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Chapter 13 The First-Order Theory Of Linear Orderings
CHAPTER 13 THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
$1. THE EHRENFEUCHT-FRAISSE GAMES AND FIRST-ORDER THEORIES In Chapters 6 and 7, where we introduced and analyzed Ehrenfeucht games, we made many references to the implications of that analysis for the first-order theory of linear orderings. We will begin this section, and this chapter, by making the connection between games and languages; that is, we will prove that A B if and only if A = B. The implications of this connection will be discussed in 913.2. Although the statement of the theorem above mentions only statements of L , its proof will proceed by induction on formulas. Thus we will need to define a version of the Ehrenfeucht game which permits parameters. As we observed in Chapter 6, the games can be played with any two structures of the same type; later, in 513.6, we will play games with k-partitioned orderings, which are just linear orderings with k unary relations. The formulation of Ehrenfeucht games we now present is sufficiently general to embrace all of these extensions of the original definitions. To simplify the notation in the coming definitions, it is convenient to separate notationally the distinguished elements of a structure from the structure itself. Thus while we have let A denote the structure ( A ; ( R i l i €I } ; ( a j l , j E J ) ) , w e w i l l n o w l e t A d e n o t e ( A ; R , , . . .,R,)anduse(A,a,, ...,a k ) or (A, a) for what was formerly denoted A. Another comment on notation : Whenever we speak of structures in general, we will use the notation A ; however, when we specialize the discussion to linear orderings, we will forget the boldface type and write A as in earlier chapters.
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DEFINITION 13.1 : Let A and B be structures of the same type 2. That is, A is ( A , R , , . . . ,R P ) and B is ( B , S , , . . . ,S,), where Ri is a z(i)-ary relation on A and Siis a z(i)-ary relation on B for each i. Let a = (a,, a 2 ,. . . ,ak) and b = ( b , ,b 2 , . . . ,b k ) be sequences of elements of A and B, respectively. Let n be a fixed natural number. A play of the game G,( (A, a), (B, b)) consists 247
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
of an ordered sequence of n repetitions of the following: PLAYER I chooses an element of either A or B and PLAYER II chooses an element of the other. The element of A selected at the rth turn is denoted a k + tand the element of B selected at the tth turn is denoted b,,,. We say that PLAYER 11 has won the play of the game if for each i, 1 I i I p , and each sequence r( l), r(2),. . . , r ( s ( i ) )of numbers between 1 and k + n,
*
* . ,ar(r(i))) E R i
if and only if
. ,br(T(i)))~Si;
otherwise we say that PLAYER I has won the play of the game. We say that PLAYER 11 has a winning strategy in the game G,( (A, a), (B,b)) if there are functions f ; ,#,, . . . ,f, such that
(i) the domain of .f, is the set of all ordered t-tuples of elements of A u B; (ii) given c 1 , c 2 , .. . ,c, E A u B (representing the choices of PLAYER I at the first t turns) f,(cl,c2,. . . , C J E A
if C , E B
f,(cl,c2, . . . , c , ) t B
if c , E A ;
and
(iii) i f c , , ~ , ,. . . , ~ , E Abandi if we define
and
for every t I n, then for every i, 1 I i Ip , and for every sequence r ( l ) , r(2),. . . , r ( z ( i ) )of numbers between 1 and k + n
..
< ~ r ( ~ ) - a r ( 2 j* ~, a r ( r ( i j ) ) E
Otherwise we say that
R,
if and only if
PLAYER I
...
9
br(,(i))) E S i .
has a winning strategy.
We observe that Definition 13.1 coincides with Definition 6.2 in the case where A and B are linear orderings and k = 0, with a slight exception. The exception is simply that the game G,(A,B) is now defined for any linear orderings A and B,and PLAYER II always has a winning strategy in G,(A,B). [Note, however, that PLAYER 11 does not have a winning strategy in G,((A,a,,a,),(B,b,,b,)) unlessa, < A a, ifandonlyifb, < B b 2 . ]
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EHRENFEUCHT-FRAYS&
GAMES AND FIRST-ORDER THEORIES
249
NOTATION : We write G,((A, a), (B, b)) E 11 if PLAYER II has a winning strategy in that game; otherwise we write G,( (A, a), (B, b)) E I.
The following lemmas are easily verified. LEMMA 13.2: (1) C,( (A, a), (B.b)) E I1 ifandonly ifC,( (B, b), (A, a)) E 11. (2) I f f is an isomorphism ,fronz A onto B and if ,f(aj)= bj f o r all j , 1 5 j 5 k, then G,((A, a), (B,b)) E 11. A LEMMA 13.3:
I f G,((A,a), (B,b))E I1 and 0 5 m < n, then G,((A, a), (B, b)) E 11.
THEOREM 13.4: G , , and
(i) for every a
E
A
(A, a), (B, b)) E I1 if and only
if
A there is a b E B such that G,((A, a, a), (B,b, b ) )E 11,
(ii) for every h E B there is an a E A such that G,( (A, a, a),(B, b, b ) )E 11.
A
Note that the form of the conditions in Theorem 13.4 is different from that in Theorem 6.6. However, it is easily seen from Definition 13.1 above that G,((A,a), (B,b ) )E I1 if and only if both G,(A'", B t b ) E I1 and G,(A'",B'b) E 11, so that the conditions here and in Theorem 6.6 agree for k = 0 when A and B are linear orderings. (When A and B are not linear orderings, the conditions of Theorem 6.6 do not make sense at all.) LEMMA 13.5 : Suppose that G,( (A, a), (B, b)) E I1 and G,( (B, b), (C,c)) E 11. Then G,( (A, a), (C,c)) E 11. A
The proofs of 13.2, 13.3, 13.4, and 13.5 are similar to those of 6.3, 6.4, 6.6, and 6.7, respectively, and are left to the reader. The upshot of the three lemmas is that we can define an equivalence relation, for each k 2 0 and each n 2 0 and each type z, on structures of the form (A, a), where A is a structure of type z and a is an ordered k-tuple of elements of A, by specifying that (A*a ) - k , n (B, b, if C,((A,a), (B, b))EII. In the case where A and B are linear orderings and k = 0, we have that A B if and only if A -,B according to Definition 6.8. (We will often suppress the k in - k , n and write -,.) As in Chapter 6,
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
we say that two structures are C-equivalent or --equivalent if they are -,-equivalent for all n. EXERCISE 13.6: Let A and B be linear orderings. Let a , < a 2 < . . . < ak and b, < b, < . . < bk be elements of A and B, respectively. Define A , = A‘“’, Ak = A’ ak and A j = [ u j , u j + for each j , 1 5 j < k, and similarly define B , = BSb1,B , = B’bk and B j = [bj, bj+l]B for each j, 1 I j < k. Show that C,((A,a), ( B ,b ) ) E I I if and only if G,(Aj,Bj)~IIfor every j 5 k. *
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The proof that A B if and only if A = B will proceed by showing that A - n B if and only if A -,, B. Intuitively, A =,, B if any statement of complexity ~n that is true in A is also true in B. The appropriate measure of complexity is not the number of quantifiers in a formula, but the depth of the quantifiers. For example, the statements (W((3Y)P(X, Y ) A (VZ)P(X, z)),
(3X)(3Y)(P(X, Y ) A ( V 4 ( 1 (P(x, 4 * P(Z, Y )1)) have the same number (and kind) of quantifiers. The first has quantifier depth 2; the second has quantifier depth 3. In the first, neither of ( 3 y ) and (Vz) is within the scope of the other; in the second, (Vz) is in the scope of (3y), which is, in turn, in the scope of (3x). (The “scope” of a quantifier is a term whose precise definition can be found references [31 and [7] of Chapter 12.) DEFINITION 13.7: We define the quantifier depth qd(4) of a formula 4 by induction on formulas d as follows:
DEFINITION 13.8: Let Lk,,be the set of formulas ofL whose free variables are among u l , u 2 , . . . ,uk and whose quantifier depth is at most n. Let A and B be interpretations of L and let u t , a 2 , . . . , a k €A and b t , b 2 ,. . . ,bkEB. We define (A,a) E k , , (B,b) if for every &u,, u 2 , . . . , u k ) E Lk.,,we have
A k 4[a1,a2,.. . ,ak]
if and only if
If k = 0, we write A -,, B insteud qf A q n B.
B k 4[bl, b2, . . . ,bk].
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THE EHRENFEUCHT-FRAi‘SSk GAMES AND FIRST-ORDER THEORIES
251
B if and only if A - n B EXERCISE 13.9: Assuming that for every n, A (Corollary 13.12),show that the statement (3x)(3y)(P(x, y) A ( V z ) ( i (P(x,z ) A P(z, y ) ) ) )considered above is not logically equivalent to any statement of quantifier depth 2. LEMMA 13.10: For each k and ii there is a finite subset @k,n of Lk,nsuch that every forniula of Lk,n is logically equivalent to a ,formula of @ k , n . Proof: It is easily verified by induction that every formula of Lk,n is a Boolean combination of formulas of Lk,n-l and of formulas of the form (Qy)+(u,,u,,. . . ,uk,y),where Q can be either 3 or V. By replacement of bound variables, all of the latter formulas are logically equivalent to formulas of the form (Quk+l)+(ul, u 2 , . . . ,tikruk+ 1), where +(ul, u 2 , . . . ,u k + 1) is in L k + 1 , n - 1.
We now proceed by induction on n to show that the desired conclusion holds for all k. For n = 0, the formulas of Lk,nare all Boolean combinations of atomic formulas in the variables u l , u,, . . . ,u k . Since the number of predicate letters is finite, there are finiteIy many such atomic formulas. Hence, by Exercise 12.5, we can define for each k an appropriate finite subset Qk.0 of Lk,o.For n > 0, we may assume that ak,,and @)k+ 1, n- have been defined and are finite; then every formula of L,’nis logically equivalent to a Boolean combination of formulas from ( D k , n - l and of formulas (Quk+l)+(ul,...,’k+~), where ~ ( U I , . . ? U k + l ) is in @ k + l , n - l . Hence, by Exercise 12.5, we can find an appropriate finite subset @)k,n of Lk,n. We are now ready to state and prove the main theorem, which relates the Ehrenfeucht games to first-order languages. Note that the proof of the lemma above depends on the assumption that the language has only finitely many predicate letters; the theorem below, which uses the lemma above, is false without this assumption, as we will show near the end of 513.4. THEOREM 13.11 : Let n 2 0 and k 2 0 be given, let A and B be structures of the same type and let a , , a,, . . . , a k €A and b , , b,, . . . ,b k E B. Then (A, a) = k , n (€3, b)
if and only if
(A, a)
-
k.n
(B, b).
We proceed by induction on n to show that the equivalence holds for all k. If n = 0, then assuming that A = ( A ,R , , . . . ,R,) and B = ( B , S,, . . . ,S,), it is clear that Go((A, a), (B,b))E 11 if and only if for each i and for each sequence r(l),r(2),. . . , r ( z ( i ) )of numbers between 1 and k
Proof:
(ar(~j,ar,,,, . . . ,ar(r(i)))E Ri
if and only if
(br(lj,br(2),. . * ,br(r,ijj> E Si
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if and only if for every formula 4(t',,u , , . . . ,U , ) E @k,O A b 4[a,,a,, . . . vak]
if and only if
A I= 4 [ U l i Q ~ ., . . -ak]
if and only if
B != +[b,, b,, . . . ,b,]. Assume that the claim has been proven for n and suppose that (A, a) -,,+ (B, b). Let 4 be in @k,,,+ ,. Then either 4 is of the form (Qu,, where or 4 is a Boolean combination of formulas in @k,,, and formulas of this form. It clearly suffices to show that B =!
4[bl, b,,
. . . ,bk]
holds for all formulas 4 in @k,n+ of the form ( Q U k + J$ since the "if and only if" is preserved under Boolean combinations of formulas. And since (Vy)$ is logically equivalent to 1 ( 3 y ) i $ , we may assume that 4 is of the form ( 3 4 , ,)$, where $ has quantifier depth at most n. Since (A,a) -,,+ (B,b), for each ~ E there A is a b E B such that (A, a, u ) -,, (B, b, b). Hence by the induction hypothesis, (A, a, a) -,, (B, b, b ) ; thus for each a E A there is a b E B such that A b $ [ a , , . . . ,a,,u]
if and only if
B 1 $ [ b l , . . , ,b,, b].
Hence if A I=4[a,,a,, . . . ,a,], then €3 k 4 [ b , , b,, . . . ,bk]. Similarly, if B k 4 [ b , ,b,, . . . ,b,], then A k 4[al, a,, . . . ,uk]. Hence, for each 4 E @ k , n + 1 , A 1 d [ a ~ , U , ,. . . >a,]
if and only if
,
B k 4 [ b l ,b,, . . . ,b k ] .
Conversely, suppose that (A,a) Q,,+ (B, b). We must show that there is a formula 4 ~ @ , , , ,such + that it is not the case that
,
A 1 4 [ a l , u 2 , . . . ,ak]
if and only if
B 1 4[b1, . . . ,b,].
Since (A, a) *,,+, (B, b), there is (without loss of generality) an element U E A such that for each b E B, (A, a, a) * ,,(B,b, b). By the induction hypothesis, there is, for each b E B, a formula qbE @)k+ ," such that A != $b[a13a2,. . . ,ak,a]
B y $b[bi,b,,. . . ,b,,b]. Since @ k + , , , , is finite, it follows that {$blbEB) is finite. Let $ be the conjunctionA{$,(bEB}.ThenA k $ [ a l , U 2 ,..., a,,a] butB y $[b1,h2,..., b,,h] for any b E B. Let 4 be ( h k + l ) $ . Then A 1 4 [a ,,a ,, . . . ,a,] and B y 4 [h ,b, , . . . ,bk], although 4 E a,.,, + This proves the converse and thus the theorem. and
,
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COROLLARY 13.12: (1) For each n, A (2) A = B ifand only if A B. W
E,, B
if and only
if
A
-,, B.
Thus two structures are elementarily equivalent if and only if they are G-equivalent.
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GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
82. THE EHRENFEUCHT GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
As a consequence of the equivalence between logical equivalence and G-equivalence, we can now interpret logically various results of Chapters 6 and 7. For example, from Corollary 6.12 it follows that o is logically equivalent to o 5 . CI for any order type CI. In particular, no statement of the first-order language distinguishes between w and w 5.
+
+
THEOREM 13.13: (Tarski [20]) an elementary class.
The class W of well-orderings is not
Proof: W is not closed under elementary equivalence, although by Exercise 12.7.2, every elementary class is closed under elementary equivalence. I
By Exercise 6.11, if A is a finite linear ordering with at least 2" - 1 elements, then A -"o o*.Hence any statement true in o o* is also true in some finite linear ordering. Moreover, given any finite set of statements true in o ID*, their conjunction, which is also true in w o*, must be true in some finite linear ordering. O n the other hand, o o* is not elementarily equivalent to any finite A since if A has m elements, a statement asserting the existence of at least m 1 distinct elements would be true in o + o*but not in A . Thus although o + w* can be distinguished from each finite A by a first-order statement, no one first-order statement distinguishes o + o*from them all. This is expressed by saying that o o*, or more properly Th(o w*), is not finitely axiomatizable.
+
+
+
+ +
+
+
+
DEFINITION 13.14: Given a theory T we say that T is finitely axiomatizable if there is a statement 4 E T such that for every 9 E T , 4 logically implies 9. The statement 4 is called an axiomatization of T. Given a structure B, we say that B is finitely axiomatizable if Th(B) is finitely axiomatizable, and if 4 is an axiomatization of Th(B), we also say that is an axiomatization of B.
Note that
4 is an axiomatization of B if and only if for every structure 4, then C = B.
C of the same type as B, if C !=
PROPOSITION 13.15: o
+ o* is not Jinitely axiomarizable.
I
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Note that 4 is an axiomatization of a theory T if and only if for every interpretation C of the language of T , C k T if and only if C b 4. Thus, for example, the conjunction 4Lof (1) ( ' J X ) ( V Y ) ( W X < y A Y < 2 --+ x < 4, (2) ( W V Y ) ( X < Y 7 ( Y < x )1, and (3) ( V X ) ( \ J Y ) ( X < Y v x = Y" Y < x ) +
is an axiomatization of the theory of linear orderings. Although w + o*is not finitely axiomatizable, w is finitely axiomatizable. For every model of the conjunction of the statements (1)
4L.t
(2) ( W 3 Y ) ( X < Y A ( V Z ) ( l ( X < z A z < Y ) ) ) , (3) (3X)(VY)(X< Y v x = Y ) , (4) (VX)((3Y)(Y< x ) --+ (3Y)(Y < x A ( W l ( Y < z A
-=x ) ) ) )
has order type w + 6 . c1 for some c1 and so is elementarily equivalent to o by Corollary 6.12 and Corollary 13.12. PROPOSITION 13.16: w is Jinitely axiomatizable.
We note that the quantifier depth of the above axiomatization of w is 3; hence, if B =30then B = w. This fact, and the fact that w is finitely axiomatizable with an axiomatization of quantifier depth 3, can also be explained combinatorially, so we make a brief digression to Ehrenfeucht games.
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DEFINITION 13.17: A structure A is said to be hitary if there is an n such that B A implies B A for all structures B of the same type. If A is finitary, then we will write f(A) = m if m is the least IZ for which B -" A implies B A for all structures B of the same type.
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Before commenting on this definition, we note that using Corollary 13.12,Theorem 6.16 says that given any linear ordering C there is a countable linear ordering elementarily equivalent to it. This conclusion, the Lowenheim-Skolem Theorem, is also true for arbitrary structures, as pointed out at the end of 812.2. By the Lowenheim-Skolem Theorem, to show that A is finitary it suffices to show that B -,, A implies B A for all countable structures B of the same type; for if C is uncountable, then C B for some countable B, so that C -,, A implies that B w nA, which implies that B A and therefore C A.
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255
EXAMPLES: (1) q is finitary, for if B m 3 q, then B must be dense and have no endpoints; hence if B is countable, it follows, by Cantor’s Theorem 2.5, that B v q. Then Lemma 6.3.2 implies that B mnq for all n, or, equivalently, that B q. O n the other hand, since 5 - 2 q , we conclude that f ( q ) > 2, so that f ( q ) = 3. (2) w is finitary, for if B m 3 w , then B must have a first element, every element of B must have an immediate successor, and every element of B, except the first, must have an immediate predecessor. Thus, if b E B is not among the first w elements of B, then cF(b)‘v 5, so that B ‘v w + 5 . a for some order type a. Since B w by Corollary 6.12, we conclude that w is finitary. On the other hand, ((D + o)-2 w so that f ( w )= 3. (3) o + w* is not finitary, for, given any n, 2” -,, (w + a*)by Exercise 6.11, but 2“ * n + (w + a*)by Exercise 6.10.2. (4) Each finite ordering is finitary, for, by Exercise 6.10.2, if k < 2” - 1 and A -” k, then A = k so that A k ; hence f(k) I n.
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EXERCISE 13.18: (1) Show that 5 is finitary and that f ( 5 ) = 3. (2) Show that w + w is finitary and that f(o+ a)= 5. (3) Show that w* ( + w is finitary and that f ( o * + 5 + w ) = 5. [Note: You will have to state and prove analogues of Corollary 6.12 for 5, w + o,and w* + 5 + w.]
+
This completes the digression to Ehrenfeucht games. We showed earlier that if B = 3 w then B = w by using a specific axiomatization of w . Alterw, so B o by Example natively, if B - 3 w, then, by Corollary 13.12, B 2 above; hence, by Corollary 13.12, B = w. Furthermore, since @ 0 , 3 is finite, &4 E @0,310 1 4} is an axiomatization of w. This is an example of the following fact. m3
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PROPOSITION 13.19 : A structure A is finitely axiomatizable if and only cf it is finitary. Furthermore, if , f ( A )= n, then A has an axiomatization of quantijier depth n but has no axiomatization of quantijier depth smaller than n. Proof : The first statement follows from Corollary 13.12. If f(A) = n, then as in the discussion above, A has an axiomatization of quantifier depth n. If A had an axiomatization of quantifier depth m < n, it would follow that f(A) I m, contrary to assumption. H
Applying Proposition 13.19 to conclude that w is finitely axiomatizable is more efficient, but does not of course give an explicit axiomatization of
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w. It does, however, guarantee that no axiomatization of o has smaller quantifier depth than @<,, .
COROLLARY 13.20:
q and
5 are Jinitely axiomatizable. A
c, +
+
EXERCISE 13.21 : ( I ) Find axiomatizations of q, o o,and o* 5 o that have minimal quantifier depth. (2) Describe an axiomatization for each n that has minimal quantifier depth.
+
Corollaries 7.10 and 7.21 have the following logical interpretation. THEOREM 13.22: (Lauchli and Leonard) (1) I f # is a sfatement true in some scattered linear ordering A , then M k 4 .for some M E A. . (2) If 4 is a statement true in some linear ordering A , then M I. # for some M E A.
Proof: Given # of quantifier depth n, use Corollary 7.10 (or 7.21) to find an M E A. (or M E . X ) such that M m n A . Then, by Corollary 13.12, M = " A , so that M k 4.
Note that the proof of Theorem 13.22 actually shows more, namely, given any (scattered) linear ordering A and any n, there is a single (scattered) in which is true every statement of quantifier linear ordering M in A (or Ao) depth at most n which is true in A . In 513.6, we will prove the following stronger version of Theorem 13.22.
THEOREM 13.23: (Amit, Myers, Schmerl, Shelah) Let A be a linear ordering. Then for each n there is a finitely axiomatizable M E Af such that A =,, M . If' A is scattered, then M can be chosen f r o m Ao. Theorem 13.22 can be used to show that various linear orderings are not finitely axiomatizable. For example, assume that cow is finitely axiomatizable. It follows from Theorem 13.22 that ow= M for some M E do since the axiomatization of coo must be true in some M E Ao.Choose M of smallest rank such that cow = M . Let M' be the subordering of M consisting of all limit points of M , that is, all elements of M that have no immediate predecessor. Now it is easily verified by induction on A. that if M E -H0 and M' is the set of limit points of M then M' E A. and rF(M') < rF(M). Hence coo f M'. Thus PLAYER I has a winning strategy in Gn(mw, M') for
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257
some n. But thenPLAYER I has the following winning strategy in G , + 2 ( ~MW) :, In his first n moves he plays only on the limit points of ww and M , which have order types ww and M ' , respectively, following his winning strategy in Gn(wW, M ' ) ; since PLAYER 11 must have chosen a non-limit point, PLAYER I can win in two more moves. This contradiction proves the following result. COROLLARY 13.24: ww is not jinitely axiomatizubfe. It is now natural to ask whether every ordinal a < ww is finitely axiomatizable. Note that we cannot automatically expect a positive answer since, as we have seen, dodoes contain non-finitely axiomatizable linear orderings. We will show that every ordinal a < owis finitely axiomatizable, and that, moreover, an axiomatization of M can be obtained effectively from a description of a. When we speak here of a description of a, we mean of course the Cantor Normal Form a = wn . a,
+ wn-' .
tE,-
1
+ . . . + w . u1 + a,,
where a, # 0. We will show how to construct an axiomatization of M from its description. To prove that the axiomatizations @ a that we present for each M really are axiomatizations, we will need lo show that if A b @a, then A = a. To do so, we will define, for each x , a class &(a) of a-like linear orderings and show that for each linear ordering A , A is G-equivalent to a if and only if A E ~ ( c c )Then, . by virtue of the equivalence of G-equivalence and elementary equivalence, to show that CD, is an axiomatization of a it will suffice to show that if A k CDa, then A E d ( x ) . We first define each d ( w " ) ,by induction on n. Then wO-likeorderings all have order type 1. For n = 1, we have already seen that A o if and only if A c o 5 . a for some a,and so we define these to be the w-like orderings. Note that both 1 and o are finitary, and, in fact, f ( w ) = 3. Now assume that we have defined, for each k < n, the class d ( w k )of wk-likeorderings, that we have shown that A E d ( w k )if and only if A cok, and that we have shown that wk is finitary and that f ( w k )5; 2k + 1. Moreover, we assume that for each k < n, A E d ( w k )if and only if A = 1 i E W ), where W is cok- I like and each 4 is w-like, and that A E d ( o k )ifand only if A = Kli E W } , where W is d i k e and each is wk-l-like. (Note that for k = 1 both of these equivalences hold.) Now suppose that A wn.L,et A' be the set of all limit points of A . Then A' must have order type in d ( ~ " for - ~if)G,2 , - l ( A ' , o " - l )E I, then, by sticking to his winning strategy in G2n-l(A',w"-1),PLAYER I can force
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I{
-
w.
I{
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
to a non-limit point and thus win in G2,,+,(A,w").Between each pair of successive limit points of A, the interval must be o-like. Thus A = x{l4(\i E W } ,where W is o"-'-like and each is o-like. Since W is on-I like, it can be written in the form W = 51j E V } , where V is o-like and each V, is wnV2-like.Thus A = c ( c { w l i E y } l j E V } ; by the induction W , ( iE 5 ) is on-'-like, so that A = c { X j ( jE V ] ,where hypothesis, each V is d i k e and each X j is w"-'-like. More explicitly, PLAYER II
I{
I{
-
where each l4(. and each W,' is w"-'-like. Thus we define &(a")to be all linear orderings of this form. Clearly, if A E &(on),then A o" by the 2n + 1. Thus the following is correct. induction hypothesis. Also, f ( w " ) I PROPOSITION 13.25: Define a sequence { d ( o " ) l n< Q } of sets of linear orderings, by induction on n, as follows:
(0) d(1) = all linear orderings of order rype 1 ; (1) d ( o )= all linear orderings of order type w + 5 . a for ci arbitrary; (n) &(on)= all linear orderinys of form
where euch
4, and each W,' A
is in &(on-'). Then, ,for each n,
E &(d)
Moreouer, f(o")= 2n + 1.
if and only if
A
-
on.
Proof : We have verified everything except that f(o")2 2n + 1. This follows from the fact that o"r r 2 , , o"+ w" (Theorem 6.18) which is not on-like. . u2 + . * . + Now let cc < owbe arbitrary and write a as w"' . a , + on2 . ak in Cantor Normal Form. We say that A is a-like if and only if A can be written in the form (link
(Wl' + . * .
+ w:;) + (W;2 + . . + w:;) + . . . + ( wl"+. . . + W:;), '
where each Wyt is an w"[-likeordering, and we let &(a) be the set of all cc-like orderings. THEOREM 13.26: For every a < ow,A over, a is jnitary.
E &(a)
i f and only i f A
-
a. More-
2.
GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
-
-
259
Proof: If A E &(a), it is clear that A a. Assume that A CI and that a = wnl . a, . . . wnk. u k .Then PLAYER I in his first c { a j l 1 I j I k } moves chooses the first elements of each of the aj copies of con, for all j. If between two successive choices of PLAYER 11 in A the interval is not co"J-like for the appropriate n j , then PLAYER 11 is punished in 2nj 1 more moves. Hence A E ~ ( c I )Moreover . f ( u ) I 2n, + 1 + c{ujI 1 Ij I k } .
+
+
+
We will now pattern our axiomatization of each a < owon the constructions above; that is to say, we will axiomatize anby saying that its limit points are w"-'-like. To do Ihis, it will be very helpful to introduce the idea of relativizing formulas of L . Relativization is useful when one wants to speak about a substructure of a given structure as a structure. Thus, for example, given a linear ordering A and an element a E A , one can write a statement about A'" that says that every element of A'" has a successor in A'"; such a statement would be simply (Vx)(3y)(x < y), since this says about any linear ordering that every element has a successor. If one wants to write something about A and a that says that A'" has this property, one would use instead the formula (vX)(X
+
( 3 y ) ( y< U
A X
< y)),
observing that A k (VX)(X < u
+
(3y)(y < u A
x < y))[a] A'" 1 (Vx)(Iy)(x< y).
if and only if
We say that the new formula is the old one relativized to the formula e(w,u ) = w < u, since every quantifier ( Q w ) is interpreted as applied to { w ( w < u } . Now for the general formulation. DEFINITION 13.27: Let O(w,vl, u 2 , . . . ,uJ be a formula of L. We define for every formula 4 of L the relativization by induction on formulas:
+@
(1) If 4 is atomic, then 4' is 4, (2) If 4 is i $then , 4' is i($@). ( 3 ) I f 4 is $1 A $ 2 , $1 " $ 2 , IC/I $ 2 , Or $1 $ 2 , then $'is $ i @ A $2', v $2', + 4b2', or $,' ++ $2', respectively. (4) If 4 is (3y)$, then 4' is (3y)(,O(y,u l , . . . ,u,) A $'). ( 5 ) If 4 is (Vy)$, then 4' is (vy)(O(y,u l , . . . ,0,) + $7. +
Thus, for example, the statement (Vx)[(Vy)(y < X ( W ( Y < < 4) (3X')(('dy)(y< x' -+ (3z)(y < Z < X')) +
+
A X
< X')]
260
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
-
is exactly the statement (Vx)(3s')(x < x') relativized to the formula (Vy)(y < w -, (3z)(y < z < w)) and thus says of a linear ordering that it has no last limit point. Similarly, the statement (3x)(Vx')(.x I x') relativized to ( V y ) ( y < w + ( 3 z ) ( y< z < w ) ) says of a linear ordering that it has a first limit point. More generally, given any statement 4, the relativization of 4 to ( V y ) ( y < w -+ ( 3 z ) ( y < z < w ) ) says of a linear ordering that its limit points have property 4. We now explain this more precisely. Given a structure A and elements a,, a 2 , . . . ,a, E A , let (A,ul, . . . ,a,)" = jblA I= 8 [ b , a , , . . . , a , ] } . Then what 4' says about ( A , a l ,. . . ,a,) is that 4 holds in ( A ,a,, . . . ,a,)'. This is a corollary of the following lemma, which is proved by induction on formulas. LEMMA 13.28 : Let @(w,v,, . . . ,c,) be a formula of L. Then for any structure A , any formula 4(x1, x 2 , . . . ,x,), any a , , . . . ,a, E A, and any b l , . . . ,b, E ( A , a , ,. . . ,a,)',
A
if and only if 4'[bl,. . . , b n r a i . .. . , a t ] (A,a,, . . . ,at)' 1 4 [ b , , . . . ,&I. A
COROLLARY 13.29: Let O(w, u l , . . . ,u,) be a formula of L . Then .for any structure A , any statement 4, and any a , , . . . ,a, E A,
A I. 4'[al, . . . .a,]
if and only
if
(A, a,, . . . ,a,)' I. 4.
A
Thus, for example, if A is a linear ordering and 4 is a true statement about the interval [a,, a 2 ] of A, then, choosing O(w, u,,u,) to be u , I w I u 2 , we see that A k @[al, a2] since 4' makes the same assertion about A that 4 makes about [a1,a 2 ] . COROLLARY 13.30: Let O ( w ) be a formula of L. Then for any structure A and any statement 4,
A b
4'
if and only if
A' k
4. A
Thus, for example, if A is a linear ordering, then 4 is a true statement about the limit points of A if and only if A I= 4', where B(w) is ( V y ) ( y< w -+ (&)(y < z < w ) ) . Note that for any formula 4, qd(4') = qd(4) qd(O), so that if 8 is quantifier-free, the quantifier depth of 4' is the same as that of 4. Also note that if 4 is a statement and O ( w , u l , . . . ,u,) is a formula, then the free variables of 4' are u,, . . . ,u, but not w.
+
2.
GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
261
The substructures of A we discussed in our treatment of relativization are all definable, so it seems appropriate to define that notion before returning to our discussion of ordinals.
DEFINITION 13.31 : Let A be a linear ordering and let B G A . We say
that B is definable if there is a formula O(w) such that b E B if and only if A k O[b]. Let c1,c2, . . . ,ct E A ; we say that B is definable over c 1 , c 2 ,. . . ,c, if there is a formula O(w,y l , y,, . . . , y,) such that b E B if and only if A k O[b,ci,cz,. . . ,cJ.
Thus, for example, the set A’ of limit points of A is definable by the formula ( V y ) ( y < w -+ ( 3 z ) ( y< z < w)), the interval [ b , , b21A is definable over b,, h, by the formula y, 5 \v 5 y,, and the interval [a, + ) A is definable over a by the formula w I y , . The results above can then be interpreted as follows. COROLLARY 13.32: (1) Let B be a dejnable subordering of A . Then, given any statement 4, we can ejj%ctively,find a statement q5A such that
Bk
4
f and only fi
A k
4A.
(2) Let B G A be dejnable over. c,, c,, . . . ,ct. Then, given any statement can efectively find a forniula 4 A ( y l ,y,, . . . ,y,) such that
4, we
Bk
4
i f and on/.) if
A k 4*[c,,c2, . . . ,c,].
Proof : Take to be @, where O defines B (over c,, c,, . . . ,c,), and apply Corollaries 13.29 and 13.30.
Given a linear ordering A, which subsets B s A are definable? Suppose, for example, that B G g and B is definable: If b , , b , ~g, then there is an automorphism f of g such that f ( b , )= b,. Hence, by Exercise 12.6, for any formula 8, q k 8[b,] if and only if q k 8[b,]. Thus B must be either @ or all of g; that is, g has no definable proper subsets. The same is true of 4, and, in fact, whenever A is transitive, A has no definable proper subsets. Which subsets of o are definable? For a complete analysis, see the exercise below. EXERCISE 13.33: (1) Show that each element of o is definable (where a € A is definable if ( a ) is definable) and that each finite subset of o is definable; show that the complement of each finite subset of o is definable. (2) Show that the only definable subsets of o are its finite and cofinite subsets. [Hint: Look at o 4.1
+
13. T H E FIRST-ORDER
262
THEORY OF LINEAR ORDERINGS
(3) Suppose that 4 is a statement which is true in infinitely many nonisomorphic finite linear orderings. Show that 4 is true in n for all but finitely many n.
Now let O1 be the axiomatization of o discussed earlier; note that qd(@,,)= 3. Let O(w) be the formula (Vy)(3z)(y < w + y < z < w ) discussed above, which says that w has no immediate predecessor. Then A k Qle if and only if A' I=a,. But A' = A', so that A b Ole if and only if A' is d i k e if is an axiomatization of 0'. and only if A E d ( w 2 ) .Thus THEOREM 13.34: For each n > 1, LteJine @, axiomatization of on.Moreover, qd(@,,)= 2n + 1.
Then @,, is an
=
Proof: By induction on n, as in the argument for n = 2. Moreover, since qd(@,) = 3 and qd(O) = 2, we can show by induction on n, using the fact that qd(de) = qd(4) qd(O), that qd(@,,) = 2n + 1 for all n. W
+
Note that the quantifier depth of the axiomatization a,,of W" is precisely
f ( d )as , predicted by Proposition 13.25, so that these axiomatizations are
optimal. . u2 + . . . + w " ~. a k , Turning now to the general case c( = w"' . a, + on2 we see that we want to axiomatize CY by saying that there are certain points such that the intervals between them are wk-like;for this we need to relativize to the formula ~(w,u,,u2) that is v 1 I w < u 2 . We thus define @(a) to be ( 3 x 1 ')
' ' '
(jXi,)(3X12)
A(X1'
' ' '
(jX:,)
' ' '
5 y)
(3Xik){(Vy)(X,' '
. . A @ e ( w . d 1 . x l 2 )A ae(w,x,2,x22)
nl
. . .A
(]Xik)
< " ' < xi, < X1' < . . . < XZ2 < . . * < X l k < . . < X i k )
A @e(w,x,',Xl2) A . A
' ' '
n1 @ ~ ( W , X , ~ . X ~ ~ )
nk
A"
A,
. .A ae(w%x:2.x,3) nz
"2
.A ( D o ( ~ ~ ~ : k _ A @e'(wdk) nk
nk
1,
where B'(w, vl) is u I I w. The following is then easily verified. THEOREM 13.35 : For each ordinal a < ww,@(a)is an axiomatization of
CY.
A
We conclude this section with several theorems of a technical nature specific to linear orderings that result from the equivalence between logical equivalence and G-equivalence. These theorems will be used in $813.4 and 13.5. Recall that for every k and n, Lk,,,consists of all formulas of L whose free variables are among u l , . . . , uk and whose quantifier depth is at most n. We showed that there is a finite subset @k,,, of Lk,,, such that every formula
2.
GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
263
of Lk,,is logically equivalent to a formula of @ k , , ; thus if 4 E a,,,, then there 4. Now let A be a structure, let is a II/ E @ k , , that is logically equivalent to i a,, . . . ,ak E A , and let @ = @(A,a , , . . . ,a,) be the conjunction of all 4 E @ k , n such that A 1 b[a,, . . . ,a,]. Then A 1 @[a,, . . . ,a,] and for any B and b , , . . . ,b, E B we have (A,a,, . . . ,ak) =,,, (B, b , , . . . ,bk) if and Only if B b @ [ b , , . . . ,b,]. We can thus find a finite set {Qi(ul, . . . ,u,)l1 I i I t } of formulas of Lk,, that completely describe all of the =,,,-equivalence classes. That is, for each structure A and elements a,, . . . , ak E A, there is a unique such that A 1 Qi[a,, . . . ,a,] and, moreover,
B k a i [ b , , . . . , bk]
if and Only if
(B,b , , . . . ,bk) - k , n (A,al, . . . ,a,).
For k = 0 this says that there is for each n a finite set {Oil1 I i I t} of statements of quantifier depth at most n that describe all =,-equivalence classes of linear orderings. That is, for each structure A, there is a unique misuch that A 1 mi and, moreover,
B 1 ai
if and only if
B =,A.
This much we can say about arbitrary structures, not just linear orderings; with linear orderings we can carry this analysis much further. Note first of all that each Q imust specify the ordering of the variables. We will work with the case where @ = mi logically implies that u , < u2 < . . . < u,; any permutation can be dealt with similarly. Let A k @ [ a , , . . . , a k ] , so that, in particular, a , < a2 < . . . < a,. Let A , , , 4 , , . . . ,A,- ,, A , be the intervals (t, a,], [a,, a,], . . . , [a,- ,,a,], [a,, +) of A . By the discussion above, we can find statements Y o , 'PI,. . . ,y k that completely describe the =,-equivalence j I k - 1, let YF be the relativizaclasses of A , , A , , . . . , A , . For each j . 1 I tion of Y jto e(w, u j , ujt = ( u j I w Iu j + ,), so that B k Y F [ b j ,b j +, ] if and only if [ b j ,b j +,] 1 Y j If and only if [ b j ,b j +, ] =, [ a j , a j + , ] . Let Y o Abe the u,) and let Y / be the relativization of relativization of Y oto O(w, uL) = (w I yk to o ( w , u k ) = (0, 5 w). Let y ( u 1 , . , . ,u k ) be the formula (01 < u 2 < ' ' ' < u k ) A YoA(u1) A YIA(ul, u,) A . . . A Y f - .1(uk- u,) A y t ( u k ) . It is easily verified that Y(u,,. . . ,u k ) characterizes the =,,,-equivalence class of @,a,, . . . ,a,); that is, A k Y [al' . . . ,ak] and, for any B and b , , . . . ,b, E B,B 1 Y [ b , , . . . ,b,] if and only if (B,b , , . . . ,b,) E k , , (.4,a,, . . . ,a,). Furthermore, since each Y j E Lk,nand relativization to quantifier-free formulas does not increase quantifier rank, we conclude that Y E Lk,,. We summarize this discussion in the following lemma.
,,
LEMMA 13.36: Let A be a linear ordering and let a , < a, < . ' ' < ak be elements of A . Then there is a ,formula Y(u,, . . . ,u k ) E L,,, such that A k Y [a . . . ,a,] and such that B 1 Y [ b , , . . . , bk] if and OnlJ) if (B,b , , . . . ,bk)-kk,,(A,al,. . . ,Uk).
264
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Furthermore, Y can be chosen to be a conjunction of v1 < vz < . . ' < v k with formulas Y ~ ~ ( U Y~ l A)( v, l rvz), . . . ,Yt- ,(uk- 1,ok)r y,"(uk) so that for any linear ordering B and elements b,, . . . , b, E B we have B
YoA[b,]
if andonly if if and only if
(+,b,] =n(+,u,], [bk, +)
B 1 Y',"[bk] and fbr each j , I 5 j 5 k - 1, B =! Y P [ b , , b j + l ]
ifandonly if
=,, [a,,
+),
[ b j , b j + J =, [ a j , a j + ] ] .
This lemma is easily modified if the elements a l , a z , . . . ,ak are in some < ' ' < an(k)for order other than a, < u2 < ' . .
THEOREM 13.37: Any formula 4 ( v l , . . . , v k ) in the language of linear orderings is logically equivalent, in the theory T of linear orderings, to a Boolean combination of ,formulas each of which has at most two free variables. Proof: Suppose that the quantifier depth of 4 is n. By the discussion above, there are a finite number of =,,,-equivalence classes to each of which is associated one of the special formulas T I ,. . . ,Yt of Lk,,. Then, for each j, either T t Y, + 4 or T 1 Y j + i 4. Let 0 be the disjunction of those Y j for which T k Y j + 4. Then clearly T t 4-0 and, since each Y j is a Boolean combination of formulas in two free variables, the same is true of
0. Another consequence of Lemma 13.36 is the following result of Rubin ~171. THEOREM 13.38: Let A be a linear ordering and let B be an interval of A. Then for each formula 4(v1,.. . ,v,) there is a formula +*(v,,. . . , v k ) such that ,for all b,, . . . ,b, E B, A
4[b], . . . ,bk]
if and only i f
B k 4 * [ b 1 , .. . ,bk].
2.
265
GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Proof : As in the proof of the preceding theorem, if 4 has quantifier depth
n, then 4 can be written as a disjunction of special formulas of Lk,n.If we find an appropriate Y * for each disjunct Y , then we can define +* to be the disjunction of the Y * . Thus we may assume that 4 is a special formula, so that, for some permutation n, & ( t i , , . . . , v k ) is (%(1,
< Un(2) < . . . < U,(k)) A
' ' '
A ykl(un(k-l)-
A
YoA(un(l)) A YlA(vn(1),%(2J
un(k))
YkA(un(k)).
Now Y k A ( t i , ( k ) ) is the relativization of Y kto w 2 where Y kis the statement completely describing the =,-equivalence class of some linear ordering C. Let P , , . . . ,P, be the statements completely describing equivalence classes whose representatives C,. C,, . . . ,C, satisfy C = n Ci A , , where A , = (a E AIB < a } . Let Yk*(v,(k)) be 1I i I r } . Similarly, YO4(u,~,)) is the relativization of Y oto w I v n ( l )where , Y ois the statement completely describing the =,-equivalence class of some linear ordering D.Let Q , , . . . , Q, be the statements completely describing those =,-equivalence classes whose representatives D1, . . . ,D, satisfy D = n A , Dj, where A , = {a E Ala < B}. Let Y c , * ( v E ( , Jbe v{QJ"5un(1)l 1 ~j I s}. To obtain +*(u,, . . .,vk)from + ( u , , . . .,u k ) we just replace 'l'oA(u,(oJ by '€'o*(vn(oJ and Y k A ( u x ( k ) ) by Yk*(u,(k)).It is then clear that for every b,, . . . ,b, E B,
+ V{PF~~~(~)I
+
A 1 4 [ b , , . . . ,bk]
if and only if
B 1 4 * [ b , , . . . ,bk].
The same result can be obtained if we allow parameters from A - B. COROLLARY 13.39: Let A be a linear ordering, let B be an interval of A, and let a , , . . . ,a, E A - B. Then for each formula 4(v1,. . . ,vk, w l , , . . , w,) there is a ,formula 4*(u1. . . , ,u k )such thnt for all b,, . . . ,h, E B.
Let aico,be the largest ai < B and let ai(,) be the smallest ai > B. Repeat the proof above, deleting from the special formula & those conjuncts discussing intervals wholly contained in A - B and replacing A , and A , by {alai(,) I u < B } and j(iIB < a I ai(,,>,respectively. Proof:
The content of Theorem 13.38 is that to determine whether or not certain elements of an interval B of A satisfy a certain formula in A, we can test to see whether they satisfy &* in B. For this reason, Rubin calls 4* the testing formula for 4 in B (relative to a,, . . . ,a,).
+
266
13. THE FIRST-ORDER
THEORY OF LINEAR ORDERINGS
Using Theorem 13.38, we see that if C is definable in A and is contained within the interval B of A, then C is definable in B. For if a E C if and only if A k $ [ a ] and 4* is the testing formula for 4 in B, then, for every a E B, a E C if and only if B != 4 * [ a ] . The following fact is true whenever B is a definable subset of A. LEMMA 13.40: Let A be a linear ordering and let B be an interval of A that is dejinable over e l , . . . , c,. Then for every formula 4 ( u l , . . . , 0,) there is a formula 4 # ( v l , .. . , v k , x l , . . . , x i ) such that for every b,, . . . ,b, E A. A if and only
4 # [ b l ‘, .
.
7
b k , cl,
. ..,
if b,, . . . ,b, E B
and
A != (P[b,,. . . ,bk].
Proof: Let O(w,x,, . . . ,xJ be a formula that defines B over c , , . . . ,ct. Then define $#(u,, . . . ,vk,xlr . . . ,x,) to be
(p(v1,.. . ~ ~ ~ ) A ~ { 6 (. .u. ,&)I1 i , x ~<, i < k } . Clearly 4’ has the desired property. W A similar conclusion is correct in general if we allow parameters from A - B. In the context of linear orderings, however, such parameters can be eliminated from the new formula (6’; this consequence of Corollary 13.39 is also due to Rubin.
THEOREM 13.41 : Let A be a linear ordering, let B be an interval of A that is dejinable over c,, . . . ,c,, and let a,, . . . ,a, E A - B. Then for every jbrmula (p(v,, . . . ,v k , w,, . . . ,w,) there is a formula 4 # ( v l , . . . ,v,, x,, . . . ,x,) such that for every b , , . . . , b, E A, A
(P#[b,,.. . ,bkrCir.. .,Ci]
if and only if b,, . . . ,b, E B
and
A k 4 [ b , , . . . ,bk,a,, . . . ,am].
Proof: Let 6 ( w , x 1 , .. . , x J be a formula that defines B over c l r . . . ,c,. Let 4*(ul,.. . ,uk) be a testing formula for 4 in B relative to a,, . . .,a,, so that for all b l , . . . ,b, in B, A k
4 [ b , , . . . ,bh,al,. . . ,am]
if and only if
B 1 +*[b,, . . . ,b,].
3.
DECIDABILITY OF FIRST-OKDHR THEORIES OF LINEAR ORDERINGS
Let $ # ( u , , . . . ,u k , x , , . . . ,xJ be the relativization of together with A{O(ui,xl, . . . , x t ) l 1 I i I k } . Then A
4*
267
to O(w, xl, . . . ,x,)
1 d # [ b , , . . . ,bk,c,, . .. ? C t ] if and only if bl, . . . ,b k E B and A 1 (4*)’[b,, . . . ,b,, cl, . . . ,c,] if and only if b l , . . . ,bk E B and B 1 4*[b,, . . . ,b k ] if and only if b l , . . . ,b, E B and A 1 4 [ b , , . . . ,b k , u , , . . . ,urn].
rn
Thus if we are interested only in which elements of the definable interval satisfy 4, we can select them out by using 4’; we thus call 4’ the selecting formula for 4 in B (relative to c i l , . . . ,urn).
53. DEClDABlLlTY OF FIRST-ORDER THEORIES OF LINEAR ORDERINGS DEFINITION 13.42: Given a language L and a theory T in the language L, we say that T is decidable if there is an algorithm that, when applied to an arbitrary statement 4 of L, will determine whether or not 4 E T ; that is, T is decidable if T is a recursive set of statements. Otherwise, we say that T is undecidable. Typically, T is either the theory Th(M) of a structure M or the theory Th (d) of a class d of structures. In this chapter, we will discuss examples of
both kinds. A useful criterion for decidability in the first case is recursive axiomatizability, which we will call axiomatizability even though that term has already been used, since. in the context of decidability, axiomatizability always means recursive axiomatizability. DEFINITION 13.43: A theory T is axiomatizable if there is a recursive set of axioms for T. That is to say, T is axiomatizable if there is a subset @ c T such that (1) If 4 E T then 0 1 4. (2) There is an effective procedure that, when applied to a statement 11, of L , determines whether or not $ E 0.
In this case we say that T is axiomatizable by 0.
268
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
If the theory T is axiomatizable by 0,then we define a “theorem-checking algorithm” for Q, as follows: Enumerate all finite sequences of statements of L, check each to see whether it is or is not a deduction from 0,and enumerate the conclusions of all deductions from Q,. By (1) and the Completeness Theorem, every statement of T will turn up on the list. Thus, if T is axiomatizable, then T is recursively enumerable. If, in addition to being axiomatizable, T is complete, then T is decidable by a “wait-for-a-proof algorithm”: To determine whether or not & E T , start the theorem-checking algorithm for Q, and wait until either & or 14 is enumerated. This algorithm works since the completeness of T guarantees & will appear. that eventually either #I or i THEOREM 13.44: I ~ f T is an axiomatizable complete theory, then T is decidable. W
This theorem is particularly useful in showing that a theory Th(M) is decidable since, for any structure M, Th(M) is complete. COROLLARY 13.45: For any structure A, Th(A) is axiomatizable i f and only if Th(A) is decidable. In particular, if Th(A) isjnitely axiomatizable, then Th(A) is decidable. Proof: Every finite set is recursive. H
Thus, for example, Th(w),Th((), and Th(q) are all decidable, as is Th(a) for each a < ow. Even if A4 is not finitely axiomatizable, Th(M) may be decidable. One need only give explicitly a complete recursive set of axioms for Th(M).Thus, for example, although M = o + o*is not finitely axiomatizable, it is axiomatizable. Thus, let Q, = C, u (4,,ln < w},where, for each n, &,, says that a model has at least n elements, and C, says that a model is a linear ordering that has a first and a last element, that every element but the last has an immediate successor, and that every element but the first has an immediate predecessor. (The subscript F is there to remind us that C, is also true in every finite linear ordering.) This set of statements is clearly recursive, and any model A of Q, must have order type w + ( * a + o*for some a, so that, by Exercise 6.11 and Corollary 13.12, A = o + w*. Thus @ is an axiomatization of Th(o + w*), so that Th(o o*)is decidable. On the other hand, one can easily give examples of linear orderings whose theories are not decidable. For example, let A be a recursively enumerable subset of N that is not recursive. (The existence of such a set is
+
3.
DECIDABILITY OF FIRST-ORDER THEORIES OF LINEAR ORDERINGS
269
demonstrated near the end of $16.1.) Let A = {a,, a,, a,, . . . } be an enumeration of A and let M = q a, q a, q + a, . . . . If T h ( M ) were decidable, then one could effectively determine whether or not n E A by checking to see whether or not M l= 9,, where 9, is the statement
+ + + +
(3X,)(3X2).
. .(3X,)[(Xl < X 2 < . '
A (vy)(Xl
AfVY)(Y
< < X,
-+
'
<
+
X,)
( y = X 2 V . . . V y = X,-
< X l + ( W Y < Z <
X,))A(~Y)(y>x,+(3z)(Y>z>x,))l,
which says that M has a maximal finite interval with n elements. For this M , Th(M) is not decidable and hence is not axiomatizable. A more interesting example is Th(o"). We know that Th(w") is not finitely axiomatizable by Corollary 13.24. Furthermore, we have not presented any axioms of Th(w"). However, for each y1, moo-,onby Theorem 6.18, so if 4 has quantifier depth n, then owk 4 if and only if w" k 4. Thus we can argue that since each w" is decidable, so is w w ;but that would be premature. The algorithm we propose for Th(o") has the following form: Given C$ calculate qd(4) and then apply algorithm to 4, where algorithm A , is the one that describes whether or not 4 E Th(o"). This proposed algorithm is actually a finite object only if the passage from a number n to the algorithm A , is recursive. That would not be the case if the algorithms A , followed no pattern. Thus to conclude above that Th(w") is decidable, we need to know not only that each Th(w") is decidable, but that there is a uniform pattern to the decision procedures. The uniform pattern is as follows: Given n, write out the axiomatization a, of ongiven in Theorem 13.34 and apply the wait-for-a-proof algorithm to @,, (Note that there is an algorithm-iterating relativizations-that, given n, will produce the statement @,.) So we can conclude that Th(w") is decidable. By Theorem 6.22, if ON is the collection of all ordinal numbers, then ON Q"', so that Th(0N) = Th(crt") and hence Th(0N) is decidable. That is, one can decide effectively whether a statement about the ordering of the ordinals is true or false. This result, due to Mostowski and Tarski [113, is proved in Doner, Mostowski, and Tarski [3] and in Ehrenfeucht c41-
-
COROLLARY 13.46: Th(o") = Th(0N) is decidable. H
Note that our discussion above also yields a complete axiomatization of Th(w"), namely, @ = u,,
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13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Tarski [ 3 ] : Let CD consist of the linear ordering axioms together with all induction axioms (3x)Cp(x) (3x)[Cp(x) ~(Vy)(Cp(y)+ x I y)] where Cp(x) is a formula whose only free variable is x. Then CD is a complete axiomatization of Th(0N). The theory of O N is discussed at length in [3]; Rosenthal [I61 discusses the models of Th(0N) and tabulates its complete extensions. For which linear orderings M is Th(M) decidable? We will see at the end of this section that Th(M) is decidable for each M E A.But for most order types M,Th(M) must be undecidable since there are only a countable number of algorithms and hence only a countable number of decidable theories. A complete answer as to when Th(M) is decidable seems to be unknown. We now consider whether Th(,X) is decidable, where Jf is a class of linear orderings. Our first example is the case for which JV is the class A!Fof finite linear orderings. We first claim that Z, is an axiomatization of Th(A,). (Note that M k C, if and only if either M is finite or M ‘v w + [ . a + w* for some order type a.) For if Cp E T h ( d F )but is false in some model of C,, then, for some order type a o , Cp is false in w + [ . a. + w * . But then, by Exercise 6.11 and Corollary 13.12, Cp is false in w + [ . rx + w* for every order t y p a ; thus C, u { i C p } is a finite axiomatization of Th(w + a*), contrary to Proposition 13.15. Hence Cp is true in every model of C,, so that, by the Completeness Theorem, X F I- Cp. (This is an example of the common phenomenon that not every model of T h ( M )necessarily belongs to .A’”.) Thus 1, is an axiomatization of Th( H,), and hence Th(A,) is recursively enumerable. To show that Th(,H,) is decidable, it suffices to show that {Cp I Cp 4 Th(&,)] is also recursively enumerable. But Cp # Th(M,) if and only if M b i C p for some finite M, so we need only recursively enumerate the set of all statements false in some finite linear ordering. But if we recursively enumerate all pairs (a,@), where n is a natural number and Cp is a statement of L, and check them one by one to see whether or not n k Cp, putting i Cp E K if n 1 Cp and Cp E K ifn y Cp, then we will have recursively enumerated K = {+I4 4 Th(,X,)). This completes the proof of the following theorem. --$
THEOREM 13.47: Th(AF)is decidable. A slightly more sophisticated argument is used to show that Th(Mw) is decidable, where .Mw is the collection of well-orderings. It follows from the proof of Proposition 7.8 and Corollary 13.12 that if a statement is true in some well-ordering, then it is also true in a well-ordering
3.
DECIDABILITY OF FIRST-ORDER THEORIES OF LINEAR ORDERINGS
271
has a finite axiomatization, there IS, for each a < ow,an algorithm for determining whether or not a 1 d. Moreover, the algorithms are uniform in a since for each r we can construct effectively the axiomatization @(a) from the Cantor Normal Form of a and then apply the wait-for-a-proof algorithm to @(a).However, checking to see whether a k 4 for all a < ow requires infinitely many applications of this algorithm; thus, although we may arrive at the conclusion 4 4 Th(.&"R,) because c( 4 for some a, we could never, by using this algorithm, arrive at the conclusion 4 E Th(AfW). But the Ehrenfeucht analysis comes to the rescue. Recall that every ordinal a is G,,-equivalent to some ordinal in the finite set
w,=
(W,.Un
+ o"-l.a,-,+ . . ' + u 2' a , + W ' f t , + a&
where each a , < 2," and a, I 1 (Exercise 6.20.8). Thus if qd(4) <2n, then Since W, is a finite set of ordinals, and furthermore since one can, given n, effectively write down all the elements a E W,,one can apply the wait-for-a-proof algorithm to @(a) for each r E W,, concluding that 4 E Th(&",) if and only if @(or) t (p for all a E W,.
4 E Th(.Aw)if and only if @(a)t- 6 for all a E W,.
THEOREM 13.48:
Th(AW)is decidable.
We now show that T h ( 9 ) is decidable, where 9 is the collection of all linear orderings. This was first proved by Ehrenfeucht [ 5 ] ; the proof we give is patterned on that of Liiuchli and Leonard [8]. (They note that Calvin found a similar proof.) By virtue of a wait-for-a-proof algorithm applied to 4L(the axiomatization of T h ( 9 ) discussed after Proposition 13.15), Th(Y)is recursively enumerable. If li/ $ Th(9), then, by Theorem 13.22, M k i$ for some M E .A?', so that it suffices to show that { ( a , 4 ) ICI is an order type in and M k 4) is recursively enumerable. THEOREM 13.49: (Liiuchli and Leonard) { ( a , d ) l a is an order type irz ..dand a 1 41 is recursively enumerable.
a theory T , in Proof: We will first define for each order type a in an expanded language L,, and show that T, is complete and logically implies Th(a),the first-order theory of a. (For technical reasons that will be apparent later, we will also define a subtheory S, G TN.)The definition of T , will be given by induction on a. Recall that when we define a theory T as a set Q, of statements, we mean that T is the set of all logical consequences of @.
272
13. THE FIRST-ORDER
THEORY OF LINEAR ORDERINGS
For a = 1, Lais just L, and T , = S, is just (Vx)(Vy)(x = y ) A 4 L .Clearly, T, is complete and logically implies the first-order theory of a. If a = a1 + a,, then La has two unary predicate symbols P , and P,, S, consists of(Vx)(P,(x)v P2(x)),(Vx)(Vy)(P,(x)A P,(y) + x < y), (3x)P,(x),and (3y)P,( y), in addition to the linear ordering axiom 4L,and Thus T, says that a model A can be partitioned into A , + A , with A l = a , and A , = a,; since any two models of T , are clearly G-equivalent, they are also logically equivalent, so that T , is complete, as a set of statements of La. Since ( a , a l , a z ) k T,, T, logically implies the first-order theory Th(a) of a. If a = /3. o,then La has a binary relation symbol and S , contains statements, in addition to 4L, that say that is an equivalence relation that partitions a model A into intervals, and that the set of intervals has order type = UJ. Also T, = S , LJ {(Vu)$ecw~")I/3k $}, where O(w,u)is u w. Thus T , says that a model A can be partitioned into o C . y intervals, for some order type y, and that each interval is E 8; since any two models of T, are G-equivalent, they are also logically equivalent, so that T , is complete, as a set of statements of La. Since (a, -) I= T,, T, logically implies the first-order theory Th(a)of a. A similar argument works in case a = /3 . a*. If a = a(F), where F = {a1,a,, . . . ,a,}, then La contains a binary relation symbol and unary relation symbols P,, P,, . . . ,P, and S , contains is an equivalence relation statements, in addition to 4L,that say that that partitions a model A into intervals, that P,, P,, . . . ,P, also partition A , that each --equivalence class is completely contained in some P i , and that between any two elements of A in different --equivalence classes there are elements belonging to each of P , , P , , . . . ,P, . Also
-
-
-
+
-
-
T, = s, u
-
u
1s i s n
((Vu)[P,(u)3
$@(W,")
1l.i'
$}3
where O(w,u) is u w. Thus T, says that a model A is obtained from the shufflea(F)by replacing each occurrence of an xi by a linear ordering logically equivalent to it; since any two models of T, are G-equivalent, they are logically equivalent, so that T, is complete, as a set of statements of L,, and, as in the other cases, T , logically implies the first-order theory Th(a) of a. We can now argue, by induction on a, that Th(a) is decidable for each aE& Thus, if. for example, if a = /3 w, then, assuming that Th(P) is decidable, we conclude that T, is a recursive and complete set of statements, so that a wait-for-a-proofalgorithm applied to T , gives a decision procedure for Th(a).The other cases are similar. This analysis is not sufficient for our purposes since it is not sufficiently uniform to draw the conclusion that ( ( a , $ ) ] $ E T,} is recursive, from
4.
MODEL THEORY AND LINEAR ORDERINGS
273
which it would follow that ( ( a ,$) I$ E Th(a)) is recursive. We turn now to showing that { ( a , $) 1 $ E T,) is recursively enumerable, which implies that it is recursive, since each T , is complete. , . . . , ( f i k , t j k ) a proof We call a finite sequence ( f i , , $ ~ ~ ) (fiz,tj2), sequence if each pi E A%, each $iE L p , , and for each i, 1 5 i I k, either
$iis a logical axiom or
$,. follows from earlier formulas
or or
$i E
or or
pi = p j . o (or p j . 0")for some,j < i and Ic/i is ( V U ) $ ~ - ~ pi = a(F)and bj E F for somej < i and $iis (Vu)[Pj(u) + $;""I.
So,
pi = ctl (or
$j
by some rule of inference
+ a2 and, for someJ < i, D,i = a1 (or a2)and $iis gl(w)
$pz"@)
Since the set of ordered pairs ( p , $ ) that occur in a proof sequence is recursively enumerable (since [ ( a , $) I $ E S,} is recursive), we need only show that (a, $> occurs in a proof sequence if $ E T,; but this is shown by a simple induction on a. THEOREM 13.50: ( 1 ) Th(M) is decidable for every M E .,A?'. (2) Th(Y) = Th(,&) is decidable. ( 3 ) Th(Y) is decidable, where 9'consists of all scattered linear orderings.
Proof : We need only prove (3), but this is immediate since, by Theorem 13.22, T h ( 9 ) = Th(,Af0)and, by the proof of Theorem 13.49 restricted to scattered linear orderings, Th(,fl[,)is decidable.
$4. MODEL THEORY AND LINEAR ORDERINGS
The model theorist studies the relationship between theories and structures; for example, if T is a theory, what can one say about the collection of models of T? In this section we will develop a number of the basic tools of model theory and use them to analyze the collection of countable models of a complete theory T of linear orderings. (By a complete theory of linear orderings we mean a complete theory that contains the theory of linear orderings; alternatively, T is a complete theory of linear orderings if there is a linear ordering A such that T = Th(A).) In the next section we will deal with the specific question of how many distinct countable models a complete theory of linear orderings can have.
274
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Whenever we speak of a theory T in this section we will assume that T is complete. We begin by presenting two basic tools-elementary substructures and n-t ypes. Although w 5 w 5 in various ways, from an “elementary” point of view one way is best since only if w is mapped onto the initial w of w 4 do the images of the elements of w in w + l, have the same first-order properties as the original elements of w . The map f : w + o ( that takes the ith element a, of o to the ith element bi of w + 4 is said to be an elementary map because, as we will see shortly, any formula satisfied by the ai in o is satisfied by the corresponding hi in w 4. The notion of an elementary substructure and the basic facts we present here are due to Tarski and Vaught [21] and Robinson [13].
+
+
+
+
DEFINITION 13.51 : Let A, B I= T and let J ’ : A-+ B. We say that f is an elementary map from A to B if for every a,, . . . ,a, E A and every formula 4 ( . X 1 . . ’ . ,X,L
A b 4[a,, . . . ,a,]
if and only if
B 1 $ [ f ( a , ) , . . . ,f(a,)].
An elementary map from A to B is also called an elementary embedding since it must be a 1-1 map. If there is an elementary map from A to B then we say that A is elementarily embeddable in B, and we write A < B. If A c B then we call A a substructure of B, denoted A E B if whenever &v,, u 2 , . . . ,0,) is an atomic formula and a l , a 2 , .. . ,a, E A, A k 4 [ a , , a 2 , . . . ,a,] if and only if B I= 4[ul, a 2 , . . . ,a,]; if A and B are linear orderings, this means that A is a subordering of B. If A G B, then we call A an elementary substructure of B if the identity map from A to B is an elementary map from A to B; this is also denoted A < B. Although using A < B to mean two different things can lead to confusion-for example, if A s B and A is elementarily embeddable in B by some map other than the identity-it will not, since in any context it will be clear which use is intended. Another possible source of confusion is our using here the same symbol, for elementary embeddings of structures, that we used in earlier chapters, for embeddings of linear orderings. To remove that confusion, we will reserve < in this chapter for elementary embeddings and use Ionly for ordinary embeddings. Note that the assumption that A and B are models of the same complete theory T is redundant since the condition in the definition with n = 0 implies that A = B. Extending this observation, we see that f is an elementary map
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MODEL THEORY AND LINEAR ORDERINGS
275
from A to B if and only if for every ( i , , . . . , a , E A
(‘%a,, . . ,a,J ’
and that if A c B, then A
= ( B , f ( a , ) ,. . . , f ( a , ) )
< B if and only if
( A ,a , , . . . , a , ) = (B,a , , . . . ,a,) for every a l , . . . ,a, E A . This observation, together with Theorem 13.11, provides the following test for A < B. THEOREM 13.52: (1) f : A if and only if
G,((A, a1,
’ ’
+
B is an elementary map from A to B
. a,), ( B ,f ( a , ) , . . . f ( a , , ) ) E) 11 3
7
.for every m and every a l , . . . ,a, E A . (2) I f A c B, then A < B if cind only
if
G , ( ( A , a l , . . . .u,J,(B,a,,.. . ,a,,))€II .for ever!] m and every a,, . . . ,a,, E A .
+
+
Thus, for example, i f f : o + cu 6 maps o onto the initial o of o 5, then, as is easily verified, f is an elementary map. We express this fact by writing o < o i.(This use of the notation presupposes that w is the o of o + 5 ; more generally, if we write A < B, where there is some ambiguity as to how A is seen as a subordering of B, then we assume that A is contained in B in the most natural way possible.) On the other hand, if y : w -+ o iis any other order preserving map, then, as the reader should verify, for some choice of rn and of a , , . . . ,u, E o,PLAYER I has a winning strategy in G , ( h a , . . . ,a,,),(0+ i,d a d , . . . ,da,,))). The following exercises likewise use Theorem 13.52.
+
+
EXERCISE 13.53: (1) Show that if A and B are dense linear orderings without endpoints, then any order-preserving f : A -+ B is an elementary em bedding. (2) Show that if A , < B , , . . . , A , < B,, then A , . . . + A , i B,
. . . +B,.
+
+
+
(3) Show that there is a unique map from o a* to (a+ w * ) . 2 that is an elementary embedding. (4) Show that if A < B and B < C , then A 4 C.
If one has an explicit description of A and B, then the game analysis is a useful tool for determining whether or not A < B, but otherwise it is inadequate. The following criterion will be useful in more abstract situations.
276
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THEOREM 13.54: Let A C_ B and assume that for euery $(x,, . . . ,x,,y j and every a , , . . . ,a, E A, if B k $[al,. . . ,an,b] for some ~ E B then , B k $ [ a , , . . . ,a,,a] for some a E A . Then A < B. Proof : Suppose otherwise and choose a formula 4(xl, . . . ,xn)that has minimum quantifier depth and such that there are elements a,, . . . ,a, E A for which B k $[al, . . . ,a,] and A F $[al, . . . ,a,]. Although 4 is in general a Boolean combination of quantifier-free formulas and formulas beginning with a quantifier, it is apparent that if there is such a 4, then there must be one that begins with a quantifier ( Q y ) since A G B implies that such a $ cannot be quantifier-free. If $(xl, . . . ,x,) is (Vy)$(x,, . . . ,x,, y), then for some a E A we have A Y $ [ a , , . . . ,u,, a] although B k $[al, . . . ,a,, a ] , contradicting the minimality condition since qd($) < qd(+). If 4(x1,. . . ,x,) is (3y)$(x,,. . . ,xnryj, so that B k $[a,, . . . ,a,, b] for some b E B, then by assumption B k $[a,, . . . .a,,a] for some a E A so that, since qd($) < qd(+), A k $ [ a , , . . . ,anra], which implies that A k + [ a , , . . . ,a,], contradicting the condition on 4. Thus no such can exist and hence A < B.
+
Note that the converse of this theorem is also true. For if A < B, a , . . . . ,a, E A, and B k $ [ a , , . . . ,a,, b] then B k (3y)+(a,, . . . ,a,, y ) , so that, since A < B, A k ( 3 y ) 4 ( a l , .. . , a,, y ) ; hence A k 4 [ a , , . . . , a n ra] for some a E A, which implies, since A < B, that B k +[al, . . . ,a,,a]. A simple application of Theorem 13.54 is the following lemma, which will be used frequently. LEMMA 13.55:
If A
G
B G C and A
< C and B 4 C , then A < B.
Proof : Assume that B k 4[a,, . . . ,a,,b], where b E B and a,, . . . ,a, E A . Since B < C, we have C k $ [ a , , . . . ,a,,b], and since A < C , we have C k 4 [ a , , . . . ,a,,a] for some a E A , which, since B < C, implies that B k $[al,. . . ,a,,a].
The method of complete diagrams, discussed below, can be used to construct an elementary extension of a given model A. DEFINITION 13.56: Let L be a given language and let A be an interpretation of L . By L ( A )we mean the language obtained from L by adding a new constant symbol g for each element a E A . By C(A) we mean the set of allstatements+(a,, . . . ,g,)ofL(A)forwhichA k @ [ a , , . . . ,a,].C(A)iscalled the complete diagram of A.
4.
277
MODEL THEORY AND LINEAR ORDERINGS
For B to be a model of C(A) means that within B we can identify elements f ( a )corresponding to the elements a E A such that B k + [ f ( a , ) ,. . . ,f(a,)] whenever A k +[a,, . . . ,an].Thus if we identify A with the substructure of B consisting of { f ( a ) 1 a E A } , we can regard A as an elementary substructure of B. We express this concisely, with some abuse of notation, in the following way. THEOREM 13.57:
A
< B if
and only if B b C(A).
The method of complete diagrams can be used to show that any infinite A has a proper elementary extension. For if we add one new constant symbol b_ to L(A) and let C = C(A) u { b _ # g l a E A } , then, by the Compactness Theorem, C has a model B since any finite subset of C has A as a model (where to b_ corresponds any element of A not mentioned in that subset of C-using the assumption that A is infinite). Then A < B since B k C(A) and A is properly contained in B since the element of B corresponding to b_ is not in (the image of) A . In the context of linear orderings, we can come to a more striking conclusion by combining the method of complete diagrams with Rubin's techniques [17]. THEOREM 13.58: Let A be a countable linear ordering with no last element. Then there is a non-empty linear ordering B such that A < A B.
+
Proof : Let C = C ( A )u { b_> glu E A } , where b_ is a new constant symbol. Since A has no last element, C has a model D by the Compactness Theorem and A < D by Theorem 13.57. We will assume, without loss of generality, that A is actually contained in D. Let D' = (d E Dld < a for some a E A } , so that D = D' B, where B is non-empty. To show that A < A + B, it suffices, by Lemma 13.55, to show that A B i D' B, and for this it suffices to show that A < D' by Exercise 13.53.2. Again using Lemma 13.55, to show that A < D', it suffices to show that D' < D' B, and that is what we now prove. Assume that D' B k + [ d , , . . . ,d,, b], where d,, . . . ,d, E D' and b E B. Choose a E A greater than d,, . . . ,d,. By Theorem 13.41, there is a selecting formula +#(x,u) for in the interval [a, +) of D' + B ; that is, for every d E D' B, we have D' B b +"[a,d] if and only if
+
+
+ +
+
+
+
+
d2a
and
D'
+ B b +[d,,
. . . ,d,,d].
278
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
+
+ +
Thus since D' B t= $ [ d , , . . . , d , , b ] and b 2 a, we have D' B k $#[a,b]. But since A < D' Band a E A, there is an a' E A such that D' B k $ " [ a , a']. This implies that we have found an element a' E D' such that
+
D
Hence D'
+ B b $[d,,
. . . ,d,,a'].
< D' + B. Thus we have shown that A
i A
+ B.
Another important application of the method of complete diagrams is the following general result. THEOREM 13.59: Let A 7hen each A, < A.
=
U{A,In < o}, where A,
< A,
i A,
< ' .. .
Proof: Let C = C(A,) u C(A,) u C(A,) u . . . . Then Z has a model, by the Compactness Theorem, since any finite subset of C is completely contained within some C(A,) and hence has that A, as a model. Let B k C. Then, by Theorem 13.57,A, < B for each k . Furthermore, since A = {An[n < w}, A !E B. By Lemma 13.55, in order to show that A, i A for each k, it suffices to show that A < B. Assume that B k $ [ a , , . . . ,a,, b], where a , , . . . ,u, E A and b E B. Choose m so that a,, . . . ,a,€ A , . Since A, i B, there is an a E A , such that B k $[al, . . . ,a,, a] ; thus we have found an a E A such that B k $[a,, . . . ,a,,a], so that we can conclude that A < B and hence that A, < A for each k. 4
u
In particular, if each Ai is a model of the theory T, then A is also a model of T ; this enables us to build new models of T from old ones. If the chain {A,ln < o}is not an elementary chain, then u{A,,In < o} need not be a model of T even if each Ai is a model of T . For example, if A , = (o+ o*) .n for each n (it being understood that if n < rn, then A, consists of the first n copies of o + o* in A,), then u(A,ln
a1though A,
+ o * ) . n = o + 6 . ( n - 1 ) + o*= o + w*
= (o
for each n. Thus if we wish to put together models of a theory T and get new models of T, it helps if suborderings are elementary suborderings. DEFINITION 13.60: Given a complete theory T, we say that a model A of T is a prime modelof T if A < B for every model B of T.
4.
279
MODEL THEORY AND LINEAR ORDERINGS
+ + + +
For example, w is a prime model of Th(w); for given any model o 5 . ct of Th(o), the map f :w + w 5 . u that maps o onto the initial w of w 5. ct is an elementary map, as is clear from a game analysis. Similarly, w w* is a prime model of its theory; for given any model w 5 . ct + o*ofTh(w a*), the map , f : o w* + o + 5 . (x + to* that maps w onto the initial o,and o* onto the terminal w*, of o 5 . (x + o* is an elementary map. If T is an arbitrary complete theory, then T need not have a prime model (see, for example, [2]), but what if T is a complete theory of linear orderings? We will return to this question later.
+
+
+
+
+
+
EXERCISE 13.61 : Let o 5 . CI and w 5 . p be models of Th(w). Show that w + [ . ct < w [ . p ifand only if ct 5 p.
+
We now introduce the notion of n-types, first discussed systematically in Vaught [22], who also provides many historical remarks on the material treated in this section. DEFINITION 13.62: Let A be a structure and let ( a l , . . . ,u,) be an ntuple of elements of A . By the n-type of ( a l , . . . ,an) we mean the set of all formulas 4 ( v l , . . . ,v,) of L such that A I= 4[ul, . . . ,a,]. We note that the n-type P of (a [ , . . . ,a,) is a consistent subset of F,(L), the set of all formulas whose free variables are among v l , . . . ,u,, and, moreover, that it is maximal among consistent subsets of F,(L) since adding any formula 4 to P would result in an inconsistency because, if 4 is not already in P, then (14) E P. (As in Chapter 12, a set Q, of formulas is consistent if no contradiction Ic/ A i ) I is deducible from Q,.) We now turn the definition around, with this observation as guide. DEFINITION 13.63: Let T be a complete theory. An n-type of T is a maximal consistent subset of F,,(L) that includes T. This definition is justified by the theorem below. THEOREM 13.64: I f P is an n-type of T then there is a countable model A of T and an n-tuple (al, . . . ,a,,> of elements of A whose n-type is P.
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Since C is consistent, it has a countable model A in which there are elements a, . . . . ,a, such that (a,, . . . ,a,) has n-type P. Note that A b T since if 4 is a statement in T, then 4 E P , so that 4 E X. H DEFINITION 13.65: Given a complete theory T, an n-type P of T , and a model A of T, we say that P is realized in A if there is an n-tuple of elements of A whose n-type is P. Otherwise we say that P is omitted in A. For example, in o each element has a different 1-type since the nth element, and only the nth element, satisfies a formula 4,(v1) that says that u , has exactly n predecessors. No element of o [ after the initial o has any of these types since any such element satisfies i$,,(vl) for every n. Do all such elements have the same 1-type? Does Th(o) have any other 1-types'? These kinds of questions are naturally discussed by connecting n-types with notions discussed earlier.
+
PROPOSITION 13.66: (1) Suppose that f is an automorphism of A such thatf(a,) = bi for 1 I iI n. Then (a,, . . . ,a,) and (b,, . . . ,b,) have the same n-type. (2) Let ( a l , . . . ,a,) be an n-tuple of elements of A and ( b l , . . . ,b,) be an n-tuple of elements of B. Then the following are equivalent: (a) (al, . . . ,a,) and (b,, . . . ,h,) have the same n-type; (b) (A, ~ 1 , ... , a,) (B, b1,. . . , b,); (c) PLAYER II has the winning strategy in G,( (A, a,, . . . ,a,), (B, b,, . . . ,b,)) j o r every m. A EXERCISE 13.67 : (1) Prove Proposition 13.66. (2) Show that all elements of o 5 after the initial o have the same 1-type. (3) Show that Th(o)has no other 1-types. (4) Show that if P' is a consistent subset of F,(L), then P' s P for some n-type P of T.
+
Given a theory T , to determine all n-types of T we need only look at all n-tuples of elements in all (countable) models of T, not a small task. The following definition will simplify matters. DEFINITION 13.68: Let P be an n-type of T and let P' G P. We say that P' generates P if for each o! E P there are formulas o!,, . . . ,clk of P' such that
T k ( V v 1 ) . . . (b'o,,)[Crl(vl, . . . ,v,)
A
'
. . A a,(v,, . . . ,v,)
+ o!(vl.
. . . ,v,)].
4. MODEL THEORY AND
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28 1
Equivalently, P' generates P if whenever an n-tuple of elements of a model of T satisfies just the formulas in P', then its n-type is automatically P. If P' generates an n-type, then we may as well call P' an n-type. Thus the l-types ofTh(o) are { {4,(uo)} In < a}and ( i 4 , ( v o ) I n < w } .
EXERCISE 13.69: (1) Assume that P' G F,(L), that A 1 T , and A b $[al, . . . ,a,] for every $ E I".Assume also that whenever B b $[b,, . . . ,b,] for every $ E P', ( b l , . . . ,b,) has the same n-type as ( u l , . . . ,uJ. Show that P' generates an n-type of T . (2) Assume that P' G F,(L), A 1 T , and for some a l , . . . ,a,E A, A 1 $ [ a , , . . . ,a,]for every $ E P'. Assume also that for any two n-tuples of.elements of A that satisfy P' there is an automorphism of A mapping one n-tuple to the other. Show that if either P' is finite or T has, up to isomorphism, a unique countable model, then P' generates an n-type of T . (3) Show 'that if T = Th(q). then the n-type of ( u l , . . . , a,) is { u , < U J U , < u,} u { U L = V j l U * = L l J . We determined above all l-types of Th(w). Let us now describe the 2-types of Th(o). For each n and m, 6,,(ul) A $ m ( ~ 2 )is a 2-type P,,m;for each n, (4,(u1)} u { i 4 m ( u z ) I m< w', isa2-type P,,,;foreachm, { i 4 , , ( u l ) l n < w } u {$ m ( u 2 ) } is a 2-type Pm,m.Every other 2-type contains ( i 4 , ( u l ) I n < w } u {i$m(~lz)I< m w } = P , , , but there are many distinct 2-types containing P,, . Thus if E,(x,y)says that x < y and there are exactly t elements such that x < t I y, then for each t 2 1, P,+ = P,, u { Et ( u 1 , u 2 ) } and P I - = P,, ~ {E , ( u , , t~ ~ ) } ar e2 - ty p es . A ls=oP~ P,, u ( u l = u2)isa2-type. Finally,
p,,, u Ch f %> u ( l E , ( u l , u 2 ) 1 t < 0)u { l ~ l ( U z , u , ) ( < t 0) is also a 2-type P*,%.That these are in fact all 2-types is easily verified by a game analysis and Proposition 13.66.2. The set S,( T )of n-types of a complete theory T has a topological structure on it that sheds light on the relationship between n-types. A basic open set in the topological space S,(T) is of the form ( P E S,(T)I 4(u1, . . . ,on)E P } for a particular formula 4(ul, . . . ,on)E F,(T). Using this definition, what we now present as a definition is actually a theorem. DEFINITION 13.70: An n-type P of T is said to be principal or isolated, if P is generated by a single formula. Otherwise it is a non-principal or limit n-type.
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Equivalently, P is a limit n-type if every formula in P also belongs to some other n-type; that is, every basic open set containing P contains other n-types. Note that if P is a principal n-type, say P = {4(v1,. . . ,u,)}, then P must be realized in every model of T for, since T is complete, T I- (3v1) . . . (3vn)4(v,,. . . ,v,).
Thus, if a model of T omits an n-type P , then P must be non-principal. The converse of this is also true; that is, if P is a non-principal type of T , then there is a countable model of T in which P is omitted. This is implied by the following theorem due to Ehrenfeucht (see Vaught [22]).
THEOREM 13.71 : Ler { P j l j < 01) be a countable set of non-principal types of' a complete the0r.y T . Then there is a countable model A of T in whic3z every P , is omitted. Proof : Add to the language of T a countable set {g,ln < w } of new constant symbols; the elements of A will be precisely this set. Enumerate all ordered pairs (a,P), where a is an ordered n-tuple of the new constant symbols, P is one of { P j l j < w } , and P is an n-type. Since there are countably many such ordered pairs, we may assume that they are enumerated as an w-sequence {(ak,Pk>I k < w } . We now define by induction an expanding sequence { T k ( k< o)of theories, each involving a finite number of additional statements and a finite number of the constant symbols (@,In < w } . We define To to be T and we assume, as the induction hypothesis, that the constants appearing in Tk are g i , 1 , , g i ( 2.) ., . We also assume that all formulas of the expanded language that have the one free variable x are enumerated in an w-sequence ( + k ( X ) / k < w } . Now given Tk we proceed as follows to define T k + l .We first add to Tk a statement of form (3x)&(x)+ &(a),where g is a constant symbol appearing neither in T , nor in 4k.X). (Such a statement says that if anything satisfies &, then some & does; this guarantees that A, consisting of {fzk I k < w } , will be a model of T . This method of proof, used in proving the Completeness Theorem, is discussed in detail in Enderton (see [31, Chapter 12). Note that this statement can be added consistently to Tk since otherWise k 1 [ ( 3 X ) b k ( X ) + 4 k ( g ) ] , so that Tk k ( 3 X ) 4 k ( X ) and Tk k 14k(g); since a appears neither in Tk nor in &(X), we can pass from T , t i4,(g) to Tk t ( v x ) i d k ( x ) , contradicting the consistency of T,. We next find a formula $ ( x l , x 2 , . . . ,x,) in Pk (assuming Pk is an n-type) such that i $ ( a k ) is consistent with Tk u { ( h ) + k ( X ) -,&(a)}. If this task is impossible, then Tk
{(3x)4k(x)
$k(!d}
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283
for all $ in Pk. Since T , has only a finite number of statements not already in T, we can write this as
T u O(ak,b) t $(ak), where b are the constants appearing in Tk u ((h)&(x) + &(a)) that are different from those in ak.This implies that
T t ()(ak,b) + $(ak), which in turn implies that
T
( 3 Y ) W k , Y)
+
$(ak)
for all $ E Pk. But this says that Pk is a principal n-type generated by
(3y)0(x 1 9 x2 . . . > x k , y), 3
contrary to the hypothesis. We now define Tk+1 to be Tk U { ( j X ) d k ( X ) + 4k(a)}U { 1$(ak),>.By putting i $(ak) into T k +1, we guarantee that the n-type of ak will not be Pk. If, as advertised, A consists of just the constant symbols {@,In < a), then no n-tuple of elements of A will have n-type Pk,so that this Pk,and all the others, will be omitted in A. Now let T* = Tklk < o) and let T** be an arbitrary complete extension of T*. We now define a structure A whose elements are the constant symbols {a,\n < w >; more precisely, we define an equivalence relation by g, -v g , if g, = g , is in T** and WG let A be the set of equivalence classes. For each relation symbol P ( x , , . . . ,xn) in the language of T , we define an n-ary relation R on A by stipulating that R(giil),. . . ,a,(,,)will hold in A if and only if P(giil),. . . , g i c E tis) in T**. It is then easily verified, by induction on formulas, that for every formula $(xl, . . . ,x,) and any elements uiil),. . . ,ai(,)of A , A 1 4[gi(l),. . . ,gi(,)] if and only if ,)&’q . . . ,gi(,)) is in T**. In particular, A 1 T and each of the types {PjIj < o}is omitted in A.
u{
It follows from Theorem 13.71 that if A is a prime model of T, then the n-type of every n-tuple of elements of A must be principal; for if a nonprincipal n-type were realized in .4 and omitted in B, then A certainly could not be embedded elementarily in B since the n-type of an n-tuple of elements of A is the same in B as it is in A if A < B. At this point, it is appropriate to explain where the terms “principal” and “non-principal” come from. Just as the terms “isolated’ and “limit” arise naturally from the topological structure of S,( T ) ,the terms “principal” and “non-principal” arise from the Boolean algebra structure of F,( T ) .That is, if we identify two formulas that are equivalent in the theory T, then
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F,( T ) becomes a Boolean algebra, denoted B,(T), and an n-type becomes a maximal (dual) ideal in this Boolean algebra. An isolated type, one that is generated by a formula, is a principal maximal (dual) ideal, and a limit type is a non-principal maximal (dual) ideal in this Boolean algebra. A fundamental fact about Boolean algebras is that every infinite Boolean algebra has a non-principal maximal (dual) ideal. Thus, for example, the Boolean algebra consisting of all finite subsets of N and all complements of finite subsets of N has infinitely many principal maximal (dual) ideals, namely, I , = { X G N I n E X} for each n E N , and exactly one non-principal maximal (dual) ideal, namely, the set of complements of finite sets. In the context of model theory, this fundamental fact translates into the statement that if there are infinitely many inequivalent formulas in F,( T ) then there is a non-principal n-type of T. To prove this statement, suppose that there are infinitely many inequivalent formulas in F,(T). It is then easily verified that there are infinitely many n-types. If all of these are principal, then there are countably many of them; call them {P,lm < o}.where each P , is generated by I),(vl, v2, . . . ,v,). Since { i I),(vl, v2, . . . ,v,) I m < o>is consistent, it is contained in an n-type P of T that is different from each P,, contrary to assumption. Thus some n-type of T must be non-principal. The following proposition is now easily verified. PROPOSITION 13.72:
The following are equivalent:
(1) There is an infinite set of formulas of F,(T) no two of which are equivalent in T. (2) B,(T) is infinite. (3) S,(T) is infinire. (4) There is a non-principal n-type of T. A Armed with the topological and algebraic interpretations of the set S,( T ) of n-types of T , we return to our consideration of the n-types of Th(o); we continue using the terminology introduced earlier. The space S,(Th(w)) of 1-types of Th(w) can be pictured as a
...
a
Po PI
p2
a
p,
7
where P, is the 1-type {4,,(v1)).and P, = {i4,,(v1)1n< o}is the limit of {Pnln w>. The space S,(Th(wj j is more complicated. However, each P,,.,, is isolated, each Pn,mis a limit of ( P , , , I rn < m}, each P,,, is a limit of { Pn,mn < o}, each P,+ isalimitof{P,,,+,In < o),andeachP,- isalimitof{P,+,,,Irn < o > .
I
4.
MODEL THEORY AND LINEAR ORDERINGS
.
285
P,'
. . . PI
I
pz
I
i l l
. . . .... . . . . . . .
...
....
0
...
........
.
p1.0
Po.0
p1.0
-
0
px,z px,l
pz,o
FIGURE1
Thus we have the picture shown in Fig. 1. Omitted in this picture is P:,, which, unlike the other 2-types of Th(o), is a limit of limit 2-types. (It This kind of analysis is also the limit of the isolated 2-types {Pt,2tlt < o}.) of S,(T), similar to the Cantor-Bendixson analysis of closed subsets of the real line (which Cantor used to show that every uncountable closed subset of R has cardinality c), was introduced by Morley [9] and has been a major tool in model theory. What about S,(Th(o))? The following result shows that, in a certain sense, the n-types of any complete theory T of linear orderings are determined by the 2-types of T. THEOREM 13.73: Let P be an n-type of a complete theory T of linear orderings and let P* be all formulas of P that have at most two free variables. Then P* generates P. Proof : If + ( u l , . . . , ~ , ) E Pthen, , by Theorem 13.37, there are formulas Il/ij such that 4 L
~ ( V I .? .
' 7
vn)
* V A Il/ij, i
j
where each t,hij has at most two free variables. Since 4 E P, there must be an i such that Ajt+hijeP.But {Il/ijl j} s P*, so that P* t 4(vl,.. . ,vJ. Hence P* generates P.
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COROLLARY 13.74: Let T be u complete theory of linear orderings. I f S,( T ) is jinite, then S,( T ) is finite ,for all n. I f S2(T ) is countable, then S,( T ) is countable ,for all n. Proof : Let P be an n-type of T and let P* be as in the proof of Theorem 13.73. For each i, j I n, let P t j be the set of formulas of P* whose free variables are among ui and u j . Then each P t j is a 2-type of T and P* = iPZjI 1 I i Ij I n } . Both of the desired conclusions follow from this.
u
The conclusions of Corollary 13.74 are not true in general. For example, it is shown in Rosenstein [15] that for each n there is a complete theory T , such that S,( T,) is finite but S,( T,) is infinite for all m > n. We asked earlier whether every complete theory T of linear orderings has a prime model. If S 2 ( T )is countable, then by Corollary 13.74, S , ( T ) is countable for all n, so that, by Theorem 13.71, there is a countable model A of T in which no non-principal type is realized. Such a model is prime, as we now show. THEOREM 13.75: Let A be a countable model of a complete theory T in which no non-principal type is realized. Then A is a prime model of T . Proof: Let B k T and let A = { a l r u 2 , a 3 ,.. .) be an enumeration of the elements of A . Since for each n the n-type P, of ( u l , a,, . . . ,a,) is principal, there is a formula +,(ul, u 2 , . . . ,u,) that generates P,. Thus it suffices to find a sequence b , , b 2 ,b 3 , . . . of elements of B such that B 1 + , [ b l , b 2 , . . . ,b,] for each n, for then the map f :A + B defined by f(a,) = b, is an elementary map. Proceeding inductively, we assume that we have defined b,, b,, . . . ,bk SO that B k (f)k[b,,bz,. . . ,hk]. Since A = ! ( b k + l [ U l , U Z , . . . ,U k , U k + I], We have also A 1 (lo,+ I ) & + ,(a,,a,, . . . , a k , ,), so that, since ( b k generates the k-type p k , T 1 + k ( U 1 , . . . u k ) ( 3 t ~ k + l ) ( b k + l ( ~ 1., . . 7 uk? u k + l ) . But then, since B k + k [ b l , .. . ,bk], there is an element b k + lE B such that B b $k+ [ b . . . ,b k ,b k +,I, and so we can continue to embed A elementarily into B. W tik+
9
+
We return now to our earlier question whether every complete theory T of linear orderings has a prime model. It follows from Theorem 13.75 that that is the case if S 2 ( T )is countable. Rubin [17] showed that if S , ( T ) is countable, then S 2 ( T )is also countable, so that in this case too T has a prime model. Thus, we consider next an example of a linear ordering A for which S,(Th(A))is uncountable; this example is motivated by the analogy between Dedekind cuts and 1-types.
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EXERCISE 13.76: Let f : Q + N be a 1-1 correspondence between Q and N and let A have order type fxq) I q E Q}. Show that Th(A) has c l-types and that A is a prime model of Th(A).
c(
This example fails since A is in fact a prime model of Th(A). We are, however, on the right track, as noted by G. Cherlin. THEOREM 13.77: prime model.
There is a linear ordering A such that Th(A) has no
Proof: Partition Q into two subsets Q1 and Q 2 each of which is dense in the other. Let f : Q 1 + N be a 1-1 correspondence between Q1 and N and let A have order type c(g(q)I q E QJ , where g(q) = f(q) if q E Ql and g(q) = 1 if q E Q 2 , Now every Dedekind cut X u Y of Q determines a l-type which asserts that z'l has no immediate predecessor or successor, and that u I is greater than all condensation classes of size f(q) for q E X and less than all condensation classes of size f(y) for q E Y. In every countable model of Th(A), infinitely many of these l-types must be realized, but none of these I-types is realized in every countable model. Thus Th(A) can have no prime model.
In Theorem 13.73 we saw that any n-type of a complete theory of linear orderings is determined by its constituent 2-types. Rubin [17] proved that, under certain circumstances, a 2-type is determined by its constituent l-types. The proof of this result, presented here as Theorem 13.80, involves a complicated induction (Lemma 13.79)and the notion of the k-n-type of a k-tuple. (Theorem 13.80 will be used in $5.) Recall that is the set of all formulas in the variables u 1 , u 2 , . . . ,uk whose quantifier depth is at most n. DEFINITION 13.78 : Let b = (6 (, 6, , . . . ,h k ) be a k-tuple of elements of B. By the k-n-type of b we mean the set of all formulas $ ( u l , u 2 , . . . ,u k ) in Lk., such that B k $ [ h , , b 2 , . . . ,hJ. The k-n-type of b is denoted P,(b). A k-n-typeof B is a maximal set of formulas of I ! , ~ , ~that is consistent withTh (B).
If we let P(b) denote the k-type of b, then P(b) = U{P,(b)In < w } . Note also that since any k-n-type of B is a finite set of formulas, the language being finite, any k-n-type of B is the k-n-type of a k-tuple of elements of B. If I is an interval of B, we let P,(I) denote the set of l-n-types of the elements of I . We introduce several notions which will simplify the statement of Lemma 13.79. A sequence Q1, Q 2 ,. . . ,Q,, of l-n-types of B is said to be realized in
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B if there are elements c1 < c2 < . . . < c, in B such that each cihas 1-n-type Qi.We let t(n) be the number of 1-n-types of B for each n and we define two sequences s(n) and u(n) by induction: s(0) = 0, u(0) = 0, s(n + I ) = 2 . s ( n ) .(t(n))s(”) 1, u(n + 1) = s(n + 1) + u(n).The formulation of the lemma is complicated because of the requirements of the induction argument.
+
LEMMA 13.79: Let B be a linear ordering and let b , < b, < ’ . . < b, und b , ’ < b2’ < * . . < b,‘ be two k-tuples of elements of B. Let B , = (t, b , ) and Bo‘ = (+-, bl’),let Bi = ( b i ,hi+,) and B,’ = (bi‘,bi+ 1) for each i, 1 I i I k - 1, and let Bk = (b,, +) and B,‘ = ( h k ‘ , +). Assume that (1) for every j , 1 I j I k, Pu(nl(bi) = PU(,)(b/); (2) for every j , 0 I j I k, P,,,- ,)(B,) = Pu(n-l)(B;)
i
j = 0 and
and either
‘u(nJB0)
= Pu(n)(Bo‘)
j = k und PU(,)(Bk) = P,(,,(Bi)
O < j < k a n d pu(n)((bj,bj+l))= Pu(,,)((bj’,b)+l))
or every sequence of 1-u(n - 1)-types of Pu(n-l , ( B j )of length s(n) is realized in B, and Bj’.
Then P , ( ( b i , b , , . . . , b J ) = Pn((b1’,b2’,. . . ,b,,’)). Proof : We proceed by induction on n to show that this conclusion holds for all k. To show that P,(b) = P,(b), where b = ( b , , b , , . . . ,bk) and b’ = (bl‘,b2’,. . . ,b i ) , it suffices to show, by Theorem 13.11, that G,((B,b), (B, b’))E 11, which is correct if the two members of each pair of corresponding intervals are both empty or both non-empty; this is guaranteed by (2). Assume now that the lemma has been proved for n and for all k. To show that P,+,(b) = Pn+l(b’),it suffices to show that G,,,((B,b),(B,b))~11.To do this, it suffices to show, by Theorem 13.4, that for each b E B there is a b ’ E B (and the reverse) such that G,((B,b, b), (B, b’,b‘))E11. Assuming that b , < b < b, (the other cases are similar), we will be able to use the induction hypotheses to conclude that
Pn((b,,b,b,, . . . bk)) = P,((bl’,b’,bz‘,. . . bk’)) 9
7
[and hence that G,( (B, b, b), (B,b’,b’))E 111 if we pick b so that b,’ < b‘ < b2’, ‘u(n)(’) = ‘u(n)(b’h P u ( n - 1 J ( b l , b ) )= Pu(n- 1)((b1’9bf)),Pu,n- 1)((b,b2))= Pucn-l)((b, b,‘)), either P,,(,,J(b,,b ) ) = PU(,)((hl’, 8 ) ) or every sequence of 1-u(n - 1)-types of P u ( n - l J ( b l , b )of ) length s(n) is realized in ( b l ,b) and (bl’,b), and either Pu(,,)((b, b,)) = PU(,)((b’,b,‘)) or every sequence of 1-u(n - 1)-types of l,((b,b , ) ) of length s(n) is realized in (b,b,) and (b‘.b,’).
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MODEL THEORY AND LINEAR ORDERINGS
There are two main cases, the first if P,(,+ ,)((b,,b,)) = P,(,+ ,)((b1', b,')), and the second if every sequence of l-u(n)-types of Pu('")((bl, b,)) of length s(n 1) is realized in (b,,b,) and ( b , ' ,b,'). In the first case, where Pu(n+l)(
+
Bb
(W(+I(bl,
x)A + Z ( % b,)
A
b , < x < b, A d44),
we also have
B b (3x)(+l(bl',x)A tJ2(x. b2')A b,' < x < b,' so that if b
E
A
$(x)),
B is such that B k t,bl(h1', b') A $2(h',b,')
A
bl' < b' < bZ' A +(b'),
then P u ( n ) ( ( b l ' > b')) = p u ( n ) ( ( b l , b ) ) and pu(n)((b'>b 2 ' ) ) = p u ( n ) ( ( b , b 2 ) ) and P,,(,,,(b') = Pu(n)(b).Note that this implies also that Pu(n-,)((bl, b))= f',('",,((b,', b'))and that Pu(n-,)( (h,h,)) = Pu("l)((b', b,')) because the existence of an element in one of these intervals with a certain l-u(n - 1)-type can be expressed as a formula of quantifier depth u(n - 1) 1 s u(n), so that the corresponding interval also has such an element, since the endpoints of the two intervals have the same 2-u(n)-type. Thus in this case we are done. In the second case, where every sequence of l-u(n)-types of PU(,,)((bl,b2)) of length s(n 1) is realized in ( b , ,b,) and (b,', b,'), there are two main subcases. Suppose first that every sequence of l-u(n - 1)-types of length s(n)that is realized in (h,,6,) is also realized in both (b,, b)and (b,b,). Let Q,, Q,, . . . ,Q, be the concatenation of all sequences of elements of Pucn-1)( (bl, b,)) of length s ( n ) ; since the number of l-u(n - 1)-types realized in (b,,b,) is at most t ( n - l), r I s(n) . (t(n - l))'(").If P IS the l-u(n)-type of b, then the sequence Q,, Q,, . . . , Q r ,P, Q , , Q,, . . . ,Q, can be considered as a sequence of l-u(n)types of length 2r 1 I 2 . s ( n ) .(r(n- 1))"'") 1 I s(n 1). Hence there are elements cl' < c,' < . . . < c,' < h' < d,' < d,' < . . . < d,' of B,' such that both c,' and d,' have l-u(n - 1)-type Q, and b' has l-u(n)-type P. Thus every sequence of l-u(n - 1)-types of Pu(n-,)((b,,b,))of length s(n) is realized in both (bl',b') and (b',b,') as well as in (b,,b) and (b,b,). Thus in this subcase we are done. Otherwise, there is a sequence of l-u(n - 1)-types of length s(n) that is realized in (b,, b,) but not in both (b,, b) and (b,b,). Assume, without loss of generality, that P,, P,, . . . ,Pscn) is realized in (blrb2)and (bl, b) but not in (b,b 2 ) . [Note that since s(n 1) 2 2s(n), the sequence P , , P , , . . . , P,(,,,,P,, P,, . . . ,P,,,, is realized in (b,, b,), so that P I ,P,, . . . ,Ps(n)must be realized in either (b,,b) or in (hb,).] The formula $(D,,U,), which says
+
+
+
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+
290
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
that u1 < u,, the 2-u(n)-type of ( u , , ~ , ) is the same as the 2-u(n)-type of ( b , b,), and P , , P,, . . . , is not realized in the interval ( v l , uz), has quantifier depth max(u(n),s(n) i- u(n - 1)) = u(n), so that, since B k (3v1)i+b(ul,b2) and Pufn+ ,)(b,) = Pufn+l)(b,’),it follows that B k $(b’,b2’) for b 2 ’ ) )= Pu(fl)((b, b,)) and, since P , , P , , . . . ,PSC,, some b’ E B. Then Pu(,,)({b’, is realized in (b,’,b,’), we must have b,’ < b’ < bz‘. To complete the proof, we must show that every sequence Q,, Qz, , . . , Qs(fl) of 1-u(n - 1)-types in P,,(,- lJ(bl,b)) is realized in both (b,,b) and (b,’,b’). But Q1, Q,, . . . , Q s f n ) , P , P I ,P , , . . . ,Pscn)is realized in both (b,, b 2 )and (bl’,bz‘),and PI,P,, . . . ,PsC,, is not realized in either (b,b,) or (b’,bz’).Hence Q,, Q,, . . . ,Qs,,)is realized in both (b,, b) and (bl’, b’). Thus in this subcase also we are done and the proof is completed. The theorem below is an immediate consequence of the lemma. THEOREM 13.80: Let B be a linear ordering and let b , < b, and b,‘ < b,’ be elements of €3. Assume that P f b , ) = P(b,’), P(b,) = P(b,’), P((b,, b,)) = P((bl’,bz’)), and every sequence of 1-types in P((b,,b,)) is realized in both ( b l , b 2 )and (b1’,b2’).Then P((bl,b2)) = P((bl’,bz’)). We now introduce the notion of a saturated model. As we will see, a countable saturated model of T contains all countable models of T as elementary subsystems if S,(T) is countable for all n.
DEFINITION 13.81 : A countable model A of T is said to be saturated if whenever (a,, . . . ,a,) has n-type P and Q is an ( n 1)-type that contains P . there is an element a,, E A such that ( a l , . . . ,a,, a,, has (n + 1)-type Q.
+
,)
We can easily show by induction on n that if A is saturated, then every n-type is realized in A ; hence if T has a countable saturated model A, then S,(T) is countable for all n. We now show that the converse is also true. THEOREM 13.82: I f S,(T) is countable for all n, then T has a countable suturated model. Proof: Add to the language of T a countable set {g,ln < w } of new constant symbols; the elements of the saturated model will be precisely this set. Enumerate all triples (a, P, Q) where a is an ordered n-tuple of the new constant symbols, P i s an n-type of T , and Q is an ( n + 1)-typeof T containing P. (Note that if n = 0, a is empty, P is T, and Q is a 1-type of T.) Since S,(T) is countable for all n, we may assume that the set of such triples is enumerated
4.
29 1
MODEL THEORY AND LINEAR ORDERINGS
as an w-sequence {(ak,Pk, Q k ) Ik < w}. We now define by induction an expanding sequence (T,lk < O ) of complete theories, each involving a finite number of the constant symbols {g,ln < w } . We define To to be T and we assume, as the induction hypothesis, that the constants appearing . . . ,ai(((,)) and that there is a t(k)-type Rk of T such that in T , are gi(l),gi(2), T , = T u {O(gi(l),gi(2), . . . ,gi(r(k)))Ifl E R k ) . If now T , u {4(ak)14E P k } is inconsistent, then we define T k + ,= T,. If, on the other hand, 7,u {4(ak)14E P k } is consistent, then we select a constant symbol a, not occurring in T , u {4(ak)l4 E P k } and we choose T k + to be any completion of T , u {$(ak,gp)I$ E Q k } in the same language. To show that T,+ is consistent, it suffices to show that T , LI { $(ak, +) I $ E Q k ) is consistent. But otherwise T u { 8(b,a,), $(ak,g p ) }is inconsistent for some O(b,a,) E T , and for some $ E Qk,where none of akor g, occur among b. Thus T k B(b, a,) -+ i $(ak,ap), so that T I- (3x)B(x,a,) -+ i ( 3 y ) $ ( a k ,y). Now (3x)O(x,v) E Pk,since otherwise (Vx)iB(x,v) E Pk, so that both (Vx)iO(x,ak) and O(b,ak) are in T , u {q5(ak)l$ E P'}, contradicting its consistency. Hence i(3y)$(v, y ) E P k ;but $(v, y) E Qk and i (3y)$(v, y ) E Qk, which contradicts the consistency of Qk. Hence Tku {$(ak,gp)l$E Q k } and T k + are consistent. We now define a structure S whose elements are the constant symbols {g,I n < a}; more precisely, we define an equivalence relation by g , g , if a, = Q, is in some T,, and we let S be the set of equivalence classes. For each relation symbol P(x,, . . . ,x,J in the language of T , we define an n-ary relation R on S by stipulating that R ( U ~ (. ~. .),ai(,,) , holds in S if and only if P(gi(l),. . . ,gi(,,))E T , for some k. It is then easily verified by induction on formulas that for every formula 4(xl, . . . ,x,) and any elements gi(l),. . . , gi(,) E S, S t= 4[gi(l),. . . ,ai(,,)]if and only if 4(ui(ll, . . . ,gi(,))E Tkfor some k. In particular, S k T, and if . . . ,gic,,) has n-type P and P G Q, then . . . ,gic,,,b) has (n 1)-type Q. there is an element b E S such that Hence S is a countable saturated model of T.
-
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We leave as an exercise the following three fundamental facts about countable saturated models. EXERCISE 13.83: (1) Show that if A is a countable saturated model of T and B is a countable model of T , then B < A. [Hint: Combine the proof of Cantor's Theorem 2.5 with the proof of Theorem 13.75.1 (2) Show that if A and B are both countable saturated models of T, then A z B. (3) Show that if A is a countable saturated model of T and the two n-tuples (al,a2, . . . ,a,) and ( b l , h,, . . . ,bJ of elements of A have the same n-type, then there is an automorphism f of A such that f ( a i ) = bi for all i.
292
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Note that if T has a countable saturated model, then T also has a prime model by Theorems 13.71 and 13.75.The converse is false, since the complete theory T of all true statements of first-order arithmetic has a prime model, namely, the natural numbers, but has no countable saturated model since T has c different 1-types, which the reader is challenged to find. Before illustrating these notions, we observe that if a, < a, < . . .
+
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EXERCISE 13.84: (1) Show that w [ . g is saturated. (2) Show that w + [ . a is weakly saturated if and only if
c1
is infinite.
An interesting theory arises if we add to the language L of linear orderings a countable list {c,(n< w ) of constant symbols and let T be the theory Th(g) together with the statements {c, < g,+ In < w ] ; this theory was first discussed by Vaught [22]. Any model of T consists of a dense linear ordering with an o-sequence (c,lrz < w } of distinguished elements. It is easily verified that, up to isomorphism, T has only 3 countable models. In one model, A , , the sequence of distinguished elements is cofinal in A , . In a second model, A,, the sequence of distinguished elements has a limit in A , . In the third model, A , , the sequence of distinguished elements is neither cofinal in A , nor has a limit in A , . These three models differ in that the elements that are greater than all the distinguished elements have order types 0, 1 g, and g in A , , A , , and A , , respectively; however, any countable model of T is isomorphic to either A , , A , , or A , . The fact that T has three non-isomorphic countable models suggests that T is not complete. Moreover, the experienced game player will see immediately that G , ( A , ,A,) E I, and G , ( A 1 , A 3 ) E I, and G , ( A , , A , ) E I, and conclude by Corollary 13.12 that A A , , and A , are not elementarily equiv-
+
,,
4.
29 3
MODEL THEORY AND LINEAR ORDERINGS
alent. That conclusion, however, would be premature, since Theorem 13.11 and Corollary 13.12 apply only in the case where the language is finite. In fact, as we will see, T is complete and A , = A , = A , , so that the assumption that the language be finite in Theorem 13.11 is necessary. We will shortly prove that T is complete, but granted that fact, let us examine the three countable models of T . We first claim that A , is a prime model. Indeed, if a,, . . . , u , E A , and each ui< c k , then the n-type of ( a , , . . . ,a,) is generated by {u,
ujlui= U j } u {Ui < < U j } u { u , =gjlui= C j } u ( u , < c j l a , < c j and j I k } u {cj< u i l c j < a,},
=
by Exercise 13.69.2 since this set is finite and, given any two n-tuples of A , satisfying these formulas, there is an automorphism of A , mapping one n-tuple to the other. Thus every n-type realized in A , is principal, so that, by Theorem 13.75, A , is a prime model. Now A , is not saturated since no element of A , satisfies all of the formulas {g, < u , In < w } , and A , is not saturated since, although the limit point c of the distinguished elements satisfies {g, < u1 In < w } , there is no element u E A , such that (c, a ) satisfies the formulas {g, < u, < u1 In < (0). Thus A , must be the saturated model. A direct proof that A , is indeed saturated (and that A , is weakly saturated) involves analyzing all n-types of T. In point of fact, the n-types of T are easily described; each n-type is of the form \u,
u I D i < I1jIui < U j } u {Vi = g j l u , = c.> J u {u, < cjlui< C j ) u { C j < vilej < a,},
= UjlUi= U j }
where a,, . . . ,a, are elements of one of the models A iof T. (To prove that these are actually n-types is not so easy, and we will leave this until after we show that T is complete.) Thus, in particular {g, < ul In < w } is a 1-type; thus the limit point of A , cannot be distinguished from any point larger than it. We turn now to a proof that T is complete. The method we will use is called the elimination of quantifiers method, and we will show both that T is complete and that it is decidable. The basic idea of the elimination of quantifiers method is to start with a formula of the form (3x)$(x, u l , . . . ,un), where 4 is quantifier-free and obtain effectively a quantifier-free formula $(u,, . . . ,u,) such that T 1 ( W . .(Vufl)[(W4(x,u 1 , . . . ,%I) '
W I , .
'.
,u,,I.
If one can also show that for each quantifier-free 4(x), either T 1 (3x)4(x) or T 1 i(3x)$(x), then T is complete; if one can also determine effectively
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
whether T t (3x)&(x) or T t i ( 3 x ) + ( x ) , then T is also decidable. For suppose that we are given a statement (D; we first find effectively a logically equivalent statement in prenex normal f i r m (Qiul)(Q2~2)
' '
. (Qk-
1%
i)(QkUk)'#)(Ui,
u25
...
9
u k - 1,
%h
where each Q is either V or 3. If k > 1 and Qk is 3, we apply the elimination ofquantifiers method to ( Q k U k ) 4 ( U I , u 2 , . . , ,uk- 1 , uk), obtaining an equivalent ). formula $ ( u l , . . . , I J ~ - ~ Then Tt
@
c1 ( Q 1 ~ 1 ) ( Q 2 ~ 2 )' '
. (Qk-
i u k - i)$(ui,
. . . > uk-
11,
so that Q, is equivalent in T to a statement with one fewer quantifier. If k > 1 and Qk is V, we apply the elimination of quantifiers method to l(Qkuk)4(uir..
.,Uk-i,uk)
obtaining an equivalent formula $ ( u , , . . . ,uk-
Then
Tt- ~ ~ ( Q i ~ i ) ( Q 2 ~ z ) . " ( Q k - i ~ k - ~ ) 1 $ ( ~ 1. ,. ,. U k - i ) r
-
so that in this case also Q, is equivalent in T to a statement with one fewer quantifier. This procedure is repeated k - 1 times until we find that T I- CD ( Q , u l ) O ( u l ) , where 8 is quantifier-free. If Q1is 3, then we use the fact that either T k (3ul)f3(ul) or T t i ( 3 u l ) 8 ( u , ) to conclude that T I- Q, or T t 10; and if Q1 is V, then we use the fact that either TI- (3vl)i8(ul) or T t i ( 3 u 1 ) i f 3 ( u 1 ) to conclude that T t -I@ or T t Q,. In any case we know or T k CD, so , that T is complete. Moreover, since all passages that T t iQ are effective, T is decidable. Thus we have to show that, in the case where T is the theory with three countable models, a quantifier can be eliminated from (3x)4(ul,.. . ,u,, x), where 4 is quantifier-free. We first put 4 in disjunctive normal form V{4;lls i l k ) and since (3x)4 is logically equivalent to v ( ( 3 ~ ) 41<~i(< k ) , we need only find a quantifier-free such that T t $i ( 3 ~ ) 4since ~, 11 5 i Ik ] . Thus we can assume that 4 is a then T t- (3x)4 t ) of atomic formulas of the language and their conjunction A(r$iI 1 I i I negations. Replace each conjunct of the form i ( u = b) by its equivalent (a < b) v (b < a) and each conjunct of the form i ( a < b) by its equivalent a = b v b < a. Having done this, we can put the resulting formula in disjunctive normal form, pass the (3x) through the disjunction, and end up facing formulas like (3x)A{+;ll Ii 5 t } , where each g5i is an atomic formula. We may assume that each 4i contains x free since otherwise that 4l can be moved out of the scope of (3x1. If x = x appears, it can be replaced by u1 = u l . If x = u j appears then ( 3 x ) A {&(ul, . . . ,un,x)I1 5 i It } is logically iI t ) . Similarly if x = cj appears. equivalent to /l{+i(ul, . . . ,u , , u j ) [ l I
- v{$i
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4.
MODEL THEORY AND LINEAR ORDERINGS
295
Thus we may assume that each conjunct is of one of the forms ui
< x,
cj < Y,
x
< ui,
x
If only the first two forms appear, then the formula (3x)$ is equivalent in T to u1 = u , ; if only the last two forms appear, the same conclusion is correct. Otherwise (3x)+ is equivalent in T to the formula A{. < hlboth u < x and x < b appear). Thus the quantifier can be eliminated from (3x)4(rl.. . . , r , , x ) if n 2 1 and 4 is quantifier-free. We now consider statements of the form (3x)4(x),where 4 is quantifierfree. As before, we need only consider the case where $(x) is a conjunction of formulas each of which is an atomic formula containing x. If x = x appears alone, then T t (3x)4(x), and otherwise it can be deleted. If x < x appears, then T t i(3x)4(x).If x = g j appears, then T t (3x)$(x) if and only if T k $ ( g j ) and, since 4(ri) just makes some assertions about the ordering of the distinguished elements, either T t 4 ( g j ) or T t i4(cj). If only formulas of the form x < g j appear or if only formulas of the form g j < x appear, then T t (3x)4(x). If formulas of both forms appear, then (3x)$(x) is equivalent t o < gjlgi< x and x < cj both appear), a statement which is easily decided by 7 .Thus for every statement (3x)4(x),where 4 is quantifier-free, either T t (3x)4(x)or T t i(3x)b(x). We summarize our conclusions about the theory T in the following theorem.
THEOREM 13.85: (Vaught) Lut T be the theory Th(q) together with the statements (c, < g,, ( n < w } . Then T is complete and decidable and has exactly three distinct countuble models. H The elimination of quantifiers method is a useful general tool for proving completeness or decidability. Some additional examples of its use are found in the exercise below. EXERCISE 13.86: (1) Add to the language of T a constant symbol a and, using the method of elimination of quantifiers, show that T u {c, < g l n < w } is complete. Deduce from this that { c, < uln < o} is a 1-type of T. (2) Similarly, prove that the it-types of T a r e all of the form
-
ujlai= U j } u { U i < Z i j I U i < U j } u { U i = cjlai = C j } LJ (q < cjlui < C j ) L! (Lj < U i ( C j< ail.
{Ui=
(3) For each m 2 1 add to the language of the theory T considered above w1 unary predicate letters U , , . . . , U , and let T , be the theory that
296
13.
THE! FIRST-ORDER THEORY OF LINEAR ORDERINGS
contains T and says that U , , . . . , U , partition the model, each is dense in the model, and every c, E U Show that T, is complete and decidable and has exactly m 2 distinct countable models (Vaught [22]). (4) Prove the decidability of Th(o)using the elimination of quantifiers method.
+
$5. THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
Given a complete theory T, how many distinct countable models can it have? For example, by Cantor’s Theorem 2.8, Th(q) has but one countable model. On the other hand, by Corollary 6.12 and Corollary 13.12, Th(w) has c distinct countable models. In this section we will present a result of Rubin [171 that says that, for linear orderings, these are the only two possibilities; that is, if T is a complete theory of linear orderings, then T has either exactly one or exactly c distinct countable models. For arbitrary complete theories T , a complete answer has not yet been obtained as to the number of distinct countable models of T. At the end of $4 we presented Vaught’s examples of theories T , that have m 2 distinct countable models for each m 2 1; on the other hand, Vaught [22] showed that no complete theory can have exactly two distinct countable models. It is also possible for a complete theory to have exactly countably many distinct countable models; for example, if T is just the theory of equality together with the statements { ci # cjl i < j < w } , then a countable model of T is determined by the number of non-distinguished elements that it has. Morley [lo] proved that if a theory has an uncountable number of countable models, then that number must be either K1 or c. If, of course, we assume the continuum hypothesis, then Morley’s result answers the question completely. Without the continuum hypothesis, there is no known example of a complete theory that has exactly K, distinct countable models, so it has been conjectured that there are none. Rubin’s result shows that the conjecture is correct for complete theories of linear orderings. The astute reader will recall that in Chapter 8 we showed that there are exactly K, countable linear orderings that are groupable; but this does not provide a counterexample since there is no set CD of first-order statements such that A is groupable if and only if A k @. For a linear ordering A to be groupable, there must exist a ternary relation on A that meets certain requirements, and the existence of such a relation cannot be expressed in a first-order language. The class of ordered groups is, on the other hand,
+
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
297
an elementary class, and there are exactly c countable ordered groups. If we ignore the group structure of these groups, the c distinct countable ordered groups fade into just K, countable linear orderings. The class of groupable linear orderings is called a pseudo-elementary class and it has K elements. Morley’s theorem applies equally well to pseudo-elementary classes, so it is conceivable, though unlikely, that a similar counterexample will be found for elementary classes. It is interesting that there is also a pseudo-elementary class containing exactly two distinct countable models.
,
THEOREM 13.87: (1) The pseudo-elementary class of linear orderings that are groupable contains exactly K, distinct countable linear orderings. (2) The pseudo-elementary class of linear orderings that are the linear orderings of models of arithmetic contains exactly two countable linear orderings.
Proof : (2) Clearly the axioms T Aof arithmetic logically imply Th(w). Thus if M is a non-standard countable model of TA, then the order type of M is w 5 CI for some order type a # 0. If 2 , is a copy of 5 in M and b E Z , , then b < b b ; if b b E Z , , then h + b = b + n for some natural number n, so that h = n, which is impossible. Hence c1 has no Iast element. Suppose that Z , < Z , are two copies of 5 in M and that a, E 2, and u2 E 2,; we may assume that a, and a, are both even, since from TA it follows that for any a E M either a is even or a 1 is even. Then a , + a, is even and a, < $(a, a2) < a,. If ;(al a2)E Z , , then $(al a,) = a , n for some natural number n, so that a, = a , + 2n, which is impossible. Similarly, +(a, az)$ Z,, so that there must be a copy Z , of 5 in A4 such that Z , < 2, < 2,. Hence c1 is a dense order type. By a similar argument, CI does not have a first element, so that a = 9. Hence every countable non-standard model of arithmetic has order type w + 5 . q. Since such models exist, the proof is complete.
+
+ 1
+
+
+
+
+
+
+
Before presenting Rubin’s analysis, we first discuss those complete theories T of linear orderings that are KO-categorical. DEFINITION 13.88: A theory T is said to be KO-categoricalif it has, up to isomorphism, exactly one countable model.
Engeler [6], Ryll-Nardzewski [18], and Svenonius [191 independently obtained a characterization of those theories that are KO-categorical. Using this characterization, Rosenstein [141 determined which linear orderings have KO-categoricaltheories.
29 8
13. THE FIRST
ORDER THEORY OF LINEAR ORDERINGS
THEOREM 13.89: A theory T is KO-categorical if and only n, B,(T) is finite.
if for every
Proof : If B,( T ) is infinite for some n, then there is a non-principal n-type P by Theorem 13.72. By Theorem 13.64, there is a countable model A of T in which P is realized and, by Theorem 13.71, there is a countable model B of T i n which P is omitted. If A 2 B, then the isomorphism would carry an n-tuple ( a l , . . . ,a,) ofelements of A that realizes P to an n-tuple ( b , , . . . ,h,) of elements of B that does not realize P. This contradicts Proposition 13.66.1, so that if B,(T) is infinite for some n, then T is not No-categorical. Suppose, on the other hand, that B,(T) is finite for each n and that A and B are both countable models of T ; we will show that A = B, so that T is N,-categorical. The proof is basically patterned on that of Cantor's Theorem 2.8 and thus proceeds by enumerating A = {a,ln < m } and B = jb, I n < w } and defining by stages an isomorphism f from A to B, specifically defining f(a,) at the even stage 2n and defining an element U E A such that f ( a ) = b, at the odd stage 2n 1. More concisely, we assume as the induction hypothesis that at the beginning of stage t = 2n we have defined f for a finite number of elements of A including at least a,, a, . . . ,a,and that for each j < n there is an such that is defined and equals b j . Furthermore we assume that if we have so far defined {f(a')l j < k } and that these elements are respectively {bjlj < k } , then the k-type of the k-tuple (a'lj < k ) in A is the same as the k-type of the k-tuple (b'lj < k) in B. We now take an element a of A and find its counterpart in B ; the reverse could be done with equal ease. Let P be the (k 1)-type of the (k + 1)-tuple (uo,a', . . . , a k - ' , a ) in A. Since B k + l ( T is ) finite, we may assume that P is finite and so may take the conjunction Y of the formulas of P . Then A I= Y [ a o , a l ,. . . , a k - ' , u ] ,so that
+
+
. ,ak-',Vk). Since (a', a ' , . . . ,u k - ' ) satisfies ( ~ L ' k ) ~ ( ~v(l ,, , . . . ,uk- 1, u k ) in A, this formula must be satisfied by (bo,h', . . . ,bk- ') in B. Thus there is a b E B such that B k Y [ b o , b l ,. . . , b k - ' , b ] .Thus P is the ( k + 1)-type of the ( k + 1)-tuple (bo,h ' , . . . ,hk-', b ) in B, so if we define f ( u ) = b, the induction hypotheses A I=
(3ti,)Y(U0,U'j..
continue to hold, so that f will be an isomorphism from A to B.
If B,( T )is finite for all n and A and B are models of T that contain k-tuples . . , a k - ' ) and (bo,bl, . . . , b k - ' ) that have the same k-type, then we can begin to define the isomorphism from A to B by mapping each a' to b'. Thus the following strengthening of Theorem 13.89 is analogous to the strengthening of Cantor's Theorem 2.8 by Theorem 2.11. (uo,ul,.
5.
THE NUMBER OF COUNTAHL,E MODELS OF A COMPLETE THEORY
299
COROLLARY 13.90: Let T be K,-categorical, let A and B be countable modeis of T and let (a'li < k ) and (b'li < k) be k-tuples of elements of A and B, respectively, that have the same k-type. Then there is an isomorphism f : A -+ B such that f ( a ' ) = b' for each i < k.
These results, when combined with the analysis of almost doubly transitive linear orderings in 98.6, yield a characterization of KO-categorical linear orderings. THEOREM 13.91: (Rosenstein) Let Al be the smallest class of order types Containing 1, containing z1 z2 whenever it contains 5 1 and z2, and containing the shufle a(F)wheneuer it contains the elements of F. Then a complete theory T of linear orderings is K,-categovical if and only (f it has a model whose order type is in A!,.
+
Proof: If T is KO-categorical and .4 is the countable model of T , then by applying Corollary 13.90 with B = A, we see that A is almost k-tuply tranIf, on the other hand, sitive for any k, so that by Theorem 8.40, 2 E A,. T has a model A whose order type is in A l , then A is almost doubly transitive by Proposition 8.37, and therefore A is almost k-tuply transitive for every k by Proposition 8.35. Using Proposition 13.66.1, it follows that the k-tuples of elements of A have only finitely many different k-types, so that B k ( T )is finite and T is KO-categorical by Theorem 13.89. EXERCISE 13.92: Define a notion of rank for KO-categorical linear orderings as follows. Let ~ t V ~ (=0 I)I } , put z 1 + T~ E A l ( n + 1) whenever T ~ T , ~ .&',(II), E and put a(F)E .&(n + 1) whenever F G &Ml(n). The rank of T E .&, is defined to be the smallest n for which T E A l ( n ) .
(1) Show that A l ( n )is finite for all n. ( 2 ) Show that if T has rank I I and if A is an interval of 0 , then A is KO-categoricaland has rank at most 2n + 1.
T
of order type
We turn now to a discussion of Rubin's theorem, which says that if T is not KO-categorical, then T has c distinct countable models. In the most familiar non-K,-categorical case, we obtain c distinct countable models of Th(w) by taking the interval I of order type [ of the model w [, which satisfies I = I . a for every countable order type a, and replacing I by each of its c different countable counterparts. What Rubin shows is that given any non-KO-categorical T for which S,(T) is countable, there is a model A of T with an interval I satisfying I = I . a for every countable order type a.
+
300
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
We will first concentrate on those linear orderings I such that I = I . a for every order type c( and later show how they can be found in models of non-KO-categoricaltheories. When does a linear ordering I have this property? As we will see, a condition just a bit stronger than I = I I is sufficient to guarantee that I = 1 * u for all CI. We first note that the condition I = 1 I is itself not sufficient, for w + w* = (w w * ) (w w * ) , although (w w * ) . o = w + [ . w = w f w + w * . That I . w = I fails is due to the fact that although the map from w w* to the first w w* in (w w * ) (w w * ) is an embedding, it is not an elementary embedding. Thus the condition we impose on I is that if I 'v f and I 'v 1, (where we assume here and elsewhere that I , is disjoint from I,), then the identity maps of I , and I , into I, I , should be elementary embeddings. As we will see, this condition implies that I = I . M for every order type a. Actually, we will prove a stronger result, one which involves elementary embeddings in the conclusion as well as in the hypothesis.
+ + + +
+
+
+
+ + +
+
+
THEOREM 13.93: Let X be a linear ordering and assume that for each x E X , I , z I . Assume that if I 2 I and I , 'v I , then the identity maps of I, and I , into I , I , are elementary embeddings. Then for any nonempty subordering Y of X , the identity map is an elementary embedding of C ( I x l x E Y } into L{I,.~X E x).
,
+
Proof: We proceed by induction on 1x1,first for X finite and then for infinite X. For 1x1 = 2, what is to be proved is exactly what has been assumed. Assume then that the result is true for 1x1= k and proceed to the case k 1, in which case we may assume that X = { n l n I k } and that x { l x l xE X } = I , I, . . . 1,. Let A = I , I, . . . 1, and let B denote x { l x l x E Y } . Suppose that A k 4 [ b , , b,, . . . ,b,,a], where b,, b 2 , . . . ,b, E B and a E A . We assume that a $ 1,; otherwise we give a similar argument using I , + I , + . . . + l , i n s t e a d o f l , + l , + . ~ ~ + I , - , .We may assumealso that Y n (0, 1,. . . , k - 1 ) # 0; for if Y = ( k } , then 1, < 1,-1 -k 1, by assumption, so that to show that 1, < A it suffices, by Exercise 13.53.4, to show that 1 k - I 1, < A , and this corresponds to the case Y = (k - 1, k}. Assume that b,, b,, . . . ,b, E 1, but b,, b,, . . . ,b,- $ 1,. By Corollary 13.39,there is a testing formula for 4 in I, . . . + lk- (relative to b,, . . . ,bn). That is, we can find 4 * ( u , , . . . , u t ) such that I , + I, + . . . + lk-, k 4*[c1,c2,. . . ,c,- ,,c] if and only if A k 4[c1,c,,. . . ,c,- b,, . . . ,b,,c] for every c,,c,, . . . , c t - l,c E I, I , . . . 1 k - I . Thus
+
+ +
+
+ +
+
,
+
+ +
1,
+ 1 , + + 1,' ' '
+
,
+
k 4*[bl,bz,
. . . ,b,- ,,a].
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
301
By the induction hypothesis, since Y n (0, 1,. . . , k - 1) # @, there exists an element b E B n ( I , 1 , . . . l k - , ) such that I , I, . . . l k - I= 4 * [ b l , b 2 , . . . ,b,- ,,b], so that A I @[b,,b,, . . . ,b,,b], as was to be proved. Assume now that X is countably infinite and let X = (x, I n < u} be an enumeration of X . For each n, let A, = C { l x ( xE { x i l i < n ) ] and let B, = IlllIx E Y n {xili < n}}. By what we have just shown, the following diagram consists of elementary embeddings :
+
+
+ +
,
+
B, 4 B2 4 B , 4 ' . .
h
h
h
A , 4 A2 < A , 4 . . . . We must show that B = U { B , ( n < o}< A = U { A , I n < o}.Assume then that A I=4 [ b , , . . . ,b,,a], where bl, . . . ,b, E B. Then for some n, we have b,, . . . ,b,, a E A,; furthermore, since, by Theorem 13.59, A , < A, we have that A , I= 4 [ b l , . . . ,b,,a]. Since h l , . . . ,b, E B , and B , 4 A,, there is an element b E B, such that A, k 4 [ b , , . . . ,b,,b]. Since A , 4 A , A I= $ [ b , , . . . ,b,,b]. Hence B 4 A. A similar proof works for any infinite set X , using, as the induction hypothesis the assumption that the theorem is correct for sets of all smaller cardinalities. This completes the proof.
Because of this theorem and its corollary, the following definition seems appropriate. DEFINITION 13.94: A linear ordering I is said to be self-additive if whenever I, 2 I and I, = I , the identity maps of I , and I, into I , I , are elementary embeddings.
+
COROLLARY 13.95: type 3.
I f I is sclf-additive, then I
=I
. CI
for every order
Is the converse true? That is, if 1 = I . a for every order type a, must I be self-additive? Perhaps the reader can find the answer to this question. As we now show, the property of being self-additive is more correctly ascribed to the theory Th(1) than to I itself. THEOREM 13.96:
(1) (2) (3) (4)
Let T = Th(I). Then the following are equivalent:
I is self-additive. Every model of T is seif-(additive. I f A b T and B 1 T , thcn A < A B and B 4 A I has no definable intervals other than I and 0.
+
+ B.
- -
302
13. THE FIRST-ORDER
THEORY OF LINEAR ORDERINGS
Proof : Obviously ( 3 )=-(2) (1). We thus need only verify (1)* (4) =-(3). We first show (1) (4). Suppose then that ( u E I1 I k 4 [ u ] )is a non-empty intervalofI.LetI, ~ l a n d l e t l , - l . S i n c e I , < I , + I , a n d l , < I , +12,
+ I,p, + 1 , k 4 [ 4 )= ( u E 1 , p , k 4 [ a ] ) u ( u E z,p2 k 4 [ a ] ) . Since ( u E I I I k b[u])is an interval of I , [ U E 1,
I k ( V . W Y , [ . X < ? ' A 4(x) A 4 ( Y ) (V'Z)(x< < Y -,4(4)1, so that C U E 1, + 12111+ 1, k + [ a ] } must be an interval of 1, + I,. Since it contains points of both I , and I,, it must contain a terminal interval of 1, and an initial interval of 1,. Since I , < I , + I,, { a E I , II, k 4 [ u ] ) is a terminal interval of I,, and, since I , i I , + I,, { u E 12112k + [ a ] ; is an +
initial interval in I , . Thus I, k (VX)[~(X) -, (Vy)(y 2 x -, 4 ( y ) ) ] and I , k (Vx)[&x) + (V'y)(y I x -+ + ( y ) ) ] ,hence both of these statements are true
in 1 and I k ( V X ) ~ ( X ) . Thus (1) * (4). We now show (4)*(3). As in the proof above, the assumption that I has no proper definable intervals implies that the same is true for every model of T. Assume that A b T and B k T ; we show that A 4 A B and B < A B. We do this by constructing a model C of T such that A 4 C, B < C, and A B G C. We let C consist of the complete diagrams C ( A ) and C ( B ) of A and B together with (a < bla E A, b E B).If C is consistent and C b X, then clearly .44 C, B 4 C, and A + B G C. If C is inconsistent, then, by the Compactness Theorem, there are elements b , , b,, . . . ,b, E B, u E A, and a statement q5(b1,b2,. . . ,b,) E C ( B ) such that
+
+
+
C ( A ) k g < b, A G < b
2 ~" r .\ g
< b,+ iq5(bl,b,,. . . ,bJ.
This implies that C ( A ) t- (~Xl)(t/X2)~ . . (VX&
< X,
A-
- . A g < X n + l+(X,,.x,,.
. . ,X,,)j
since b,, . . . , b, do not appear in C(A). On the other hand,
W )1 4(bl, b 2 , . . . ,b,), which implies that A k (3x,) . . . (3x,,)4(xl, . . . ,x,,); but A has no first element (since that would yield a definable interval of A ) , so that {CE
A1A k (jx,). . . ( ~ x , ) ( c< x1 A " . A C < x , A ~ ( x .~. ,. ,x,))
is a non-empty initial interval of A. By assumption it must be all of A, contradicting the conclusion above that a is not in the interval. Thus C is consistent. Let C k C and divide C into C, + C, C,, where C, = ( c E Clc Iu for some a E A ) and C, = {c E Clc 2 b for some b E B } , so that A G C,
+
5.
303
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
and B G C,. Let us for the moment assume that A 4A + C 2 + B and B 4 A C, B and show that A + B < A + C, + B. By Lemma 13.55, this will imply that A 4A + B and B < A + B. Assume, therefore, that A + C’, + B k Cp[d,,d,, . . . ,dk,c], where d,, d,, . . . ,dk E A u B and c E C,. Let a be the largest of the di in A and let b be the smallest of the d , in B ; if there are no d, in A or in B a slight modification is necessary. By Theorem 13.41, there is a selecting formula 4’ for Cp in [a,b], that is, a formula ~ # ( x 1 , x 2 , usuch ) that A + C, + B 1 ~$”[a,b,d] if and only if d E [a,b] and A + C 2 + B 1 4 [ d , , . . . ,d,,d]. It thus suffices to find a c’ E A such that A + C2 t B 1 4#[a,b, c’], given that A + C, + B k 4 ” [ a ,b,c]. Now there is, by Corollary 13.39, a testing formula 4#* for Cp“ in A (relative to b); that is, there is a formula d)#*(x,,o) such that A k ##*[a’,c’] if and only if A + (’, + B k Cp#[a’,b,c’], for all a’, c’ E A. It thus suffices to find an element c‘ E A such that A k 4#*[a, c’]. Suppose, on the contrary, that A 1 i ( 3 z ) $ # * ( a , z); then, since A < A + C, + B, we have
+ +
A
+ C, + B k
1(32)4#*(U,Z)A
Cp#[U,b,C].
Hence Af
c 2
f B k (3x,)(3x,)[x,
< .Y2 5 C
A1(3Z)~#*(Xl,Z)A~#(X,,b,X,)].
The same formula is satisfied by every element of A + C, + B that is larger than c ; we claim that in fact it must be satisfied by all elements ofA + C, B. For if not. then
+
which implies that the same is true in B since B < A is an element b‘ E B such that B
F@X1)(3X2)[X1
+ C, + B. Thus there
< X2 2 I J ’ A1 ( 3 5 ) ~ ’ * ( x , , z ) ~ ~ # ( x , , b , x , ) ] ,
+ +
which implies that the same is correct in A C, B, which is impossible since every element of A + C, + B larger than c satisfies this formula, including b‘. Thus A
+ c, + B k (vU)(3X1)(3X2)[X1
< X2
0 A 1(3Z)4#*(X1,2)
In particular, if e E A, then there is an a ,
E
A $#(XI,
b,x,)].
A and an a2 E A such that
whichisimpossiblesinceifA + C, + B k 4#[a,,b,a,],thenA k Cp#*[a,,a,] by definition of +#*, and if A + C 2 5 k i(3z)Cp”*(al,z), then, since A < A + C, + B, we also have ’4 k -I (3z)4#*(al,z). Hence we must conclude that A k (3z)Cp#*(a,z),so that for some c’ E A, A k ~ # * [ u , c ’ ]as, was
+
304
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
to be shown. Thus, assuming that A < A + C, + B and B < A + C2 + B, we have shown that A + B < A + C, + B. It thus remains to be shown that A < A + C, + B and that B < A + C2 B. Recall that C = C, + C, C, where A c C,, B G C,, A < C , + C 2 C , , a n d B i C , + C, C,.WewillshowthatC,
+
+
+ +
+
+
+
+
,,
+
+ +
+
+
+ + +
c1 + c2 + c3 k 4 [ a , , .
’ ’
, a,,a’],
as was to be shown. The argument for C3 4 C, the proof is therefore complete. I
+ C, + C, is similar, and
EXERCISE 13.97: Show that if I is self-additive and +(x) is a formula, then (a E I ( I 1 + [ a ] ) is either empty or unbounded.
We would now like to show that if T = Th(Z) is self-additive, then T has c distinct countable models I . CI,where cx ranges over all countable order types. Unfortunately this is false; in fact, it is possible that all these models are isomorphic, as happens when I is q or is any other KO-categorical self-additive linear ordering. Even if I is not KO-categorical, it is possible that all these models are isomorphic, although Th(Z) does have c countable models; an example of this occurs when I = o . g. We will see later that these are essentially the only kinds of exceptions, and so we focus our attention on how to show that I . cx and I . p are different when CI # B. We first define a condensation cD on self-additive linear orderings such that any isomorphism carries condensation classes onto condensation classes. DEFINITION 13.98: Given a self-additive linear ordering A and an element a E A we define c,(a) to be the union of all bounded intervals of A containing a which are definable over a. Thus b E c,(a) if and only if there is a formula +(x,y) such that A 1 +[a,b] and { d A1~ A k + [ a , d ] ) is a bounded interval of A .
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
305
Note that, as with other condensations, although cD(a)is a union of bounded intervals, it itself may be unbounded; thus, for example, c,(u) = Z for every a E Z . Note also that if we wish cD to be a condensation, then A cannot have definable intervals. For example, in Z + N, since each element of N is definable, c,(u) = Z N for every a E Z , but cD(a)= N for every a E N . [For the interval from a to the nth element of N is definable by the formula (Vz)($Jz) + x i y I z), where $,,(z) defines the nth element of N . On the other hand if a E N and {d b $ [ a , d ] } were bounded, then it could contain no elements of 2 since given any d,, d, E Z there is an automorphism of Z N that fixes each element of N and carries d , to d,.] To be a condensation, C, must satisfy c,(b) = c,(u) whenever b E c,(a). Hence in order for cD to be a condensation, A must be self-additive; we now show that this condition is also sufficient.
+
+
THEOREM 13.99: l f A is self-udditive, then cD is u condensation of A .
Proof : We must verify that if b E c,(a), then c,(b) = c,(u). Assume that a < b ; the other case is similar. Then there is a formula $(x,y) such that A != $[a,b] and such that {d E A1-4 k $ [ a , d ] } is a bounded interval of A whose minimum is a. We focus our attention first on {c E A I c > b}. Suppose that b < c and c E cD(a).Choose a formula $(x, v ) such that A k $[a,c] and such that { d l A b @ [ u , d ] } is a bounded [nterval of A with minimum a. By Theorem 13.41, there is a selecting formula 4” for $ in {d E Ald 2 b ) relative to a ; that is, there is a formula ~ ” ( c , J ) such that A k 4 ” [ b , d ] if and only if d 2 b and A b 4 [ a , d ] . Then ( d l A b 4”[b,d]} is a bounded interval of A containing both b and c, so that L‘ E c,(b). Suppose on the other hand that b < c and c E c,(b). Choose a formula 4(x,y) such that A b 4[b,c] and such that { d l A I= 4 [ b , d ] ) is a bounded interval of A with minimum b. If we replace 4(x, y) by the formula 4’(x, y), so that X
y
A (vZ)(X
5ZI y + 4(X,
Z)) A
(3Z)(Z > X A
1 4 ( X , Z)),
then { d l A 1 @ [ b , d ] ) = { d l A 1 4 [ b , d ] ) ; moreover, for every c E A, { d l A I= @[c,d ] } is either empty or IS a bounded interval of A with minimum c. We may thus assume that for every c E A, {dl A != +[c,d ] } is either empty or is a bounded interval of A with minimum c. Consider the formula @(x,y) that is (3z)($(x, z) A 4(z, y ) ) v $(x. y). Then { d E A I A I= e[a, d ] } is an interval of A whose minimum is a and that includes c. It suffices to show that this interval is bounded, for then c E c,(a), as desired. But otherwise, A I= (Vy)(a I y + e(a,y)). Let C consist of C ( A ) together with v o > g and
13.
306
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
{ i d ( c , u o ) I Ab + [ u , c ] } . Since for each ~ E A {, d l A t= d [ c , d ] ) is either empty or is a bounded interval of A with minimum c, any finite union of such intervals is bounded, so that any finite subset of C is satisfied by some element of A . Hence C is consistent and has a model C = C , + C , , where C , = { d l d I a for some a E A } . As in the last paragraph of the proof of the previous theorem, since A < C, C,, C, < C , C , , so that A < C , ; this implies A + C , < C , C,, by Exercise 13.53.2, so that A < A C , . Since A C , b C, there is an element d E A C, such that i + ( c , d ) holds for all c such that A b +[a,c]. But since A < A C,, A C, 1 (Vv)(a5 y + &a, y)), which is a contradiction. We have thus shown that if b E c,(u) and a < b, then for all c > b, c E c,(u) if and only if c E CD(b).Analogously, if b E c,(u) and b < a, then for all c < b, c E c,(a) if and only if c E c,(b). To complete the proof that c , is a condensation we need only show that ifb E c,(u), then a E c,(b). For clearly, ifb E c,(a), then for all d, a I d Ib, we have both d E c , p ) and d E c,(b). But also, if b E c,(a) and a < b, then a E c,(b) and a < b, so by the second statement above with a and b reversed, for all c < a, c E c,(a) if and only if c E c,(b). Thus for all c E A , c E c,(a) if and only if c E c,(b). Thus suppose that a < b and b E c,(u); we will show that a E c,(b). We may assume that c,(u) is bounded above, for otherwise we could replace A by A A , , where A , Y A, and, since A < A A , , c,(a) in A A , is identical to what it is in A , so that c,(u) is bounded above in A + A , . (Recall that c,(u) in A is a union of bounded intervals). We may also assume, as we did two paragraphs earlier, that for every a' E A, { d l A b + [ a ' , d ] } is either empty or is a bounded interval of A with minimum a'. We show that A Y (Vy)(3z)(z < y A $(z, h)).Otherwise, since A is self-additive, it follows from Exercise 13.97 that there is element b' > cD(a)such that
+
+
+
+
+ +
+
+
+
+
+
A b (VY)(~Z)< ( Z Y A +(z,b')); hence for some a' < a, A != +[u',b], so that b' E c,(u'). Since a' < a < b , by the first part of the proof, we conclude that b' E c,(a') if and only if b' E c,(u), which is a contradiction. Thus A y (Vy)(3z)(z < y A $(z, b ) ) . Choose an element e E A such that A k $ [ c , b ] for no c < e. Then (dl A I= b [ b , d ] } , where +(x, uo) is the formula x 2 uo A ( 3 y ) ( y5 uo A $(y, x)), is bounded below by e and above by b and contains both a and b, so that a E c,(b), as was to be shown. W Suppose now that f : A + B is an isomorphism, where A is self-additive, and that a E A . Then, since for any a' E A and any formula 4 ( x ,y), A b 4 [a, u'] if and only if B b $[f(a),f(a')], it is clear that f maps cD(a)onto c , ( ~ ( u ) ) , so that isomorphisms preserve condensation classes. We are now ready to
5.
THE NUMBER OF COUNTABLEMODELS OF A COMPLETE THEORY
307
prove that T = Th(l), where I is self-additive, has c distinct countable models in the case where S,( T ) is infinite.
THEOREM 13.100: (Rubin) Assume that I is self-additive and that S,(T) is injinite, where T = Th(Z). Then T has c distinct countable models.
Proof : Since S,(T) is infinite, there is a non-principal type P. Choose a model I , of T in which P is omitted and another model I , of T in which P is realized by an element e and let B = I , cD(e) I , , where cD(e)is taken in I , and I, = I,. We will show that B k T, that the 1-type of e in B is P, cD(e)is a condensation class in B, and cD(e)is the only condensation class of B in which P is realized. Then if a , and u2 are order types and f is an isomorphism from B . a, onto B . a,, then, since b has type P if and only if f ( b )has type P and f [ c D ( b ) ]= c , ( f ( b ) )for every b, f maps the condensation classes of B . a , in which P is realized onto the condensation classes of B . a, in which P is realized. Thus ,f establishes an isomorphism from a1 onto a 2 , so that a1 = u,. Hence if a, and u2 are distinct order types, then B . a, and B . a, are not isomorphic and T has c distinct countable models. We claim that it suffices to show that B < I , 1, I,. For this implies that B k Tand, since I, < I , + 1 , t I , , the 1-typeofein I , I , + I , isP, so that, since B < I , 1, I , , the 1-type of e in B is also P. The condensation class cD(e) of 1, is also a condensation class of B since any formula defining over e an unbounded interval of I 2 will also define an unbounded interval of I , 1, 1, and hence also of B. Furthermore, the I-type of any element of either I , or 1, in I , + I, 1, is the same in B as it is in I , or I , and I , I , I,, as are the condensation classes of such elements. Thus cD(e)is the only condensation class in which P is realized. Hence to complete the proof, we need only show that R < II I 2 + I,. Let A = I , 1, 1 3 .We must show that if A k $ [ b , , . . . , b k ,a ] , where each bj E B, then there is a b E B such that A k $ [ b , , . . . ,b,, b ] . We may assume that a E Z 2 - cD(e)and a > e. and that b , < b , < . . . < b, with b p the first bi E I,. Choose b' E cD(e) such that b ' 2 b p P l ;then b' < a < b,. By Theorem 13.41, there is a selecting formula $" for $ in [b',b,] (relative to b , , . . . ,bk),that is, a formula $ # ( x L , x 2 v) , such that A k $"[b, b,, c] if and only if b ' < c < b p and A k $ [ b , . . . . ,b,,c]. If we find b e B such that A k $ " [ b , b , , b ] , then we will have a b E B such that A k $ [ b , , . . . ,b,, b ] . Thus we may assume that we are given A k q@" b,, a], where b < a
+
+
+ +
+ +
+ + + +
+ +
+
+
+
308
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
A b +[b’,c,,c] for all c, c p E I , . Hence if there is no c E I , such that A 1 c$[b’,b p ,c ] , then I , k i ( 3 u ) 4 * ( b p u). , Since I , < A, we would have
A 1 4[b’,b,,a]
A
1(344*(b,,u).
Consider this as a formula satisfied by b and modify the formula so that it will be satisfied by all points in an interval including b‘. Let O(u,w ) be the formula W
2
UA
(3X)(3y)(X2
WA
2
W A 4 ( U , X, V ) A
1(3U)4*(X,
U)).
Now (dl A ’k O[b, d ] 3 is an interval of A whose initial point is 6’and which contains asince A k O[b’,a].SinceI,
and
A ’k
i (3u)4*(b,‘, u).
Hence I , b i(3u)+*(b,‘,u) since I , < A, so that in particular I , k i +*(bp), a’). But this means that A 1 i 4[b’,b,‘, a’], a contradiction. What we have shown is that if I is self-additive and S , ( T ) is infinite, where T = Th(l), then there is another countable model I’ = I such that I‘ .cx and I’ . fl are different whenever cx # fl. The example I = 0 . q shows that the same result could not be obtained for I itself; for this example, the model I ’ obtained in Theorem 13.100 is (w . q ) + (o+ 5) (w . q). The proof of Theorem 13.100 fails when S , ( T ) is finite since there are then no non-principal 1-types. The analysis for the case when S , ( T ) is finite hinges on the fact that every 1-type is principal.
+
LEMMA 13.101 : Assume that I is self-additive and S , ( T ) is jinite, where T = Th(1). Then either c,(a) < I ,fbr every a E I or c,(u) is definable ouer a for euery a E I . In the latter case, c , [ l ] has dense order type and for each 1-type P the set of condensation classes in which it is realized is dense in c , [ l ] .
Proof : If c,(a) = I for some (and hence every) a E I, then clearly c,(a) < I for every a E I . Suppose then that c,(a) is definable over a for every a E I and c d a ) # I for any n E I . We will show that in this case the second possibility holds; thus, we must show that if c,(a) < c,(6) and P is a 1-type of I , then there is an element c that realizes P and satisfies c,(a) < c < c,(b). Suppose not. Let $(a, y ) define { d ~ c , ( o ) l d2 a } and let $’(b, y ) define { d E C,(b)ld I b } ; let cx(x) generate the 1-type P and let B(x) generate the
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
309
1-type of b. We now define an interval with initial point a in which P is not realized. Let +(u, w ) be the formula W
2 U A (jX)(X 2
W A
P(X) A (vy)(O 5 y 5
X
-+
$(U,
y) V $’(X,
J’) V
la( y))).
Then { d I I k 4 [a, d ] ) is an interval whose initial point is a and that contains b. Since b $c,(a), { d l l k + [ a , d ] } must contain [a, -). In particular, no condensation class past c,(u) can contain an element whose 1-type is P ; this is only possible, according to Exercise 13.97, if c,(a) is the last condensation class of I . But then c,(u) is a definable interval of I ; for if + ( a , y ) defines c,(u) over a, then W[(VY)((Y
’
x
-
44x7 Y ) ) A +(x,
A]
defines c,(u). This is impossible since I is self-additive and cD(a)# I . Thus if c,(u) is definable over a for every a E I , then c,(u) has dense order type and for each 1-type P the set of condensation classes in which it is realized is dense in ~ ~ ( 1 ) . Suppose, on the other hand, that c,(u) is not definable over a for some a E I . We will show that c,(u) < I . Of course we continue to assume that cD(a) # I . We first show that c,(a) is not definable over any a’ E c,(a). Suppose, to the contrary, that c,(a) is definable over a’ and that I != +[a’, c] if and only if c E c,(u). Note that c,(a) cannot be either the first or the last condensation class in I , because then, as above, cD(a)would be definable, contradicting the self-additivity of I . Suppose that u ( x ) generates the 1-type of a’. Then we claim that (3x)(u(x) A 4(x, a) A +(x, y ) ) defines c,(a) over a. For clearly, if b E c,(a), then I != (3x)(a(x) A 4(x,a) A +(x, b)). O n the other hand, since I
( 3 x 1 ) ( W ( v ~ ) ( 4 ( aY’ ,) + x1 I Y I x2)
and u generates the 1-type of a‘, whenever I != ~ [ c ] then , {dlI b d [ c , d ] ) is contained within a bounded interval of I that is definable over c, and hence must be contained within c,(c); hence { d E I 1 I != (gx)(a(x) A +(x, a) A 4(x, d ) ) } =
tJ{
{ ~ J1I+ [ c , d ] ) ( cE c,(u) and I k .[c])
c
cD(ff).
Thus (3x)(a(x)A +(x, a) A +(x, y ) ) defines c,(a) over a. Hence if c,(u) is not definable over a, then c,(a) is not definable over any a’ E c,(u). Suppose now that I != + [ a , , . . . ,a,, b], where a,, . . . ,a, E c,(u) and b > c,(a). To show that c,(a) 4 I , we must find an element c E c,(u) such that I k 4[a,, . . . ,a,, c ] . Assuming that a’ is the largest among a,, . . . ,a,, we let d* be a testing formula for 4 in [a’, -); that is, I != +*[.’,dl if and only if I k +[a,, . . . ,a,,d]. Let u ( x ) generate the 1-type of a’. If there is no
13.
310
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
d E c,(u) such that I b $*[a’,d], then ( c ( 1 k $ , [ a ’ , c ] ) , where $l(x, y ) is x I y A (Vu)(x I uIy +i $*(x, u ) ) , defines a bounded interval including (c E c,(a)lc 2 a’} and hence equals { c E c,(a)lc 2 a’}. If cD(a)is the first condensation class in I, then (c E c,(a)lc I a’} = { c E I1 c Ia’} is definable over a’, so that cD(u)is definable over a’, contrary to what was shown above. Thus c,(a) is not the first condensation class in I. Hence, by Exercise 13.97, there are elements d < c,(u) whose l-type is P. Since a(x) generates P and I b (3z)$*(a:z), we must also have I 1 (3z)$*(d,z). If for every d < cD(a) whose l-type is P, there is no e between d and a’ such that I b $*[d, el, then the formula (vU)(U
X A %(U) +
(yZ)((4*(V,Z)A
Z
> U ) + 2 > X))
defines a proper interval of I . Hence for some d < c,(a) whose l-type is P, there is an e between d and a’ such that I k $*[d,e]. Hence the formula &a‘. Y), y<
U’A(’JtL1)((yI U < U ’ A C ( ( V ) ) - + ( \ J Z ) ( ( $ * ( U , Z ) A Z >
V)+Z>U’))
defines a bounded interval whose last point is a’, and must therefore be contained within c,(u). This interval cannot be all of (+-,a’] n c,(u) for in that case c,(u) would be definable over a’, contrary to assumption. Hence there is an element c E c,(u) such that c < a’ but c is not in this interval. This a’ in c,(a) such implies that there is a d‘ E c,(u) whose l-type is P and an e‘ I that I 1 $*[d’, e’]. Since d‘ E c,(u) and a‘ E c,(a), also a’ E c,(d’), so that there is a formula t+b(x,y)such that ( b l l k $ [ d ’ , b ] } is a bounded interval of I containing a’, d‘, and e‘ and hence contained within cD(a).But then I b ( 3 z ) ( 4 * ( d ‘ , z ) ~ $ ( d ’ , zHence, )). since a generates the l-type of both d‘ and a’, we have I k (3z)(4*(a’,z ) A $(a’, z ) ) ; for the same reason { b I I k $[a’, b ] ) is a bounded interval of I containing a’ and is therefore contained within c,(n). Thus, despite what we first assumed, there is a c E cD(u)such that I 1 $*[a’,c]. Hence c,(a) < I . Now let b E I - c,(u). Since c,(u) i I, every l-type of T is realized in c,(u), so that there is an element b‘ E c,(u) such that (I,b ) = (I,b‘). If c,(b) is definable over b, then cD(b‘)would be definable over b’, contrary to assumption. Hence C,(b) < I for every b E I. Thus I is a sum of elementarily equivalent self-additive models. 4 Let us elaborate on Lemma 13.101. Suppose that I is self-additive, that S , ( T ) is finite, where T = Th(I), and that c,(a) is definable over a for every a E I. Let a E I have l-type P and let e(a,y ) define c,(a) over a ; we claim that if I‘ is a model of T and a’ E I ’ has l-type P then &a’, y) defines c,(a’) over a’. Indeed, if { b E I / I’ k 4 [ a ’ , b ] } is a bounded interval of I containing a’,
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
31 1
then that fact is recorded in P, so that { b E III k 4[a, b ] } 2 { b E I11 1 6[a, b]}. But since (Vy)(+(x,yj + 6 ( x ,y ) ) is in P, ( b E I'I I' k +[a',b]} G { b E I'IZ' !F 6[a', b ] 1, so the claim is verified. Moreover, if a E I and a' E I ' have the same 1-type, then c,(u) = c,(a'). It follows that for any a E I and a' E I' the set of 1-types of T realized in cD(a)and c,(a') are either identical or disjoint. Since for each 1-type of I the set of condensation classes in which it is realized is dense, it follows that there is a finite set D,,D,, . . . ,D,of linear orderings such that I = o ( D , , D , , . . . ,D J . To summarize, T has a model that is a shuffle a ( D , , D , , . . . ,DJ such that each copy of D,is definable over each of its elements; these definitions are uniform in that if two elements have the same 1-types, then the copies of D, containing them are definable by the same formula. Turning to the other case, if cD(a)< I for every a E 1, then, by the discussion above, the same must be true for every model I' of T . Furthermore, there is a countable model of T , namely, c,(a), that consists of one condensation class. We now complete our discussion of the self-additive case by showing that if S , ( T ) is finite, where T = Th(l), then T has exactly one or exactly c distinct countable models. This together with Theorem 13.100 shows that the same conclusion is true for every self-additive I . Although we need only prove the result below for self-additive models, the proof requires that we deal simultaneously with non-self-additive linear orderings since a selfadditive I can be a finite shuffle of non-self-additive linear orderings. THEOREM 13.102: (Rubin) Let T be a complete theory of linear orderings ,for which S,( T j is Jinite. Then T has exactly one or exactly c distinct countable models. Proof: We proceed by induction on the size of S , ( T ) . If IS,(T)I = 1 and A !F T , then either +(a) z 1 for all u E A , in which case A 'v 1 or A N 1, or cF(a)2: 5 for all a E A , in which case A = 5. Thus in this case T has exactly one or exactly c distinct countable models. We proceed now to the case where IS,(T)I = n 1. If a model A k T is not self-additive, then by Theorem 13.96, there is a formula that defines a proper interval of A . Thus there is a formula 4 ( x j such that A = A , + A , , where A , = { U E AIA k + [ a ] ) and A , = { a E AIA 1 i 4 [ a ] } . Since there are 1-types that contain each of +(x) and i 4 ( x ) , IS,(Th(A,))I I n for each i. If both A , and A , are No-categorical, then so is A , A , = A by Theorem 13.91. If A , is not No-categorical, then, by the induction hypothesis, there are c distinct countable models {B,~cI< c> that are elementarily equivalent to A . Let D, = B, A , for each CI < c. Clearly D, = A for every a; and if { d E D,I D, !F 4 [ d ] } = B, for every a,then any isomorphism from D, to D,
+
+
,
+
312
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
would map B, onto B,, which is only possible when M = p. So let d E B, and let m = qd(4). Since G,+ l(B,, A E 11, there is an element a E A such that G,((B,, d ) , ( A l , a ) )E 11. Since G,(A,, A , ) E 11, we have G,((D,, d ) , ( A , a ) )E 11. Hence (D,,d ) = (A, a ) by Theorem 13.11, so that D, 1 4 [ d ] ifandonlyifA t= +[a].SinceA 1 4 [ a ] for all a E A , , D, I= 4 [ d ] foralld E B,. Similarly, if d E D, - B,, then D, 1 i 4 [ d ] .Hence { d E D,ID, t= 4 [d] 1 = B,. This implies that T has c distinct countable models. We now turn to the case in which every model of T is self-additive. According to Lemma 13.101, there are two cases to consider. If c,(a) < I for every a E A, then there is a countable model, namely, c,(u), that consists of one condensation class. Thus c,(u) . M has exactly c1 condensation classes, so that c,(u). M and c,(u). b are not isomorphic unless M = fi, so that T has c distinct countable models. Otherwise, by the discussion following Lemma 13.101, we can find a finite set of linear orderings D , , D , , . . . ,D,such that A = a@,, D,, . . . ,DJ. Thus A = A , I q E Q } , where each A , = D,for some j and, for each a E A,, c,(u) = A,. Now if every D,has an KO-categorical theory, then so does A by Theorem 13.91. If, on the other hand, some D iis not KO-categorical,then there are two possibilities. If Sl(Th(Di))I n, then Th(Di) has c distinct countable models {B,la< c } . If S,(Th(Di))= n + 1, then every 1-type of T is realized in each copy of Di,so that A = D, . q. Since A $ D i , D, is not self-additive, so that Th(Di) has c distinct countable models { B, I M < c f . In either case, if we let I' be the result of replacing some A , = Di by B, and all other A , = Di by Bi,then I' . GI is different from I' . b whenever M # p, so that T has c distinct countable models.
c{
COROLLARY 13.103: (Rubin) If T is selfadditive, then either T is KO-categorical or there is a countable model I of T such that I . c1 is diflerent from I ' p whenever c1 # p. In particular, if T is self-additive, then either T has exactly one or exactly c distinct countable models.
This completes our discussion of self-additive linear orderings, and we are now prepared to return to the general case. Before doing so, however, we will present another result of Rubin that provides a characterization of those complete theories T of linear orderings for which S , ( T )is finite. DEFINITION 13.104: Let M 2 be the smallest class of order types containing 1, containing T~ + t 2 whenever it contains z1 and T,, containing 5 . 5 whenever it contains T, and containing a(F) whenever it contains the elements of F.
5.
313
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
The theorem we will present, which is analogous to Theorem 13.91, is that S , ( T ) is finite if and only if T has a model whose order type is in A 2 . Before doing so, however, we define the following notion of rank for order types of DEFINITION 13.105: We define inductively a sequence { A , ( n ) l n < w } of subsets of A2by setting A,(01 = { l } and defining A 2 ( n + 1) to include A',(n) and to contain z1 z, if z l , z 2 E A 2 ( n ) ,to contain z . 5 if z E A 2 ( n ) , and to contain a(F) if F E A z ( n ) . Clearly, U { A 2 ( n ) In < w } = A 2 .We define the rank of an element z E A'2to be the smallest n such that z E A 2 ( n ) .
+
THEOREM 13.106: (Rubin) Let T be a complete theory of linear orderings. Then S , ( T ) is ,finite zf [inti only if T has u model whose order type is in .A2. Moreover, iJ' S,(T ) is Jiniri?,then T is Jinitely axiomatizable.
Proof : It is easily verified, using Ehrenfeucht games, that if z E A 2 then , Sl(Th(7))is finite. We will show, by induction on n, that if IS,(T)I I n, then T is finitely axiomatizable and has a model in A , ( 2 n - 1). If (S,(T)I = 1 then, as we pointed out earlier, T is either Th(l), Th(q), or Th([), and in each of these cases T is finitely axiomatizable and has a 1) as required. Proceeding with the induction, we assume that model in A2( if ISl(T)1 I n, then T is finitely axiomatizable and has a model in A , ( 2 n - 1) and we turn to the case where ISl(T)I = n 1. If T is not self-additive, then there is a countable model A of T and a formula 4(x) such that A = A , A , , where A , = { u E AIA b 4 [ a ] }and .4, = {a E AIA 1 iq5[a]}. Since each of 4 and 14 belong to 1-types of T , IS,(Th(A,))I I n and IS,(Th(A,))( In, so that, by induction hypothesis, each Th(Ai) is finitely axiomatizable and has a model Bi whose order type is in A1(2n - 1). Then T has a model, namely, B , B,, whose order type is in A,(2n). Moreover, if 8, and e2 are axiomatizations of Th(A,) and Th(,4,), then T is axiomatized by
+
+
+
(vX)(vy)(4(X) A 1$(y ) + X < y ) A 8f'"') A 8;'(uo).
We may thus assume that T is self-additive and apply Lemma 13.101 and the discussion appended to it. We consider first the case in which T has a There are then two possibilities. If IS,(Th(D,))(In model a(D,, D,, . . . ,Dk). for each i, then, using the induction hypothesis, there are Bi = D iwhose order types are in A , ( 2 n - I); then a(Bl, I?,, - . . ,Bk)= a(D1,D2,.. . ,Dk) and has order type in A2(2n).Moreover, if for each i we choose a 1-type Pi realized in D i , a formula cl/i(x)that generates Pi, and a formula 6 J x , y )
314
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
that defines over any element whose 1-type is Pi the copy of Di to which it belongs, and if we use the hypothesis to give us an axiomatization Bi of Th(D,), then we can construct the following axiomatization of T : (3x)+i(x) A
A ( V X ~ ) ( V X ~ V) ( + i ( X t ) A X 1 < X
Ai (vx)(lCli(x)
-+
1
~f~(x,uo)
~ 7A4 i ( x I 9 ~ 2 ) )
( i --t
A ( ~ x ) (
< ~ 2 ~ + j ( 1x 4) j ~ (x
>~1)~14j(x,~2))
j
A (~Y)(3Xl)(3X2)(Xl
< Y < x 2 A lClI(X1) A lCll(X2)
A 1 4 1 ( X l , Y ) A 741(XZ,Y)).
On the other hand, if IS,(Th(D,))I= n + 1 for each i, then Di = Dj for all i and j , so that Di . g is a model of T. Since each Di is definable over each of its elements, Di # Di . g, so that Di is not self-additive. But we have already shown that in the non-self-additive case, Th(D,) is finitely axiomatizable and has a model B in A 2 ( 2 n ) .Thus T has a model, namely, B . g, in A 2 ( 2 n + 1) and, as in the case above, T is finitely axiomatizable. We now turn to the other case in which T has a model A that consists of one condensation class. We intend to show that A = T . [ for some z E A 2 ( 2 n ) . Hence we first define a suitable Z-sequence of points of A. Let P be a 1-type of T and let a E A have 1-type P. Since P is principal and A is self-additive, it follows from Exercise 13.97 that { b E AIP(b) = P } is unbounded in B ; moreover, since S , ( T ) is finite, there is an element b > u such that b has 1-type P and such that every 1-type of T is realized in [a, b]. We claim that such an element b may be chosen so that for no element c between a and b whose 1-type is P is it the case that every 1-type of P is realized in both [u, c ] and [c, b ] . Using this claim, and the reverse claim, we can start with u = ao, let b = a,, and generate a [-sequence { a z l z E Z> of elements of A . To verify the claim, we note that otherwise we can find a sequence co > c1 > c2 > * . . of elements above a such that each ci has 1-type P and such that every 1-type of T is realized in both [a,ci+, ] and [ci+ c i ] . Since c,(a) = A, there is a formula $(u, w) such that ( d j A k 4 [ u , d ] ) is a bounded interval of A including co. Since A is self-additive, A < A + A’, where A‘ = A . In A A’ we can find an element c whose 1-type is P and such that every 1-type of T is realized between { d l A k + [ a , d ] } and c. Since there are only a finite number of 1-types and all of them are principal, this can be expressed by a formula with parameter a that, since A < A + A’, is also true in A . Thus there is an element c - E~ A such that { d l A k $[a,(il) <
+
5.
THE NUMBER OF COUNTAHLE MODELS OF A COMPLETE THEORY
315
c P I and every l-type of T is realized in [C,,C_~].Since P(a) = P(co) = P ( c - ,) = P and every sequence of l-types of T is realized in both [a, co] and [a,c-,], we can apply Theorem 13.80 to conclude that (a,co) and ( a , chave the same 2-type in A . But this is impossible because A b 4 [a, co] but A k i 4 [ a , c - ,I. This contradiction substantiates the claim. We have thus generated a [-sequence {a,lz E Z } of elements of A such that each a, has l-type P, every l-type of A is realized in each [a,, a,, and for every b E (u,,a,+,) whose l-type is P there is a l-type of T not realized in either [a,, b] or [b,a,+,]. If now a < a, for all z, then the con< a - < a,, would lead to the same contradiction figuration a < . . . < as before. Hence the sequence {a,la E Z } is coinitial in A, and, similarly, it is cofinal in A . For each z E Z we let A,' = {clfor some b of l-type P, a, I c I b, and not every l-type of T is realized in [a,,b]}; we define A,similarly and we set A , = A , AZt. Then A , c (a,- a, + ,). To show that A , does not overlap with A , + , , it suffices to show that there is a l-type of T that is not realized in any A,, for then A, u A , + , could not include [a,, a,+ ,] since every l-type of T is realized in [a,, a,, ,I. Even if we show that there is a l-type Q of T that is not realized in some A,, this will be sufficient; for since A , is definable over a,, the fact that no element of A , has l-type Q is expressed by a formula in P and hence no A, has an element whose l-type is Q. Suppose, on the other hand, that every l-type of T is realized in A , . Then, in particular, there are elements b,,b, E A , of l-type P such that b , I a, I b, and every l-type of T is realized in [b,, b,]. If some l-type of T is not realized in [b,,~,], then the interval Z(b,) = { d 2 b, lnot every l-type of T is realized in [b,,d]} is definable over b,, and contains [b,, a,]. Similarly, if some l-type of T is not realized in [a,, b,], then the interval I(a,) is definable over a, and contains [a,, bJ. Since a, and b, both have l-type P , if Q is not realized in [b,,~,], then Q is also not realized in [a,, b,] ;this is impossible since every l-type is realized in [b,, b,]. Hence either every l-type of T is realized in [b,,a,] or every l-type of T is realized in [a,,b,]. This contradicts the definition of A , . Hence some l-type Q of P is not realized in A , , and hence, as pointed out above, Q is not realized in any A , . Thus A , u A; does not include [a,, a,+ ,], so that there are non-empty linear orderings {B,(z E Z } such that A = C { A , B,lz E Z } . Note that P is not realized in any B,; for if a, < b < a, + where b E B , and b has l-type P, then either some l-type of T is not realized in [a,, b]-in some l-type is not realized in [b,a,+,]-in which which case b E A,-or case we would have b E A, + . To show that A = z * 5 for some z E A 2 ( 2 n ) ,we will show that every A, = for some T~ E A 2 ( 2 n - 1) and that every B, = z2 for some z 2 E ,4,(2n - 1). Then t, + t 2 E A 2 ( 2 a ) , so that (z, z,). [ E A 2 ( 2 n 1). Since A = (zl+ t2). [, this will complete the proof.
,)
,I,
,
+
,,
,
,,
+
,
'I,
+
+
316
13. THE! FIRST-ORDER
THEORY OF LINEAR ORDERINGS
That A,, = A,, for every zl, z , E Z is clear since each A , is definable over a, by the same formula, so that given any statement true in A, its relativization is found in P. We now show that lS,(Th(A,))( < IS,(Th(A))I, so that we can apply the induction hypothesis. Let Qf(a,x)say that there is a y of type P such that x lies between a and y and not every 1-type of T is realized between a and y. Then { b E A I ~ [ u , , b ] } = A,. We claim that for any a E A , , { b E AIQf[a,b]} = A , . Suppose, for example, that a, < a < b < c and not every 1-type of T is realized in (a, c) but b $ A,; then we can choose an element a‘ that is in A , and has 1-type P such that a, < a < a’ < b < c; but then not every 1-type of T is realized in [a,,a’] and not every 1-type of T is realized in [a’, c], so that, as shown several paragraphs ago, not every 1-type of T is realized in [a,, c]-so that b E A , after all. Thus if two elements b , and b, of A , have the same 1-type in A , then in playing, and winning, the game G,((A,b,),( A , b 2 ) )for n > qd(4), PLAYER II confines his choices to A , whenever PLAYER I does so. Hence the 1-types of b , and b, in A , are the same. Thus IS,(Th(A,)) < IS,(T)l since Q is not realized in A , . Hence, by the induction hypothesis, Th(A,) has a model z1 in A2(2n - 1). Similarly, since P is not realized in any B, and elements realizing P are coinitial and cofinal in each A , , for any b E B,, B, is exactly the set of all elements c such that nothing between b and c has 1-type P. Thus B,, = B,, if zl, z , E Z , and if two elements of B, have the same 1-type in A , then they have the same 1-type in B,. Thus, as above, (S,(Th(B,))I < ISl(T)I, so that Th(B,) has a model t 2E A,(2n - 1). It follows that Th(A) has a model, namely, (51 t 2 ). c in A2(2n l), as was to be shown. The axiomatization of (7, + t 2 ). [ we leave to the reader.
I
+
+
A special case of Theorem 13.106 is that every KO-categorical linear ordering is finitely axiomatizable. This result is due to Lauchli and to Rosenstein [14]. We return now to our discussion of an arbitrary complete theory T of linear orderings. There are now two possibilities: Either S,(T) is countable for each n or, for some n, S,(T) is uncountable. We will first dispose of the second case by showing that the cardinality of S,(T) must actually be c if it is uncountable. This will in fact dispose of this case because if S,(T) has cardinality c, then, since each n-type of T must be realized in some countable model of T and only countably many n-types of T can be realized in any single countable model of T , it follows that T has exactly c distinct countable models. [The fact that S,(T) has cardinality c if it is uncountable is true more generally for the collection of maximal ideals in any Boolean algebra, and the proof given below also works in the general case.] LEMMA 13.107: Zf S,(T) is uncountable, then it has cardinality C.
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
317
Proof : We will first show that if II/ belongs to uncountably many n-types of T, then there is a formula 6’ in F,( T ) such that both $ A 8 and I,+ A (1 0) belong to uncountably many n-types of T. Suppose not, so that given any f3E F,( T ) either $ A 0 or $ A (1 0) belongs to only countably many n-types of T. Let { O j l j < o}be an enumeration of all formulas in F,(T). We define a sequence ( $ j l j < o}of formulas of F,(T) such that each $ j is either d j or i B j and such that for each m, ( t j j l j I rn} LJ {$) is in uncountably many j < rn) u {$, i $,} is in at most countably many n-types. n-types but Assume that the { $ j I j < rn) have already been defined and satisfy the cone,) are in uncountably many ditions above. Since not both $ A 0, and $ A (1 n-types, it follows that not both ( $ j l j < m } u ($.em} and ( $ j l j < rn} LJ { $ A (10,)) are in uncountably many n-types; on the other hand, since {tjjlj < m } u ($1 are in uncountably many n-types, each of which contains either 8, or lo,, it follows that one of the two sets is contained in at most countably many n-types and the other is in uncountably many n-types. Define I),,, to be 8, if { t,hj I j < rn} u { $, 0,) is in uncountably many n-types and otherwise ,,)I is lo,,,.Note that the induction hypothesis continues to hold. Since each finite subset of (qb,l,j < o}is in uncountably many n-types, the set itself is certainly consistent and is therefore contained in an n-type. Moreover, since for each O j E F,,(T) either O j or i O j is II/j, it follows that {t,bjl j < a} is an n-type of T. On the other hand, as we defined the sequence j < o}, at each stage we sloughed off only countably many n-types that j < ID}, $ is only in countably many n-types contained $. Thus besides of T. This contradicts the original assumption that $ is in uncountably many n-types of T. Thus we have proved that if $ belongs to uncountably many n-types of T, then there is a formula f3 in F,( T ) such that both t,b A f3 and $ A (1 19) belong to uncountably many n-types of T. Using this fact, we now define for each finite sequence s of 0 s and 1’s a formula 4sof F,(T) that is in uncountably many n-types of T. For the empty finite sequence ( ), $( is just u1 = u1 A u2 = u2 A . . . A v, = v,,, which is in all of the uncountably many n-types of T . Proceeding inductively, if 4sis defined and is in uncountably 0) many n-types of T, then we find a formula 8 so that both rfJs A 0 and 4sA (1 are in uncountably many l-types of T and we let one of these be 4soand the other rfJsl. Now i f f : N -, (0, l} is an arbitrary sequence of 0’s and 1’s and S , is its set of finite initial sequences, then (4,Is E S,} is contained in an n-type of T. Since different sequences result in different n-types and since there are c such sequences, it follows that SdT) has cardinality c. W
{$jl
{$jl
($jl
Thus we can confine ourselves to the case in which S,(T) is countable for each n. We introduce a new condensation cE to analyze the models of such a theory.
31 8
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
DEFINITION 13.108: Let 4(x) be a formula of L. We say that d, is an interval formula of T if T t (3x)+(x)A i(Vx)$(x) and
T I-
(~X)(VY)(VZ)((X
< y < z A 4(x) A 4b))
+
4(Y)).
Thus 4 is an interval formula of T if and only if for every model A of T, { a E A I A k 4 [ u ] } is a proper interval of A. The set of interval formulas of T is denoted i(T).For any model A of T and any a E A we define c E ( a )= (blfor every interval formula #(x) of T , A k 4[b] if and only if A k 4 [ a ] } . Note that cE is a condensation since, as is easily verified, if b E c,(u), then C E ( b ) = C E ( ~ ) . DEFINITION 13.109: An interval formula 4(x) of T is called a terminal formula of T if T t (Vx)(Vy)((x < y A 4 ( x ) )+ & y ) ) . Similarly, #(x) is an initial formula of T if T t (Vx)(Vy)((x < y A 4( y ) ) -+ &x)).
Given an interval formula #(x) of T , we let +'(x) denote (Vy)(&y) -+ y < x) and c$-(x) denote ( V y ) ( + ( y ) --* y > x). Then $'(x) is a terminal formula of T unless +(x) is a terminal formula of T , and &(x) is an initial formula of T unless $(x) is an initial formula of T . If @x) is an interval formula of T , then any model A of T can be written as Cap k
4-CaN +
4C4 +b
( A
4"aIh
with the first (last) term absent if 4 is an initial (terminal) formula of T. Given a model A of T and a E A , we let @a be the set of all interval forthen either 4+(x) E or +-(x) E Oa, mulas of T satisfied by a. If +(x) $ so that 4(x) cannot be consistently added to Thus @ a is a maximal consistent set of interval formulas of T. Conversely, if Q, is a maximal consistent set of interval formulas of T , then @ s P for some 1-type P by Exercise 13.67.4 since @ G F,(T),and hence there is a countable model A k T and an element a E A whose 1-type is P, and therefore @a = @. DEFINITION 13.110: A maximal consistent subset of the set I ( T ) of interval formulas of T is called an interval type of T. The collection of interval types of T is denoted ZT(T).If @ is an interval type of T and A i= T, we let A(@)= { u E A ( @ ,= @}. Note that A(@) is an interval of A if it is non-empty. If A ( @ )# $3,we say that @ is realized in A, and otherwise we say that @ is omitted in A . We let IT(A) denote the set of interval types of T realized in A .
Note that for any model A k T, there is a natural correspondence between c,-condensation classes cE(u)of A and interval types @a realized in
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
319
A ; thus b E cE(aJif and only if Qb = @a, so that A(@,) = cE(a).This correspondence induces a linear ordering on IT(A) by @, < I T ( A ) @ b if and only if cE(u)< cE(b);thus I T ( A ) = c E [ A ] . Let and be two interval types of T and assume that $ ( x ) E 0,but + ( x ) $ 0,.We may assume, without loss of generality, that $ J + ( x E) 0,. From this we see that if Ol,Q2 E I T ( A )n I T ( B ) , then < I T ( A ) Q2 if and only if O1 m2. Thus the various linear orderings induced on the subsets IT(A) of I T ( T ) by the correspondences with c E [ A ]are all compatible. Since S,,(T)is assumed to be countable for all n, by Theorem 13.82 there is a countable saturated model S of T . Since each interval type is contained in a l-type and since each l-type is realized in S, it follows that every interval type is realized in S , so that I T ( S )= I T ( T )and I T ( T )2 c E [ S ] ;in particular, I T ( T ) is countable. DEFINITION 13.111: An interval type CD is called a principal interval type if there is a formula $(x) E @ that generates 0;that is T k (Vx)[$(x) -, $(x)] for all 4(x) E 0.Otherwise, @ is called a non-principa1,or limit, interval
type. We now obtain a characterization of the principal interval types. LEMMA 13.112: Suppose that 4(x) is an interval formula of T. Then {@ E I T ( T ) )$(x) E @} has a first mid last element.
Proof: Suppose that it has no last element. Then there is a countable < < a2< . . . of elements of I T ( T ) that is cofinal in sequence {@ E I T ( T ) l 4 ( x )E 0.). For each n, choose a formula 4,,(x) with 4,,(x)E @, and $n'(x) E On+1 . Let C = {$(x)} u {$,,+(x)In < 01). This is a consistent set of interval formulas of T since any finite subset is contained in some @,,. Hence C c P for some 1-type P of T ; but the set of interval formulas of T in P forms an interval type, so that C c Q, for some @ E ZT(T). But each On< CD and 4(x) E Q, contradicting the assumption that the sequence {a,,In < w } is cofinal in ICD E I T (T )1 + ( x ) E a}. H COROLLARY 13.113: Let @ E I T ( T ) . Then @ is a principal interval type fi and only i f @ has an immediute predecessor and an immediate successor
in ZT(T). Proof : If @ has an immediate predecessor
and an immediate successor in I T ( T ) ,then there are formulas Cpl(x)E CDl and 4 2 ( x )E O 2 neither of which is in @. Then 4 1 + ( x )A $ J 2 - ( . x ) E @ and clearly generates
Q2
320
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Conversely, if $(x) generates @, then, by Lemma 13.112, {@ E
w T ) I $ - ( x ) E @}
has a last element, which is the immediate predecessor of 0,and
{a E ZT(T)($+(X)E @] has a first element, which is the immediate successor of 0 in ZT(T). The proof of Lemma 13.112 can be adapted to yield the following result. THEOREM 13.114:
I T ( T )is complete.
+
Proof : Let IT( T ) = A B. We must show that either A has a last element or B has a first element. Otherwise, we can choose a sequence (@,,In < o) cofinal in A and a sequence {'PI,/n < o} coinitial in B. If +,,(x)E @, and +,,'(x) E a,,+ for each n, and $,,(x) E Y,,and $,,-(x) E Y,,+ for each n, then {+,,+(zc))n < w } u {$"-(x)In < o}is consistent and hence is contained in some @ E I T ( T ) which is bigger than each a,, and less than each Y,,. This contradiction implies that IT( T ) is complete.
COROLLARY 13.115: I T ( T )is scattered. Proof: If q 5 ZT(T),then ZT(T)has c cuts, by Exercise 2.33.4. But since IT( T )is countable and complete, this is impossible.
We now obtain critical information about the intervals S(@) of the saturated model S of T. THEOREM 13.116: Let S be a saturated model of T and let @ be an interval type of T. Then either S(@) is a single element of S or S(@) is self-additive. Proof: We first observe that if +(x) is any formula such that S k + [ a ] for some a E S(@), then { b E S(@)JSb (P[b]) is cofinal in S(@). For suppose that C E SPD). If there is a formula $(x)E@ such that S Y +[b] whenever b 2 c and S k $ [ b ] , then the interval formula (3y)(xI y~ +( y) A $(y)) is satisfied by all elements of S(@)up to a, but by no elements of S(@) beyond c. Since CD is an interval type, this is impossible. Hence for every $(x) E @ there is a b 2 c such that S k $[b] and S I=+ [ b ] . Thus X = {$(v2)1$(x)E @} u {&vz), v2 2 u , } is consistent with the 1-type of
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
321
c, and hence, since S is saturated, there is an element b E S that, together with satisfies all formulas of C. Thus h 2 c, S I: 4 [ b ] , and b E S(@), as was to be shown. From this it follows that for any 1-type P of T that is realized in S(@), the set of elements of S ( 0 ) whose 1-type is P is cofinal in S(@).For if c E S(@), then P u {c I u } is finitely satisfiable by the argument above, and hence u> is satisfied is satisfied in S, since S is saturated; but since @ G P, P u { c I by some element of S(@). We use this to show that S(@)has no proper definable intervals. For if A is a proper interval of S(@)that is, without loss of generality, not a terminal interval, we choose a E A and, using the conclusion above, we choose b > A so that a and b have the same 1-type in S. Since S is saturated, there is an automorphism of S that maps a to b ; but any automorphism of S induces an automorphism of S(@).Thus a and b have the same 1-type in S(@),so that A cannot be a definable interval of S(@).Thus S(@)is either self-additive or S(@)has exactly one element. c,
EXERCISE 13.117: Show that if S is a saturated model of T , @ is an interval type of T ,and S(@)has a single element,then that element is definable.
LEMMA 13.118: I f @ is a principal interval type, then S(@)is a saturated model of Th(S(@)). Proof : Let P be an n-type of Th(S(@))and Q be an (n such that P c Q. Let @ be generated by O(x).Then
+ 1)-typeof Th(S(0))
Qe(w)= {4(xl, . . . , X , , X , + ~ ) ' ( ~ ) (E~Q ) 2 Pecw) = {4(Xl>..
1
>x,)e(w)]4 E PI.
If ( a l , . . . , a , ) is an n-tuple of elements of S(@)whose n-type is P, then S I: 4 e ( w ) [ a l. ,. . ,a,] for every 4 E P, so that, since S is saturated, there is an element b E S such that S k @('"')[a,,. . . ,a,*,b] for every 4 E Q and such that S 1 O[b]. Hence S(@)is a saturated model of Th(S(@)). We now have all the information we need, so we proceed to prove Rubin's theorem. First consider the possibility that for some isolated cb E I T ( T ) , S(@)is not KO-categorical.By Theorem 13.116, S(@)is self-additive, so that, by Corollary 13.103, Th(S(0))has c countable models; by Lemma 13.118, S(@)is saturated, so that, by Exercise 13.83.1, there are distinct models (A al a< c} such that each A , < S(@). Write S = S, + S(@)+ S,. Then, for each a < c, S, = S, + A , + S , 4 S and { a E SalsaI: O[a]} = A,, where e(x) generates @. Hence these c countable models are all distinct models of T, since any isomorphism from S, to S, must carry A , onto A , .
13.
322
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Thus we may assume that S(@) is KO-categoricalfor every isolated @ E I T ( T ) .If ZT(T)is finite, then S = S , . . . S,, where each Th(S,) is KO-categorical.By Theorem 13.91, each Si- M ifor some Mi € A l ,so that S = M , . . . M , E A , ; hence T is K,-categorical. Thus we may also assume that ZT(T) is infinite. Since it is scattered, it has an interval of order type w or o*;we assume the former, with no loss of generality. Since I T ( T ) is complete by Theorem 13.114, it has an interval of order type o 1. Thus there are interval types @ 0 , @ 1 , @ 2 r . . . ,@ such that each S(@,) is KO-categoricaland S(@,) S(a1)+ S(@,) . . . S ( @ ) is an interval of S. Let us pause to consider a particular example A = a(l,2) a(2,3) + a(3,4) + . . . of this case. It is easily verified that the c,-condensation classes of A are precisely {a(.,. 1)ln < w } , so that ZT(A) has order type o.Since Z T ( T ) is complete, it must have order type o 1; thus ZT(T) consists of @, < 0,< Q3 < . . . < 0 and S((D,) = a(n,n + 1) for each n < w since a(n,n 1) is KO-categorical.What is S(@)? We observed that cF(a)cannot be finite for any a E S(@) since there is no maximal finite interval of length n in A after any maximal finite interval of length n + 2. Furthermore, given any element a E S(Q there must be a first element b 2 a with no immediate successor and a last element c I a with no immediate predecessor. Thus S ( 0 ) is a sum of intervals each of which has order type w [ . CI o* for some order type a. Moreover, since the elements of A with no immediate predecessor have order type q, we conclude that S(@) = (w w * ) . q. Thus c distinct countable models of T h ( A )can be obtained by replacing one copy of o o*in A (o w * ) . q by each possible w + [ . c t o*. In this example, S(@,) f S(@,) whenever n # m. The procedure we now describe is reminiscent of the example and works whenever (S(@,)) n < w > contains infinitely many non-isomorphic linear orderings. We need the following lemma.
+
+
+
+
+
+
+
+
+
+
+
+
+ +
+
+ + +
+
LEMMA 13.119: Let JV be an infinite collection of C.C,-categorical order n o} of order types such that each types. Then there is a sequence { ~ , \ < zi is the order t y p e of an interval of some z E N and such that each z, is isomorphic to a proper interval of z,+ 1.
Proof : We define the notion of predecessor for KO-categoricalorder types by induction on rank (see Exercise 13.92). Thus if z has rank n + 1 and z = z1 T ~ where , both z1 and z 2 have rank at most n, then the predecessors of z consist of z,, z2, and the predecessors of z1 and 7,; similarly, if t has rank n + 1 and T = a(F), where each z’ E F has rank at most n, then the predecessors of z consists of F together with the predecessors of each T’ E F.
+
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
323
Now given N , we let A"* consist of Jli together with all predecessors of order types in N.We now define by induction a sequence {z,(n < o) of elements of N * such that each z, has rank n. We will assume, as part of the induction hypothesis, that the set T , of elements of N* that have z, as a predecessor is infinite. We define zo to be 1 and, assuming that z, is defined and T , is infinite, we enumerate the elements ol,.. . , ( T ~of T , that have rank n + 1. This set is finite by Exercise 13.92. Moreover, if T E T,, then, as is easily verified by an induction on rank, some ( T ~is a predecessor of z. Thus if we define T"+ to be the first oithat is a predecessor of infinitely many elements of T,, then the induction hypothesis continues to hold and we obtain a sequence {z,ln < w ) , as desired. Suppose now that { S ( 0 , ) 1 n < o} contains infinitely many nonisomorphic KO-categorical linear orderings. Then, by applying Lemma 13.119, we can find a sequence (t,ln < co} of order types such that each T~ is a predecessor of some S(@,,,J; a subsequence {a,ln < w } can easily be selected so that each (T, is a predecessor of some S((Dq(,)),where the subscripts { q ( n )\ ti < o}form an increasing sequence. Let C, be an interval of S(aq(,,J of order type B, for each n. let g, map C, onto an interval of C,+, for each n, and let (T be the order type of C = U{C,[ n < o},where in this union each C , is identified with g,[C,]. In the example discussed earlier, each C, has order type n and the order type of C is either w, w*, or ( depending on the functions {gnln < ( 1 ) ) chosen. We now show that S(@) has an interval of order type rr. Indeed, if we choose elements {a,lz E Z ) G C so that z1 5 z2 implies aZlI azz and so that { a , , l n N ~ } is cofinal in C and { a , / n ~ Z N ) is coinitial in C, then C = U { [ a _ , , , a , , ] I nN~) ; since each [a-,,a,] is an interval of some S(@"J, each [a_,,,a,] is KO-categorical by Exercise 13.92, and hence is finitely axiomatizable by Theorem 13.106. Let $, be an axiomatization of [a-.,a,] for each n. Add to the language of linear orderings a (-sequence (a,] z E Z } of constant symbols and let C = C ( S ) u ($,'In < wI1 u { g z , I gz2/z1 5 z 2 } u {Cp(%)(zE 2 and Cp E @ } , where t+bn' is the relativization of $, to the interval [a-,, 4.Any finite subset of C can be realized in S by selecting elements of some S(0.,) to correspond to the constant symbols. Hence 'c is consistent and is realized in some countcontains elements { c, I z E Z ) able model s' of T. Since s' < s,we see that s(@) such that each interval [c - ,,c,] Y [ u - ,, a,] ;hence S ( 0 )contains an interval of order type rr. Now B itself cannot be KO-categoricalsince if its rank were k, then every interval of it would have rank at most 2k + 1 by Exercise 13.92; this is impossible since the sequence { rr, 1 YE < o}of intervals of (T all have different
324
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
ranks. Since o is an interval of S(@), it follows that S(@) cannot be KOcategorical. Since by Theorem 13.116, S(@)is self-additive, it follows from Corollary 13.103 that there is a countable model I of T such that I . CI is different from I . fl whenever ct # p. Let S’(@)be a saturated model of Th(S(@))and let S’ be obtained from S by replacing S(@)by S’(@).Since S(@) is saturated, it follows that each I . a < S’(@). If we let S, be the result of replacing S‘(@)by I . c( in S‘, then each S , < S’ and { u E S,I S, k 4 [ u ] for all 4 E @} = I . a. Hence these c countable models are all distinct models of T since any isomorphism from S, to S, must carry I, onto I , . Thus if the sequence {S(@,)ln < a}contains infinitely many distinct KO-categoricallinear orderings, then T has c distinct countable models. We thus need only consider the case in which the sequence (S(@,)ln < 01) contains only finitely many distinct KO-categorical linear orderings. As an example of this case we can take A = A , + 3 + A , + 3 + A , + 3 + . . . , where A, = (q 2 * q ) . n. Here again I T ( A ) has order type w and the condensation classes have order type q, 2 . q, and 3. Thus I T ( T ) consists of @, < Q2 < Q 3 < . . . < 0,and we must again calculate S(@).We leave this calculation to the reader.
+
+ +
EXERCISE 13.120: Verify that if A = A , + 3 A , + 3 + . . . , where A , = ( q 2 . q) . n, then S ( @ ) = (3 ( q 2 . q)(w o*)) . 5 * q.
+
+ +
The phenomenon that occurs in this example, namely, the existence of c . . . , occurs in all cases in which ( S ( @ , ) l n < o)contains only finitely many different models. That is, we claim that there is an increasing sequence { i(n)ln < u}of natural numbers such that i(n 1) 2 i(n) n and such that S(Qicn,) N S(QiCO,) for all n 2 0, S(@i(n)+l) ‘v S(Qi(,)+ 1) for all n 2 1, and, more generally, S(Qi(,)+;)= S(@i(,l+,)for all n 2 j and all j < a. If we now define A , = S(@,(,,)+ S(Qi(,,)+ + . . . S(@‘i(n)+n+ ,), then each A, is isomorphic to an initial interval of A,+ 1. To prove the existence of {i(n)ln< o},we define it inductively together with a contracting sequence {Z(n)ln < w ) of infinite subsets of N . Let B,, B,, . . . ,B,-, be the finitely many distinct models in the sequence { S ( @ , ) l n < w ) , and let Ij(0)= { nI S(@,)N B j } for each j < k. Since { Ij(0)I j < p } is a partition of N , some Ij(0)must be infinite; call it I(0)and let i(0)be its smallest element. Proceeding inductively, we assume that i(0) < i(1) < . . . < i(k) have been defined, i(t + 1) 2 i(r) + t for every t < k, and for every t 5 k and every n E I(t), s(Qn) S(@i(O),S(Qn+ 1) L s ( Q i ( i ) + 11, . . - S ( Q n + t ) S(@i(i)+t). We now partition I(k) - (0, 1 , 2 , . . . , i(k) + k - 1) into { I j @ + 1)1j < p ) , where I j ( k + 1) = { n 2 i ( k ) + klS(@,+k+ ,) N Biand n E I ( k ) } .Then some Ij(k + 1) must be infinite, so we define that one to be I ( k + 1) and we define its smallest A E A, EA,
+
+
3
+
6.
FINITELY AXIOMATIZABLE LINEAR ORDERINGS
325
element to be i(k + I). The induction hypotheses continue to hold; hence the existence of (i,,1 n < w } is substantiated. We now show that the sequence {A,ln < w } of KO-categorical linear orderings is pairwise non-isomorphic. For suppose that n < m and that A, 'v A,. Now A , is isomorphic to the initial interval C of A,. Since C is a finite sum of S ( m j ) ,each of which is definable, it follows that C is definable, and, by Theorem 13.38, C is definable in A,. It follows that more 1-types are realized in A , than in C, so that C. and hence A,, cannot be isomorphic to A,. Hence { A , In < w } is pairwise non-isomorphic. Let cr be the order type of A = U{A,I n < 01, where in this union each A , is identified with the corresponding initial interval of A,, Then, as in the preceding case, S(@)has an interval of order type cr and, since cr cannot be KO-categorical,S(@)is not KO-categorical.Then, as in the preceding case, since S(@) is self-additive, T has c distinct countable models. This completes the proof of the main result of this section. THEOREM 13.121: (Rubin) Let T be a complete theory of linear urderings. Then T is either N,-cutegorical or has exactly c distinct countable models.
Using Corollary 13.103, the proof we have given of Theorem 13.121 can be modified to obtain the following result. COROLLARY 13.122: (Rubin) Let T be a complete theory of linear orderings. Then either T is KO-categorical,S,(T) is uncountable for some n, or there is a countable model A of T that has a self-additive interval I such that if A = A , + I + A,, then, whenever CI # p, the linear orderings A , + I ' C I + A , and A , + 1 . /3 + A , are not isomorphic.
56. FINITELY AXIOMATIZABLE LINEAR ORDERINGS
Our goal in this section is to prove that a statement that is true in some linear ordering is true in some finitely axiomatizable linear ordering. This was mentioned earlier as Theorem 13.23: Given any linear ordering A and M ; i f A is any n, there is a finitely axiomatizable M E A? such that A scattered, then M can be chosen from d oThis . was conjectured by Lauchli and Leonard [8] and was finally proved by Myers [12], Shelah, and Schmerl (independently); our proof is patterned on that of Amit [l].
326
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We restrict our attention for now to scattered linear orderings. The general case will be considered subsequently. Since we know, by Corollary 7.10, that any countable scattered linear ordering is G,-equivalent to an element of Ao,it suffices to prove, by the transitivity of -,, that every element of ,Aois G,,-equivalent to some finitary linear ordering. Thus, proceeding inductively, we could assume that A -,F, where F is finitary, conclude by Lemma 6.5.2 that A . w -,F . w, and then try to show that F . w is finitary. The difficulty with this approach is that F . w need not be finitary even when F is finitary; moreover, conditions under which F . w must be finitary are elusive. We digress for a moment from our goal and elaborate on the sentence above. An example of a finitary F for which F . w is not finitary is provided by F = w* + 5 + w (shown finitary in Exercise 13.18.3). To verify that
F.
= O*
+ 5 + (O + 0") + 5 + (w + o*)+ . . .
is not finitary, we observe that F . w (and even F o + w* in the following technical sense.
+ F) contains a copy of
DEFINITION 13.123: Let I be an interval of a linear ordering A. We say that 1 is a copy of w w* if 1 has order type w + 5 . CI + w* for some order type u, and if the first element of I has no immediate predecessor and the last element of I has no immediate successor. If A has an interval I that is a copy of w + w*, then we say that A contains a copy of w + w * .
+
+
+
Note that 5 5 does not contain a copy of w + a*,although F F does, where F = m* 5 + w. We will show later, in Corollary 13.142, that any A that contains a copy of w + w* is not finitary. For now, we will show that F . w is not finitary; a similar argument works for F + F. It follows from Exercise 6.1 1 and Lemma 6.5.2 that
+
F .
-,,
W*
+ (5 + 2").0
for any n; so it suffices to show that for every n. PLAYER I chooses, in his first two turns, the endpoints of a copy of 2" in w* + (5 + 2"). w ; PLAYER II must then choose a point of F . w with no immediate predecessor and a point of F . w with no immediate successor (or lose in two more turns). The interval of F . w thus determined must have order type w F . m w* for some m 2 0. But, by Exercise 6.10.2, PLAYER I has the winning strategy in C,(2" - 2, w + F . m + w * ) and hence in G , , 2(w* (5 Z n ) . w, F . 0).Thus F . w is not finitary. This completes the digression.
+ + +
+
6.
FINITELY AXIOMATIZABLE LINEAR ORDERINGS
327
The approach in [l] is to use the appropriate condensation map. Thus we will define a condensation c of each linear ordering A E .Aoso that each c(a)is finitary and so that c[A] has lower rank than A , and hence is G,-equivalent to a finitary F . We will then replace the sum A
=
c ( c ( a ) c[A]} E
by a sum that is indexed by F , is (3,-equivalent to A , and is finitary. In order to replace A by a sum indexed by F , we need to think of c[A] as not just a linear ordering, but as a labeled ordering-where each element c(a) E c [ A ] is labeled by the G,-equivalence class of c(a). If it turns out that c [ A ] -,F not only as linear orderings but as labeled orderings, then we can easily take a sum indexed by F that is G,-equivalent to A. As we will see, each c(a) can be taken to be G,,-equivalent to one of a j n i t e set of linear orderings, so that the labeling of c[A] determines a finite partition of c[A]. (We have been speaking of “labelings’’because of similarities to the discussion in 58.6, but it will be more convenient to talk about partitions.) Because of these considerations we need to enlarge our scope so that we consider partitioned orderings, as well as ordinary linear orderings, as fair game for the Ehrenfeucht analysis. DEFINITION 13.124: Let A be a linear ordering and let A , , A l , . . . , A,be an ordered sequence of k subsets of A that partition A , where k is a fixed natural number. The structure ( A ; & , A , , . . . ,A k - 1 ) is called a partitioned ordering. In some contexts, we will use A to denote this structure. If there is a possibility for confusing the linear ordering A with the partitioned ordering A, then we will use the notation A k for the latter. If we wish to emphasize the number k, then we will speak of A k as a k-partitioned ordering. (We will not introduce the notion of partitioned order types, although that too can be easily defined.)
Given two k-partitioned orderings A k = ( A ; A , , A , , . . . , Ak-l)and Bk= ( B ; B , , B , , . . . ,Bk-,),thedefinitionoftheEhrenfeuchtgameG,(Ak,Bk) is included in Definition 13.1. In order for PLAYER II to win, he must so select his moves that, if a, and b, denote the elements of A and B chosen at the tth turn, not only must a, < A a,
if and only if
b, < B b,
if and only if
b, E B,
for all s, t 5 n, but also a, E A j
for all s I n and all j < k. Thus PLAYER 11 must match PLAYER 1’s moves both with respect to the linear orderings and with respect to the partitions.
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
If he can do so, then PLAYER 11 has the winning strategy in G,(Ak,B k ) ;if he cannot, then PLAYER I has the winning strategy. For k = 1 these notions coincide with the definitions given for ordinary linear orderings. The exercise below, involving the case k = 2, gives some indication of what happens at the next level. (Assume that G,-equivalence has already been discussed for k-partitioned orderings-it will be soon, and this exercise will thus serve as motivation.) EXERCISE 13.125 : Corollary 6.9 can be interpreted, for n = 2, as saying that every finite linear ordering with more than 3 elements is G,-equivalent to a finite linear ordering with no more than 3 elements. This is the situation for k = 1; the statements below concern the case where k = 2. (1) Show that there is a finite ( A ; A , , A , ) where A has 8 elements which is not G,-equivalent to any finite ( B ; B 1 , B 2 ) ,where B has fewer than 8 elements. (2) Show that every finite ( A ;A 1, A , ) where A has more than 8 elements is G,-equivalent to a finite ( B ; B1, B , ) , where B has no more than 8 elements.
We now proceed by reviewing the analysis of Ehrenfeucht games on linear orderings presented in Chapter 6 and determining to what extent it applies to Ehrenfeucht games on k-partitioned orderings. (We will be referring to this discussion when, later on, we speak of the "adapted" versions of the statements of Chapter 6.) If we define Ak Bk to mean that there is an isomorphism f : A -+ B such that f [ A j ] = B j for j < k, then Lemma 6.3 still holds. If we define Ak + Bk (for A and B disjoint) to be (A
+ B ; A0 U B o , . . . , Ak-1
U Bk-1)
and similarly for generalized sums (of partitioned orderings over an index set that is an ordinary linear ordering) and shuffles, then Lemma 6.5 and Exercise 7.18 still hold. If we define I k to be ( I ; A , n I , . . . ,Ak-1 n Z)
for any interval I of A, then the following form of Theorem 6.6 is correct: G , , ,(Ak,B k )E I1
if and only if (i) for every j < k and every a E A j there is a b e Bj such that C,((A'")k,(B'b)k)E I1 and G,((A<")k,(B
6.
FINITELY AXIOMATIZABLE LINEAR ORDERINGS
3 29
Corollary 6.7 and Definition 6.8 can be given as before, so that we can speak of G,-equivalence and G-equivalence of k-partitioned orderings. The careful calculations of Corollary 6.9 and Corollary 6.12 are completely upset, but the proof of Theorem 6.13 is correct without any changes except that f(1) = 2k instead of 2. Thus, for each k and each n, there are only a finite number of G,-equivalence classes of k-partitioned orderings. Consequently, the following version of Corollary 6.9 is correct: For each k and each n, there is a natural number g(k, n) such that every finite k-partitioned ordering with more than g(k,n) elements is G,-equivalent to one with no more than g ( k , n) elements. Corollary 6.9 thus says that g( 1, n) = 2” - 1; the difficulty in calculating explicit values of g(k,n) for k > 1 will be apparent to the reader who tries to show that g(2,2) = 8 in Exercise 13.125. Corollary 6.12 holds if the partitions are “periodic”; that is to say, if F is a finite kpartitioned ordering and a is any order type, then F.o-,F.(o+[.a) for every n. The proof of this adapted version of Corollary 6.12 uses an adapted version of Exercise 6.1 1, where the function g(1,n) = 2” - 1 has to be replaced by a function h(k, n) defined inductively by h(k, 1) = g(k, 1) and h(k, n + 1) = max{2h(k,n) 1, h(k,n) g ( k , n) l}. (Note that in products A . B, as in generalized sums, the index set B is taken to be an ordinary linear ordering, so that A . B can be thought of as B copies of the partitioned ordering A.) We define a k-partitioned ordering to be finitary as before; that is, Ak is finitary if, for some n, whenever Bk is a countable k-partitioned ordering and Bk -” Ak, then Bk Ak. The following result is fundamental for the subsequent analysis. (We noted earlier that for arbitrary finitary linear orderings F, F . o need not be finitary.)
+
+
+
-
LEMMA 13.126: Zf F is a jinite k-partitioned ordering, then F . o is finitary.
+
Proof : Suppose that F has m elements. We claim that f ( F . u)2 2m 1. To prove this we must show that if B is any countable k-partitioned ordering and if B N~,,,+ F * o,then B F . o.But since 2m + 1 2 3, we know that the order type of B, ignoring its partition structure, is o 4 . a for some order type a (since the order type of F . o is 0).It thus suffices to show that the first w elements of B are actually F . o and that each (-sequence in B is actually F * 5, for then B 2: F . (o+ 5 * a), which by Corollary 6.12 (adapted) is G-equivalent to F . w .
-
+
13.
330
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
If PLAYER I chooses the first m elements of F . w, then he would win in the game G,, l(B,F . w ) unless B begins with an F. Proceeding inductively, assume that B begins with n copies of F. If PLAYER I chooses the m elements of the nth copy of F in B and then chooses the next m elements of B, then PLAYER 11 must pick 2m successive elements of F . w, lest PLAYER I win on his (2m 1)st turn. By the periodicity of F . w, the second group of m elements of F . w must be exactly like the first group, so the same must be true of the 2m elements of B ; hence B begins with n 1 copies of F, so, by induction, B begins with F . w. Similarly, for any copy of [ in B, PLAYER I can choose 2m - 1 consecutive elements in that [, and, since PLAYER 11’s selections in F * w must contain a copy of F , we conclude that PLAYER I’S selection from B also contains a copy of F. This copy of F propagates copies of F in both directions, as in the preceding paragraph. Thus each copy of 5 in B is actually F .[, so that B 2 F . (w 5 . a), as required.
+
+
+
COROLLARY 13.127: If F , , F , , and F , are j n i t e k-partitioned orderings, than F , F , . w, F , . w* F,, and F , . w* F , F , . w arealljinitary.
+
+
+ +
DEFINITION 13.128: A basic (finitary) k-partitioned ordering is a kpartitioned ordering that is in one of the following forms, where F , , F,, and F 3 are finite: F , , F , + F , . o , F 1 . w * + F , , F ; o * + F , + F , . w .
We use these notions in the following way.
-
DEFINITION 13.129 : Define a condensation c, of a k-partitioned ordering A by stipulating that b c if every subinterval [x, y) of [b, c] is G,#-equivaIent to a finite k-partitioned ordering. (Recall that when condensations are under discussion, the notation [b, c] is to be interpreted as [c, b] if c < b.)
This is clearly a condensation map for each n. Furthermore, assuming that n 2 3, as we will do throughout this section, the order type of each subinterval [x, y) of an interval [b,c] within c,(a) must be either finite or of the form w + 6 . a + a*for some a. (If A F, where F is finite, then A has a first and last element, and all elements have immediate successors and predecessors.) Thus the order type of c,(a) is either finite or is of one of the following forms: w + 4 . a + w * , c . a + w * , w + 5 . a, 5 . a. The use of Ramsey’s Theorem in the proof below is reminiscent of the proof of Proposition 7.8.
-,
6.
FINITELY AXIOMATIZABLE LINEAR ORDERINGS
33 1
PROPOSITION 13.130 : Let Ak be a k-partitioned ordering. Then for each a E A, c,(a) is G,-equivalent to a basic (finitary)k-partitioned ordering. Proof: Suppose that c,(a) has no last element. Choose an o-sequence a = a, < a, < a, < . . . that is cofinal in c,(a). Then each interval [ a i , a j ) is G,-equivalent to a finite k-partitioned ordering. By Corollary 6.9 (adapted), we may assume that there are a finite number of finite k-partitioned orderings such that each [ai,aj)is G,-equivalent to precisely one of them. Thus we can partition [ N I 2 into a finite number of sets in such a way that {i,j} and {k,l) are in the same set if and only if i < j , k < 1, and [ a i , a j )w n [ak,a,). Applying Ramsey’s Theorem (Theorem 7.4), we can choose a,, < a,, < ai2< . . . so that, for each r, [u,,,u,,+~)is G,-equivalent to a fixed finite k-partitioned ordering F. Thus {b E c,(a)(b 2 a } -,[a, a,,,) + F . o. But [a, a,,) is G,equivalent to some finite G, so that { b E c,(a)lb 2 a} -,G F o. If c,(a) has P last element c, then { b E c,(a)l b 2 a } = [a, c], which is G,-equivalent to a finite k-partitioned ordering. Similarly, {b E c,(a) b < a } is G,-equivalent to either a finite k-partitioned ordering or to F , . o* G I for some finite F , and G,. Thus in every case c,(a) is G,-equivalent to a basic k-partitioned ordering. H
+
I
+
Note that c,(a)is naturally G,-equivalent to a basic k-partitioned ordering in the following sense. DEFINITION 13.131 : We say that A is naturally G,-equivalent to F , F , . o,where F , and F , are finite, if A can be written in the form A = B c { B i l i < a}, where B -,F , and B, -,F , for all i. (The definition is analogous for other basic finitary k-partitioned orderings.)
+
+
EXERCISE 13.132: Show that if A m 4 D, where D is a basic finitary k-partitioned ordering and q 2 f ( D ) . then A is naturally G,-equivalent to D.
Now, by Corollary 6.9 (adapted), there is a finite set of finite k-partitioned orderings such that each finite k-partitioned ordering is G,equivalent to one of these. Hence there is a finite set of basic k-partitioned orderings such that each basic k-partitioned ordering is naturally G,-equivalent to one of these. Thus we conclude that there is a fixed finite ordered sequence D o , D,, . . . ,D, - of distinct basic finitary k-partitioned orderings such that, for every Ak and every a E A, c,(u) is naturally G,-equivalent to some D,. [Note that it is possible that c,(a) is naturally G,-equivalent to more than one D,; to arrange for uniqueness seems troublesome and is unnecessary
332
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
since, in the definition below, we insist on a uniform and optimal method of selecting a Di corresponding to each possible c,(a).] DEFINITION 13.133: Given a k-partitioned ordering Ak, we define, for each c,(a), a basic ordering D(c,(a))from the sequence Do, D,, . . . , D,by the following three-step procedure: First, select those Di such that c,(a) is naturally G,-equivalent to D i ; second, eliminate Dj if c,(a) w q D j but c,(a) -q Di for some Di and some q ; and third, let D(c,(a)) be the first remaining Di. We then define ( C , [ A ~ ]to) ~be the s-partitioned ordering ( c , [ A k ] ;Po,!‘,, . . . ,Pspl),where Pi = {c,,(a))~D(c,,(a)) = Di).
,
Note that the third step in the procedure for defining D(c,(a))guarantees uniformity. That is, if c,(a) rr c,(b), then D(c,(a))= D(c,(b));consequently, if Ak Y Bk, then (c,[Ak])”‘v (c,[Bk]l,”.The second step is a condition of optimality; no Di is more similar to c,(a) than D(c,(a)).The first step, of course, guarantees that D(c,(a)) will be naturally G,-equivalent to c,(a). Following the program we outlined at the beginning of this section, we now want to show that for A E Ao,c,[A] is also in A. and has lower rank. Then, in trying to show that A is G,-equivalent to something finitary, we ] } argue inductively that A is G,can resolve A into ~ { c , ( a ) ~ c , [ Aand equivalent to a finitary sum of pieces each of which, by Proposition 13.1 30, is also finitary. Before we do so, we must of course define the set of k-partitioned orderings corresponding to .A?’, since that is where A really lives. At the same time, we must present the notion of rank that is most useful for our purposes. (As observed earlier, we have not introduced partitioned consisted of order types, d o will k consist order types, so that, although &lo of partitioned orderings.) DEFINITION 13.134: We will define, by induction on n, a class .MOk(n) of scattered k-partitioned orderings of rank I n . For n = 0, A o k ( 0 consists ) of all finite k-partitioned orderings. Assuming that A o k ( n )has been defined, we define A o k ( n + 1) to consist of all finite sums A I k+ A,k + . . . + A / , where each A: is either in .Aok(n)or is w or w* copies of an element of -&‘:(n). We define -A?‘: = U,A’,k(n). If Ak E A:, we define the rank of Ak to be the smallest n such that Ak E .Mok(n).
LEMMA 13.135: Let Ak E d oand k let 1 be an interval ofA. Then l kE dtok and the rank of’I k is no more than the rank of Ak. then Ak can be written as A I k+ A Z k+ . . + Amk,where (2) l f Ak E dok, each Aik is either jinite or is w or a* copies of a k-partitioned ordering of lower rank than Ak.
6.
333
FINITELY AXIOMATIZABLE LlNEAR ORDERINGS
Proof : By induction on the rank of Ak.
A
PROPOSITION 13.136: For each Ak E d o k and each n, ( c n [ A k ] )E” dos and has lower rank than Ak. Proof : We will first show that A k = A l k + . . . + Amkcan be rewritten as Ak = M I k . . . M,k, where for each i, either
+
+
(i) c,(a) = M t for all a E M t , or (ii) M: consists of o or o*copies of something of lower rank than A , each of which contains c,(a) whenever it contains a.
We may assume, by Lemma 13.135.2, that each A,k is either finite or is o or o*copies of something of lower rank. So assume that AIk= Bi I i < o}, where each Bi v B and B has lower rank than A. Suppose that for some a, E B , and b , E B , we have c,(a,) = c,(b,). If a , is the element of B, corresponding (under the assumed isomorphism) to a, and if c,(a,) = c,(a,), then c,(a,) contains all of Atk. Otherwise b , < a,, so that if we set T o = ( a E Bola E c,(ao)} and I , = (a E B , l a E c,(a,)}, then c,?(a0)= T o + I , . Using the isomorphisms between the B i , we can write
I{
+ N o + T o )+ (I, + N , + T,) + (I, + N , + T 2 )+ . . . , where c,(a) = T i + I i + , for each a E Ti + Ii+, and for each i. Setting Ci = N i + T i+ I i + , for each i, we get A,k = I , + c { C i l i < a}.Note that AIk= ( I ,
each Nj, Ti, and I i is in &lok, so that each Cj E Aok. Also the rank of C iis no more than the maximum of the ranks of Ni,T i , and I i , each of which, by Lemma 13.135.1, is no more than the rank of B i , which is less than the rank of A . Finally, for a E I,, c,(a) 2 I,. Examining the various cases of the preceding paragraph, we see that A,k can be decomposed into at most two pieces, one of which is completely included in c,(a) for any of its elements a, and the other satisfies (ii). By decomposing each A,k in this way, we obtain a decomposition of Ak into at most 2rn pieces; by combining adjacent pieces in the same condensation class, we arrive at a decomposition Ak = M I k+ . . . + M,k that has the required properties. We now consider c,[M?] for each i. If c,(a) = M t for a E M i ,then ( C , [ M ; ] ) E~ ,Ao‘and has rank 0. Otherwise M t = x { M i j lj < o},where each M i j = M and is a union of condensation classes. But then ( c , [ M t ] ) ”= ~ ( ( c , [ M , , ] ) “< / jo}; by the induction hypothesis, each ( c , [ M i j ] f Shas rank smaller than M i j and is in A,“. By uniformity, (c,[Mij])” (cn[Mij.])sfor and its rank is smaller than everyj, j’ < (0.This implies that (c,[M,“])”E A?’,‘ that of MF. Thus, in any case, c,[M?] E A’,‘ and has rank smaller than M/‘.
-
334
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
+ +
+
But c,[Ak] = c , [ M , ~ ] c , [ M ~ ~ ]. . . c,[M,k], so that c,[Ak] E dos and has rank equal to the maximum of the ranks of the c n [ M t ] ,which is smaller than the maximum of the ranks of the M b ; hence the rank of c n [ A k ] is smaller than the rank of Ak. We have thus defined, for each n, a condensation c, of each k-partitioned and we have shown that each c,(a) is finitary and that ordering Ak E AOk, cn[Ak]has lower rank than Ak.To prove that Ak is G,-equivalent to a finitary k-partitioned ordering, we will assume that c , [ A ] is G,-equivalent to a finitary s-partitioned ordering F and replace A = ~ { c , ( aE)c , [ A ] ) by a sum of D;s indexed by F.
DEFINITION 13.137 : Given an s-partitioned ordering F S = ( F ; F o , F , ,...,F 5 - l ) ,
we define the sum FDk to be the k-partitioned ordering ~ ( D “ ( a )E ~Fa} , where Dk(a)2: Di if a E Fi. The following lemma allows us to make the replacement described above. LEMMA 13.138: Assume that ( C ” [ A ~-, ] ) F“. ~ Then Ak -,FDk.
A
Now all of the pieces are in place and we can proceed to the proof of the main theorem. THEOREM 13.139: (Amit, Schmerl, Shelah) Every Ak E .,dokis G,equivalent to a finitary k-partitioned ordering. Proof : We proceed by induction on the rank of Ak.If Ak has rank 0, then Ak is finite and there is nothing further to say. If Ak has positive rank, we apply c, to Ak and write Ak = ~ { c , ( aE)c , [ A k ] } . By Proposition 13.136. (c,[Ak])”E .A/ and has lower rank than Ak.Hence by the induction hypoth-
esis, (c,[Ak])”-,F S for some finitary s-partitioned ordering F’. Therefore, by Lemma 13.138, Ak -,FDk. Also, by Proposition 13.130, we can assume that each c,(a) is naturally G,-equivalent to D(c,(a)),which is on the list D o , D,, . . . ,D,- of basic finitary k-partitioned orderings. We thus need only show that FDk is finitary. We first show that, for each element d E FDk, c d d ) is precisely the copy of D j to which d belongs. Since each copy of D j has order type it is clear that each c,(d) includes the copy of D j to which it belongs. On the other hand, those D j that are included
<[,
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335
FINITELY AXIOMATIZABLE LINEAR ORDERINGS
in c,(d) determine an interval of F”;if that interval is non-trivial, it must contain some element together with its successor since F” is scattered. Thus F” contains two successive elements, one in Fi and its successor in F j , and c,(d) 2 D , + D,, where D, and D, are consecutive copies of Di and D j in FDk. Since Fs -,( C , [ A ] ) ~we, conclude that c , [ A ] has successive elements c,(b,) and c,(b,), where c,(b,) and c,(b,) are naturally G,-equivalent to Di and D j , respectively. (To make this conclusion, we must assume that n 2 3, but the theorem for n 2 3 implies the same result for n < 3.) Now either c,(b,) has no last element or c,(b,) has no first element since otherwise c,(b,) = c,(b,). Hence either D, has no last element or D, has no first element. There are now three cases, which, of course, are treated similarly. Suppose that D, has no last element and that D , has no first element. Then D , ends in F , . w and D , begins with F , . w * . Since c,(b,) and c,(b,) are naturally G,-equivalent to D, and D,, respectively, we may assume that c,(b,) ends in [ c t , c, ,), where each [c,, c, + ,) -,F , , and that c,(b2) begins with x - r [ d - , - l , L I ) , where each [ L - , , d - J -,F,. But if x ~ c , ( b , )and Y E c,(b,), then
1,
[x, Y ) = [Ix,ci) +
-,,
1[ci+r c, 7
I
F
+ r t 1)
+ 1[ d j - t --f
+ F , . + F,
. O*
1, d j - t )
+ [dj, y )
+ G.
But F , . w + F , . w* is an interval of D, + D, and therefore is G,-equivalent to a finite ordering. Hence [ X , J J ) is G,-equivalent to something finite for every x E c,(b,) and y E c,(b2),so that c,(b,) = c,(b,), which is a clear contradiction. Hence for each d E FDk,c,(d) is precisely the copy of D j to which d belongs. Now by assumption, Fs is finitary; let p = f ( F s ) ) .Each Di is finitary; let q = max{{f(Di)(i< s} u { i i + 1 ) ) . We will show that ,f(FDk)5 Y, where r = p + q + n + 2. Assume that Bk -,FDk. We will show that if we write Bk = x { c , ( b ) E c , ~ [ B ~ ]then } , c,[Bk] - p F“ (and hence cfl[Bk] F”)), and that if D(c,(b))= Di [so that c,(b) O i l , then c,(b) m 4 Di [and hence c,(b) O i l . This will imply that Bk FDk. showing that FDk is finitary, and completing the proof of the theorem. If PLAYER I has a winning strategy in Gp(c,[Bk],FS), then PLAYER I will also have the following winning strategy in G,(Bk,FDk),contrary to assumption. For his first p moves, PLAYER I pretends he is pursuing his winning strategy in the game G,(c,[Bk], F’). choosing, at each of his turns, elements from different condensation classes. At some point, to avoid losing in p turns, PLAYER 11 must either choose a second element from some condensation class or choose, in response to PLAYER 1’s choice of x, an element y such that D(c,,(x))# D(c,(y)).In the first case, PLAYER I can win in n + 2 further moves-let x, < x2 be the two elements in the same condensation
-
-
-
336
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
class, and let y 1 < y 2 be the corresponding elements in different classes. first chooses the end points z1 and z2 of a bounded interval within [y,,y,] that is not G,-equivalent to any finite k-partitioned ordering-which PLAYER 11 cannot duplicate within [xl, x,]-and demonstrates that difference in n more moves. In the second case, PLAYER I can win in q + n + 2 further moves-for if PLAYER 11’s choice x did not satisfy D(c,,(x))= D(c,( y)), where y was PLAYER 1’s choice, then PLAYER I, in his next q moves, demonstrates that fact and wins unless PLAYER 11 goes outside of c,(x) or c,(y), in which case PLAYER I wins in at most n + 2 further moves, as in the preceding case. Similarly, suppose that PLAYER I has the winning strategy in G,(c,(b),Di), where D(c,(b))= Di, so that c,(b) -,D ibut c,(b) N, Di.We claim that it follows that PLAYER I also has winning strategies in G,(c,(b),c,(d) ) for all d E FDk. For suppose, to the contrary, that c,(b) c,(d) for some d E FDk. Then, by Exercise 13.132, since c , ( d ) = D j for somej # i and since q 2 .f(Dj), it follows that c,(b) is naturally G,-equivalent to c,(d); since q 2 n + 1, this implies that D(c,(b)) could not have been Di. Hence PLAYER I has the winning strategy in G,(c,(b),c,(d))for all d E FDk. But then PLAYER I has the following winning strategy in G q + , +3(Bk,FDk), contrary to assumption. PLAYER I first picks an element b E B such that for no ti E FDk is c,(b) c,,(d).Assuming PLAYER 11 picks a particular element d E FDk, PLAYER I uses his winning strategy in G,(c,(b),c,(d)) to achieve victory in q more moves, unless PLAYER 11 picks, on one of his turns, an element outside of c,(b) or c,(d). Then, as in the argument above, PLAYER I wins in ra + 2 additional moves, so that he has the winning strategy in Gq+n+3(Bkr F D ~ ) contrary , to assumption, since q + n + 3 I r. PLAYER I
-,
-,
The reader who is so inclined can replace the second half of the proof of Theorem 13.139 by an appeal to Corollary 13.12 and thereby obtain an axiomatization of FDk using the given axiomatizations of Fs and D o . D1, . . . ,DsCOROLLARY 13.140 : Every scattered linear ordering is G,-equivalent to a jinitary linear ordering, for every n. Proof : If A is scattered, then by Corollary 7.10 there is an M E A0such that A -,, M. By Theorem 13.139, M is G,-equivalent to a finitary linear ordering.
If # is a statement true in a scattered linear ordering and # has quantifier depth n, then 4 is also true in a finitary linear ordering that, by Proposition 13.19, is finitely axiomatizable.
6.
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FINITELY AXIOMATIZABLE LINEAR ORDERINGS
COROLLARY 13.141 : Any statenterit trite in a scattered linear ordering is true in a jinitely axiomatizable scuttered linear ordering. W
Before turning to the non-scattered case, we make several additional remarks concerning scattered linear orderings. COROLLARY 13.142: Let A be (1 scattered linear ordering that contains u copy ofo o*.Then A is not ,finitary.
+
-
Proof: Assume that A is finitary and that f ( A ) = n. By Corollary 7.10, there is a B E A0such that A -,, B ; B is then finitary, f ( B ) I n and A B.
By Exercise 7.7.2, there is a number k that is the size of the largest maximal finite interval of B, k = max { Ic,(b)((b E B and (cF(b)lis finite}. Since A B, A has no larger maximal finite intervals. Also since A B,o o* occurs in B ; otherwise, PLAYER I could win the game G,+ 3(A,B), where p 2 n, k, by choosing the first and last elements of a copy of o o*in A. Now replace each copy of o o* in B by 2p to get C. Then, by Exercise 6.11, B - p C, so that B C . But since 2p occurs in C but not in B, B * p + 2 C . This contradicts the assumption that A is finitary.
-
-
+
+
+
The proof of Corollary 13.142depends only on the statement of Corollary 7.10 and nowhere else uses the assumption that A is scattered. Thus, using Corollary 7.21, we can conclude immediately that no linear ordering containing a copy of w o*is finitary. After verifying the corollary and the above observations, Fellner [7] conjectured that the following converse is correct: Let A be a linear ordering of A. that does not contain o + o* copies of any linear ordering, except possibly <; then A is finitary. (To say that A contains o o* copies of B is to say that there is an interval I of A that can be written
+
+
z=C{B,JiEW+[.a+o*},
where each Bi is G-equivalent to B, and that no interval of A containing I has this property.) Fellner [7] verified this conjecture for certain subclasses of .ko.A related conjecture [7] is that for finitary F , F . o is finitary if and only if F F is finitary. We return now to the general case, which, as we will see, can be handled in much the same manner. Our goal is to show that every linear ordering is G,-equivalent to some finitary linear ordering. As before, because of Corollary 7.21, we need only prove this for order types in A. We will define a class A!* of partitioned orderings (whose 1-partitioned orderings will have precisely the order types in A') and show, by induction on rank, that
+
338
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
every k-partitioned ordering in A’* is G,-equivalent to a finitary k-partitioned ordering. The desired conclusion will then be a corollary. To carry out the main argument we will define a condensation c,* so that, as in the scattered case, (a) If A E A*,then c,*[A] E .,ti!* and has smaller rank than A. (b) Each c,*(a) is G,-equivalent to something finitary. Then we can resolve A into x(c,*(a) E c,*[A]], argue inductively that A is G,-equivalent to a finitary sum of pieces each of which is finitary, and then conclude that, because of the special character of these pieces, their sum is finitary. Because of the similarities to the scattered case, we will be brief. DEFINITION 13.143: We define, by induction on n, a class J k ( n ) of k-partitioned orderings. For n = 0, A k ( 0 )consists of all finite k-partitioned orderings. Assuming that A k ( n ) has been defined, we define A k ( n 1 ) to consist of all finite sums A I k Azk . . * Amk,where each A: is either in .Lk(n),or is o or w* copies of an element of A k ( n ) ,or is a finite shuffle c{B , , . . . , B r ; of elements of A k ( n ) . We define Ak= u , J k ( n ) , and we define A’* = U k A k . If Ak E Ak, we define the rank of Ak to be the smallest n such that Ak E , A k ( n ) .
+
+
+
+
LEMMA 13.144: (1) Let Ak E = d kand let I be an interval of A . Then I kE .Ak and the rank of I kis no more than the rank of A‘. ( 2 ) i f A k E J k ,then Ak can be written as A l k A z k . . . Amk, where each A: is eitherjnite, or is w or w* copies of a k-partitioned ordering of lower rank than Ak, or is a j n i t e shufle of k-partitioned orderings of lower rank than Ak. A
+
+
+
We will define the condensation c,* so that each c,*(a) is either precisely c,(a), and hence is G,-equivalent to a basic k-partitioned ordering, or else is G,-equivalent to a finite shuffle of basic k-partitioned orderings. The following lemma will thus imply that each c,*(a) is G,-equivalent to a finitary k-partitioned ordering. DEFINITION 13.145: An elementary k-partitioned ordering is either a basic k-partitioned ordering or a finite shuffle of basic k-partitioned orderings. LEMMA 13.146 : Let G = u { G G, , . . . ,G,} be an elementary k-partitioned ordering. Then G is @fury. A
6.
339
FINITELY AXIOMATIZABLE LINEAR ORDERINGS
EXERCISE 13.147: Lemma 13.126.1
Prove Lemma 13.146. [Hint: Review the proof of
Referring back to Definition 13.129 and Proposition 13.130, we see that the condensation map c,, is well defined for arbitrary k-partitioned orderings and that for any element of an arbitrary k-partitioned ordering Ak,c,(u) is G,-equivalent to a basic (finitary) k-partitioned ordering, whether or not Ak is scattered. Thus we may assume that for every Ak and every a E A, c,(a) is G,-equivalent to some member of the list Do, D,, . . . ,Dsp1 of basic k-partitioned orderings. Define D(c,,(a)) for each a E Ak and then define (c,[Ak])” exactly as in Definition 13.133. The definitions below, which lead up to the definition of c,*(a), are reminiscent of Definitions 8.38 and 8.39. DEFINITION 13.148: An interval I of cn[A] of order type q or 1 is said to be homogeneous if there is a subset 9 c {Do,D1,.. . ,Ds,} such that if c,(a) E I , then D(c,(a)) E 9,and if, whenever c,(a) << c,(b) are elements of I , then, for each D j E $3, there is an element dj E A such that c,(a) << c,(dj) << c,(b) and D(c,(dj)) = Dj.Given a E A, there is a maximal homogeneous interval I, of c,,[A] that contains c,(a). We define c,*(a) = x{c,(b) E I,}. PROPOSITION 13.149: Let Ak be a k-partitioned ordering. Then for each a E A , c,*(a) is G,-equivalent to a n elementary jinitary k-partitioned ordering. Proof: By Proposition 13.130. c,(a) -,G j for some basic G j . If I, = {c,(u)), then c,*(a) = c,(a). If I , has order type q, then, by Exercise 7.18 (adapted), c,,*(a) is G,-equivalent to a finite shuffle of basic k-partitioned orderings, which, by Lemma 13.146, is an elementary finitary k-partitioned ordering.
We observe that if c,*(a) = Clc,,(b) E l a } ,then c,*(a) is naturally G,equivalent to an elementary k-partitioned ordering in the sense that, in the elementary ordering, each interval c,(b) of c,*(a) is replaced by a basic ordering to which it is naturally (2,-equivalent. We let D o , D,,. . . , D t - be a list of elementary k-partitioned orderings such that D o , D , , . . . ,D,- are basic (as before) and D,,D,+ .. .,Dtare all the finite shuffles of subsets of (Do,D1,. . . ,D,-l>.
,,
DEFINITION 13.150 : Given A k and having defined (c,[Ak])S, we associate with each c,*(a) an element D*(c,,*(a)) from the list Do,D1, . . . , D t - , in
340
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
the following way. If I, is a singleton, then D*(c,*(a))= D(c,(a)). If I , is not a singleton and the subset of ( D o ,D , , . . . ,D,- associated with I , is 9, then D*(c,*(a))is a(9), which is among D,, D S c 1 , . . , D,- We define (c,*[Ak])*to be the t-partitioned ordering ( c , * [ A k ] ;B,,B,, . . . ,B,- ,), where B j = {c,*(a)lD*(c,*(a))= Dj).
,.
Then, as in Definition 13.133, this definition is uniform. That is, ifc,*(a) 2 c,*(b), then D*(c,*(a))= D*(c,*(b)); and consequently, if Ak rr Bk, then (C,*[Ak])f= (C,*[Bk])'. PROPOSITION 13.151 : For each Ak E ,Ak and each n, (c,*[Ak])'E At
and has lower rank than Ak.
Proof: As in the proof of Proposition 13.136, we need to show that A k = A + . . . + Amk can be rewritten Ak = M , + . . . + M,!, where, for each i, either c,*(a) = M f for all a E Mf,or M f is w or a*copies of something of lower rank than A (each copy of which contains c,(a) whenever it contains a), or M f is a finite shuffle of orderings of lower rank than Ak (each copy of which contains c,(a) whenever it contains a). Treating each A,k separately, and proceeding as in the proof of Proposition 13.136, we see that if Ajk is o or w* copies of something of lower rank, then the analysis is still correct. If, on the other hand, A; is a shuffle, then it is clear (by the remarks preceding Definition 13.150) that if b E c,*(a) and b is not in the same copy as a, then c,*(a) 2 A:, whereas otherwise Ajk already satisfies the desired condition. Having represented Ak properly, we now, as in Proposition 13.136, need only analyze c,*[Mjk] for each j , and in particular for those cases where Mjk is a finite shuffle (and c,*[Mjk] is not trivial). So suppose that Mjk = a ( B , , . . . ,B4) and that each copy of Bicontains c,*(a) whenever it contains a. Then cn*[M;] = t ~ ( c , * [ B ~. ]. ,. , c,*[B,]), and since each cn*[Bi] has lower rank than B i , it follows that c,,*[M:] has lower rank than Mjk. DEFINITION 13.152 : Given a t-partitioned ordering
F' = ( F ; F o , F , , . . . ,F,-,), we define the sum F*Dk to be the k-partitioned ordering z { D k ( a ) l aE F } , where Dk(a)= D iif a E Fi. LEMMA 13.153: Assume that (c,*[Ak])'-,F'. Then Ak -,F*Dk.
A
6.
341
FINITELY AXIOMATIZABLE LINEAR ORDERINGS
THEOREM 13.154 : (Amit, Schmerl, Shelah) Every Ak E Mkis G,-equivalent to a finitary k-partitioned ordering. Proof: As in the proof of Theorem 13.139, the only fact needing to be verified is that F*Dk is finitary for a suitably chosen finitary F'. The first step, again, is to show that for each element d E F*Dk, c,*(d) is precisely the copy of D j to which d belongs. To do this, we first rewrite F*Dk as a sum of Di where i < s and, condensing these Di, obtain an s-partitioned ordering that we denote FS. We first show that for each d E F*Dk, c,(d) is precisely the copy of Di to which d belongs, where i < s. Otherwise, those D,included in c,(d) determine a non-trivial interval of FS. If that interval contains some element and its successor, we proceed exactly as in Theorem 13.139. Otherwise, it is dense, so by Theorem 7.17, there is a subinterval of that interval of FS to which there corresponds in F*Dk a shuffle of some subset of { D o , D , , . . . , D,-l}. If that shuffle, or some non-trivial part of it, is contained within some Dk(a) for some a E I', then, as in Theorem 13.139, we could argue that for certain distinct c,(a) and c,(b) in cn[Ak] we should have c,(a) = c,(b); otherwise we could argue that the homogeneous interval I,, for some a E A , was not really maximal. Thus, in any case, c,(d) is precisely the copy of Di to which d belongs, where i < s. It then follows that c,*(d) is precisely the copy of D j to which d belongs, where j < t. By assumption, F' is finitary; let p = f ( F ' ) . Each D j is finitary; let q = max((f(D,)li < t } u { n 1)). We will show that f ( F * D k )5 r, where r = maxip 2n 5, q n 3}. So assume that Bk .vr F*Dk. We will show that if we write Bk = x{c,,*(b)E c,*[Bk]}, then c,*[Bk] F' (and hence c,*[Bk] F r ) , and that if D(c,(b)) = D i [so that c,(b) -,Oil, then c,(b) Dj [and hence c,(b) Oil. This latter implies that if D*(c,*(b)) = D j , then c,*(b) D j , which, together with the above, yields Bk F*Dk, showing that F*Dk is finitary and completing the proof of the theorem. If PLAYER I has a winning strategy in Gp(c,*[Bk],F'), then PLAYER I will also have a winning strategy in G,, 2n+5(Bk, F*Dk), contrary to assumption. For his first p moves, PLAYER I pretends he is pursuing his winning strategy in the game G,(c,*[Bk], F')%choosing, at each of his turns, elements from different c,*-condensation classes. At some point, to avoid losing in p turns, PLAYER 11 must either choose a second element from some c,*condensation class or choose an element of a c,*-condensation class that is not G,-equivalent to the c,*-condensation class of PLAYER 1's last move. In the first case, there are two possibilities. Either the two elements are also in the same c,-condensation class, in which case PLAYER I wins in n 2 further moves, as in Theorem 13.139, or the c,*-condensation class of the two elements must be a shuffle whereas the c,*-condensation classes of the other two elements, since they are distinct, are not part of a similar shuffle.
+ +
-
+ + +
-
-,
-
+
342
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Then, in three more moves, PLAYER I can choose an element whose c,condensation class cannot be duplicated by PLAYER 11’s choice. Hence, as in Theorem 13.139, PLAYER I wins in at most 2n + 1 further moves. In the second case, where PLAYER 11 has chosen an element of a c,*-condensation class that is not G,-equivalent to the c,*-condensation class of PLAYER 1’s last move, there are again two possibilities. Either both of these c,*-equivalence classes are also c,-equivalence classes, in which case PLAYER I wins in 2n + 1 more moves, or at least one is a shuffle, in which case, by choosing an appropriate element of the shuffle, PLAYER I can force PLAYER 11 into the first-case situation and so win in at most 2n + 4 more moves. In this latter case, PLAYER I thus wins in p 2n + 5 moves. The proof that if D(c,(b)) = D i , then c,(b) -4 Di proceeds exactly as in Theorem 13.139, by showing that otherwise PLAYER I has a winning strategy in G,+,+3(Bk,F*Dk).This completes the proof.
+
=
COROLLARY 13.155: (Amit, Myers, Schmerl, Shelah) Every linear ordering is G,-equivalent to a jnitary linear ordering, for each n.
The corollary below can be interpreted as saying that the Boolean algebra of statements of T h ( 9 ) is atomic. COROLLARY 13.156: Any statement true in a linear ordering is also [rue in ajnitely axiomatizable linear ordering. REFERENCES Amit, R., and Shelah, S., The complete finitely axiomatized theories of order are dense, Israel J . Math. 23 (1976). 200-208.[MR 58, #5162] [2] Baldwin, J. T., Blass. A. R.. Glass. A. M. W., and Kueker, D W., A “natural” theory without a prime model. Algebra Ui~irersalis3 (1973). 152-155. [ 3 ] Doner, J., Mostowski, A,, and Tarski. A,, The elementary theory of well-ordering-a metamathematical study, Loyic Colloquium 77 (Proc. Cons. Wroclaw, 1977). I -54. Studies in Logic and Foundations of Mathematics 96, Amsterdam: North-Holland Publ.. 1978. [ M R 8Od: 03027a1 [4] Ehrenfeucht, A,, An application of games to the completeness problem for formalized theories, Fund. Math. 49 (1961), 129-141. [ M R 23, #3666] [5] Ehrenfeucht, A., Decidability of the theory of the linear ordering relation. Notices Amer. Math. Soc. 6 (19591. 556-38. [6] Engeler, E., A characterization of theories with isomorphic denumerable models, Notices Amer. Math. SOC.6 (1959), 161. [7] Fellner, S., Recursiveness and finite axiomatizability of linear orderings. Thesis, Rutgers University, 1976. [XI Lauchli, H., and Leonard, J., On the elementary theory of linear order, Fund. Moth. 59 (1966), 109-1 16. [ M R 33. #7258] [l]
REFERENCES
343
Morley, M., Categoricity in power., rrans. Amer. Math. Soc. 114 (1965), 514-538. [ M R 31, # 5 8 ] Morley, M., The number of countable models, J . Symbolic Logic 35 (1970), 14-18. [ M R 44,#5213] Mostowski, A,. and Tarski, A,, Arithmetically definable classes and types of well-ordered systems. Bull. Amer. Marh. Soc. 55 (1949), 65, 1192. Myers, D., The Boolean algebra of the theory of linear orders, Israel J . Math. 35 (1980), 234-256. [ M R 81i: 030381 r131 Robinson, A,, Introduction to Model Theory and to the Metamathematics of Algebra, Amsterdam: North-Holland, 1963. Rosenstein. J. G.. KO-categoricity of linear orderings, Fund. Math. 64 (1969), 1-5. [ M R 39. ~ 3 9 8 2 1 Rosenstein. J . G., Theories which are not KO-categorical,in Proceedings Summer School in Logicat b e d s f967. pp. 273 278. Berlin: Springer, 1968. Rosenthal, J. W . . Models of T ~ ( ( w "<)), , Nutre Dame J . Formal Logic 15 (1974), 122132. [ M R 49, 4'88531 Rubin, M., Theories of linear order, IsraelJ. Math. 17 (1974). 392-443. [ M R 50, #I8711 Ryll-Nardzewski. C., On the categoricity in power IN,, Bull. Acad. Polon. Sci. Ser. Sci. Math. Asrro. Phys. 7 (1959),545-548, Svenonius, L., KO-categoricityin first-order predicate calculus, Theoria 25 (1959), 82 -94. Tarski, A,. Grundziige des Systemenkalkiils, Fund. Math. 26 (1936), 301. Tarski, A,. and Vaught, R., Arithmetical extensions of relational systems, Compositio Math. 13(1957), 81-102. Vaught, R. L., Denumerable models of complete theories, in InJnitistic Methods (Symposium on Foundutions ofMathmiutic.i. Warsaw, 1959), 1961, 303-321. [ M R 32, #4011]
$1. THE EHRENFEUCHT-FRAISSE GAMES AND FIRST-ORDER THEORIES In Chapters 6 and 7, where we introduced and analyzed Ehrenfeucht games, we made many references to the implications of that analysis for the first-order theory of linear orderings. We will begin this section, and this chapter, by making the connection between games and languages; that is, we will prove that A B if and only if A = B. The implications of this connection will be discussed in 913.2. Although the statement of the theorem above mentions only statements of L , its proof will proceed by induction on formulas. Thus we will need to define a version of the Ehrenfeucht game which permits parameters. As we observed in Chapter 6, the games can be played with any two structures of the same type; later, in 513.6, we will play games with k-partitioned orderings, which are just linear orderings with k unary relations. The formulation of Ehrenfeucht games we now present is sufficiently general to embrace all of these extensions of the original definitions. To simplify the notation in the coming definitions, it is convenient to separate notationally the distinguished elements of a structure from the structure itself. Thus while we have let A denote the structure ( A ; ( R i l i €I } ; ( a j l , j E J ) ) , w e w i l l n o w l e t A d e n o t e ( A ; R , , . . .,R,)anduse(A,a,, ...,a k ) or (A, a) for what was formerly denoted A. Another comment on notation : Whenever we speak of structures in general, we will use the notation A ; however, when we specialize the discussion to linear orderings, we will forget the boldface type and write A as in earlier chapters.
-
DEFINITION 13.1 : Let A and B be structures of the same type 2. That is, A is ( A , R , , . . . ,R P ) and B is ( B , S , , . . . ,S,), where Ri is a z(i)-ary relation on A and Siis a z(i)-ary relation on B for each i. Let a = (a,, a 2 ,. . . ,ak) and b = ( b , ,b 2 , . . . ,b k ) be sequences of elements of A and B, respectively. Let n be a fixed natural number. A play of the game G,( (A, a), (B, b)) consists 247
13.
248
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
of an ordered sequence of n repetitions of the following: PLAYER I chooses an element of either A or B and PLAYER II chooses an element of the other. The element of A selected at the rth turn is denoted a k + tand the element of B selected at the tth turn is denoted b,,,. We say that PLAYER 11 has won the play of the game if for each i, 1 I i I p , and each sequence r( l), r(2),. . . , r ( s ( i ) )of numbers between 1 and k + n,
*
* . ,ar(r(i))) E R i
if and only if
. ,br(T(i)))~Si;
otherwise we say that PLAYER I has won the play of the game. We say that PLAYER 11 has a winning strategy in the game G,( (A, a), (B,b)) if there are functions f ; ,#,, . . . ,f, such that
(i) the domain of .f, is the set of all ordered t-tuples of elements of A u B; (ii) given c 1 , c 2 , .. . ,c, E A u B (representing the choices of PLAYER I at the first t turns) f,(cl,c2,. . . , C J E A
if C , E B
f,(cl,c2, . . . , c , ) t B
if c , E A ;
and
(iii) i f c , , ~ , ,. . . , ~ , E Abandi if we define
and
for every t I n, then for every i, 1 I i Ip , and for every sequence r ( l ) , r(2),. . . , r ( z ( i ) )of numbers between 1 and k + n
..
< ~ r ( ~ ) - a r ( 2 j* ~, a r ( r ( i j ) ) E
Otherwise we say that
R,
if and only if
PLAYER I
...
9
br(,(i))) E S i .
has a winning strategy.
We observe that Definition 13.1 coincides with Definition 6.2 in the case where A and B are linear orderings and k = 0, with a slight exception. The exception is simply that the game G,(A,B) is now defined for any linear orderings A and B,and PLAYER II always has a winning strategy in G,(A,B). [Note, however, that PLAYER 11 does not have a winning strategy in G,((A,a,,a,),(B,b,,b,)) unlessa, < A a, ifandonlyifb, < B b 2 . ]
1.
THE
EHRENFEUCHT-FRAYS&
GAMES AND FIRST-ORDER THEORIES
249
NOTATION : We write G,((A, a), (B, b)) E 11 if PLAYER II has a winning strategy in that game; otherwise we write G,( (A, a), (B, b)) E I.
The following lemmas are easily verified. LEMMA 13.2: (1) C,( (A, a), (B.b)) E I1 ifandonly ifC,( (B, b), (A, a)) E 11. (2) I f f is an isomorphism ,fronz A onto B and if ,f(aj)= bj f o r all j , 1 5 j 5 k, then G,((A, a), (B,b)) E 11. A LEMMA 13.3:
I f G,((A,a), (B,b))E I1 and 0 5 m < n, then G,((A, a), (B, b)) E 11.
THEOREM 13.4: G , , and
(i) for every a
E
A
(A, a), (B, b)) E I1 if and only
if
A there is a b E B such that G,((A, a, a), (B,b, b ) )E 11,
(ii) for every h E B there is an a E A such that G,( (A, a, a),(B, b, b ) )E 11.
A
Note that the form of the conditions in Theorem 13.4 is different from that in Theorem 6.6. However, it is easily seen from Definition 13.1 above that G,((A,a), (B,b ) )E I1 if and only if both G,(A'", B t b ) E I1 and G,(A'",B'b) E 11, so that the conditions here and in Theorem 6.6 agree for k = 0 when A and B are linear orderings. (When A and B are not linear orderings, the conditions of Theorem 6.6 do not make sense at all.) LEMMA 13.5 : Suppose that G,( (A, a), (B, b)) E I1 and G,( (B, b), (C,c)) E 11. Then G,( (A, a), (C,c)) E 11. A
The proofs of 13.2, 13.3, 13.4, and 13.5 are similar to those of 6.3, 6.4, 6.6, and 6.7, respectively, and are left to the reader. The upshot of the three lemmas is that we can define an equivalence relation, for each k 2 0 and each n 2 0 and each type z, on structures of the form (A, a), where A is a structure of type z and a is an ordered k-tuple of elements of A, by specifying that (A*a ) - k , n (B, b, if C,((A,a), (B, b))EII. In the case where A and B are linear orderings and k = 0, we have that A B if and only if A -,B according to Definition 6.8. (We will often suppress the k in - k , n and write -,.) As in Chapter 6,
250
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
we say that two structures are C-equivalent or --equivalent if they are -,-equivalent for all n. EXERCISE 13.6: Let A and B be linear orderings. Let a , < a 2 < . . . < ak and b, < b, < . . < bk be elements of A and B, respectively. Define A , = A‘“’, Ak = A’ ak and A j = [ u j , u j + for each j , 1 5 j < k, and similarly define B , = BSb1,B , = B’bk and B j = [bj, bj+l]B for each j, 1 I j < k. Show that C,((A,a), ( B ,b ) ) E I I if and only if G,(Aj,Bj)~IIfor every j 5 k. *
-
The proof that A B if and only if A = B will proceed by showing that A - n B if and only if A -,, B. Intuitively, A =,, B if any statement of complexity ~n that is true in A is also true in B. The appropriate measure of complexity is not the number of quantifiers in a formula, but the depth of the quantifiers. For example, the statements (W((3Y)P(X, Y ) A (VZ)P(X, z)),
(3X)(3Y)(P(X, Y ) A ( V 4 ( 1 (P(x, 4 * P(Z, Y )1)) have the same number (and kind) of quantifiers. The first has quantifier depth 2; the second has quantifier depth 3. In the first, neither of ( 3 y ) and (Vz) is within the scope of the other; in the second, (Vz) is in the scope of (3y), which is, in turn, in the scope of (3x). (The “scope” of a quantifier is a term whose precise definition can be found references [31 and [7] of Chapter 12.) DEFINITION 13.7: We define the quantifier depth qd(4) of a formula 4 by induction on formulas d as follows:
DEFINITION 13.8: Let Lk,,be the set of formulas ofL whose free variables are among u l , u 2 , . . . ,uk and whose quantifier depth is at most n. Let A and B be interpretations of L and let u t , a 2 , . . . , a k €A and b t , b 2 ,. . . ,bkEB. We define (A,a) E k , , (B,b) if for every &u,, u 2 , . . . , u k ) E Lk.,,we have
A k 4[a1,a2,.. . ,ak]
if and only if
If k = 0, we write A -,, B insteud qf A q n B.
B k 4[bl, b2, . . . ,bk].
1.
THE EHRENFEUCHT-FRAi‘SSk GAMES AND FIRST-ORDER THEORIES
251
B if and only if A - n B EXERCISE 13.9: Assuming that for every n, A (Corollary 13.12),show that the statement (3x)(3y)(P(x, y) A ( V z ) ( i (P(x,z ) A P(z, y ) ) ) )considered above is not logically equivalent to any statement of quantifier depth 2. LEMMA 13.10: For each k and ii there is a finite subset @k,n of Lk,nsuch that every forniula of Lk,n is logically equivalent to a ,formula of @ k , n . Proof: It is easily verified by induction that every formula of Lk,n is a Boolean combination of formulas of Lk,n-l and of formulas of the form (Qy)+(u,,u,,. . . ,uk,y),where Q can be either 3 or V. By replacement of bound variables, all of the latter formulas are logically equivalent to formulas of the form (Quk+l)+(ul, u 2 , . . . ,tikruk+ 1), where +(ul, u 2 , . . . ,u k + 1) is in L k + 1 , n - 1.
We now proceed by induction on n to show that the desired conclusion holds for all k. For n = 0, the formulas of Lk,nare all Boolean combinations of atomic formulas in the variables u l , u,, . . . ,u k . Since the number of predicate letters is finite, there are finiteIy many such atomic formulas. Hence, by Exercise 12.5, we can define for each k an appropriate finite subset Qk.0 of Lk,o.For n > 0, we may assume that ak,,and @)k+ 1, n- have been defined and are finite; then every formula of L,’nis logically equivalent to a Boolean combination of formulas from ( D k , n - l and of formulas (Quk+l)+(ul,...,’k+~), where ~ ( U I , . . ? U k + l ) is in @ k + l , n - l . Hence, by Exercise 12.5, we can find an appropriate finite subset @)k,n of Lk,n. We are now ready to state and prove the main theorem, which relates the Ehrenfeucht games to first-order languages. Note that the proof of the lemma above depends on the assumption that the language has only finitely many predicate letters; the theorem below, which uses the lemma above, is false without this assumption, as we will show near the end of 513.4. THEOREM 13.11 : Let n 2 0 and k 2 0 be given, let A and B be structures of the same type and let a , , a,, . . . , a k €A and b , , b,, . . . ,b k E B. Then (A, a) = k , n (€3, b)
if and only if
(A, a)
-
k.n
(B, b).
We proceed by induction on n to show that the equivalence holds for all k. If n = 0, then assuming that A = ( A ,R , , . . . ,R,) and B = ( B , S,, . . . ,S,), it is clear that Go((A, a), (B,b))E 11 if and only if for each i and for each sequence r(l),r(2),. . . , r ( z ( i ) )of numbers between 1 and k
Proof:
(ar(~j,ar,,,, . . . ,ar(r(i)))E Ri
if and only if
(br(lj,br(2),. . * ,br(r,ijj> E Si
252
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINCS
if and only if for every formula 4(t',,u , , . . . ,U , ) E @k,O A b 4[a,,a,, . . . vak]
if and only if
A I= 4 [ U l i Q ~ ., . . -ak]
if and only if
B != +[b,, b,, . . . ,b,]. Assume that the claim has been proven for n and suppose that (A, a) -,,+ (B, b). Let 4 be in @k,,,+ ,. Then either 4 is of the form (Qu,, where or 4 is a Boolean combination of formulas in @k,,, and formulas of this form. It clearly suffices to show that B =!
4[bl, b,,
. . . ,bk]
holds for all formulas 4 in @k,n+ of the form ( Q U k + J$ since the "if and only if" is preserved under Boolean combinations of formulas. And since (Vy)$ is logically equivalent to 1 ( 3 y ) i $ , we may assume that 4 is of the form ( 3 4 , ,)$, where $ has quantifier depth at most n. Since (A,a) -,,+ (B,b), for each ~ E there A is a b E B such that (A, a, u ) -,, (B, b, b). Hence by the induction hypothesis, (A, a, a) -,, (B, b, b ) ; thus for each a E A there is a b E B such that A b $ [ a , , . . . ,a,,u]
if and only if
B 1 $ [ b l , . . , ,b,, b].
Hence if A I=4[a,,a,, . . . ,a,], then €3 k 4 [ b , , b,, . . . ,bk]. Similarly, if B k 4 [ b , ,b,, . . . ,b,], then A k 4[al, a,, . . . ,uk]. Hence, for each 4 E @ k , n + 1 , A 1 d [ a ~ , U , ,. . . >a,]
if and only if
,
B k 4 [ b l ,b,, . . . ,b k ] .
Conversely, suppose that (A,a) Q,,+ (B, b). We must show that there is a formula 4 ~ @ , , , ,such + that it is not the case that
,
A 1 4 [ a l , u 2 , . . . ,ak]
if and only if
B 1 4[b1, . . . ,b,].
Since (A, a) *,,+, (B, b), there is (without loss of generality) an element U E A such that for each b E B, (A, a, a) * ,,(B,b, b). By the induction hypothesis, there is, for each b E B, a formula qbE @)k+ ," such that A != $b[a13a2,. . . ,ak,a]
B y $b[bi,b,,. . . ,b,,b]. Since @ k + , , , , is finite, it follows that {$blbEB) is finite. Let $ be the conjunctionA{$,(bEB}.ThenA k $ [ a l , U 2 ,..., a,,a] butB y $[b1,h2,..., b,,h] for any b E B. Let 4 be ( h k + l ) $ . Then A 1 4 [a ,,a ,, . . . ,a,] and B y 4 [h ,b, , . . . ,bk], although 4 E a,.,, + This proves the converse and thus the theorem. and
,
-
COROLLARY 13.12: (1) For each n, A (2) A = B ifand only if A B. W
E,, B
if and only
if
A
-,, B.
Thus two structures are elementarily equivalent if and only if they are G-equivalent.
2.
253
GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
82. THE EHRENFEUCHT GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
As a consequence of the equivalence between logical equivalence and G-equivalence, we can now interpret logically various results of Chapters 6 and 7. For example, from Corollary 6.12 it follows that o is logically equivalent to o 5 . CI for any order type CI. In particular, no statement of the first-order language distinguishes between w and w 5.
+
+
THEOREM 13.13: (Tarski [20]) an elementary class.
The class W of well-orderings is not
Proof: W is not closed under elementary equivalence, although by Exercise 12.7.2, every elementary class is closed under elementary equivalence. I
By Exercise 6.11, if A is a finite linear ordering with at least 2" - 1 elements, then A -"o o*.Hence any statement true in o o* is also true in some finite linear ordering. Moreover, given any finite set of statements true in o ID*, their conjunction, which is also true in w o*, must be true in some finite linear ordering. O n the other hand, o o* is not elementarily equivalent to any finite A since if A has m elements, a statement asserting the existence of at least m 1 distinct elements would be true in o + o*but not in A . Thus although o + w* can be distinguished from each finite A by a first-order statement, no one first-order statement distinguishes o + o*from them all. This is expressed by saying that o o*, or more properly Th(o w*), is not finitely axiomatizable.
+
+
+
+ +
+
+
+
DEFINITION 13.14: Given a theory T we say that T is finitely axiomatizable if there is a statement 4 E T such that for every 9 E T , 4 logically implies 9. The statement 4 is called an axiomatization of T. Given a structure B, we say that B is finitely axiomatizable if Th(B) is finitely axiomatizable, and if 4 is an axiomatization of Th(B), we also say that is an axiomatization of B.
Note that
4 is an axiomatization of B if and only if for every structure 4, then C = B.
C of the same type as B, if C !=
PROPOSITION 13.15: o
+ o* is not Jinitely axiomarizable.
I
13.
254
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Note that 4 is an axiomatization of a theory T if and only if for every interpretation C of the language of T , C k T if and only if C b 4. Thus, for example, the conjunction 4Lof (1) ( ' J X ) ( V Y ) ( W X < y A Y < 2 --+ x < 4, (2) ( W V Y ) ( X < Y 7 ( Y < x )1, and (3) ( V X ) ( \ J Y ) ( X < Y v x = Y" Y < x ) +
is an axiomatization of the theory of linear orderings. Although w + o*is not finitely axiomatizable, w is finitely axiomatizable. For every model of the conjunction of the statements (1)
4L.t
(2) ( W 3 Y ) ( X < Y A ( V Z ) ( l ( X < z A z < Y ) ) ) , (3) (3X)(VY)(X< Y v x = Y ) , (4) (VX)((3Y)(Y< x ) --+ (3Y)(Y < x A ( W l ( Y < z A
-=x ) ) ) )
has order type w + 6 . c1 for some c1 and so is elementarily equivalent to o by Corollary 6.12 and Corollary 13.12. PROPOSITION 13.16: w is Jinitely axiomatizable.
We note that the quantifier depth of the above axiomatization of w is 3; hence, if B =30then B = w. This fact, and the fact that w is finitely axiomatizable with an axiomatization of quantifier depth 3, can also be explained combinatorially, so we make a brief digression to Ehrenfeucht games.
-
DEFINITION 13.17: A structure A is said to be hitary if there is an n such that B A implies B A for all structures B of the same type. If A is finitary, then we will write f(A) = m if m is the least IZ for which B -" A implies B A for all structures B of the same type.
-
Before commenting on this definition, we note that using Corollary 13.12,Theorem 6.16 says that given any linear ordering C there is a countable linear ordering elementarily equivalent to it. This conclusion, the Lowenheim-Skolem Theorem, is also true for arbitrary structures, as pointed out at the end of 812.2. By the Lowenheim-Skolem Theorem, to show that A is finitary it suffices to show that B -,, A implies B A for all countable structures B of the same type; for if C is uncountable, then C B for some countable B, so that C -,, A implies that B w nA, which implies that B A and therefore C A.
-
- -
-
2.
GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
255
EXAMPLES: (1) q is finitary, for if B m 3 q, then B must be dense and have no endpoints; hence if B is countable, it follows, by Cantor’s Theorem 2.5, that B v q. Then Lemma 6.3.2 implies that B mnq for all n, or, equivalently, that B q. O n the other hand, since 5 - 2 q , we conclude that f ( q ) > 2, so that f ( q ) = 3. (2) w is finitary, for if B m 3 w , then B must have a first element, every element of B must have an immediate successor, and every element of B, except the first, must have an immediate predecessor. Thus, if b E B is not among the first w elements of B, then cF(b)‘v 5, so that B ‘v w + 5 . a for some order type a. Since B w by Corollary 6.12, we conclude that w is finitary. On the other hand, ((D + o)-2 w so that f ( w )= 3. (3) o + w* is not finitary, for, given any n, 2” -,, (w + a*)by Exercise 6.11, but 2“ * n + (w + a*)by Exercise 6.10.2. (4) Each finite ordering is finitary, for, by Exercise 6.10.2, if k < 2” - 1 and A -” k, then A = k so that A k ; hence f(k) I n.
-
-
-
EXERCISE 13.18: (1) Show that 5 is finitary and that f ( 5 ) = 3. (2) Show that w + w is finitary and that f(o+ a)= 5. (3) Show that w* ( + w is finitary and that f ( o * + 5 + w ) = 5. [Note: You will have to state and prove analogues of Corollary 6.12 for 5, w + o,and w* + 5 + w.]
+
This completes the digression to Ehrenfeucht games. We showed earlier that if B = 3 w then B = w by using a specific axiomatization of w . Alterw, so B o by Example natively, if B - 3 w, then, by Corollary 13.12, B 2 above; hence, by Corollary 13.12, B = w. Furthermore, since @ 0 , 3 is finite, &4 E @0,310 1 4} is an axiomatization of w. This is an example of the following fact. m3
-
PROPOSITION 13.19 : A structure A is finitely axiomatizable if and only cf it is finitary. Furthermore, if , f ( A )= n, then A has an axiomatization of quantijier depth n but has no axiomatization of quantijier depth smaller than n. Proof : The first statement follows from Corollary 13.12. If f(A) = n, then as in the discussion above, A has an axiomatization of quantifier depth n. If A had an axiomatization of quantifier depth m < n, it would follow that f(A) I m, contrary to assumption. H
Applying Proposition 13.19 to conclude that w is finitely axiomatizable is more efficient, but does not of course give an explicit axiomatization of
256
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
w. It does, however, guarantee that no axiomatization of o has smaller quantifier depth than @<,, .
COROLLARY 13.20:
q and
5 are Jinitely axiomatizable. A
c, +
+
EXERCISE 13.21 : ( I ) Find axiomatizations of q, o o,and o* 5 o that have minimal quantifier depth. (2) Describe an axiomatization for each n that has minimal quantifier depth.
+
Corollaries 7.10 and 7.21 have the following logical interpretation. THEOREM 13.22: (Lauchli and Leonard) (1) I f # is a sfatement true in some scattered linear ordering A , then M k 4 .for some M E A. . (2) If 4 is a statement true in some linear ordering A , then M I. # for some M E A.
Proof: Given # of quantifier depth n, use Corollary 7.10 (or 7.21) to find an M E A. (or M E . X ) such that M m n A . Then, by Corollary 13.12, M = " A , so that M k 4.
Note that the proof of Theorem 13.22 actually shows more, namely, given any (scattered) linear ordering A and any n, there is a single (scattered) in which is true every statement of quantifier linear ordering M in A (or Ao) depth at most n which is true in A . In 513.6, we will prove the following stronger version of Theorem 13.22.
THEOREM 13.23: (Amit, Myers, Schmerl, Shelah) Let A be a linear ordering. Then for each n there is a finitely axiomatizable M E Af such that A =,, M . If' A is scattered, then M can be chosen f r o m Ao. Theorem 13.22 can be used to show that various linear orderings are not finitely axiomatizable. For example, assume that cow is finitely axiomatizable. It follows from Theorem 13.22 that ow= M for some M E do since the axiomatization of coo must be true in some M E Ao.Choose M of smallest rank such that cow = M . Let M' be the subordering of M consisting of all limit points of M , that is, all elements of M that have no immediate predecessor. Now it is easily verified by induction on A. that if M E -H0 and M' is the set of limit points of M then M' E A. and rF(M') < rF(M). Hence coo f M'. Thus PLAYER I has a winning strategy in Gn(mw, M') for
2.
GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
257
some n. But thenPLAYER I has the following winning strategy in G , + 2 ( ~MW) :, In his first n moves he plays only on the limit points of ww and M , which have order types ww and M ' , respectively, following his winning strategy in Gn(wW, M ' ) ; since PLAYER 11 must have chosen a non-limit point, PLAYER I can win in two more moves. This contradiction proves the following result. COROLLARY 13.24: ww is not jinitely axiomatizubfe. It is now natural to ask whether every ordinal a < ww is finitely axiomatizable. Note that we cannot automatically expect a positive answer since, as we have seen, dodoes contain non-finitely axiomatizable linear orderings. We will show that every ordinal a < owis finitely axiomatizable, and that, moreover, an axiomatization of M can be obtained effectively from a description of a. When we speak here of a description of a, we mean of course the Cantor Normal Form a = wn . a,
+ wn-' .
tE,-
1
+ . . . + w . u1 + a,,
where a, # 0. We will show how to construct an axiomatization of M from its description. To prove that the axiomatizations @ a that we present for each M really are axiomatizations, we will need lo show that if A b @a, then A = a. To do so, we will define, for each x , a class &(a) of a-like linear orderings and show that for each linear ordering A , A is G-equivalent to a if and only if A E ~ ( c c )Then, . by virtue of the equivalence of G-equivalence and elementary equivalence, to show that CD, is an axiomatization of a it will suffice to show that if A k CDa, then A E d ( x ) . We first define each d ( w " ) ,by induction on n. Then wO-likeorderings all have order type 1. For n = 1, we have already seen that A o if and only if A c o 5 . a for some a,and so we define these to be the w-like orderings. Note that both 1 and o are finitary, and, in fact, f ( w ) = 3. Now assume that we have defined, for each k < n, the class d ( w k )of wk-likeorderings, that we have shown that A E d ( w k )if and only if A cok, and that we have shown that wk is finitary and that f ( w k )5; 2k + 1. Moreover, we assume that for each k < n, A E d ( w k )if and only if A = 1 i E W ), where W is cok- I like and each 4 is w-like, and that A E d ( o k )ifand only if A = Kli E W } , where W is d i k e and each is wk-l-like. (Note that for k = 1 both of these equivalences hold.) Now suppose that A wn.L,et A' be the set of all limit points of A . Then A' must have order type in d ( ~ " for - ~if)G,2 , - l ( A ' , o " - l )E I, then, by sticking to his winning strategy in G2n-l(A',w"-1),PLAYER I can force
-
+
N
I{
-
w.
I{
13.
258
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
to a non-limit point and thus win in G2,,+,(A,w").Between each pair of successive limit points of A, the interval must be o-like. Thus A = x{l4(\i E W } ,where W is o"-'-like and each is o-like. Since W is on-I like, it can be written in the form W = 51j E V } , where V is o-like and each V, is wnV2-like.Thus A = c ( c { w l i E y } l j E V } ; by the induction W , ( iE 5 ) is on-'-like, so that A = c { X j ( jE V ] ,where hypothesis, each V is d i k e and each X j is w"-'-like. More explicitly, PLAYER II
I{
I{
-
where each l4(. and each W,' is w"-'-like. Thus we define &(a")to be all linear orderings of this form. Clearly, if A E &(on),then A o" by the 2n + 1. Thus the following is correct. induction hypothesis. Also, f ( w " ) I PROPOSITION 13.25: Define a sequence { d ( o " ) l n< Q } of sets of linear orderings, by induction on n, as follows:
(0) d(1) = all linear orderings of order rype 1 ; (1) d ( o )= all linear orderings of order type w + 5 . a for ci arbitrary; (n) &(on)= all linear orderinys of form
where euch
4, and each W,' A
is in &(on-'). Then, ,for each n,
E &(d)
Moreouer, f(o")= 2n + 1.
if and only if
A
-
on.
Proof : We have verified everything except that f(o")2 2n + 1. This follows from the fact that o"r r 2 , , o"+ w" (Theorem 6.18) which is not on-like. . u2 + . * . + Now let cc < owbe arbitrary and write a as w"' . a , + on2 . ak in Cantor Normal Form. We say that A is a-like if and only if A can be written in the form (link
(Wl' + . * .
+ w:;) + (W;2 + . . + w:;) + . . . + ( wl"+. . . + W:;), '
where each Wyt is an w"[-likeordering, and we let &(a) be the set of all cc-like orderings. THEOREM 13.26: For every a < ow,A over, a is jnitary.
E &(a)
i f and only i f A
-
a. More-
2.
GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
-
-
259
Proof: If A E &(a), it is clear that A a. Assume that A CI and that a = wnl . a, . . . wnk. u k .Then PLAYER I in his first c { a j l 1 I j I k } moves chooses the first elements of each of the aj copies of con, for all j. If between two successive choices of PLAYER 11 in A the interval is not co"J-like for the appropriate n j , then PLAYER 11 is punished in 2nj 1 more moves. Hence A E ~ ( c I )Moreover . f ( u ) I 2n, + 1 + c{ujI 1 Ij I k } .
+
+
+
We will now pattern our axiomatization of each a < owon the constructions above; that is to say, we will axiomatize anby saying that its limit points are w"-'-like. To do Ihis, it will be very helpful to introduce the idea of relativizing formulas of L . Relativization is useful when one wants to speak about a substructure of a given structure as a structure. Thus, for example, given a linear ordering A and an element a E A , one can write a statement about A'" that says that every element of A'" has a successor in A'"; such a statement would be simply (Vx)(3y)(x < y), since this says about any linear ordering that every element has a successor. If one wants to write something about A and a that says that A'" has this property, one would use instead the formula (vX)(X
+
( 3 y ) ( y< U
A X
< y)),
observing that A k (VX)(X < u
+
(3y)(y < u A
x < y))[a] A'" 1 (Vx)(Iy)(x< y).
if and only if
We say that the new formula is the old one relativized to the formula e(w,u ) = w < u, since every quantifier ( Q w ) is interpreted as applied to { w ( w < u } . Now for the general formulation. DEFINITION 13.27: Let O(w,vl, u 2 , . . . ,uJ be a formula of L. We define for every formula 4 of L the relativization by induction on formulas:
+@
(1) If 4 is atomic, then 4' is 4, (2) If 4 is i $then , 4' is i($@). ( 3 ) I f 4 is $1 A $ 2 , $1 " $ 2 , IC/I $ 2 , Or $1 $ 2 , then $'is $ i @ A $2', v $2', + 4b2', or $,' ++ $2', respectively. (4) If 4 is (3y)$, then 4' is (3y)(,O(y,u l , . . . ,u,) A $'). ( 5 ) If 4 is (Vy)$, then 4' is (vy)(O(y,u l , . . . ,0,) + $7. +
Thus, for example, the statement (Vx)[(Vy)(y < X ( W ( Y < < 4) (3X')(('dy)(y< x' -+ (3z)(y < Z < X')) +
+
A X
< X')]
260
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
-
is exactly the statement (Vx)(3s')(x < x') relativized to the formula (Vy)(y < w -, (3z)(y < z < w)) and thus says of a linear ordering that it has no last limit point. Similarly, the statement (3x)(Vx')(.x I x') relativized to ( V y ) ( y < w + ( 3 z ) ( y< z < w ) ) says of a linear ordering that it has a first limit point. More generally, given any statement 4, the relativization of 4 to ( V y ) ( y < w -+ ( 3 z ) ( y < z < w ) ) says of a linear ordering that its limit points have property 4. We now explain this more precisely. Given a structure A and elements a,, a 2 , . . . ,a, E A , let (A,ul, . . . ,a,)" = jblA I= 8 [ b , a , , . . . , a , ] } . Then what 4' says about ( A , a l ,. . . ,a,) is that 4 holds in ( A ,a,, . . . ,a,)'. This is a corollary of the following lemma, which is proved by induction on formulas. LEMMA 13.28 : Let @(w,v,, . . . ,c,) be a formula of L. Then for any structure A , any formula 4(x1, x 2 , . . . ,x,), any a , , . . . ,a, E A, and any b l , . . . ,b, E ( A , a , ,. . . ,a,)',
A
if and only if 4'[bl,. . . , b n r a i . .. . , a t ] (A,a,, . . . ,at)' 1 4 [ b , , . . . ,&I. A
COROLLARY 13.29: Let O(w, u l , . . . ,u,) be a formula of L . Then .for any structure A , any statement 4, and any a , , . . . ,a, E A,
A I. 4'[al, . . . .a,]
if and only
if
(A, a,, . . . ,a,)' I. 4.
A
Thus, for example, if A is a linear ordering and 4 is a true statement about the interval [a,, a 2 ] of A, then, choosing O(w, u,,u,) to be u , I w I u 2 , we see that A k @[al, a2] since 4' makes the same assertion about A that 4 makes about [a1,a 2 ] . COROLLARY 13.30: Let O ( w ) be a formula of L. Then for any structure A and any statement 4,
A b
4'
if and only if
A' k
4. A
Thus, for example, if A is a linear ordering, then 4 is a true statement about the limit points of A if and only if A I= 4', where B(w) is ( V y ) ( y< w -+ (&)(y < z < w ) ) . Note that for any formula 4, qd(4') = qd(4) qd(O), so that if 8 is quantifier-free, the quantifier depth of 4' is the same as that of 4. Also note that if 4 is a statement and O ( w , u l , . . . ,u,) is a formula, then the free variables of 4' are u,, . . . ,u, but not w.
+
2.
GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
261
The substructures of A we discussed in our treatment of relativization are all definable, so it seems appropriate to define that notion before returning to our discussion of ordinals.
DEFINITION 13.31 : Let A be a linear ordering and let B G A . We say
that B is definable if there is a formula O(w) such that b E B if and only if A k O[b]. Let c1,c2, . . . ,ct E A ; we say that B is definable over c 1 , c 2 ,. . . ,c, if there is a formula O(w,y l , y,, . . . , y,) such that b E B if and only if A k O[b,ci,cz,. . . ,cJ.
Thus, for example, the set A’ of limit points of A is definable by the formula ( V y ) ( y < w -+ ( 3 z ) ( y< z < w)), the interval [ b , , b21A is definable over b,, h, by the formula y, 5 \v 5 y,, and the interval [a, + ) A is definable over a by the formula w I y , . The results above can then be interpreted as follows. COROLLARY 13.32: (1) Let B be a dejnable subordering of A . Then, given any statement 4, we can ejj%ctively,find a statement q5A such that
Bk
4
f and only fi
A k
4A.
(2) Let B G A be dejnable over. c,, c,, . . . ,ct. Then, given any statement can efectively find a forniula 4 A ( y l ,y,, . . . ,y,) such that
4, we
Bk
4
i f and on/.) if
A k 4*[c,,c2, . . . ,c,].
Proof : Take to be @, where O defines B (over c,, c,, . . . ,c,), and apply Corollaries 13.29 and 13.30.
Given a linear ordering A, which subsets B s A are definable? Suppose, for example, that B G g and B is definable: If b , , b , ~g, then there is an automorphism f of g such that f ( b , )= b,. Hence, by Exercise 12.6, for any formula 8, q k 8[b,] if and only if q k 8[b,]. Thus B must be either @ or all of g; that is, g has no definable proper subsets. The same is true of 4, and, in fact, whenever A is transitive, A has no definable proper subsets. Which subsets of o are definable? For a complete analysis, see the exercise below. EXERCISE 13.33: (1) Show that each element of o is definable (where a € A is definable if ( a ) is definable) and that each finite subset of o is definable; show that the complement of each finite subset of o is definable. (2) Show that the only definable subsets of o are its finite and cofinite subsets. [Hint: Look at o 4.1
+
13. T H E FIRST-ORDER
262
THEORY OF LINEAR ORDERINGS
(3) Suppose that 4 is a statement which is true in infinitely many nonisomorphic finite linear orderings. Show that 4 is true in n for all but finitely many n.
Now let O1 be the axiomatization of o discussed earlier; note that qd(@,,)= 3. Let O(w) be the formula (Vy)(3z)(y < w + y < z < w ) discussed above, which says that w has no immediate predecessor. Then A k Qle if and only if A' I=a,. But A' = A', so that A b Ole if and only if A' is d i k e if is an axiomatization of 0'. and only if A E d ( w 2 ) .Thus THEOREM 13.34: For each n > 1, LteJine @, axiomatization of on.Moreover, qd(@,,)= 2n + 1.
Then @,, is an
=
Proof: By induction on n, as in the argument for n = 2. Moreover, since qd(@,) = 3 and qd(O) = 2, we can show by induction on n, using the fact that qd(de) = qd(4) qd(O), that qd(@,,) = 2n + 1 for all n. W
+
Note that the quantifier depth of the axiomatization a,,of W" is precisely
f ( d )as , predicted by Proposition 13.25, so that these axiomatizations are
optimal. . u2 + . . . + w " ~. a k , Turning now to the general case c( = w"' . a, + on2 we see that we want to axiomatize CY by saying that there are certain points such that the intervals between them are wk-like;for this we need to relativize to the formula ~(w,u,,u2) that is v 1 I w < u 2 . We thus define @(a) to be ( 3 x 1 ')
' ' '
(jXi,)(3X12)
A(X1'
' ' '
(jX:,)
' ' '
5 y)
(3Xik){(Vy)(X,' '
. . A @ e ( w . d 1 . x l 2 )A ae(w,x,2,x22)
nl
. . .A
(]Xik)
< " ' < xi, < X1' < . . . < XZ2 < . . * < X l k < . . < X i k )
A @e(w,x,',Xl2) A . A
' ' '
n1 @ ~ ( W , X , ~ . X ~ ~ )
nk
A"
A,
. .A ae(w%x:2.x,3) nz
"2
.A ( D o ( ~ ~ ~ : k _ A @e'(wdk) nk
nk
1,
where B'(w, vl) is u I I w. The following is then easily verified. THEOREM 13.35 : For each ordinal a < ww,@(a)is an axiomatization of
CY.
A
We conclude this section with several theorems of a technical nature specific to linear orderings that result from the equivalence between logical equivalence and G-equivalence. These theorems will be used in $813.4 and 13.5. Recall that for every k and n, Lk,,,consists of all formulas of L whose free variables are among u l , . . . , uk and whose quantifier depth is at most n. We showed that there is a finite subset @k,,, of Lk,,, such that every formula
2.
GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
263
of Lk,,is logically equivalent to a formula of @ k , , ; thus if 4 E a,,,, then there 4. Now let A be a structure, let is a II/ E @ k , , that is logically equivalent to i a,, . . . ,ak E A , and let @ = @(A,a , , . . . ,a,) be the conjunction of all 4 E @ k , n such that A 1 b[a,, . . . ,a,]. Then A 1 @[a,, . . . ,a,] and for any B and b , , . . . ,b, E B we have (A,a,, . . . ,ak) =,,, (B, b , , . . . ,bk) if and Only if B b @ [ b , , . . . ,b,]. We can thus find a finite set {Qi(ul, . . . ,u,)l1 I i I t } of formulas of Lk,, that completely describe all of the =,,,-equivalence classes. That is, for each structure A and elements a,, . . . , ak E A, there is a unique such that A 1 Qi[a,, . . . ,a,] and, moreover,
B k a i [ b , , . . . , bk]
if and Only if
(B,b , , . . . ,bk) - k , n (A,al, . . . ,a,).
For k = 0 this says that there is for each n a finite set {Oil1 I i I t} of statements of quantifier depth at most n that describe all =,-equivalence classes of linear orderings. That is, for each structure A, there is a unique misuch that A 1 mi and, moreover,
B 1 ai
if and only if
B =,A.
This much we can say about arbitrary structures, not just linear orderings; with linear orderings we can carry this analysis much further. Note first of all that each Q imust specify the ordering of the variables. We will work with the case where @ = mi logically implies that u , < u2 < . . . < u,; any permutation can be dealt with similarly. Let A k @ [ a , , . . . , a k ] , so that, in particular, a , < a2 < . . . < a,. Let A , , , 4 , , . . . ,A,- ,, A , be the intervals (t, a,], [a,, a,], . . . , [a,- ,,a,], [a,, +) of A . By the discussion above, we can find statements Y o , 'PI,. . . ,y k that completely describe the =,-equivalence j I k - 1, let YF be the relativizaclasses of A , , A , , . . . , A , . For each j . 1 I tion of Y jto e(w, u j , ujt = ( u j I w Iu j + ,), so that B k Y F [ b j ,b j +, ] if and only if [ b j ,b j +,] 1 Y j If and only if [ b j ,b j +, ] =, [ a j , a j + , ] . Let Y o Abe the u,) and let Y / be the relativization of relativization of Y oto O(w, uL) = (w I yk to o ( w , u k ) = (0, 5 w). Let y ( u 1 , . , . ,u k ) be the formula (01 < u 2 < ' ' ' < u k ) A YoA(u1) A YIA(ul, u,) A . . . A Y f - .1(uk- u,) A y t ( u k ) . It is easily verified that Y(u,,. . . ,u k ) characterizes the =,,,-equivalence class of @,a,, . . . ,a,); that is, A k Y [al' . . . ,ak] and, for any B and b , , . . . ,b, E B,B 1 Y [ b , , . . . ,b,] if and only if (B,b , , . . . ,b,) E k , , (.4,a,, . . . ,a,). Furthermore, since each Y j E Lk,nand relativization to quantifier-free formulas does not increase quantifier rank, we conclude that Y E Lk,,. We summarize this discussion in the following lemma.
,,
LEMMA 13.36: Let A be a linear ordering and let a , < a, < . ' ' < ak be elements of A . Then there is a ,formula Y(u,, . . . ,u k ) E L,,, such that A k Y [a . . . ,a,] and such that B 1 Y [ b , , . . . , bk] if and OnlJ) if (B,b , , . . . ,bk)-kk,,(A,al,. . . ,Uk).
264
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Furthermore, Y can be chosen to be a conjunction of v1 < vz < . . ' < v k with formulas Y ~ ~ ( U Y~ l A)( v, l rvz), . . . ,Yt- ,(uk- 1,ok)r y,"(uk) so that for any linear ordering B and elements b,, . . . , b, E B we have B
YoA[b,]
if andonly if if and only if
(+,b,] =n(+,u,], [bk, +)
B 1 Y',"[bk] and fbr each j , I 5 j 5 k - 1, B =! Y P [ b , , b j + l ]
ifandonly if
=,, [a,,
+),
[ b j , b j + J =, [ a j , a j + ] ] .
This lemma is easily modified if the elements a l , a z , . . . ,ak are in some < ' ' < an(k)for order other than a, < u2 < ' . .
THEOREM 13.37: Any formula 4 ( v l , . . . , v k ) in the language of linear orderings is logically equivalent, in the theory T of linear orderings, to a Boolean combination of ,formulas each of which has at most two free variables. Proof: Suppose that the quantifier depth of 4 is n. By the discussion above, there are a finite number of =,,,-equivalence classes to each of which is associated one of the special formulas T I ,. . . ,Yt of Lk,,. Then, for each j, either T t Y, + 4 or T 1 Y j + i 4. Let 0 be the disjunction of those Y j for which T k Y j + 4. Then clearly T t 4-0 and, since each Y j is a Boolean combination of formulas in two free variables, the same is true of
0. Another consequence of Lemma 13.36 is the following result of Rubin ~171. THEOREM 13.38: Let A be a linear ordering and let B be an interval of A. Then for each formula 4(v1,.. . ,v,) there is a formula +*(v,,. . . , v k ) such that ,for all b,, . . . ,b, E B, A
4[b], . . . ,bk]
if and only i f
B k 4 * [ b 1 , .. . ,bk].
2.
265
GAMES AND THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Proof : As in the proof of the preceding theorem, if 4 has quantifier depth
n, then 4 can be written as a disjunction of special formulas of Lk,n.If we find an appropriate Y * for each disjunct Y , then we can define +* to be the disjunction of the Y * . Thus we may assume that 4 is a special formula, so that, for some permutation n, & ( t i , , . . . , v k ) is (%(1,
< Un(2) < . . . < U,(k)) A
' ' '
A ykl(un(k-l)-
A
YoA(un(l)) A YlA(vn(1),%(2J
un(k))
YkA(un(k)).
Now Y k A ( t i , ( k ) ) is the relativization of Y kto w 2 where Y kis the statement completely describing the =,-equivalence class of some linear ordering C. Let P , , . . . ,P, be the statements completely describing equivalence classes whose representatives C,. C,, . . . ,C, satisfy C = n Ci A , , where A , = (a E AIB < a } . Let Yk*(v,(k)) be 1I i I r } . Similarly, YO4(u,~,)) is the relativization of Y oto w I v n ( l )where , Y ois the statement completely describing the =,-equivalence class of some linear ordering D.Let Q , , . . . , Q, be the statements completely describing those =,-equivalence classes whose representatives D1, . . . ,D, satisfy D = n A , Dj, where A , = {a E Ala < B}. Let Y c , * ( v E ( , Jbe v{QJ"5un(1)l 1 ~j I s}. To obtain +*(u,, . . .,vk)from + ( u , , . . .,u k ) we just replace 'l'oA(u,(oJ by '€'o*(vn(oJ and Y k A ( u x ( k ) ) by Yk*(u,(k)).It is then clear that for every b,, . . . ,b, E B,
+ V{PF~~~(~)I
+
A 1 4 [ b , , . . . ,bk]
if and only if
B 1 4 * [ b , , . . . ,bk].
The same result can be obtained if we allow parameters from A - B. COROLLARY 13.39: Let A be a linear ordering, let B be an interval of A, and let a , , . . . ,a, E A - B. Then for each formula 4(v1,. . . ,vk, w l , , . . , w,) there is a ,formula 4*(u1. . . , ,u k )such thnt for all b,, . . . ,h, E B.
Let aico,be the largest ai < B and let ai(,) be the smallest ai > B. Repeat the proof above, deleting from the special formula & those conjuncts discussing intervals wholly contained in A - B and replacing A , and A , by {alai(,) I u < B } and j(iIB < a I ai(,,>,respectively. Proof:
The content of Theorem 13.38 is that to determine whether or not certain elements of an interval B of A satisfy a certain formula in A, we can test to see whether they satisfy &* in B. For this reason, Rubin calls 4* the testing formula for 4 in B (relative to a,, . . . ,a,).
+
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13. THE FIRST-ORDER
THEORY OF LINEAR ORDERINGS
Using Theorem 13.38, we see that if C is definable in A and is contained within the interval B of A, then C is definable in B. For if a E C if and only if A k $ [ a ] and 4* is the testing formula for 4 in B, then, for every a E B, a E C if and only if B != 4 * [ a ] . The following fact is true whenever B is a definable subset of A. LEMMA 13.40: Let A be a linear ordering and let B be an interval of A that is dejinable over e l , . . . , c,. Then for every formula 4 ( u l , . . . , 0,) there is a formula 4 # ( v l , .. . , v k , x l , . . . , x i ) such that for every b,, . . . ,b, E A. A if and only
4 # [ b l ‘, .
.
7
b k , cl,
. ..,
if b,, . . . ,b, E B
and
A != (P[b,,. . . ,bk].
Proof: Let O(w,x,, . . . ,xJ be a formula that defines B over c , , . . . ,ct. Then define $#(u,, . . . ,vk,xlr . . . ,x,) to be
(p(v1,.. . ~ ~ ~ ) A ~ { 6 (. .u. ,&)I1 i , x ~<, i < k } . Clearly 4’ has the desired property. W A similar conclusion is correct in general if we allow parameters from A - B. In the context of linear orderings, however, such parameters can be eliminated from the new formula (6’; this consequence of Corollary 13.39 is also due to Rubin.
THEOREM 13.41 : Let A be a linear ordering, let B be an interval of A that is dejinable over c,, . . . ,c,, and let a,, . . . ,a, E A - B. Then for every jbrmula (p(v,, . . . ,v k , w,, . . . ,w,) there is a formula 4 # ( v l , . . . ,v,, x,, . . . ,x,) such that for every b , , . . . , b, E A, A
(P#[b,,.. . ,bkrCir.. .,Ci]
if and only if b,, . . . ,b, E B
and
A k 4 [ b , , . . . ,bk,a,, . . . ,am].
Proof: Let 6 ( w , x 1 , .. . , x J be a formula that defines B over c l r . . . ,c,. Let 4*(ul,.. . ,uk) be a testing formula for 4 in B relative to a,, . . .,a,, so that for all b l , . . . ,b, in B, A k
4 [ b , , . . . ,bh,al,. . . ,am]
if and only if
B 1 +*[b,, . . . ,b,].
3.
DECIDABILITY OF FIRST-OKDHR THEORIES OF LINEAR ORDERINGS
Let $ # ( u , , . . . ,u k , x , , . . . ,xJ be the relativization of together with A{O(ui,xl, . . . , x t ) l 1 I i I k } . Then A
4*
267
to O(w, xl, . . . ,x,)
1 d # [ b , , . . . ,bk,c,, . .. ? C t ] if and only if bl, . . . ,b k E B and A 1 (4*)’[b,, . . . ,b,, cl, . . . ,c,] if and only if b l , . . . ,bk E B and B 1 4*[b,, . . . ,b k ] if and only if b l , . . . ,b, E B and A 1 4 [ b , , . . . ,b k , u , , . . . ,urn].
rn
Thus if we are interested only in which elements of the definable interval satisfy 4, we can select them out by using 4’; we thus call 4’ the selecting formula for 4 in B (relative to c i l , . . . ,urn).
53. DEClDABlLlTY OF FIRST-ORDER THEORIES OF LINEAR ORDERINGS DEFINITION 13.42: Given a language L and a theory T in the language L, we say that T is decidable if there is an algorithm that, when applied to an arbitrary statement 4 of L, will determine whether or not 4 E T ; that is, T is decidable if T is a recursive set of statements. Otherwise, we say that T is undecidable. Typically, T is either the theory Th(M) of a structure M or the theory Th (d) of a class d of structures. In this chapter, we will discuss examples of
both kinds. A useful criterion for decidability in the first case is recursive axiomatizability, which we will call axiomatizability even though that term has already been used, since. in the context of decidability, axiomatizability always means recursive axiomatizability. DEFINITION 13.43: A theory T is axiomatizable if there is a recursive set of axioms for T. That is to say, T is axiomatizable if there is a subset @ c T such that (1) If 4 E T then 0 1 4. (2) There is an effective procedure that, when applied to a statement 11, of L , determines whether or not $ E 0.
In this case we say that T is axiomatizable by 0.
268
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
If the theory T is axiomatizable by 0,then we define a “theorem-checking algorithm” for Q, as follows: Enumerate all finite sequences of statements of L, check each to see whether it is or is not a deduction from 0,and enumerate the conclusions of all deductions from Q,. By (1) and the Completeness Theorem, every statement of T will turn up on the list. Thus, if T is axiomatizable, then T is recursively enumerable. If, in addition to being axiomatizable, T is complete, then T is decidable by a “wait-for-a-proof algorithm”: To determine whether or not & E T , start the theorem-checking algorithm for Q, and wait until either & or 14 is enumerated. This algorithm works since the completeness of T guarantees & will appear. that eventually either #I or i THEOREM 13.44: I ~ f T is an axiomatizable complete theory, then T is decidable. W
This theorem is particularly useful in showing that a theory Th(M) is decidable since, for any structure M, Th(M) is complete. COROLLARY 13.45: For any structure A, Th(A) is axiomatizable i f and only if Th(A) is decidable. In particular, if Th(A) isjnitely axiomatizable, then Th(A) is decidable. Proof: Every finite set is recursive. H
Thus, for example, Th(w),Th((), and Th(q) are all decidable, as is Th(a) for each a < ow. Even if A4 is not finitely axiomatizable, Th(M) may be decidable. One need only give explicitly a complete recursive set of axioms for Th(M).Thus, for example, although M = o + o*is not finitely axiomatizable, it is axiomatizable. Thus, let Q, = C, u (4,,ln < w},where, for each n, &,, says that a model has at least n elements, and C, says that a model is a linear ordering that has a first and a last element, that every element but the last has an immediate successor, and that every element but the first has an immediate predecessor. (The subscript F is there to remind us that C, is also true in every finite linear ordering.) This set of statements is clearly recursive, and any model A of Q, must have order type w + ( * a + o*for some a, so that, by Exercise 6.11 and Corollary 13.12, A = o + w*. Thus @ is an axiomatization of Th(o + w*), so that Th(o o*)is decidable. On the other hand, one can easily give examples of linear orderings whose theories are not decidable. For example, let A be a recursively enumerable subset of N that is not recursive. (The existence of such a set is
+
3.
DECIDABILITY OF FIRST-ORDER THEORIES OF LINEAR ORDERINGS
269
demonstrated near the end of $16.1.) Let A = {a,, a,, a,, . . . } be an enumeration of A and let M = q a, q a, q + a, . . . . If T h ( M ) were decidable, then one could effectively determine whether or not n E A by checking to see whether or not M l= 9,, where 9, is the statement
+ + + +
(3X,)(3X2).
. .(3X,)[(Xl < X 2 < . '
A (vy)(Xl
AfVY)(Y
< < X,
-+
'
<
+
X,)
( y = X 2 V . . . V y = X,-
< X l + ( W Y < Z <
X,))A(~Y)(y>x,+(3z)(Y>z>x,))l,
which says that M has a maximal finite interval with n elements. For this M , Th(M) is not decidable and hence is not axiomatizable. A more interesting example is Th(o"). We know that Th(w") is not finitely axiomatizable by Corollary 13.24. Furthermore, we have not presented any axioms of Th(w"). However, for each y1, moo-,onby Theorem 6.18, so if 4 has quantifier depth n, then owk 4 if and only if w" k 4. Thus we can argue that since each w" is decidable, so is w w ;but that would be premature. The algorithm we propose for Th(o") has the following form: Given C$ calculate qd(4) and then apply algorithm to 4, where algorithm A , is the one that describes whether or not 4 E Th(o"). This proposed algorithm is actually a finite object only if the passage from a number n to the algorithm A , is recursive. That would not be the case if the algorithms A , followed no pattern. Thus to conclude above that Th(w") is decidable, we need to know not only that each Th(w") is decidable, but that there is a uniform pattern to the decision procedures. The uniform pattern is as follows: Given n, write out the axiomatization a, of ongiven in Theorem 13.34 and apply the wait-for-a-proof algorithm to @,, (Note that there is an algorithm-iterating relativizations-that, given n, will produce the statement @,.) So we can conclude that Th(w") is decidable. By Theorem 6.22, if ON is the collection of all ordinal numbers, then ON Q"', so that Th(0N) = Th(crt") and hence Th(0N) is decidable. That is, one can decide effectively whether a statement about the ordering of the ordinals is true or false. This result, due to Mostowski and Tarski [113, is proved in Doner, Mostowski, and Tarski [3] and in Ehrenfeucht c41-
-
COROLLARY 13.46: Th(o") = Th(0N) is decidable. H
Note that our discussion above also yields a complete axiomatization of Th(w"), namely, @ = u,,
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13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Tarski [ 3 ] : Let CD consist of the linear ordering axioms together with all induction axioms (3x)Cp(x) (3x)[Cp(x) ~(Vy)(Cp(y)+ x I y)] where Cp(x) is a formula whose only free variable is x. Then CD is a complete axiomatization of Th(0N). The theory of O N is discussed at length in [3]; Rosenthal [I61 discusses the models of Th(0N) and tabulates its complete extensions. For which linear orderings M is Th(M) decidable? We will see at the end of this section that Th(M) is decidable for each M E A.But for most order types M,Th(M) must be undecidable since there are only a countable number of algorithms and hence only a countable number of decidable theories. A complete answer as to when Th(M) is decidable seems to be unknown. We now consider whether Th(,X) is decidable, where Jf is a class of linear orderings. Our first example is the case for which JV is the class A!Fof finite linear orderings. We first claim that Z, is an axiomatization of Th(A,). (Note that M k C, if and only if either M is finite or M ‘v w + [ . a + w* for some order type a.) For if Cp E T h ( d F )but is false in some model of C,, then, for some order type a o , Cp is false in w + [ . a. + w * . But then, by Exercise 6.11 and Corollary 13.12, Cp is false in w + [ . rx + w* for every order t y p a ; thus C, u { i C p } is a finite axiomatization of Th(w + a*), contrary to Proposition 13.15. Hence Cp is true in every model of C,, so that, by the Completeness Theorem, X F I- Cp. (This is an example of the common phenomenon that not every model of T h ( M )necessarily belongs to .A’”.) Thus 1, is an axiomatization of Th( H,), and hence Th(A,) is recursively enumerable. To show that Th(,H,) is decidable, it suffices to show that {Cp I Cp 4 Th(&,)] is also recursively enumerable. But Cp # Th(M,) if and only if M b i C p for some finite M, so we need only recursively enumerate the set of all statements false in some finite linear ordering. But if we recursively enumerate all pairs (a,@), where n is a natural number and Cp is a statement of L, and check them one by one to see whether or not n k Cp, putting i Cp E K if n 1 Cp and Cp E K ifn y Cp, then we will have recursively enumerated K = {+I4 4 Th(,X,)). This completes the proof of the following theorem. --$
THEOREM 13.47: Th(AF)is decidable. A slightly more sophisticated argument is used to show that Th(Mw) is decidable, where .Mw is the collection of well-orderings. It follows from the proof of Proposition 7.8 and Corollary 13.12 that if a statement is true in some well-ordering, then it is also true in a well-ordering
3.
DECIDABILITY OF FIRST-ORDER THEORIES OF LINEAR ORDERINGS
271
has a finite axiomatization, there IS, for each a < ow,an algorithm for determining whether or not a 1 d. Moreover, the algorithms are uniform in a since for each r we can construct effectively the axiomatization @(a) from the Cantor Normal Form of a and then apply the wait-for-a-proof algorithm to @(a).However, checking to see whether a k 4 for all a < ow requires infinitely many applications of this algorithm; thus, although we may arrive at the conclusion 4 4 Th(.&"R,) because c( 4 for some a, we could never, by using this algorithm, arrive at the conclusion 4 E Th(AfW). But the Ehrenfeucht analysis comes to the rescue. Recall that every ordinal a is G,,-equivalent to some ordinal in the finite set
w,=
(W,.Un
+ o"-l.a,-,+ . . ' + u 2' a , + W ' f t , + a&
where each a , < 2," and a, I 1 (Exercise 6.20.8). Thus if qd(4) <2n, then Since W, is a finite set of ordinals, and furthermore since one can, given n, effectively write down all the elements a E W,,one can apply the wait-for-a-proof algorithm to @(a) for each r E W,, concluding that 4 E Th(&",) if and only if @(or) t (p for all a E W,.
4 E Th(.Aw)if and only if @(a)t- 6 for all a E W,.
THEOREM 13.48:
Th(AW)is decidable.
We now show that T h ( 9 ) is decidable, where 9 is the collection of all linear orderings. This was first proved by Ehrenfeucht [ 5 ] ; the proof we give is patterned on that of Liiuchli and Leonard [8]. (They note that Calvin found a similar proof.) By virtue of a wait-for-a-proof algorithm applied to 4L(the axiomatization of T h ( 9 ) discussed after Proposition 13.15), Th(Y)is recursively enumerable. If li/ $ Th(9), then, by Theorem 13.22, M k i$ for some M E .A?', so that it suffices to show that { ( a , 4 ) ICI is an order type in and M k 4) is recursively enumerable. THEOREM 13.49: (Liiuchli and Leonard) { ( a , d ) l a is an order type irz ..dand a 1 41 is recursively enumerable.
a theory T , in Proof: We will first define for each order type a in an expanded language L,, and show that T, is complete and logically implies Th(a),the first-order theory of a. (For technical reasons that will be apparent later, we will also define a subtheory S, G TN.)The definition of T , will be given by induction on a. Recall that when we define a theory T as a set Q, of statements, we mean that T is the set of all logical consequences of @.
272
13. THE FIRST-ORDER
THEORY OF LINEAR ORDERINGS
For a = 1, Lais just L, and T , = S, is just (Vx)(Vy)(x = y ) A 4 L .Clearly, T, is complete and logically implies the first-order theory of a. If a = a1 + a,, then La has two unary predicate symbols P , and P,, S, consists of(Vx)(P,(x)v P2(x)),(Vx)(Vy)(P,(x)A P,(y) + x < y), (3x)P,(x),and (3y)P,( y), in addition to the linear ordering axiom 4L,and Thus T, says that a model A can be partitioned into A , + A , with A l = a , and A , = a,; since any two models of T , are clearly G-equivalent, they are also logically equivalent, so that T , is complete, as a set of statements of La. Since ( a , a l , a z ) k T,, T, logically implies the first-order theory Th(a) of a. If a = /3. o,then La has a binary relation symbol and S , contains statements, in addition to 4L, that say that is an equivalence relation that partitions a model A into intervals, and that the set of intervals has order type = UJ. Also T, = S , LJ {(Vu)$ecw~")I/3k $}, where O(w,u)is u w. Thus T , says that a model A can be partitioned into o C . y intervals, for some order type y, and that each interval is E 8; since any two models of T, are G-equivalent, they are also logically equivalent, so that T , is complete, as a set of statements of La. Since (a, -) I= T,, T, logically implies the first-order theory Th(a)of a. A similar argument works in case a = /3 . a*. If a = a(F), where F = {a1,a,, . . . ,a,}, then La contains a binary relation symbol and unary relation symbols P,, P,, . . . ,P, and S , contains is an equivalence relation statements, in addition to 4L,that say that that partitions a model A into intervals, that P,, P,, . . . ,P, also partition A , that each --equivalence class is completely contained in some P i , and that between any two elements of A in different --equivalence classes there are elements belonging to each of P , , P , , . . . ,P, . Also
-
-
-
+
-
-
T, = s, u
-
u
1s i s n
((Vu)[P,(u)3
$@(W,")
1l.i'
$}3
where O(w,u) is u w. Thus T, says that a model A is obtained from the shufflea(F)by replacing each occurrence of an xi by a linear ordering logically equivalent to it; since any two models of T, are G-equivalent, they are logically equivalent, so that T, is complete, as a set of statements of L,, and, as in the other cases, T , logically implies the first-order theory Th(a) of a. We can now argue, by induction on a, that Th(a) is decidable for each aE& Thus, if. for example, if a = /3 w, then, assuming that Th(P) is decidable, we conclude that T, is a recursive and complete set of statements, so that a wait-for-a-proofalgorithm applied to T , gives a decision procedure for Th(a).The other cases are similar. This analysis is not sufficient for our purposes since it is not sufficiently uniform to draw the conclusion that ( ( a , $ ) ] $ E T,} is recursive, from
4.
MODEL THEORY AND LINEAR ORDERINGS
273
which it would follow that ( ( a ,$) I$ E Th(a)) is recursive. We turn now to showing that { ( a , $) 1 $ E T,) is recursively enumerable, which implies that it is recursive, since each T , is complete. , . . . , ( f i k , t j k ) a proof We call a finite sequence ( f i , , $ ~ ~ ) (fiz,tj2), sequence if each pi E A%, each $iE L p , , and for each i, 1 5 i I k, either
$iis a logical axiom or
$,. follows from earlier formulas
or or
$i E
or or
pi = p j . o (or p j . 0")for some,j < i and Ic/i is ( V U ) $ ~ - ~ pi = a(F)and bj E F for somej < i and $iis (Vu)[Pj(u) + $;""I.
So,
pi = ctl (or
$j
by some rule of inference
+ a2 and, for someJ < i, D,i = a1 (or a2)and $iis gl(w)
$pz"@)
Since the set of ordered pairs ( p , $ ) that occur in a proof sequence is recursively enumerable (since [ ( a , $) I $ E S,} is recursive), we need only show that (a, $> occurs in a proof sequence if $ E T,; but this is shown by a simple induction on a. THEOREM 13.50: ( 1 ) Th(M) is decidable for every M E .,A?'. (2) Th(Y) = Th(,&) is decidable. ( 3 ) Th(Y) is decidable, where 9'consists of all scattered linear orderings.
Proof : We need only prove (3), but this is immediate since, by Theorem 13.22, T h ( 9 ) = Th(,Af0)and, by the proof of Theorem 13.49 restricted to scattered linear orderings, Th(,fl[,)is decidable.
$4. MODEL THEORY AND LINEAR ORDERINGS
The model theorist studies the relationship between theories and structures; for example, if T is a theory, what can one say about the collection of models of T? In this section we will develop a number of the basic tools of model theory and use them to analyze the collection of countable models of a complete theory T of linear orderings. (By a complete theory of linear orderings we mean a complete theory that contains the theory of linear orderings; alternatively, T is a complete theory of linear orderings if there is a linear ordering A such that T = Th(A).) In the next section we will deal with the specific question of how many distinct countable models a complete theory of linear orderings can have.
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13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Whenever we speak of a theory T in this section we will assume that T is complete. We begin by presenting two basic tools-elementary substructures and n-t ypes. Although w 5 w 5 in various ways, from an “elementary” point of view one way is best since only if w is mapped onto the initial w of w 4 do the images of the elements of w in w + l, have the same first-order properties as the original elements of w . The map f : w + o ( that takes the ith element a, of o to the ith element bi of w + 4 is said to be an elementary map because, as we will see shortly, any formula satisfied by the ai in o is satisfied by the corresponding hi in w 4. The notion of an elementary substructure and the basic facts we present here are due to Tarski and Vaught [21] and Robinson [13].
+
+
+
+
DEFINITION 13.51 : Let A, B I= T and let J ’ : A-+ B. We say that f is an elementary map from A to B if for every a,, . . . ,a, E A and every formula 4 ( . X 1 . . ’ . ,X,L
A b 4[a,, . . . ,a,]
if and only if
B 1 $ [ f ( a , ) , . . . ,f(a,)].
An elementary map from A to B is also called an elementary embedding since it must be a 1-1 map. If there is an elementary map from A to B then we say that A is elementarily embeddable in B, and we write A < B. If A c B then we call A a substructure of B, denoted A E B if whenever &v,, u 2 , . . . ,0,) is an atomic formula and a l , a 2 , .. . ,a, E A, A k 4 [ a , , a 2 , . . . ,a,] if and only if B I= 4[ul, a 2 , . . . ,a,]; if A and B are linear orderings, this means that A is a subordering of B. If A G B, then we call A an elementary substructure of B if the identity map from A to B is an elementary map from A to B; this is also denoted A < B. Although using A < B to mean two different things can lead to confusion-for example, if A s B and A is elementarily embeddable in B by some map other than the identity-it will not, since in any context it will be clear which use is intended. Another possible source of confusion is our using here the same symbol, for elementary embeddings of structures, that we used in earlier chapters, for embeddings of linear orderings. To remove that confusion, we will reserve < in this chapter for elementary embeddings and use Ionly for ordinary embeddings. Note that the assumption that A and B are models of the same complete theory T is redundant since the condition in the definition with n = 0 implies that A = B. Extending this observation, we see that f is an elementary map
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MODEL THEORY AND LINEAR ORDERINGS
275
from A to B if and only if for every ( i , , . . . , a , E A
(‘%a,, . . ,a,J ’
and that if A c B, then A
= ( B , f ( a , ) ,. . . , f ( a , ) )
< B if and only if
( A ,a , , . . . , a , ) = (B,a , , . . . ,a,) for every a l , . . . ,a, E A . This observation, together with Theorem 13.11, provides the following test for A < B. THEOREM 13.52: (1) f : A if and only if
G,((A, a1,
’ ’
+
B is an elementary map from A to B
. a,), ( B ,f ( a , ) , . . . f ( a , , ) ) E) 11 3
7
.for every m and every a l , . . . ,a, E A . (2) I f A c B, then A < B if cind only
if
G , ( ( A , a l , . . . .u,J,(B,a,,.. . ,a,,))€II .for ever!] m and every a,, . . . ,a,, E A .
+
+
Thus, for example, i f f : o + cu 6 maps o onto the initial o of o 5, then, as is easily verified, f is an elementary map. We express this fact by writing o < o i.(This use of the notation presupposes that w is the o of o + 5 ; more generally, if we write A < B, where there is some ambiguity as to how A is seen as a subordering of B, then we assume that A is contained in B in the most natural way possible.) On the other hand, if y : w -+ o iis any other order preserving map, then, as the reader should verify, for some choice of rn and of a , , . . . ,u, E o,PLAYER I has a winning strategy in G , ( h a , . . . ,a,,),(0+ i,d a d , . . . ,da,,))). The following exercises likewise use Theorem 13.52.
+
+
EXERCISE 13.53: (1) Show that if A and B are dense linear orderings without endpoints, then any order-preserving f : A -+ B is an elementary em bedding. (2) Show that if A , < B , , . . . , A , < B,, then A , . . . + A , i B,
. . . +B,.
+
+
+
(3) Show that there is a unique map from o a* to (a+ w * ) . 2 that is an elementary embedding. (4) Show that if A < B and B < C , then A 4 C.
If one has an explicit description of A and B, then the game analysis is a useful tool for determining whether or not A < B, but otherwise it is inadequate. The following criterion will be useful in more abstract situations.
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THEOREM 13.54: Let A C_ B and assume that for euery $(x,, . . . ,x,,y j and every a , , . . . ,a, E A, if B k $[al,. . . ,an,b] for some ~ E B then , B k $ [ a , , . . . ,a,,a] for some a E A . Then A < B. Proof : Suppose otherwise and choose a formula 4(xl, . . . ,xn)that has minimum quantifier depth and such that there are elements a,, . . . ,a, E A for which B k $[al, . . . ,a,] and A F $[al, . . . ,a,]. Although 4 is in general a Boolean combination of quantifier-free formulas and formulas beginning with a quantifier, it is apparent that if there is such a 4, then there must be one that begins with a quantifier ( Q y ) since A G B implies that such a $ cannot be quantifier-free. If $(xl, . . . ,x,) is (Vy)$(x,, . . . ,x,, y), then for some a E A we have A Y $ [ a , , . . . ,u,, a] although B k $[al, . . . ,a,, a ] , contradicting the minimality condition since qd($) < qd(+). If 4(x1,. . . ,x,) is (3y)$(x,,. . . ,xnryj, so that B k $[a,, . . . ,a,, b] for some b E B, then by assumption B k $[a,, . . . .a,,a] for some a E A so that, since qd($) < qd(+), A k $ [ a , , . . . ,anra], which implies that A k + [ a , , . . . ,a,], contradicting the condition on 4. Thus no such can exist and hence A < B.
+
Note that the converse of this theorem is also true. For if A < B, a , . . . . ,a, E A, and B k $ [ a , , . . . ,a,, b] then B k (3y)+(a,, . . . ,a,, y ) , so that, since A < B, A k ( 3 y ) 4 ( a l , .. . , a,, y ) ; hence A k 4 [ a , , . . . , a n ra] for some a E A, which implies, since A < B, that B k +[al, . . . ,a,,a]. A simple application of Theorem 13.54 is the following lemma, which will be used frequently. LEMMA 13.55:
If A
G
B G C and A
< C and B 4 C , then A < B.
Proof : Assume that B k 4[a,, . . . ,a,,b], where b E B and a,, . . . ,a, E A . Since B < C, we have C k $ [ a , , . . . ,a,,b], and since A < C , we have C k 4 [ a , , . . . ,a,,a] for some a E A , which, since B < C, implies that B k $[al,. . . ,a,,a].
The method of complete diagrams, discussed below, can be used to construct an elementary extension of a given model A. DEFINITION 13.56: Let L be a given language and let A be an interpretation of L . By L ( A )we mean the language obtained from L by adding a new constant symbol g for each element a E A . By C(A) we mean the set of allstatements+(a,, . . . ,g,)ofL(A)forwhichA k @ [ a , , . . . ,a,].C(A)iscalled the complete diagram of A.
4.
277
MODEL THEORY AND LINEAR ORDERINGS
For B to be a model of C(A) means that within B we can identify elements f ( a )corresponding to the elements a E A such that B k + [ f ( a , ) ,. . . ,f(a,)] whenever A k +[a,, . . . ,an].Thus if we identify A with the substructure of B consisting of { f ( a ) 1 a E A } , we can regard A as an elementary substructure of B. We express this concisely, with some abuse of notation, in the following way. THEOREM 13.57:
A
< B if
and only if B b C(A).
The method of complete diagrams can be used to show that any infinite A has a proper elementary extension. For if we add one new constant symbol b_ to L(A) and let C = C(A) u { b _ # g l a E A } , then, by the Compactness Theorem, C has a model B since any finite subset of C has A as a model (where to b_ corresponds any element of A not mentioned in that subset of C-using the assumption that A is infinite). Then A < B since B k C(A) and A is properly contained in B since the element of B corresponding to b_ is not in (the image of) A . In the context of linear orderings, we can come to a more striking conclusion by combining the method of complete diagrams with Rubin's techniques [17]. THEOREM 13.58: Let A be a countable linear ordering with no last element. Then there is a non-empty linear ordering B such that A < A B.
+
Proof : Let C = C ( A )u { b_> glu E A } , where b_ is a new constant symbol. Since A has no last element, C has a model D by the Compactness Theorem and A < D by Theorem 13.57. We will assume, without loss of generality, that A is actually contained in D. Let D' = (d E Dld < a for some a E A } , so that D = D' B, where B is non-empty. To show that A < A + B, it suffices, by Lemma 13.55, to show that A B i D' B, and for this it suffices to show that A < D' by Exercise 13.53.2. Again using Lemma 13.55, to show that A < D', it suffices to show that D' < D' B, and that is what we now prove. Assume that D' B k + [ d , , . . . ,d,, b], where d,, . . . ,d, E D' and b E B. Choose a E A greater than d,, . . . ,d,. By Theorem 13.41, there is a selecting formula +#(x,u) for in the interval [a, +) of D' + B ; that is, for every d E D' B, we have D' B b +"[a,d] if and only if
+
+
+ +
+
+
+
+
d2a
and
D'
+ B b +[d,,
. . . ,d,,d].
278
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
+
+ +
Thus since D' B t= $ [ d , , . . . , d , , b ] and b 2 a, we have D' B k $#[a,b]. But since A < D' Band a E A, there is an a' E A such that D' B k $ " [ a , a']. This implies that we have found an element a' E D' such that
+
D
Hence D'
+ B b $[d,,
. . . ,d,,a'].
< D' + B. Thus we have shown that A
i A
+ B.
Another important application of the method of complete diagrams is the following general result. THEOREM 13.59: Let A 7hen each A, < A.
=
U{A,In < o}, where A,
< A,
i A,
< ' .. .
Proof: Let C = C(A,) u C(A,) u C(A,) u . . . . Then Z has a model, by the Compactness Theorem, since any finite subset of C is completely contained within some C(A,) and hence has that A, as a model. Let B k C. Then, by Theorem 13.57,A, < B for each k . Furthermore, since A = {An[n < w}, A !E B. By Lemma 13.55, in order to show that A, i A for each k, it suffices to show that A < B. Assume that B k $ [ a , , . . . ,a,, b], where a , , . . . ,u, E A and b E B. Choose m so that a,, . . . ,a,€ A , . Since A, i B, there is an a E A , such that B k $[al, . . . ,a,, a] ; thus we have found an a E A such that B k $[a,, . . . ,a,,a], so that we can conclude that A < B and hence that A, < A for each k. 4
u
In particular, if each Ai is a model of the theory T, then A is also a model of T ; this enables us to build new models of T from old ones. If the chain {A,ln < o}is not an elementary chain, then u{A,,In < o} need not be a model of T even if each Ai is a model of T . For example, if A , = (o+ o*) .n for each n (it being understood that if n < rn, then A, consists of the first n copies of o + o* in A,), then u(A,ln
a1though A,
+ o * ) . n = o + 6 . ( n - 1 ) + o*= o + w*
= (o
for each n. Thus if we wish to put together models of a theory T and get new models of T, it helps if suborderings are elementary suborderings. DEFINITION 13.60: Given a complete theory T, we say that a model A of T is a prime modelof T if A < B for every model B of T.
4.
279
MODEL THEORY AND LINEAR ORDERINGS
+ + + +
For example, w is a prime model of Th(w); for given any model o 5 . ct of Th(o), the map f :w + w 5 . u that maps o onto the initial w of w 5. ct is an elementary map, as is clear from a game analysis. Similarly, w w* is a prime model of its theory; for given any model w 5 . ct + o*ofTh(w a*), the map , f : o w* + o + 5 . (x + to* that maps w onto the initial o,and o* onto the terminal w*, of o 5 . (x + o* is an elementary map. If T is an arbitrary complete theory, then T need not have a prime model (see, for example, [2]), but what if T is a complete theory of linear orderings? We will return to this question later.
+
+
+
+
+
+
EXERCISE 13.61 : Let o 5 . CI and w 5 . p be models of Th(w). Show that w + [ . ct < w [ . p ifand only if ct 5 p.
+
We now introduce the notion of n-types, first discussed systematically in Vaught [22], who also provides many historical remarks on the material treated in this section. DEFINITION 13.62: Let A be a structure and let ( a l , . . . ,u,) be an ntuple of elements of A . By the n-type of ( a l , . . . ,an) we mean the set of all formulas 4 ( v l , . . . ,v,) of L such that A I= 4[ul, . . . ,a,]. We note that the n-type P of (a [ , . . . ,a,) is a consistent subset of F,(L), the set of all formulas whose free variables are among v l , . . . ,u,, and, moreover, that it is maximal among consistent subsets of F,(L) since adding any formula 4 to P would result in an inconsistency because, if 4 is not already in P, then (14) E P. (As in Chapter 12, a set Q, of formulas is consistent if no contradiction Ic/ A i ) I is deducible from Q,.) We now turn the definition around, with this observation as guide. DEFINITION 13.63: Let T be a complete theory. An n-type of T is a maximal consistent subset of F,,(L) that includes T. This definition is justified by the theorem below. THEOREM 13.64: I f P is an n-type of T then there is a countable model A of T and an n-tuple (al, . . . ,a,,> of elements of A whose n-type is P.
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13.
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Since C is consistent, it has a countable model A in which there are elements a, . . . . ,a, such that (a,, . . . ,a,) has n-type P. Note that A b T since if 4 is a statement in T, then 4 E P , so that 4 E X. H DEFINITION 13.65: Given a complete theory T, an n-type P of T , and a model A of T, we say that P is realized in A if there is an n-tuple of elements of A whose n-type is P. Otherwise we say that P is omitted in A. For example, in o each element has a different 1-type since the nth element, and only the nth element, satisfies a formula 4,(v1) that says that u , has exactly n predecessors. No element of o [ after the initial o has any of these types since any such element satisfies i$,,(vl) for every n. Do all such elements have the same 1-type? Does Th(o) have any other 1-types'? These kinds of questions are naturally discussed by connecting n-types with notions discussed earlier.
+
PROPOSITION 13.66: (1) Suppose that f is an automorphism of A such thatf(a,) = bi for 1 I iI n. Then (a,, . . . ,a,) and (b,, . . . ,b,) have the same n-type. (2) Let ( a l , . . . ,a,) be an n-tuple of elements of A and ( b l , . . . ,b,) be an n-tuple of elements of B. Then the following are equivalent: (a) (al, . . . ,a,) and (b,, . . . ,h,) have the same n-type; (b) (A, ~ 1 , ... , a,) (B, b1,. . . , b,); (c) PLAYER II has the winning strategy in G,( (A, a,, . . . ,a,), (B, b,, . . . ,b,)) j o r every m. A EXERCISE 13.67 : (1) Prove Proposition 13.66. (2) Show that all elements of o 5 after the initial o have the same 1-type. (3) Show that Th(o)has no other 1-types. (4) Show that if P' is a consistent subset of F,(L), then P' s P for some n-type P of T.
+
Given a theory T , to determine all n-types of T we need only look at all n-tuples of elements in all (countable) models of T, not a small task. The following definition will simplify matters. DEFINITION 13.68: Let P be an n-type of T and let P' G P. We say that P' generates P if for each o! E P there are formulas o!,, . . . ,clk of P' such that
T k ( V v 1 ) . . . (b'o,,)[Crl(vl, . . . ,v,)
A
'
. . A a,(v,, . . . ,v,)
+ o!(vl.
. . . ,v,)].
4. MODEL THEORY AND
LINEAR ORDERINGS
28 1
Equivalently, P' generates P if whenever an n-tuple of elements of a model of T satisfies just the formulas in P', then its n-type is automatically P. If P' generates an n-type, then we may as well call P' an n-type. Thus the l-types ofTh(o) are { {4,(uo)} In < a}and ( i 4 , ( v o ) I n < w } .
EXERCISE 13.69: (1) Assume that P' G F,(L), that A 1 T , and A b $[al, . . . ,a,] for every $ E I".Assume also that whenever B b $[b,, . . . ,b,] for every $ E P', ( b l , . . . ,b,) has the same n-type as ( u l , . . . ,uJ. Show that P' generates an n-type of T . (2) Assume that P' G F,(L), A 1 T , and for some a l , . . . ,a,E A, A 1 $ [ a , , . . . ,a,]for every $ E P'. Assume also that for any two n-tuples of.elements of A that satisfy P' there is an automorphism of A mapping one n-tuple to the other. Show that if either P' is finite or T has, up to isomorphism, a unique countable model, then P' generates an n-type of T . (3) Show 'that if T = Th(q). then the n-type of ( u l , . . . , a,) is { u , < U J U , < u,} u { U L = V j l U * = L l J . We determined above all l-types of Th(w). Let us now describe the 2-types of Th(o). For each n and m, 6,,(ul) A $ m ( ~ 2 )is a 2-type P,,m;for each n, (4,(u1)} u { i 4 m ( u z ) I m< w', isa2-type P,,,;foreachm, { i 4 , , ( u l ) l n < w } u {$ m ( u 2 ) } is a 2-type Pm,m.Every other 2-type contains ( i 4 , ( u l ) I n < w } u {i$m(~lz)I< m w } = P , , , but there are many distinct 2-types containing P,, . Thus if E,(x,y)says that x < y and there are exactly t elements such that x < t I y, then for each t 2 1, P,+ = P,, u { Et ( u 1 , u 2 ) } and P I - = P,, ~ {E , ( u , , t~ ~ ) } ar e2 - ty p es . A ls=oP~ P,, u ( u l = u2)isa2-type. Finally,
p,,, u Ch f %> u ( l E , ( u l , u 2 ) 1 t < 0)u { l ~ l ( U z , u , ) ( < t 0) is also a 2-type P*,%.That these are in fact all 2-types is easily verified by a game analysis and Proposition 13.66.2. The set S,( T )of n-types of a complete theory T has a topological structure on it that sheds light on the relationship between n-types. A basic open set in the topological space S,(T) is of the form ( P E S,(T)I 4(u1, . . . ,on)E P } for a particular formula 4(ul, . . . ,on)E F,(T). Using this definition, what we now present as a definition is actually a theorem. DEFINITION 13.70: An n-type P of T is said to be principal or isolated, if P is generated by a single formula. Otherwise it is a non-principal or limit n-type.
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Equivalently, P is a limit n-type if every formula in P also belongs to some other n-type; that is, every basic open set containing P contains other n-types. Note that if P is a principal n-type, say P = {4(v1,. . . ,u,)}, then P must be realized in every model of T for, since T is complete, T I- (3v1) . . . (3vn)4(v,,. . . ,v,).
Thus, if a model of T omits an n-type P , then P must be non-principal. The converse of this is also true; that is, if P is a non-principal type of T , then there is a countable model of T in which P is omitted. This is implied by the following theorem due to Ehrenfeucht (see Vaught [22]).
THEOREM 13.71 : Ler { P j l j < 01) be a countable set of non-principal types of' a complete the0r.y T . Then there is a countable model A of T in whic3z every P , is omitted. Proof : Add to the language of T a countable set {g,ln < w } of new constant symbols; the elements of A will be precisely this set. Enumerate all ordered pairs (a,P), where a is an ordered n-tuple of the new constant symbols, P is one of { P j l j < w } , and P is an n-type. Since there are countably many such ordered pairs, we may assume that they are enumerated as an w-sequence {(ak,Pk>I k < w } . We now define by induction an expanding sequence { T k ( k< o)of theories, each involving a finite number of additional statements and a finite number of the constant symbols (@,In < w } . We define To to be T and we assume, as the induction hypothesis, that the constants appearing in Tk are g i , 1 , , g i ( 2.) ., . We also assume that all formulas of the expanded language that have the one free variable x are enumerated in an w-sequence ( + k ( X ) / k < w } . Now given Tk we proceed as follows to define T k + l .We first add to Tk a statement of form (3x)&(x)+ &(a),where g is a constant symbol appearing neither in T , nor in 4k.X). (Such a statement says that if anything satisfies &, then some & does; this guarantees that A, consisting of {fzk I k < w } , will be a model of T . This method of proof, used in proving the Completeness Theorem, is discussed in detail in Enderton (see [31, Chapter 12). Note that this statement can be added consistently to Tk since otherWise k 1 [ ( 3 X ) b k ( X ) + 4 k ( g ) ] , so that Tk k ( 3 X ) 4 k ( X ) and Tk k 14k(g); since a appears neither in Tk nor in &(X), we can pass from T , t i4,(g) to Tk t ( v x ) i d k ( x ) , contradicting the consistency of T,. We next find a formula $ ( x l , x 2 , . . . ,x,) in Pk (assuming Pk is an n-type) such that i $ ( a k ) is consistent with Tk u { ( h ) + k ( X ) -,&(a)}. If this task is impossible, then Tk
{(3x)4k(x)
$k(!d}
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283
for all $ in Pk. Since T , has only a finite number of statements not already in T, we can write this as
T u O(ak,b) t $(ak), where b are the constants appearing in Tk u ((h)&(x) + &(a)) that are different from those in ak.This implies that
T t ()(ak,b) + $(ak), which in turn implies that
T
( 3 Y ) W k , Y)
+
$(ak)
for all $ E Pk. But this says that Pk is a principal n-type generated by
(3y)0(x 1 9 x2 . . . > x k , y), 3
contrary to the hypothesis. We now define Tk+1 to be Tk U { ( j X ) d k ( X ) + 4k(a)}U { 1$(ak),>.By putting i $(ak) into T k +1, we guarantee that the n-type of ak will not be Pk. If, as advertised, A consists of just the constant symbols {@,In < a), then no n-tuple of elements of A will have n-type Pk,so that this Pk,and all the others, will be omitted in A. Now let T* = Tklk < o) and let T** be an arbitrary complete extension of T*. We now define a structure A whose elements are the constant symbols {a,\n < w >; more precisely, we define an equivalence relation by g, -v g , if g, = g , is in T** and WG let A be the set of equivalence classes. For each relation symbol P ( x , , . . . ,xn) in the language of T , we define an n-ary relation R on A by stipulating that R(giil),. . . ,a,(,,)will hold in A if and only if P(giil),. . . , g i c E tis) in T**. It is then easily verified, by induction on formulas, that for every formula $(xl, . . . ,x,) and any elements uiil),. . . ,ai(,)of A , A 1 4[gi(l),. . . ,gi(,)] if and only if ,)&’q . . . ,gi(,)) is in T**. In particular, A 1 T and each of the types {PjIj < o}is omitted in A.
u{
It follows from Theorem 13.71 that if A is a prime model of T, then the n-type of every n-tuple of elements of A must be principal; for if a nonprincipal n-type were realized in .4 and omitted in B, then A certainly could not be embedded elementarily in B since the n-type of an n-tuple of elements of A is the same in B as it is in A if A < B. At this point, it is appropriate to explain where the terms “principal” and “non-principal” come from. Just as the terms “isolated’ and “limit” arise naturally from the topological structure of S,( T ) ,the terms “principal” and “non-principal” arise from the Boolean algebra structure of F,( T ) .That is, if we identify two formulas that are equivalent in the theory T, then
284
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F,( T ) becomes a Boolean algebra, denoted B,(T), and an n-type becomes a maximal (dual) ideal in this Boolean algebra. An isolated type, one that is generated by a formula, is a principal maximal (dual) ideal, and a limit type is a non-principal maximal (dual) ideal in this Boolean algebra. A fundamental fact about Boolean algebras is that every infinite Boolean algebra has a non-principal maximal (dual) ideal. Thus, for example, the Boolean algebra consisting of all finite subsets of N and all complements of finite subsets of N has infinitely many principal maximal (dual) ideals, namely, I , = { X G N I n E X} for each n E N , and exactly one non-principal maximal (dual) ideal, namely, the set of complements of finite sets. In the context of model theory, this fundamental fact translates into the statement that if there are infinitely many inequivalent formulas in F,( T ) then there is a non-principal n-type of T. To prove this statement, suppose that there are infinitely many inequivalent formulas in F,(T). It is then easily verified that there are infinitely many n-types. If all of these are principal, then there are countably many of them; call them {P,lm < o}.where each P , is generated by I),(vl, v2, . . . ,v,). Since { i I),(vl, v2, . . . ,v,) I m < o>is consistent, it is contained in an n-type P of T that is different from each P,, contrary to assumption. Thus some n-type of T must be non-principal. The following proposition is now easily verified. PROPOSITION 13.72:
The following are equivalent:
(1) There is an infinite set of formulas of F,(T) no two of which are equivalent in T. (2) B,(T) is infinite. (3) S,(T) is infinire. (4) There is a non-principal n-type of T. A Armed with the topological and algebraic interpretations of the set S,( T ) of n-types of T , we return to our consideration of the n-types of Th(o); we continue using the terminology introduced earlier. The space S,(Th(w)) of 1-types of Th(w) can be pictured as a
...
a
Po PI
p2
a
p,
7
where P, is the 1-type {4,,(v1)).and P, = {i4,,(v1)1n< o}is the limit of {Pnln w>. The space S,(Th(wj j is more complicated. However, each P,,.,, is isolated, each Pn,mis a limit of ( P , , , I rn < m}, each P,,, is a limit of { Pn,mn < o}, each P,+ isalimitof{P,,,+,In < o),andeachP,- isalimitof{P,+,,,Irn < o > .
I
4.
MODEL THEORY AND LINEAR ORDERINGS
.
285
P,'
. . . PI
I
pz
I
i l l
. . . .... . . . . . . .
...
....
0
...
........
.
p1.0
Po.0
p1.0
-
0
px,z px,l
pz,o
FIGURE1
Thus we have the picture shown in Fig. 1. Omitted in this picture is P:,, which, unlike the other 2-types of Th(o), is a limit of limit 2-types. (It This kind of analysis is also the limit of the isolated 2-types {Pt,2tlt < o}.) of S,(T), similar to the Cantor-Bendixson analysis of closed subsets of the real line (which Cantor used to show that every uncountable closed subset of R has cardinality c), was introduced by Morley [9] and has been a major tool in model theory. What about S,(Th(o))? The following result shows that, in a certain sense, the n-types of any complete theory T of linear orderings are determined by the 2-types of T. THEOREM 13.73: Let P be an n-type of a complete theory T of linear orderings and let P* be all formulas of P that have at most two free variables. Then P* generates P. Proof : If + ( u l , . . . , ~ , ) E Pthen, , by Theorem 13.37, there are formulas Il/ij such that 4 L
~ ( V I .? .
' 7
vn)
* V A Il/ij, i
j
where each t,hij has at most two free variables. Since 4 E P, there must be an i such that Ajt+hijeP.But {Il/ijl j} s P*, so that P* t 4(vl,.. . ,vJ. Hence P* generates P.
286
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COROLLARY 13.74: Let T be u complete theory of linear orderings. I f S,( T ) is jinite, then S,( T ) is finite ,for all n. I f S2(T ) is countable, then S,( T ) is countable ,for all n. Proof : Let P be an n-type of T and let P* be as in the proof of Theorem 13.73. For each i, j I n, let P t j be the set of formulas of P* whose free variables are among ui and u j . Then each P t j is a 2-type of T and P* = iPZjI 1 I i Ij I n } . Both of the desired conclusions follow from this.
u
The conclusions of Corollary 13.74 are not true in general. For example, it is shown in Rosenstein [15] that for each n there is a complete theory T , such that S,( T,) is finite but S,( T,) is infinite for all m > n. We asked earlier whether every complete theory T of linear orderings has a prime model. If S 2 ( T )is countable, then by Corollary 13.74, S , ( T ) is countable for all n, so that, by Theorem 13.71, there is a countable model A of T in which no non-principal type is realized. Such a model is prime, as we now show. THEOREM 13.75: Let A be a countable model of a complete theory T in which no non-principal type is realized. Then A is a prime model of T . Proof: Let B k T and let A = { a l r u 2 , a 3 ,.. .) be an enumeration of the elements of A . Since for each n the n-type P, of ( u l , a,, . . . ,a,) is principal, there is a formula +,(ul, u 2 , . . . ,u,) that generates P,. Thus it suffices to find a sequence b , , b 2 ,b 3 , . . . of elements of B such that B 1 + , [ b l , b 2 , . . . ,b,] for each n, for then the map f :A + B defined by f(a,) = b, is an elementary map. Proceeding inductively, we assume that we have defined b,, b,, . . . ,bk SO that B k (f)k[b,,bz,. . . ,hk]. Since A = ! ( b k + l [ U l , U Z , . . . ,U k , U k + I], We have also A 1 (lo,+ I ) & + ,(a,,a,, . . . , a k , ,), so that, since ( b k generates the k-type p k , T 1 + k ( U 1 , . . . u k ) ( 3 t ~ k + l ) ( b k + l ( ~ 1., . . 7 uk? u k + l ) . But then, since B k + k [ b l , .. . ,bk], there is an element b k + lE B such that B b $k+ [ b . . . ,b k ,b k +,I, and so we can continue to embed A elementarily into B. W tik+
9
+
We return now to our earlier question whether every complete theory T of linear orderings has a prime model. It follows from Theorem 13.75 that that is the case if S 2 ( T )is countable. Rubin [17] showed that if S , ( T ) is countable, then S 2 ( T )is also countable, so that in this case too T has a prime model. Thus, we consider next an example of a linear ordering A for which S,(Th(A))is uncountable; this example is motivated by the analogy between Dedekind cuts and 1-types.
4. MODEL THEORY
AND LINEAR ORDERINGS
287
EXERCISE 13.76: Let f : Q + N be a 1-1 correspondence between Q and N and let A have order type fxq) I q E Q}. Show that Th(A) has c l-types and that A is a prime model of Th(A).
c(
This example fails since A is in fact a prime model of Th(A). We are, however, on the right track, as noted by G. Cherlin. THEOREM 13.77: prime model.
There is a linear ordering A such that Th(A) has no
Proof: Partition Q into two subsets Q1 and Q 2 each of which is dense in the other. Let f : Q 1 + N be a 1-1 correspondence between Q1 and N and let A have order type c(g(q)I q E QJ , where g(q) = f(q) if q E Ql and g(q) = 1 if q E Q 2 , Now every Dedekind cut X u Y of Q determines a l-type which asserts that z'l has no immediate predecessor or successor, and that u I is greater than all condensation classes of size f(q) for q E X and less than all condensation classes of size f(y) for q E Y. In every countable model of Th(A), infinitely many of these l-types must be realized, but none of these I-types is realized in every countable model. Thus Th(A) can have no prime model.
In Theorem 13.73 we saw that any n-type of a complete theory of linear orderings is determined by its constituent 2-types. Rubin [17] proved that, under certain circumstances, a 2-type is determined by its constituent l-types. The proof of this result, presented here as Theorem 13.80, involves a complicated induction (Lemma 13.79)and the notion of the k-n-type of a k-tuple. (Theorem 13.80 will be used in $5.) Recall that is the set of all formulas in the variables u 1 , u 2 , . . . ,uk whose quantifier depth is at most n. DEFINITION 13.78 : Let b = (6 (, 6, , . . . ,h k ) be a k-tuple of elements of B. By the k-n-type of b we mean the set of all formulas $ ( u l , u 2 , . . . ,u k ) in Lk., such that B k $ [ h , , b 2 , . . . ,hJ. The k-n-type of b is denoted P,(b). A k-n-typeof B is a maximal set of formulas of I ! , ~ , ~that is consistent withTh (B).
If we let P(b) denote the k-type of b, then P(b) = U{P,(b)In < w } . Note also that since any k-n-type of B is a finite set of formulas, the language being finite, any k-n-type of B is the k-n-type of a k-tuple of elements of B. If I is an interval of B, we let P,(I) denote the set of l-n-types of the elements of I . We introduce several notions which will simplify the statement of Lemma 13.79. A sequence Q1, Q 2 ,. . . ,Q,, of l-n-types of B is said to be realized in
288
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B if there are elements c1 < c2 < . . . < c, in B such that each cihas 1-n-type Qi.We let t(n) be the number of 1-n-types of B for each n and we define two sequences s(n) and u(n) by induction: s(0) = 0, u(0) = 0, s(n + I ) = 2 . s ( n ) .(t(n))s(”) 1, u(n + 1) = s(n + 1) + u(n).The formulation of the lemma is complicated because of the requirements of the induction argument.
+
LEMMA 13.79: Let B be a linear ordering and let b , < b, < ’ . . < b, und b , ’ < b2’ < * . . < b,‘ be two k-tuples of elements of B. Let B , = (t, b , ) and Bo‘ = (+-, bl’),let Bi = ( b i ,hi+,) and B,’ = (bi‘,bi+ 1) for each i, 1 I i I k - 1, and let Bk = (b,, +) and B,‘ = ( h k ‘ , +). Assume that (1) for every j , 1 I j I k, Pu(nl(bi) = PU(,)(b/); (2) for every j , 0 I j I k, P,,,- ,)(B,) = Pu(n-l)(B;)
i
j = 0 and
and either
‘u(nJB0)
= Pu(n)(Bo‘)
j = k und PU(,)(Bk) = P,(,,(Bi)
O < j < k a n d pu(n)((bj,bj+l))= Pu(,,)((bj’,b)+l))
or every sequence of 1-u(n - 1)-types of Pu(n-l , ( B j )of length s(n) is realized in B, and Bj’.
Then P , ( ( b i , b , , . . . , b J ) = Pn((b1’,b2’,. . . ,b,,’)). Proof : We proceed by induction on n to show that this conclusion holds for all k. To show that P,(b) = P,(b), where b = ( b , , b , , . . . ,bk) and b’ = (bl‘,b2’,. . . ,b i ) , it suffices to show, by Theorem 13.11, that G,((B,b), (B, b’))E 11, which is correct if the two members of each pair of corresponding intervals are both empty or both non-empty; this is guaranteed by (2). Assume now that the lemma has been proved for n and for all k. To show that P,+,(b) = Pn+l(b’),it suffices to show that G,,,((B,b),(B,b))~11.To do this, it suffices to show, by Theorem 13.4, that for each b E B there is a b ’ E B (and the reverse) such that G,((B,b, b), (B, b’,b‘))E11. Assuming that b , < b < b, (the other cases are similar), we will be able to use the induction hypotheses to conclude that
Pn((b,,b,b,, . . . bk)) = P,((bl’,b’,bz‘,. . . bk’)) 9
7
[and hence that G,( (B, b, b), (B,b’,b’))E 111 if we pick b so that b,’ < b‘ < b2’, ‘u(n)(’) = ‘u(n)(b’h P u ( n - 1 J ( b l , b ) )= Pu(n- 1)((b1’9bf)),Pu,n- 1)((b,b2))= Pucn-l)((b, b,‘)), either P,,(,,J(b,,b ) ) = PU(,)((hl’, 8 ) ) or every sequence of 1-u(n - 1)-types of P u ( n - l J ( b l , b )of ) length s(n) is realized in ( b l ,b) and (bl’,b), and either Pu(,,)((b, b,)) = PU(,)((b’,b,‘)) or every sequence of 1-u(n - 1)-types of l,((b,b , ) ) of length s(n) is realized in (b,b,) and (b‘.b,’).
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MODEL THEORY AND LINEAR ORDERINGS
There are two main cases, the first if P,(,+ ,)((b,,b,)) = P,(,+ ,)((b1', b,')), and the second if every sequence of l-u(n)-types of Pu('")((bl, b,)) of length s(n 1) is realized in (b,,b,) and ( b , ' ,b,'). In the first case, where Pu(n+l)(
+
Bb
(W(+I(bl,
x)A + Z ( % b,)
A
b , < x < b, A d44),
we also have
B b (3x)(+l(bl',x)A tJ2(x. b2')A b,' < x < b,' so that if b
E
A
$(x)),
B is such that B k t,bl(h1', b') A $2(h',b,')
A
bl' < b' < bZ' A +(b'),
then P u ( n ) ( ( b l ' > b')) = p u ( n ) ( ( b l , b ) ) and pu(n)((b'>b 2 ' ) ) = p u ( n ) ( ( b , b 2 ) ) and P,,(,,,(b') = Pu(n)(b).Note that this implies also that Pu(n-,)((bl, b))= f',('",,((b,', b'))and that Pu(n-,)( (h,h,)) = Pu("l)((b', b,')) because the existence of an element in one of these intervals with a certain l-u(n - 1)-type can be expressed as a formula of quantifier depth u(n - 1) 1 s u(n), so that the corresponding interval also has such an element, since the endpoints of the two intervals have the same 2-u(n)-type. Thus in this case we are done. In the second case, where every sequence of l-u(n)-types of PU(,,)((bl,b2)) of length s(n 1) is realized in ( b , ,b,) and (b,', b,'), there are two main subcases. Suppose first that every sequence of l-u(n - 1)-types of length s(n)that is realized in (h,,6,) is also realized in both (b,, b)and (b,b,). Let Q,, Q,, . . . ,Q, be the concatenation of all sequences of elements of Pucn-1)( (bl, b,)) of length s ( n ) ; since the number of l-u(n - 1)-types realized in (b,,b,) is at most t ( n - l), r I s(n) . (t(n - l))'(").If P IS the l-u(n)-type of b, then the sequence Q,, Q,, . . . , Q r ,P, Q , , Q,, . . . ,Q, can be considered as a sequence of l-u(n)types of length 2r 1 I 2 . s ( n ) .(r(n- 1))"'") 1 I s(n 1). Hence there are elements cl' < c,' < . . . < c,' < h' < d,' < d,' < . . . < d,' of B,' such that both c,' and d,' have l-u(n - 1)-type Q, and b' has l-u(n)-type P. Thus every sequence of l-u(n - 1)-types of Pu(n-,)((b,,b,))of length s(n) is realized in both (bl',b') and (b',b,') as well as in (b,,b) and (b,b,). Thus in this subcase we are done. Otherwise, there is a sequence of l-u(n - 1)-types of length s(n) that is realized in (b,, b,) but not in both (b,, b) and (b,b,). Assume, without loss of generality, that P,, P,, . . . ,Pscn) is realized in (blrb2)and (bl, b) but not in (b,b 2 ) . [Note that since s(n 1) 2 2s(n), the sequence P , , P , , . . . , P,(,,,,P,, P,, . . . ,P,,,, is realized in (b,, b,), so that P I ,P,, . . . ,Ps(n)must be realized in either (b,,b) or in (hb,).] The formula $(D,,U,), which says
+
+
+
+
+
+
290
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
that u1 < u,, the 2-u(n)-type of ( u , , ~ , ) is the same as the 2-u(n)-type of ( b , b,), and P , , P,, . . . , is not realized in the interval ( v l , uz), has quantifier depth max(u(n),s(n) i- u(n - 1)) = u(n), so that, since B k (3v1)i+b(ul,b2) and Pufn+ ,)(b,) = Pufn+l)(b,’),it follows that B k $(b’,b2’) for b 2 ’ ) )= Pu(fl)((b, b,)) and, since P , , P , , . . . ,PSC,, some b’ E B. Then Pu(,,)({b’, is realized in (b,’,b,’), we must have b,’ < b’ < bz‘. To complete the proof, we must show that every sequence Q,, Qz, , . . , Qs(fl) of 1-u(n - 1)-types in P,,(,- lJ(bl,b)) is realized in both (b,,b) and (b,’,b’). But Q1, Q,, . . . , Q s f n ) , P , P I ,P , , . . . ,Pscn)is realized in both (b,, b 2 )and (bl’,bz‘),and PI,P,, . . . ,PsC,, is not realized in either (b,b,) or (b’,bz’).Hence Q,, Q,, . . . ,Qs,,)is realized in both (b,, b) and (bl’, b’). Thus in this subcase also we are done and the proof is completed. The theorem below is an immediate consequence of the lemma. THEOREM 13.80: Let B be a linear ordering and let b , < b, and b,‘ < b,’ be elements of €3. Assume that P f b , ) = P(b,’), P(b,) = P(b,’), P((b,, b,)) = P((bl’,bz’)), and every sequence of 1-types in P((b,,b,)) is realized in both ( b l , b 2 )and (b1’,b2’).Then P((bl,b2)) = P((bl’,bz’)). We now introduce the notion of a saturated model. As we will see, a countable saturated model of T contains all countable models of T as elementary subsystems if S,(T) is countable for all n.
DEFINITION 13.81 : A countable model A of T is said to be saturated if whenever (a,, . . . ,a,) has n-type P and Q is an ( n 1)-type that contains P . there is an element a,, E A such that ( a l , . . . ,a,, a,, has (n + 1)-type Q.
+
,)
We can easily show by induction on n that if A is saturated, then every n-type is realized in A ; hence if T has a countable saturated model A, then S,(T) is countable for all n. We now show that the converse is also true. THEOREM 13.82: I f S,(T) is countable for all n, then T has a countable suturated model. Proof: Add to the language of T a countable set {g,ln < w } of new constant symbols; the elements of the saturated model will be precisely this set. Enumerate all triples (a, P, Q) where a is an ordered n-tuple of the new constant symbols, P i s an n-type of T , and Q is an ( n + 1)-typeof T containing P. (Note that if n = 0, a is empty, P is T, and Q is a 1-type of T.) Since S,(T) is countable for all n, we may assume that the set of such triples is enumerated
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MODEL THEORY AND LINEAR ORDERINGS
as an w-sequence {(ak,Pk, Q k ) Ik < w}. We now define by induction an expanding sequence (T,lk < O ) of complete theories, each involving a finite number of the constant symbols {g,ln < w } . We define To to be T and we assume, as the induction hypothesis, that the constants appearing . . . ,ai(((,)) and that there is a t(k)-type Rk of T such that in T , are gi(l),gi(2), T , = T u {O(gi(l),gi(2), . . . ,gi(r(k)))Ifl E R k ) . If now T , u {4(ak)14E P k } is inconsistent, then we define T k + ,= T,. If, on the other hand, 7,u {4(ak)14E P k } is consistent, then we select a constant symbol a, not occurring in T , u {4(ak)l4 E P k } and we choose T k + to be any completion of T , u {$(ak,gp)I$ E Q k } in the same language. To show that T,+ is consistent, it suffices to show that T , LI { $(ak, +) I $ E Q k ) is consistent. But otherwise T u { 8(b,a,), $(ak,g p ) }is inconsistent for some O(b,a,) E T , and for some $ E Qk,where none of akor g, occur among b. Thus T k B(b, a,) -+ i $(ak,ap), so that T I- (3x)B(x,a,) -+ i ( 3 y ) $ ( a k ,y). Now (3x)O(x,v) E Pk,since otherwise (Vx)iB(x,v) E Pk, so that both (Vx)iO(x,ak) and O(b,ak) are in T , u {q5(ak)l$ E P'}, contradicting its consistency. Hence i(3y)$(v, y ) E P k ;but $(v, y) E Qk and i (3y)$(v, y ) E Qk, which contradicts the consistency of Qk. Hence Tku {$(ak,gp)l$E Q k } and T k + are consistent. We now define a structure S whose elements are the constant symbols {g,I n < a}; more precisely, we define an equivalence relation by g , g , if a, = Q, is in some T,, and we let S be the set of equivalence classes. For each relation symbol P(x,, . . . ,x,J in the language of T , we define an n-ary relation R on S by stipulating that R ( U ~ (. ~. .),ai(,,) , holds in S if and only if P(gi(l),. . . ,gi(,,))E T , for some k. It is then easily verified by induction on formulas that for every formula 4(xl, . . . ,x,) and any elements gi(l),. . . , gi(,) E S, S t= 4[gi(l),. . . ,ai(,,)]if and only if 4(ui(ll, . . . ,gi(,))E Tkfor some k. In particular, S k T, and if . . . ,gic,,) has n-type P and P G Q, then . . . ,gic,,,b) has (n 1)-type Q. there is an element b E S such that Hence S is a countable saturated model of T.
-
+
We leave as an exercise the following three fundamental facts about countable saturated models. EXERCISE 13.83: (1) Show that if A is a countable saturated model of T and B is a countable model of T , then B < A. [Hint: Combine the proof of Cantor's Theorem 2.5 with the proof of Theorem 13.75.1 (2) Show that if A and B are both countable saturated models of T, then A z B. (3) Show that if A is a countable saturated model of T and the two n-tuples (al,a2, . . . ,a,) and ( b l , h,, . . . ,bJ of elements of A have the same n-type, then there is an automorphism f of A such that f ( a i ) = bi for all i.
292
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Note that if T has a countable saturated model, then T also has a prime model by Theorems 13.71 and 13.75.The converse is false, since the complete theory T of all true statements of first-order arithmetic has a prime model, namely, the natural numbers, but has no countable saturated model since T has c different 1-types, which the reader is challenged to find. Before illustrating these notions, we observe that if a, < a, < . . .
+
+
EXERCISE 13.84: (1) Show that w [ . g is saturated. (2) Show that w + [ . a is weakly saturated if and only if
c1
is infinite.
An interesting theory arises if we add to the language L of linear orderings a countable list {c,(n< w ) of constant symbols and let T be the theory Th(g) together with the statements {c, < g,+ In < w ] ; this theory was first discussed by Vaught [22]. Any model of T consists of a dense linear ordering with an o-sequence (c,lrz < w } of distinguished elements. It is easily verified that, up to isomorphism, T has only 3 countable models. In one model, A , , the sequence of distinguished elements is cofinal in A , . In a second model, A,, the sequence of distinguished elements has a limit in A , . In the third model, A , , the sequence of distinguished elements is neither cofinal in A , nor has a limit in A , . These three models differ in that the elements that are greater than all the distinguished elements have order types 0, 1 g, and g in A , , A , , and A , , respectively; however, any countable model of T is isomorphic to either A , , A , , or A , . The fact that T has three non-isomorphic countable models suggests that T is not complete. Moreover, the experienced game player will see immediately that G , ( A , ,A,) E I, and G , ( A 1 , A 3 ) E I, and G , ( A , , A , ) E I, and conclude by Corollary 13.12 that A A , , and A , are not elementarily equiv-
+
,,
4.
29 3
MODEL THEORY AND LINEAR ORDERINGS
alent. That conclusion, however, would be premature, since Theorem 13.11 and Corollary 13.12 apply only in the case where the language is finite. In fact, as we will see, T is complete and A , = A , = A , , so that the assumption that the language be finite in Theorem 13.11 is necessary. We will shortly prove that T is complete, but granted that fact, let us examine the three countable models of T . We first claim that A , is a prime model. Indeed, if a,, . . . , u , E A , and each ui< c k , then the n-type of ( a , , . . . ,a,) is generated by {u,
ujlui= U j } u {Ui < < U j } u { u , =gjlui= C j } u ( u , < c j l a , < c j and j I k } u {cj< u i l c j < a,},
=
by Exercise 13.69.2 since this set is finite and, given any two n-tuples of A , satisfying these formulas, there is an automorphism of A , mapping one n-tuple to the other. Thus every n-type realized in A , is principal, so that, by Theorem 13.75, A , is a prime model. Now A , is not saturated since no element of A , satisfies all of the formulas {g, < u , In < w } , and A , is not saturated since, although the limit point c of the distinguished elements satisfies {g, < u1 In < w } , there is no element u E A , such that (c, a ) satisfies the formulas {g, < u, < u1 In < (0). Thus A , must be the saturated model. A direct proof that A , is indeed saturated (and that A , is weakly saturated) involves analyzing all n-types of T. In point of fact, the n-types of T are easily described; each n-type is of the form \u,
u I D i < I1jIui < U j } u {Vi = g j l u , = c.> J u {u, < cjlui< C j ) u { C j < vilej < a,},
= UjlUi= U j }
where a,, . . . ,a, are elements of one of the models A iof T. (To prove that these are actually n-types is not so easy, and we will leave this until after we show that T is complete.) Thus, in particular {g, < ul In < w } is a 1-type; thus the limit point of A , cannot be distinguished from any point larger than it. We turn now to a proof that T is complete. The method we will use is called the elimination of quantifiers method, and we will show both that T is complete and that it is decidable. The basic idea of the elimination of quantifiers method is to start with a formula of the form (3x)$(x, u l , . . . ,un), where 4 is quantifier-free and obtain effectively a quantifier-free formula $(u,, . . . ,u,) such that T 1 ( W . .(Vufl)[(W4(x,u 1 , . . . ,%I) '
W I , .
'.
,u,,I.
If one can also show that for each quantifier-free 4(x), either T 1 (3x)4(x) or T 1 i(3x)$(x), then T is complete; if one can also determine effectively
294
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
whether T t (3x)&(x) or T t i ( 3 x ) + ( x ) , then T is also decidable. For suppose that we are given a statement (D; we first find effectively a logically equivalent statement in prenex normal f i r m (Qiul)(Q2~2)
' '
. (Qk-
1%
i)(QkUk)'#)(Ui,
u25
...
9
u k - 1,
%h
where each Q is either V or 3. If k > 1 and Qk is 3, we apply the elimination ofquantifiers method to ( Q k U k ) 4 ( U I , u 2 , . . , ,uk- 1 , uk), obtaining an equivalent ). formula $ ( u l , . . . , I J ~ - ~ Then Tt
@
c1 ( Q 1 ~ 1 ) ( Q 2 ~ 2 )' '
. (Qk-
i u k - i)$(ui,
. . . > uk-
11,
so that Q, is equivalent in T to a statement with one fewer quantifier. If k > 1 and Qk is V, we apply the elimination of quantifiers method to l(Qkuk)4(uir..
.,Uk-i,uk)
obtaining an equivalent formula $ ( u , , . . . ,uk-
Then
Tt- ~ ~ ( Q i ~ i ) ( Q 2 ~ z ) . " ( Q k - i ~ k - ~ ) 1 $ ( ~ 1. ,. ,. U k - i ) r
-
so that in this case also Q, is equivalent in T to a statement with one fewer quantifier. This procedure is repeated k - 1 times until we find that T I- CD ( Q , u l ) O ( u l ) , where 8 is quantifier-free. If Q1is 3, then we use the fact that either T k (3ul)f3(ul) or T t i ( 3 u l ) 8 ( u , ) to conclude that T I- Q, or T t 10; and if Q1 is V, then we use the fact that either TI- (3vl)i8(ul) or T t i ( 3 u 1 ) i f 3 ( u 1 ) to conclude that T t -I@ or T t Q,. In any case we know or T k CD, so , that T is complete. Moreover, since all passages that T t iQ are effective, T is decidable. Thus we have to show that, in the case where T is the theory with three countable models, a quantifier can be eliminated from (3x)4(ul,.. . ,u,, x), where 4 is quantifier-free. We first put 4 in disjunctive normal form V{4;lls i l k ) and since (3x)4 is logically equivalent to v ( ( 3 ~ ) 41<~i(< k ) , we need only find a quantifier-free such that T t $i ( 3 ~ ) 4since ~, 11 5 i Ik ] . Thus we can assume that 4 is a then T t- (3x)4 t ) of atomic formulas of the language and their conjunction A(r$iI 1 I i I negations. Replace each conjunct of the form i ( u = b) by its equivalent (a < b) v (b < a) and each conjunct of the form i ( a < b) by its equivalent a = b v b < a. Having done this, we can put the resulting formula in disjunctive normal form, pass the (3x) through the disjunction, and end up facing formulas like (3x)A{+;ll Ii 5 t } , where each g5i is an atomic formula. We may assume that each 4i contains x free since otherwise that 4l can be moved out of the scope of (3x1. If x = x appears, it can be replaced by u1 = u l . If x = u j appears then ( 3 x ) A {&(ul, . . . ,un,x)I1 5 i It } is logically iI t ) . Similarly if x = cj appears. equivalent to /l{+i(ul, . . . ,u , , u j ) [ l I
- v{$i
-
4.
MODEL THEORY AND LINEAR ORDERINGS
295
Thus we may assume that each conjunct is of one of the forms ui
< x,
cj < Y,
x
< ui,
x
If only the first two forms appear, then the formula (3x)$ is equivalent in T to u1 = u , ; if only the last two forms appear, the same conclusion is correct. Otherwise (3x)+ is equivalent in T to the formula A{. < hlboth u < x and x < b appear). Thus the quantifier can be eliminated from (3x)4(rl.. . . , r , , x ) if n 2 1 and 4 is quantifier-free. We now consider statements of the form (3x)4(x),where 4 is quantifierfree. As before, we need only consider the case where $(x) is a conjunction of formulas each of which is an atomic formula containing x. If x = x appears alone, then T t (3x)4(x), and otherwise it can be deleted. If x < x appears, then T t i(3x)4(x).If x = g j appears, then T t (3x)$(x) if and only if T k $ ( g j ) and, since 4(ri) just makes some assertions about the ordering of the distinguished elements, either T t 4 ( g j ) or T t i4(cj). If only formulas of the form x < g j appear or if only formulas of the form g j < x appear, then T t (3x)4(x). If formulas of both forms appear, then (3x)$(x) is equivalent t o < gjlgi< x and x < cj both appear), a statement which is easily decided by 7 .Thus for every statement (3x)4(x),where 4 is quantifier-free, either T t (3x)4(x)or T t i(3x)b(x). We summarize our conclusions about the theory T in the following theorem.
THEOREM 13.85: (Vaught) Lut T be the theory Th(q) together with the statements (c, < g,, ( n < w } . Then T is complete and decidable and has exactly three distinct countuble models. H The elimination of quantifiers method is a useful general tool for proving completeness or decidability. Some additional examples of its use are found in the exercise below. EXERCISE 13.86: (1) Add to the language of T a constant symbol a and, using the method of elimination of quantifiers, show that T u {c, < g l n < w } is complete. Deduce from this that { c, < uln < o} is a 1-type of T. (2) Similarly, prove that the it-types of T a r e all of the form
-
ujlai= U j } u { U i < Z i j I U i < U j } u { U i = cjlai = C j } LJ (q < cjlui < C j ) L! (Lj < U i ( C j< ail.
{Ui=
(3) For each m 2 1 add to the language of the theory T considered above w1 unary predicate letters U , , . . . , U , and let T , be the theory that
296
13.
THE! FIRST-ORDER THEORY OF LINEAR ORDERINGS
contains T and says that U , , . . . , U , partition the model, each is dense in the model, and every c, E U Show that T, is complete and decidable and has exactly m 2 distinct countable models (Vaught [22]). (4) Prove the decidability of Th(o)using the elimination of quantifiers method.
+
$5. THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
Given a complete theory T, how many distinct countable models can it have? For example, by Cantor’s Theorem 2.8, Th(q) has but one countable model. On the other hand, by Corollary 6.12 and Corollary 13.12, Th(w) has c distinct countable models. In this section we will present a result of Rubin [171 that says that, for linear orderings, these are the only two possibilities; that is, if T is a complete theory of linear orderings, then T has either exactly one or exactly c distinct countable models. For arbitrary complete theories T , a complete answer has not yet been obtained as to the number of distinct countable models of T. At the end of $4 we presented Vaught’s examples of theories T , that have m 2 distinct countable models for each m 2 1; on the other hand, Vaught [22] showed that no complete theory can have exactly two distinct countable models. It is also possible for a complete theory to have exactly countably many distinct countable models; for example, if T is just the theory of equality together with the statements { ci # cjl i < j < w } , then a countable model of T is determined by the number of non-distinguished elements that it has. Morley [lo] proved that if a theory has an uncountable number of countable models, then that number must be either K1 or c. If, of course, we assume the continuum hypothesis, then Morley’s result answers the question completely. Without the continuum hypothesis, there is no known example of a complete theory that has exactly K, distinct countable models, so it has been conjectured that there are none. Rubin’s result shows that the conjecture is correct for complete theories of linear orderings. The astute reader will recall that in Chapter 8 we showed that there are exactly K, countable linear orderings that are groupable; but this does not provide a counterexample since there is no set CD of first-order statements such that A is groupable if and only if A k @. For a linear ordering A to be groupable, there must exist a ternary relation on A that meets certain requirements, and the existence of such a relation cannot be expressed in a first-order language. The class of ordered groups is, on the other hand,
+
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
297
an elementary class, and there are exactly c countable ordered groups. If we ignore the group structure of these groups, the c distinct countable ordered groups fade into just K, countable linear orderings. The class of groupable linear orderings is called a pseudo-elementary class and it has K elements. Morley’s theorem applies equally well to pseudo-elementary classes, so it is conceivable, though unlikely, that a similar counterexample will be found for elementary classes. It is interesting that there is also a pseudo-elementary class containing exactly two distinct countable models.
,
THEOREM 13.87: (1) The pseudo-elementary class of linear orderings that are groupable contains exactly K, distinct countable linear orderings. (2) The pseudo-elementary class of linear orderings that are the linear orderings of models of arithmetic contains exactly two countable linear orderings.
Proof : (2) Clearly the axioms T Aof arithmetic logically imply Th(w). Thus if M is a non-standard countable model of TA, then the order type of M is w 5 CI for some order type a # 0. If 2 , is a copy of 5 in M and b E Z , , then b < b b ; if b b E Z , , then h + b = b + n for some natural number n, so that h = n, which is impossible. Hence c1 has no Iast element. Suppose that Z , < Z , are two copies of 5 in M and that a, E 2, and u2 E 2,; we may assume that a, and a, are both even, since from TA it follows that for any a E M either a is even or a 1 is even. Then a , + a, is even and a, < $(a, a2) < a,. If ;(al a2)E Z , , then $(al a,) = a , n for some natural number n, so that a, = a , + 2n, which is impossible. Similarly, +(a, az)$ Z,, so that there must be a copy Z , of 5 in A4 such that Z , < 2, < 2,. Hence c1 is a dense order type. By a similar argument, CI does not have a first element, so that a = 9. Hence every countable non-standard model of arithmetic has order type w + 5 . q. Since such models exist, the proof is complete.
+
+ 1
+
+
+
+
+
+
+
Before presenting Rubin’s analysis, we first discuss those complete theories T of linear orderings that are KO-categorical. DEFINITION 13.88: A theory T is said to be KO-categoricalif it has, up to isomorphism, exactly one countable model.
Engeler [6], Ryll-Nardzewski [18], and Svenonius [191 independently obtained a characterization of those theories that are KO-categorical. Using this characterization, Rosenstein [141 determined which linear orderings have KO-categoricaltheories.
29 8
13. THE FIRST
ORDER THEORY OF LINEAR ORDERINGS
THEOREM 13.89: A theory T is KO-categorical if and only n, B,(T) is finite.
if for every
Proof : If B,( T ) is infinite for some n, then there is a non-principal n-type P by Theorem 13.72. By Theorem 13.64, there is a countable model A of T in which P is realized and, by Theorem 13.71, there is a countable model B of T i n which P is omitted. If A 2 B, then the isomorphism would carry an n-tuple ( a l , . . . ,a,) ofelements of A that realizes P to an n-tuple ( b , , . . . ,h,) of elements of B that does not realize P. This contradicts Proposition 13.66.1, so that if B,(T) is infinite for some n, then T is not No-categorical. Suppose, on the other hand, that B,(T) is finite for each n and that A and B are both countable models of T ; we will show that A = B, so that T is N,-categorical. The proof is basically patterned on that of Cantor's Theorem 2.8 and thus proceeds by enumerating A = {a,ln < m } and B = jb, I n < w } and defining by stages an isomorphism f from A to B, specifically defining f(a,) at the even stage 2n and defining an element U E A such that f ( a ) = b, at the odd stage 2n 1. More concisely, we assume as the induction hypothesis that at the beginning of stage t = 2n we have defined f for a finite number of elements of A including at least a,, a, . . . ,a,and that for each j < n there is an such that is defined and equals b j . Furthermore we assume that if we have so far defined {f(a')l j < k } and that these elements are respectively {bjlj < k } , then the k-type of the k-tuple (a'lj < k ) in A is the same as the k-type of the k-tuple (b'lj < k) in B. We now take an element a of A and find its counterpart in B ; the reverse could be done with equal ease. Let P be the (k 1)-type of the (k + 1)-tuple (uo,a', . . . , a k - ' , a ) in A. Since B k + l ( T is ) finite, we may assume that P is finite and so may take the conjunction Y of the formulas of P . Then A I= Y [ a o , a l ,. . . , a k - ' , u ] ,so that
+
+
. ,ak-',Vk). Since (a', a ' , . . . ,u k - ' ) satisfies ( ~ L ' k ) ~ ( ~v(l ,, , . . . ,uk- 1, u k ) in A, this formula must be satisfied by (bo,h', . . . ,bk- ') in B. Thus there is a b E B such that B k Y [ b o , b l ,. . . , b k - ' , b ] .Thus P is the ( k + 1)-type of the ( k + 1)-tuple (bo,h ' , . . . ,hk-', b ) in B, so if we define f ( u ) = b, the induction hypotheses A I=
(3ti,)Y(U0,U'j..
continue to hold, so that f will be an isomorphism from A to B.
If B,( T )is finite for all n and A and B are models of T that contain k-tuples . . , a k - ' ) and (bo,bl, . . . , b k - ' ) that have the same k-type, then we can begin to define the isomorphism from A to B by mapping each a' to b'. Thus the following strengthening of Theorem 13.89 is analogous to the strengthening of Cantor's Theorem 2.8 by Theorem 2.11. (uo,ul,.
5.
THE NUMBER OF COUNTAHL,E MODELS OF A COMPLETE THEORY
299
COROLLARY 13.90: Let T be K,-categorical, let A and B be countable modeis of T and let (a'li < k ) and (b'li < k) be k-tuples of elements of A and B, respectively, that have the same k-type. Then there is an isomorphism f : A -+ B such that f ( a ' ) = b' for each i < k.
These results, when combined with the analysis of almost doubly transitive linear orderings in 98.6, yield a characterization of KO-categorical linear orderings. THEOREM 13.91: (Rosenstein) Let Al be the smallest class of order types Containing 1, containing z1 z2 whenever it contains 5 1 and z2, and containing the shufle a(F)wheneuer it contains the elements of F. Then a complete theory T of linear orderings is K,-categovical if and only (f it has a model whose order type is in A!,.
+
Proof: If T is KO-categorical and .4 is the countable model of T , then by applying Corollary 13.90 with B = A, we see that A is almost k-tuply tranIf, on the other hand, sitive for any k, so that by Theorem 8.40, 2 E A,. T has a model A whose order type is in A l , then A is almost doubly transitive by Proposition 8.37, and therefore A is almost k-tuply transitive for every k by Proposition 8.35. Using Proposition 13.66.1, it follows that the k-tuples of elements of A have only finitely many different k-types, so that B k ( T )is finite and T is KO-categorical by Theorem 13.89. EXERCISE 13.92: Define a notion of rank for KO-categorical linear orderings as follows. Let ~ t V ~ (=0 I)I } , put z 1 + T~ E A l ( n + 1) whenever T ~ T , ~ .&',(II), E and put a(F)E .&(n + 1) whenever F G &Ml(n). The rank of T E .&, is defined to be the smallest n for which T E A l ( n ) .
(1) Show that A l ( n )is finite for all n. ( 2 ) Show that if T has rank I I and if A is an interval of 0 , then A is KO-categoricaland has rank at most 2n + 1.
T
of order type
We turn now to a discussion of Rubin's theorem, which says that if T is not KO-categorical, then T has c distinct countable models. In the most familiar non-K,-categorical case, we obtain c distinct countable models of Th(w) by taking the interval I of order type [ of the model w [, which satisfies I = I . a for every countable order type a, and replacing I by each of its c different countable counterparts. What Rubin shows is that given any non-KO-categorical T for which S,(T) is countable, there is a model A of T with an interval I satisfying I = I . a for every countable order type a.
+
300
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
We will first concentrate on those linear orderings I such that I = I . a for every order type c( and later show how they can be found in models of non-KO-categoricaltheories. When does a linear ordering I have this property? As we will see, a condition just a bit stronger than I = I I is sufficient to guarantee that I = 1 * u for all CI. We first note that the condition I = 1 I is itself not sufficient, for w + w* = (w w * ) (w w * ) , although (w w * ) . o = w + [ . w = w f w + w * . That I . w = I fails is due to the fact that although the map from w w* to the first w w* in (w w * ) (w w * ) is an embedding, it is not an elementary embedding. Thus the condition we impose on I is that if I 'v f and I 'v 1, (where we assume here and elsewhere that I , is disjoint from I,), then the identity maps of I , and I , into I, I , should be elementary embeddings. As we will see, this condition implies that I = I . M for every order type a. Actually, we will prove a stronger result, one which involves elementary embeddings in the conclusion as well as in the hypothesis.
+ + + +
+
+
+
+ + +
+
+
THEOREM 13.93: Let X be a linear ordering and assume that for each x E X , I , z I . Assume that if I 2 I and I , 'v I , then the identity maps of I, and I , into I , I , are elementary embeddings. Then for any nonempty subordering Y of X , the identity map is an elementary embedding of C ( I x l x E Y } into L{I,.~X E x).
,
+
Proof: We proceed by induction on 1x1,first for X finite and then for infinite X. For 1x1 = 2, what is to be proved is exactly what has been assumed. Assume then that the result is true for 1x1= k and proceed to the case k 1, in which case we may assume that X = { n l n I k } and that x { l x l xE X } = I , I, . . . 1,. Let A = I , I, . . . 1, and let B denote x { l x l x E Y } . Suppose that A k 4 [ b , , b,, . . . ,b,,a], where b,, b 2 , . . . ,b, E B and a E A . We assume that a $ 1,; otherwise we give a similar argument using I , + I , + . . . + l , i n s t e a d o f l , + l , + . ~ ~ + I , - , .We may assumealso that Y n (0, 1,. . . , k - 1 ) # 0; for if Y = ( k } , then 1, < 1,-1 -k 1, by assumption, so that to show that 1, < A it suffices, by Exercise 13.53.4, to show that 1 k - I 1, < A , and this corresponds to the case Y = (k - 1, k}. Assume that b,, b,, . . . ,b, E 1, but b,, b,, . . . ,b,- $ 1,. By Corollary 13.39,there is a testing formula for 4 in I, . . . + lk- (relative to b,, . . . ,bn). That is, we can find 4 * ( u , , . . . , u t ) such that I , + I, + . . . + lk-, k 4*[c1,c2,. . . ,c,- ,,c] if and only if A k 4[c1,c,,. . . ,c,- b,, . . . ,b,,c] for every c,,c,, . . . , c t - l,c E I, I , . . . 1 k - I . Thus
+
+ +
+
+ +
+
,
+
+ +
1,
+ 1 , + + 1,' ' '
+
,
+
k 4*[bl,bz,
. . . ,b,- ,,a].
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
301
By the induction hypothesis, since Y n (0, 1,. . . , k - 1) # @, there exists an element b E B n ( I , 1 , . . . l k - , ) such that I , I, . . . l k - I= 4 * [ b l , b 2 , . . . ,b,- ,,b], so that A I @[b,,b,, . . . ,b,,b], as was to be proved. Assume now that X is countably infinite and let X = (x, I n < u} be an enumeration of X . For each n, let A, = C { l x ( xE { x i l i < n ) ] and let B, = IlllIx E Y n {xili < n}}. By what we have just shown, the following diagram consists of elementary embeddings :
+
+
+ +
,
+
B, 4 B2 4 B , 4 ' . .
h
h
h
A , 4 A2 < A , 4 . . . . We must show that B = U { B , ( n < o}< A = U { A , I n < o}.Assume then that A I=4 [ b , , . . . ,b,,a], where bl, . . . ,b, E B. Then for some n, we have b,, . . . ,b,, a E A,; furthermore, since, by Theorem 13.59, A , < A, we have that A , I= 4 [ b l , . . . ,b,,a]. Since h l , . . . ,b, E B , and B , 4 A,, there is an element b E B, such that A, k 4 [ b , , . . . ,b,,b]. Since A , 4 A , A I= $ [ b , , . . . ,b,,b]. Hence B 4 A. A similar proof works for any infinite set X , using, as the induction hypothesis the assumption that the theorem is correct for sets of all smaller cardinalities. This completes the proof.
Because of this theorem and its corollary, the following definition seems appropriate. DEFINITION 13.94: A linear ordering I is said to be self-additive if whenever I, 2 I and I, = I , the identity maps of I , and I, into I , I , are elementary embeddings.
+
COROLLARY 13.95: type 3.
I f I is sclf-additive, then I
=I
. CI
for every order
Is the converse true? That is, if 1 = I . a for every order type a, must I be self-additive? Perhaps the reader can find the answer to this question. As we now show, the property of being self-additive is more correctly ascribed to the theory Th(1) than to I itself. THEOREM 13.96:
(1) (2) (3) (4)
Let T = Th(I). Then the following are equivalent:
I is self-additive. Every model of T is seif-(additive. I f A b T and B 1 T , thcn A < A B and B 4 A I has no definable intervals other than I and 0.
+
+ B.
- -
302
13. THE FIRST-ORDER
THEORY OF LINEAR ORDERINGS
Proof : Obviously ( 3 )=-(2) (1). We thus need only verify (1)* (4) =-(3). We first show (1) (4). Suppose then that ( u E I1 I k 4 [ u ] )is a non-empty intervalofI.LetI, ~ l a n d l e t l , - l . S i n c e I , < I , + I , a n d l , < I , +12,
+ I,p, + 1 , k 4 [ 4 )= ( u E 1 , p , k 4 [ a ] ) u ( u E z,p2 k 4 [ a ] ) . Since ( u E I I I k b[u])is an interval of I , [ U E 1,
I k ( V . W Y , [ . X < ? ' A 4(x) A 4 ( Y ) (V'Z)(x< < Y -,4(4)1, so that C U E 1, + 12111+ 1, k + [ a ] } must be an interval of 1, + I,. Since it contains points of both I , and I,, it must contain a terminal interval of 1, and an initial interval of 1,. Since I , < I , + I,, { a E I , II, k 4 [ u ] ) is a terminal interval of I,, and, since I , i I , + I,, { u E 12112k + [ a ] ; is an +
initial interval in I , . Thus I, k (VX)[~(X) -, (Vy)(y 2 x -, 4 ( y ) ) ] and I , k (Vx)[&x) + (V'y)(y I x -+ + ( y ) ) ] ,hence both of these statements are true
in 1 and I k ( V X ) ~ ( X ) . Thus (1) * (4). We now show (4)*(3). As in the proof above, the assumption that I has no proper definable intervals implies that the same is true for every model of T. Assume that A b T and B k T ; we show that A 4 A B and B < A B. We do this by constructing a model C of T such that A 4 C, B < C, and A B G C. We let C consist of the complete diagrams C ( A ) and C ( B ) of A and B together with (a < bla E A, b E B).If C is consistent and C b X, then clearly .44 C, B 4 C, and A + B G C. If C is inconsistent, then, by the Compactness Theorem, there are elements b , , b,, . . . ,b, E B, u E A, and a statement q5(b1,b2,. . . ,b,) E C ( B ) such that
+
+
+
C ( A ) k g < b, A G < b
2 ~" r .\ g
< b,+ iq5(bl,b,,. . . ,bJ.
This implies that C ( A ) t- (~Xl)(t/X2)~ . . (VX&
< X,
A-
- . A g < X n + l+(X,,.x,,.
. . ,X,,)j
since b,, . . . , b, do not appear in C(A). On the other hand,
W )1 4(bl, b 2 , . . . ,b,), which implies that A k (3x,) . . . (3x,,)4(xl, . . . ,x,,); but A has no first element (since that would yield a definable interval of A ) , so that {CE
A1A k (jx,). . . ( ~ x , ) ( c< x1 A " . A C < x , A ~ ( x .~. ,. ,x,))
is a non-empty initial interval of A. By assumption it must be all of A, contradicting the conclusion above that a is not in the interval. Thus C is consistent. Let C k C and divide C into C, + C, C,, where C, = ( c E Clc Iu for some a E A ) and C, = {c E Clc 2 b for some b E B } , so that A G C,
+
5.
303
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
and B G C,. Let us for the moment assume that A 4A + C 2 + B and B 4 A C, B and show that A + B < A + C, + B. By Lemma 13.55, this will imply that A 4A + B and B < A + B. Assume, therefore, that A + C’, + B k Cp[d,,d,, . . . ,dk,c], where d,, d,, . . . ,dk E A u B and c E C,. Let a be the largest of the di in A and let b be the smallest of the d , in B ; if there are no d, in A or in B a slight modification is necessary. By Theorem 13.41, there is a selecting formula 4’ for Cp in [a,b], that is, a formula ~ # ( x 1 , x 2 , usuch ) that A + C, + B 1 ~$”[a,b,d] if and only if d E [a,b] and A + C 2 + B 1 4 [ d , , . . . ,d,,d]. It thus suffices to find a c’ E A such that A + C2 t B 1 4#[a,b, c’], given that A + C, + B k 4 ” [ a ,b,c]. Now there is, by Corollary 13.39, a testing formula 4#* for Cp“ in A (relative to b); that is, there is a formula d)#*(x,,o) such that A k ##*[a’,c’] if and only if A + (’, + B k Cp#[a’,b,c’], for all a’, c’ E A. It thus suffices to find an element c‘ E A such that A k 4#*[a, c’]. Suppose, on the contrary, that A 1 i ( 3 z ) $ # * ( a , z); then, since A < A + C, + B, we have
+ +
A
+ C, + B k
1(32)4#*(U,Z)A
Cp#[U,b,C].
Hence Af
c 2
f B k (3x,)(3x,)[x,
< .Y2 5 C
A1(3Z)~#*(Xl,Z)A~#(X,,b,X,)].
The same formula is satisfied by every element of A + C, + B that is larger than c ; we claim that in fact it must be satisfied by all elements ofA + C, B. For if not. then
+
which implies that the same is true in B since B < A is an element b‘ E B such that B
F@X1)(3X2)[X1
+ C, + B. Thus there
< X2 2 I J ’ A1 ( 3 5 ) ~ ’ * ( x , , z ) ~ ~ # ( x , , b , x , ) ] ,
+ +
which implies that the same is correct in A C, B, which is impossible since every element of A + C, + B larger than c satisfies this formula, including b‘. Thus A
+ c, + B k (vU)(3X1)(3X2)[X1
< X2
0 A 1(3Z)4#*(X1,2)
In particular, if e E A, then there is an a ,
E
A $#(XI,
b,x,)].
A and an a2 E A such that
whichisimpossiblesinceifA + C, + B k 4#[a,,b,a,],thenA k Cp#*[a,,a,] by definition of +#*, and if A + C 2 5 k i(3z)Cp”*(al,z), then, since A < A + C, + B, we also have ’4 k -I (3z)4#*(al,z). Hence we must conclude that A k (3z)Cp#*(a,z),so that for some c’ E A, A k ~ # * [ u , c ’ ]as, was
+
304
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
to be shown. Thus, assuming that A < A + C, + B and B < A + C2 + B, we have shown that A + B < A + C, + B. It thus remains to be shown that A < A + C, + B and that B < A + C2 B. Recall that C = C, + C, C, where A c C,, B G C,, A < C , + C 2 C , , a n d B i C , + C, C,.WewillshowthatC,
+
+
+ +
+
+
+
+
,,
+
+ +
+
+
+ + +
c1 + c2 + c3 k 4 [ a , , .
’ ’
, a,,a’],
as was to be shown. The argument for C3 4 C, the proof is therefore complete. I
+ C, + C, is similar, and
EXERCISE 13.97: Show that if I is self-additive and +(x) is a formula, then (a E I ( I 1 + [ a ] ) is either empty or unbounded.
We would now like to show that if T = Th(Z) is self-additive, then T has c distinct countable models I . CI,where cx ranges over all countable order types. Unfortunately this is false; in fact, it is possible that all these models are isomorphic, as happens when I is q or is any other KO-categorical self-additive linear ordering. Even if I is not KO-categorical, it is possible that all these models are isomorphic, although Th(Z) does have c countable models; an example of this occurs when I = o . g. We will see later that these are essentially the only kinds of exceptions, and so we focus our attention on how to show that I . cx and I . p are different when CI # B. We first define a condensation cD on self-additive linear orderings such that any isomorphism carries condensation classes onto condensation classes. DEFINITION 13.98: Given a self-additive linear ordering A and an element a E A we define c,(a) to be the union of all bounded intervals of A containing a which are definable over a. Thus b E c,(a) if and only if there is a formula +(x,y) such that A 1 +[a,b] and { d A1~ A k + [ a , d ] ) is a bounded interval of A .
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
305
Note that, as with other condensations, although cD(a)is a union of bounded intervals, it itself may be unbounded; thus, for example, c,(u) = Z for every a E Z . Note also that if we wish cD to be a condensation, then A cannot have definable intervals. For example, in Z + N, since each element of N is definable, c,(u) = Z N for every a E Z , but cD(a)= N for every a E N . [For the interval from a to the nth element of N is definable by the formula (Vz)($Jz) + x i y I z), where $,,(z) defines the nth element of N . On the other hand if a E N and {d b $ [ a , d ] } were bounded, then it could contain no elements of 2 since given any d,, d, E Z there is an automorphism of Z N that fixes each element of N and carries d , to d,.] To be a condensation, C, must satisfy c,(b) = c,(u) whenever b E c,(a). Hence in order for cD to be a condensation, A must be self-additive; we now show that this condition is also sufficient.
+
+
THEOREM 13.99: l f A is self-udditive, then cD is u condensation of A .
Proof : We must verify that if b E c,(a), then c,(b) = c,(u). Assume that a < b ; the other case is similar. Then there is a formula $(x,y) such that A != $[a,b] and such that {d E A1-4 k $ [ a , d ] } is a bounded interval of A whose minimum is a. We focus our attention first on {c E A I c > b}. Suppose that b < c and c E cD(a).Choose a formula $(x, v ) such that A k $[a,c] and such that { d l A b @ [ u , d ] } is a bounded [nterval of A with minimum a. By Theorem 13.41, there is a selecting formula 4” for $ in {d E Ald 2 b ) relative to a ; that is, there is a formula ~ ” ( c , J ) such that A k 4 ” [ b , d ] if and only if d 2 b and A b 4 [ a , d ] . Then ( d l A b 4”[b,d]} is a bounded interval of A containing both b and c, so that L‘ E c,(b). Suppose on the other hand that b < c and c E c,(b). Choose a formula 4(x,y) such that A b 4[b,c] and such that { d l A I= 4 [ b , d ] ) is a bounded interval of A with minimum b. If we replace 4(x, y) by the formula 4’(x, y), so that X
y
A (vZ)(X
5ZI y + 4(X,
Z)) A
(3Z)(Z > X A
1 4 ( X , Z)),
then { d l A 1 @ [ b , d ] ) = { d l A 1 4 [ b , d ] ) ; moreover, for every c E A, { d l A I= @[c,d ] } is either empty or IS a bounded interval of A with minimum c. We may thus assume that for every c E A, {dl A != +[c,d ] } is either empty or is a bounded interval of A with minimum c. Consider the formula @(x,y) that is (3z)($(x, z) A 4(z, y ) ) v $(x. y). Then { d E A I A I= e[a, d ] } is an interval of A whose minimum is a and that includes c. It suffices to show that this interval is bounded, for then c E c,(a), as desired. But otherwise, A I= (Vy)(a I y + e(a,y)). Let C consist of C ( A ) together with v o > g and
13.
306
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
{ i d ( c , u o ) I Ab + [ u , c ] } . Since for each ~ E A {, d l A t= d [ c , d ] ) is either empty or is a bounded interval of A with minimum c, any finite union of such intervals is bounded, so that any finite subset of C is satisfied by some element of A . Hence C is consistent and has a model C = C , + C , , where C , = { d l d I a for some a E A } . As in the last paragraph of the proof of the previous theorem, since A < C, C,, C, < C , C , , so that A < C , ; this implies A + C , < C , C,, by Exercise 13.53.2, so that A < A C , . Since A C , b C, there is an element d E A C, such that i + ( c , d ) holds for all c such that A b +[a,c]. But since A < A C,, A C, 1 (Vv)(a5 y + &a, y)), which is a contradiction. We have thus shown that if b E c,(u) and a < b, then for all c > b, c E c,(u) if and only if c E CD(b).Analogously, if b E c,(u) and b < a, then for all c < b, c E c,(a) if and only if c E c,(b). To complete the proof that c , is a condensation we need only show that ifb E c,(u), then a E c,(b). For clearly, ifb E c,(a), then for all d, a I d Ib, we have both d E c , p ) and d E c,(b). But also, if b E c,(a) and a < b, then a E c,(b) and a < b, so by the second statement above with a and b reversed, for all c < a, c E c,(a) if and only if c E c,(b). Thus for all c E A , c E c,(a) if and only if c E c,(b). Thus suppose that a < b and b E c,(u); we will show that a E c,(b). We may assume that c,(u) is bounded above, for otherwise we could replace A by A A , , where A , Y A, and, since A < A A , , c,(a) in A A , is identical to what it is in A , so that c,(u) is bounded above in A + A , . (Recall that c,(u) in A is a union of bounded intervals). We may also assume, as we did two paragraphs earlier, that for every a' E A, { d l A b + [ a ' , d ] } is either empty or is a bounded interval of A with minimum a'. We show that A Y (Vy)(3z)(z < y A $(z, h)).Otherwise, since A is self-additive, it follows from Exercise 13.97 that there is element b' > cD(a)such that
+
+
+
+
+ +
+
+
+
+
+
A b (VY)(~Z)< ( Z Y A +(z,b')); hence for some a' < a, A != +[u',b], so that b' E c,(u'). Since a' < a < b , by the first part of the proof, we conclude that b' E c,(a') if and only if b' E c,(u), which is a contradiction. Thus A y (Vy)(3z)(z < y A $(z, b ) ) . Choose an element e E A such that A k $ [ c , b ] for no c < e. Then (dl A I= b [ b , d ] } , where +(x, uo) is the formula x 2 uo A ( 3 y ) ( y5 uo A $(y, x)), is bounded below by e and above by b and contains both a and b, so that a E c,(b), as was to be shown. W Suppose now that f : A + B is an isomorphism, where A is self-additive, and that a E A . Then, since for any a' E A and any formula 4 ( x ,y), A b 4 [a, u'] if and only if B b $[f(a),f(a')], it is clear that f maps cD(a)onto c , ( ~ ( u ) ) , so that isomorphisms preserve condensation classes. We are now ready to
5.
THE NUMBER OF COUNTABLEMODELS OF A COMPLETE THEORY
307
prove that T = Th(l), where I is self-additive, has c distinct countable models in the case where S,( T ) is infinite.
THEOREM 13.100: (Rubin) Assume that I is self-additive and that S,(T) is injinite, where T = Th(Z). Then T has c distinct countable models.
Proof : Since S,(T) is infinite, there is a non-principal type P. Choose a model I , of T in which P is omitted and another model I , of T in which P is realized by an element e and let B = I , cD(e) I , , where cD(e)is taken in I , and I, = I,. We will show that B k T, that the 1-type of e in B is P, cD(e)is a condensation class in B, and cD(e)is the only condensation class of B in which P is realized. Then if a , and u2 are order types and f is an isomorphism from B . a, onto B . a,, then, since b has type P if and only if f ( b )has type P and f [ c D ( b ) ]= c , ( f ( b ) )for every b, f maps the condensation classes of B . a , in which P is realized onto the condensation classes of B . a, in which P is realized. Thus ,f establishes an isomorphism from a1 onto a 2 , so that a1 = u,. Hence if a, and u2 are distinct order types, then B . a, and B . a, are not isomorphic and T has c distinct countable models. We claim that it suffices to show that B < I , 1, I,. For this implies that B k Tand, since I, < I , + 1 , t I , , the 1-typeofein I , I , + I , isP, so that, since B < I , 1, I , , the 1-type of e in B is also P. The condensation class cD(e) of 1, is also a condensation class of B since any formula defining over e an unbounded interval of I 2 will also define an unbounded interval of I , 1, 1, and hence also of B. Furthermore, the I-type of any element of either I , or 1, in I , + I, 1, is the same in B as it is in I , or I , and I , I , I,, as are the condensation classes of such elements. Thus cD(e)is the only condensation class in which P is realized. Hence to complete the proof, we need only show that R < II I 2 + I,. Let A = I , 1, 1 3 .We must show that if A k $ [ b , , . . . , b k ,a ] , where each bj E B, then there is a b E B such that A k $ [ b , , . . . ,b,, b ] . We may assume that a E Z 2 - cD(e)and a > e. and that b , < b , < . . . < b, with b p the first bi E I,. Choose b' E cD(e) such that b ' 2 b p P l ;then b' < a < b,. By Theorem 13.41, there is a selecting formula $" for $ in [b',b,] (relative to b , , . . . ,bk),that is, a formula $ # ( x L , x 2 v) , such that A k $"[b, b,, c] if and only if b ' < c < b p and A k $ [ b , . . . . ,b,,c]. If we find b e B such that A k $ " [ b , b , , b ] , then we will have a b E B such that A k $ [ b , , . . . ,b,, b ] . Thus we may assume that we are given A k q@" b,, a], where b < a
+
+
+ +
+ +
+ + + +
+ +
+
+
+
308
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
A b +[b’,c,,c] for all c, c p E I , . Hence if there is no c E I , such that A 1 c$[b’,b p ,c ] , then I , k i ( 3 u ) 4 * ( b p u). , Since I , < A, we would have
A 1 4[b’,b,,a]
A
1(344*(b,,u).
Consider this as a formula satisfied by b and modify the formula so that it will be satisfied by all points in an interval including b‘. Let O(u,w ) be the formula W
2
UA
(3X)(3y)(X2
WA
2
W A 4 ( U , X, V ) A
1(3U)4*(X,
U)).
Now (dl A ’k O[b, d ] 3 is an interval of A whose initial point is 6’and which contains asince A k O[b’,a].SinceI,
and
A ’k
i (3u)4*(b,‘, u).
Hence I , b i(3u)+*(b,‘,u) since I , < A, so that in particular I , k i +*(bp), a’). But this means that A 1 i 4[b’,b,‘, a’], a contradiction. What we have shown is that if I is self-additive and S , ( T ) is infinite, where T = Th(l), then there is another countable model I’ = I such that I‘ .cx and I’ . fl are different whenever cx # fl. The example I = 0 . q shows that the same result could not be obtained for I itself; for this example, the model I ’ obtained in Theorem 13.100 is (w . q ) + (o+ 5) (w . q). The proof of Theorem 13.100 fails when S , ( T ) is finite since there are then no non-principal 1-types. The analysis for the case when S , ( T ) is finite hinges on the fact that every 1-type is principal.
+
LEMMA 13.101 : Assume that I is self-additive and S , ( T ) is jinite, where T = Th(1). Then either c,(a) < I ,fbr every a E I or c,(u) is definable ouer a for euery a E I . In the latter case, c , [ l ] has dense order type and for each 1-type P the set of condensation classes in which it is realized is dense in c , [ l ] .
Proof : If c,(a) = I for some (and hence every) a E I, then clearly c,(a) < I for every a E I . Suppose then that c,(a) is definable over a for every a E I and c d a ) # I for any n E I . We will show that in this case the second possibility holds; thus, we must show that if c,(a) < c,(6) and P is a 1-type of I , then there is an element c that realizes P and satisfies c,(a) < c < c,(b). Suppose not. Let $(a, y ) define { d ~ c , ( o ) l d2 a } and let $’(b, y ) define { d E C,(b)ld I b } ; let cx(x) generate the 1-type P and let B(x) generate the
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
309
1-type of b. We now define an interval with initial point a in which P is not realized. Let +(u, w ) be the formula W
2 U A (jX)(X 2
W A
P(X) A (vy)(O 5 y 5
X
-+
$(U,
y) V $’(X,
J’) V
la( y))).
Then { d I I k 4 [a, d ] ) is an interval whose initial point is a and that contains b. Since b $c,(a), { d l l k + [ a , d ] } must contain [a, -). In particular, no condensation class past c,(u) can contain an element whose 1-type is P ; this is only possible, according to Exercise 13.97, if c,(a) is the last condensation class of I . But then c,(u) is a definable interval of I ; for if + ( a , y ) defines c,(u) over a, then W[(VY)((Y
’
x
-
44x7 Y ) ) A +(x,
A]
defines c,(u). This is impossible since I is self-additive and cD(a)# I . Thus if c,(u) is definable over a for every a E I , then c,(u) has dense order type and for each 1-type P the set of condensation classes in which it is realized is dense in ~ ~ ( 1 ) . Suppose, on the other hand, that c,(u) is not definable over a for some a E I . We will show that c,(u) < I . Of course we continue to assume that cD(a) # I . We first show that c,(a) is not definable over any a’ E c,(a). Suppose, to the contrary, that c,(a) is definable over a’ and that I != +[a’, c] if and only if c E c,(u). Note that c,(a) cannot be either the first or the last condensation class in I , because then, as above, cD(a)would be definable, contradicting the self-additivity of I . Suppose that u ( x ) generates the 1-type of a’. Then we claim that (3x)(u(x) A 4(x, a) A +(x, y ) ) defines c,(a) over a. For clearly, if b E c,(a), then I != (3x)(a(x) A 4(x,a) A +(x, b)). O n the other hand, since I
( 3 x 1 ) ( W ( v ~ ) ( 4 ( aY’ ,) + x1 I Y I x2)
and u generates the 1-type of a‘, whenever I != ~ [ c ] then , {dlI b d [ c , d ] ) is contained within a bounded interval of I that is definable over c, and hence must be contained within c,(c); hence { d E I 1 I != (gx)(a(x) A +(x, a) A 4(x, d ) ) } =
tJ{
{ ~ J1I+ [ c , d ] ) ( cE c,(u) and I k .[c])
c
cD(ff).
Thus (3x)(a(x)A +(x, a) A +(x, y ) ) defines c,(a) over a. Hence if c,(u) is not definable over a, then c,(a) is not definable over any a’ E c,(u). Suppose now that I != + [ a , , . . . ,a,, b], where a,, . . . ,a, E c,(u) and b > c,(a). To show that c,(a) 4 I , we must find an element c E c,(u) such that I k 4[a,, . . . ,a,, c ] . Assuming that a’ is the largest among a,, . . . ,a,, we let d* be a testing formula for 4 in [a’, -); that is, I != +*[.’,dl if and only if I k +[a,, . . . ,a,,d]. Let u ( x ) generate the 1-type of a’. If there is no
13.
310
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
d E c,(u) such that I b $*[a’,d], then ( c ( 1 k $ , [ a ’ , c ] ) , where $l(x, y ) is x I y A (Vu)(x I uIy +i $*(x, u ) ) , defines a bounded interval including (c E c,(a)lc 2 a’} and hence equals { c E c,(a)lc 2 a’}. If cD(a)is the first condensation class in I, then (c E c,(a)lc I a’} = { c E I1 c Ia’} is definable over a’, so that cD(u)is definable over a’, contrary to what was shown above. Thus c,(a) is not the first condensation class in I. Hence, by Exercise 13.97, there are elements d < c,(u) whose l-type is P. Since a(x) generates P and I b (3z)$*(a:z), we must also have I 1 (3z)$*(d,z). If for every d < cD(a) whose l-type is P, there is no e between d and a’ such that I b $*[d, el, then the formula (vU)(U
X A %(U) +
(yZ)((4*(V,Z)A
Z
> U ) + 2 > X))
defines a proper interval of I . Hence for some d < c,(a) whose l-type is P, there is an e between d and a’ such that I k $*[d,e]. Hence the formula &a‘. Y), y<
U’A(’JtL1)((yI U < U ’ A C ( ( V ) ) - + ( \ J Z ) ( ( $ * ( U , Z ) A Z >
V)+Z>U’))
defines a bounded interval whose last point is a’, and must therefore be contained within c,(u). This interval cannot be all of (+-,a’] n c,(u) for in that case c,(u) would be definable over a’, contrary to assumption. Hence there is an element c E c,(u) such that c < a’ but c is not in this interval. This a’ in c,(a) such implies that there is a d‘ E c,(u) whose l-type is P and an e‘ I that I 1 $*[d’, e’]. Since d‘ E c,(u) and a‘ E c,(a), also a’ E c,(d’), so that there is a formula t+b(x,y)such that ( b l l k $ [ d ’ , b ] } is a bounded interval of I containing a’, d‘, and e‘ and hence contained within cD(a).But then I b ( 3 z ) ( 4 * ( d ‘ , z ) ~ $ ( d ’ , zHence, )). since a generates the l-type of both d‘ and a’, we have I k (3z)(4*(a’,z ) A $(a’, z ) ) ; for the same reason { b I I k $[a’, b ] ) is a bounded interval of I containing a’ and is therefore contained within c,(n). Thus, despite what we first assumed, there is a c E cD(u)such that I 1 $*[a’,c]. Hence c,(a) < I . Now let b E I - c,(u). Since c,(u) i I, every l-type of T is realized in c,(u), so that there is an element b‘ E c,(u) such that (I,b ) = (I,b‘). If c,(b) is definable over b, then cD(b‘)would be definable over b’, contrary to assumption. Hence C,(b) < I for every b E I. Thus I is a sum of elementarily equivalent self-additive models. 4 Let us elaborate on Lemma 13.101. Suppose that I is self-additive, that S , ( T ) is finite, where T = Th(I), and that c,(a) is definable over a for every a E I. Let a E I have l-type P and let e(a,y ) define c,(a) over a ; we claim that if I‘ is a model of T and a’ E I ’ has l-type P then &a’, y) defines c,(a’) over a’. Indeed, if { b E I / I’ k 4 [ a ’ , b ] } is a bounded interval of I containing a’,
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
31 1
then that fact is recorded in P, so that { b E III k 4[a, b ] } 2 { b E I11 1 6[a, b]}. But since (Vy)(+(x,yj + 6 ( x ,y ) ) is in P, ( b E I'I I' k +[a',b]} G { b E I'IZ' !F 6[a', b ] 1, so the claim is verified. Moreover, if a E I and a' E I ' have the same 1-type, then c,(u) = c,(a'). It follows that for any a E I and a' E I' the set of 1-types of T realized in cD(a)and c,(a') are either identical or disjoint. Since for each 1-type of I the set of condensation classes in which it is realized is dense, it follows that there is a finite set D,,D,, . . . ,D,of linear orderings such that I = o ( D , , D , , . . . ,D J . To summarize, T has a model that is a shuffle a ( D , , D , , . . . ,DJ such that each copy of D,is definable over each of its elements; these definitions are uniform in that if two elements have the same 1-types, then the copies of D, containing them are definable by the same formula. Turning to the other case, if cD(a)< I for every a E 1, then, by the discussion above, the same must be true for every model I' of T . Furthermore, there is a countable model of T , namely, c,(a), that consists of one condensation class. We now complete our discussion of the self-additive case by showing that if S , ( T ) is finite, where T = Th(l), then T has exactly one or exactly c distinct countable models. This together with Theorem 13.100 shows that the same conclusion is true for every self-additive I . Although we need only prove the result below for self-additive models, the proof requires that we deal simultaneously with non-self-additive linear orderings since a selfadditive I can be a finite shuffle of non-self-additive linear orderings. THEOREM 13.102: (Rubin) Let T be a complete theory of linear orderings ,for which S,( T j is Jinite. Then T has exactly one or exactly c distinct countable models. Proof: We proceed by induction on the size of S , ( T ) . If IS,(T)I = 1 and A !F T , then either +(a) z 1 for all u E A , in which case A 'v 1 or A N 1, or cF(a)2: 5 for all a E A , in which case A = 5. Thus in this case T has exactly one or exactly c distinct countable models. We proceed now to the case where IS,(T)I = n 1. If a model A k T is not self-additive, then by Theorem 13.96, there is a formula that defines a proper interval of A . Thus there is a formula 4 ( x j such that A = A , + A , , where A , = { U E AIA k + [ a ] ) and A , = { a E AIA 1 i 4 [ a ] } . Since there are 1-types that contain each of +(x) and i 4 ( x ) , IS,(Th(A,))I I n for each i. If both A , and A , are No-categorical, then so is A , A , = A by Theorem 13.91. If A , is not No-categorical, then, by the induction hypothesis, there are c distinct countable models {B,~cI< c> that are elementarily equivalent to A . Let D, = B, A , for each CI < c. Clearly D, = A for every a; and if { d E D,I D, !F 4 [ d ] } = B, for every a,then any isomorphism from D, to D,
+
+
,
+
312
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
would map B, onto B,, which is only possible when M = p. So let d E B, and let m = qd(4). Since G,+ l(B,, A E 11, there is an element a E A such that G,((B,, d ) , ( A l , a ) )E 11. Since G,(A,, A , ) E 11, we have G,((D,, d ) , ( A , a ) )E 11. Hence (D,,d ) = (A, a ) by Theorem 13.11, so that D, 1 4 [ d ] ifandonlyifA t= +[a].SinceA 1 4 [ a ] for all a E A , , D, I= 4 [ d ] foralld E B,. Similarly, if d E D, - B,, then D, 1 i 4 [ d ] .Hence { d E D,ID, t= 4 [d] 1 = B,. This implies that T has c distinct countable models. We now turn to the case in which every model of T is self-additive. According to Lemma 13.101, there are two cases to consider. If c,(a) < I for every a E A, then there is a countable model, namely, c,(u), that consists of one condensation class. Thus c,(u) . M has exactly c1 condensation classes, so that c,(u). M and c,(u). b are not isomorphic unless M = fi, so that T has c distinct countable models. Otherwise, by the discussion following Lemma 13.101, we can find a finite set of linear orderings D , , D , , . . . ,D,such that A = a@,, D,, . . . ,DJ. Thus A = A , I q E Q } , where each A , = D,for some j and, for each a E A,, c,(u) = A,. Now if every D,has an KO-categorical theory, then so does A by Theorem 13.91. If, on the other hand, some D iis not KO-categorical,then there are two possibilities. If Sl(Th(Di))I n, then Th(Di) has c distinct countable models {B,la< c } . If S,(Th(Di))= n + 1, then every 1-type of T is realized in each copy of Di,so that A = D, . q. Since A $ D i , D, is not self-additive, so that Th(Di) has c distinct countable models { B, I M < c f . In either case, if we let I' be the result of replacing some A , = Di by B, and all other A , = Di by Bi,then I' . GI is different from I' . b whenever M # p, so that T has c distinct countable models.
c{
COROLLARY 13.103: (Rubin) If T is selfadditive, then either T is KO-categorical or there is a countable model I of T such that I . c1 is diflerent from I ' p whenever c1 # p. In particular, if T is self-additive, then either T has exactly one or exactly c distinct countable models.
This completes our discussion of self-additive linear orderings, and we are now prepared to return to the general case. Before doing so, however, we will present another result of Rubin that provides a characterization of those complete theories T of linear orderings for which S , ( T )is finite. DEFINITION 13.104: Let M 2 be the smallest class of order types containing 1, containing T~ + t 2 whenever it contains z1 and T,, containing 5 . 5 whenever it contains T, and containing a(F) whenever it contains the elements of F.
5.
313
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
The theorem we will present, which is analogous to Theorem 13.91, is that S , ( T ) is finite if and only if T has a model whose order type is in A 2 . Before doing so, however, we define the following notion of rank for order types of DEFINITION 13.105: We define inductively a sequence { A , ( n ) l n < w } of subsets of A2by setting A,(01 = { l } and defining A 2 ( n + 1) to include A',(n) and to contain z1 z, if z l , z 2 E A 2 ( n ) ,to contain z . 5 if z E A 2 ( n ) , and to contain a(F) if F E A z ( n ) . Clearly, U { A 2 ( n ) In < w } = A 2 .We define the rank of an element z E A'2to be the smallest n such that z E A 2 ( n ) .
+
THEOREM 13.106: (Rubin) Let T be a complete theory of linear orderings. Then S , ( T ) is ,finite zf [inti only if T has u model whose order type is in .A2. Moreover, iJ' S,(T ) is Jiniri?,then T is Jinitely axiomatizable.
Proof : It is easily verified, using Ehrenfeucht games, that if z E A 2 then , Sl(Th(7))is finite. We will show, by induction on n, that if IS,(T)I I n, then T is finitely axiomatizable and has a model in A , ( 2 n - 1). If (S,(T)I = 1 then, as we pointed out earlier, T is either Th(l), Th(q), or Th([), and in each of these cases T is finitely axiomatizable and has a 1) as required. Proceeding with the induction, we assume that model in A2( if ISl(T)1 I n, then T is finitely axiomatizable and has a model in A , ( 2 n - 1) and we turn to the case where ISl(T)I = n 1. If T is not self-additive, then there is a countable model A of T and a formula 4(x) such that A = A , A , , where A , = { u E AIA b 4 [ a ] }and .4, = {a E AIA 1 iq5[a]}. Since each of 4 and 14 belong to 1-types of T , IS,(Th(A,))I I n and IS,(Th(A,))( In, so that, by induction hypothesis, each Th(Ai) is finitely axiomatizable and has a model Bi whose order type is in A1(2n - 1). Then T has a model, namely, B , B,, whose order type is in A,(2n). Moreover, if 8, and e2 are axiomatizations of Th(A,) and Th(,4,), then T is axiomatized by
+
+
+
(vX)(vy)(4(X) A 1$(y ) + X < y ) A 8f'"') A 8;'(uo).
We may thus assume that T is self-additive and apply Lemma 13.101 and the discussion appended to it. We consider first the case in which T has a There are then two possibilities. If IS,(Th(D,))(In model a(D,, D,, . . . ,Dk). for each i, then, using the induction hypothesis, there are Bi = D iwhose order types are in A , ( 2 n - I); then a(Bl, I?,, - . . ,Bk)= a(D1,D2,.. . ,Dk) and has order type in A2(2n).Moreover, if for each i we choose a 1-type Pi realized in D i , a formula cl/i(x)that generates Pi, and a formula 6 J x , y )
314
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
that defines over any element whose 1-type is Pi the copy of Di to which it belongs, and if we use the hypothesis to give us an axiomatization Bi of Th(D,), then we can construct the following axiomatization of T : (3x)+i(x) A
A ( V X ~ ) ( V X ~ V) ( + i ( X t ) A X 1 < X
Ai (vx)(lCli(x)
-+
1
~f~(x,uo)
~ 7A4 i ( x I 9 ~ 2 ) )
( i --t
A ( ~ x ) (
< ~ 2 ~ + j ( 1x 4) j ~ (x
>~1)~14j(x,~2))
j
A (~Y)(3Xl)(3X2)(Xl
< Y < x 2 A lClI(X1) A lCll(X2)
A 1 4 1 ( X l , Y ) A 741(XZ,Y)).
On the other hand, if IS,(Th(D,))I= n + 1 for each i, then Di = Dj for all i and j , so that Di . g is a model of T. Since each Di is definable over each of its elements, Di # Di . g, so that Di is not self-additive. But we have already shown that in the non-self-additive case, Th(D,) is finitely axiomatizable and has a model B in A 2 ( 2 n ) .Thus T has a model, namely, B . g, in A 2 ( 2 n + 1) and, as in the case above, T is finitely axiomatizable. We now turn to the other case in which T has a model A that consists of one condensation class. We intend to show that A = T . [ for some z E A 2 ( 2 n ) . Hence we first define a suitable Z-sequence of points of A. Let P be a 1-type of T and let a E A have 1-type P. Since P is principal and A is self-additive, it follows from Exercise 13.97 that { b E AIP(b) = P } is unbounded in B ; moreover, since S , ( T ) is finite, there is an element b > u such that b has 1-type P and such that every 1-type of T is realized in [a, b]. We claim that such an element b may be chosen so that for no element c between a and b whose 1-type is P is it the case that every 1-type of P is realized in both [u, c ] and [c, b ] . Using this claim, and the reverse claim, we can start with u = ao, let b = a,, and generate a [-sequence { a z l z E Z> of elements of A . To verify the claim, we note that otherwise we can find a sequence co > c1 > c2 > * . . of elements above a such that each ci has 1-type P and such that every 1-type of T is realized in both [a,ci+, ] and [ci+ c i ] . Since c,(a) = A, there is a formula $(u, w) such that ( d j A k 4 [ u , d ] ) is a bounded interval of A including co. Since A is self-additive, A < A + A’, where A‘ = A . In A A’ we can find an element c whose 1-type is P and such that every 1-type of T is realized between { d l A k + [ a , d ] } and c. Since there are only a finite number of 1-types and all of them are principal, this can be expressed by a formula with parameter a that, since A < A + A’, is also true in A . Thus there is an element c - E~ A such that { d l A k $[a,(il) <
+
5.
THE NUMBER OF COUNTAHLE MODELS OF A COMPLETE THEORY
315
c P I and every l-type of T is realized in [C,,C_~].Since P(a) = P(co) = P ( c - ,) = P and every sequence of l-types of T is realized in both [a, co] and [a,c-,], we can apply Theorem 13.80 to conclude that (a,co) and ( a , chave the same 2-type in A . But this is impossible because A b 4 [a, co] but A k i 4 [ a , c - ,I. This contradiction substantiates the claim. We have thus generated a [-sequence {a,lz E Z } of elements of A such that each a, has l-type P, every l-type of A is realized in each [a,, a,, and for every b E (u,,a,+,) whose l-type is P there is a l-type of T not realized in either [a,, b] or [b,a,+,]. If now a < a, for all z, then the con< a - < a,, would lead to the same contradiction figuration a < . . . < as before. Hence the sequence {a,la E Z } is coinitial in A, and, similarly, it is cofinal in A . For each z E Z we let A,' = {clfor some b of l-type P, a, I c I b, and not every l-type of T is realized in [a,,b]}; we define A,similarly and we set A , = A , AZt. Then A , c (a,- a, + ,). To show that A , does not overlap with A , + , , it suffices to show that there is a l-type of T that is not realized in any A,, for then A, u A , + , could not include [a,, a,+ ,] since every l-type of T is realized in [a,, a,, ,I. Even if we show that there is a l-type Q of T that is not realized in some A,, this will be sufficient; for since A , is definable over a,, the fact that no element of A , has l-type Q is expressed by a formula in P and hence no A, has an element whose l-type is Q. Suppose, on the other hand, that every l-type of T is realized in A , . Then, in particular, there are elements b,,b, E A , of l-type P such that b , I a, I b, and every l-type of T is realized in [b,, b,]. If some l-type of T is not realized in [b,,~,], then the interval Z(b,) = { d 2 b, lnot every l-type of T is realized in [b,,d]} is definable over b,, and contains [b,, a,]. Similarly, if some l-type of T is not realized in [a,, b,], then the interval I(a,) is definable over a, and contains [a,, bJ. Since a, and b, both have l-type P , if Q is not realized in [b,,~,], then Q is also not realized in [a,, b,] ;this is impossible since every l-type is realized in [b,, b,]. Hence either every l-type of T is realized in [b,,a,] or every l-type of T is realized in [a,,b,]. This contradicts the definition of A , . Hence some l-type Q of P is not realized in A , , and hence, as pointed out above, Q is not realized in any A , . Thus A , u A; does not include [a,, a,+ ,], so that there are non-empty linear orderings {B,(z E Z } such that A = C { A , B,lz E Z } . Note that P is not realized in any B,; for if a, < b < a, + where b E B , and b has l-type P, then either some l-type of T is not realized in [a,, b]-in some l-type is not realized in [b,a,+,]-in which which case b E A,-or case we would have b E A, + . To show that A = z * 5 for some z E A 2 ( 2 n ) ,we will show that every A, = for some T~ E A 2 ( 2 n - 1) and that every B, = z2 for some z 2 E ,4,(2n - 1). Then t, + t 2 E A 2 ( 2 a ) , so that (z, z,). [ E A 2 ( 2 n 1). Since A = (zl+ t2). [, this will complete the proof.
,)
,I,
,
+
,,
,
,,
+
,
'I,
+
+
316
13. THE! FIRST-ORDER
THEORY OF LINEAR ORDERINGS
That A,, = A,, for every zl, z , E Z is clear since each A , is definable over a, by the same formula, so that given any statement true in A, its relativization is found in P. We now show that lS,(Th(A,))( < IS,(Th(A))I, so that we can apply the induction hypothesis. Let Qf(a,x)say that there is a y of type P such that x lies between a and y and not every 1-type of T is realized between a and y. Then { b E A I ~ [ u , , b ] } = A,. We claim that for any a E A , , { b E AIQf[a,b]} = A , . Suppose, for example, that a, < a < b < c and not every 1-type of T is realized in (a, c) but b $ A,; then we can choose an element a‘ that is in A , and has 1-type P such that a, < a < a’ < b < c; but then not every 1-type of T is realized in [a,,a’] and not every 1-type of T is realized in [a’, c], so that, as shown several paragraphs ago, not every 1-type of T is realized in [a,, c]-so that b E A , after all. Thus if two elements b , and b, of A , have the same 1-type in A , then in playing, and winning, the game G,((A,b,),( A , b 2 ) )for n > qd(4), PLAYER II confines his choices to A , whenever PLAYER I does so. Hence the 1-types of b , and b, in A , are the same. Thus IS,(Th(A,)) < IS,(T)l since Q is not realized in A , . Hence, by the induction hypothesis, Th(A,) has a model z1 in A2(2n - 1). Similarly, since P is not realized in any B, and elements realizing P are coinitial and cofinal in each A , , for any b E B,, B, is exactly the set of all elements c such that nothing between b and c has 1-type P. Thus B,, = B,, if zl, z , E Z , and if two elements of B, have the same 1-type in A , then they have the same 1-type in B,. Thus, as above, (S,(Th(B,))I < ISl(T)I, so that Th(B,) has a model t 2E A,(2n - 1). It follows that Th(A) has a model, namely, (51 t 2 ). c in A2(2n l), as was to be shown. The axiomatization of (7, + t 2 ). [ we leave to the reader.
I
+
+
A special case of Theorem 13.106 is that every KO-categorical linear ordering is finitely axiomatizable. This result is due to Lauchli and to Rosenstein [14]. We return now to our discussion of an arbitrary complete theory T of linear orderings. There are now two possibilities: Either S,(T) is countable for each n or, for some n, S,(T) is uncountable. We will first dispose of the second case by showing that the cardinality of S,(T) must actually be c if it is uncountable. This will in fact dispose of this case because if S,(T) has cardinality c, then, since each n-type of T must be realized in some countable model of T and only countably many n-types of T can be realized in any single countable model of T , it follows that T has exactly c distinct countable models. [The fact that S,(T) has cardinality c if it is uncountable is true more generally for the collection of maximal ideals in any Boolean algebra, and the proof given below also works in the general case.] LEMMA 13.107: Zf S,(T) is uncountable, then it has cardinality C.
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
317
Proof : We will first show that if II/ belongs to uncountably many n-types of T, then there is a formula 6’ in F,( T ) such that both $ A 8 and I,+ A (1 0) belong to uncountably many n-types of T. Suppose not, so that given any f3E F,( T ) either $ A 0 or $ A (1 0) belongs to only countably many n-types of T. Let { O j l j < o}be an enumeration of all formulas in F,(T). We define a sequence ( $ j l j < o}of formulas of F,(T) such that each $ j is either d j or i B j and such that for each m, ( t j j l j I rn} LJ {$) is in uncountably many j < rn) u {$, i $,} is in at most countably many n-types. n-types but Assume that the { $ j I j < rn) have already been defined and satisfy the cone,) are in uncountably many ditions above. Since not both $ A 0, and $ A (1 n-types, it follows that not both ( $ j l j < m } u ($.em} and ( $ j l j < rn} LJ { $ A (10,)) are in uncountably many n-types; on the other hand, since {tjjlj < m } u ($1 are in uncountably many n-types, each of which contains either 8, or lo,, it follows that one of the two sets is contained in at most countably many n-types and the other is in uncountably many n-types. Define I),,, to be 8, if { t,hj I j < rn} u { $, 0,) is in uncountably many n-types and otherwise ,,)I is lo,,,.Note that the induction hypothesis continues to hold. Since each finite subset of (qb,l,j < o}is in uncountably many n-types, the set itself is certainly consistent and is therefore contained in an n-type. Moreover, since for each O j E F,,(T) either O j or i O j is II/j, it follows that {t,bjl j < a} is an n-type of T. On the other hand, as we defined the sequence j < o}, at each stage we sloughed off only countably many n-types that j < ID}, $ is only in countably many n-types contained $. Thus besides of T. This contradicts the original assumption that $ is in uncountably many n-types of T. Thus we have proved that if $ belongs to uncountably many n-types of T, then there is a formula f3 in F,( T ) such that both t,b A f3 and $ A (1 19) belong to uncountably many n-types of T. Using this fact, we now define for each finite sequence s of 0 s and 1’s a formula 4sof F,(T) that is in uncountably many n-types of T. For the empty finite sequence ( ), $( is just u1 = u1 A u2 = u2 A . . . A v, = v,,, which is in all of the uncountably many n-types of T . Proceeding inductively, if 4sis defined and is in uncountably 0) many n-types of T, then we find a formula 8 so that both rfJs A 0 and 4sA (1 are in uncountably many l-types of T and we let one of these be 4soand the other rfJsl. Now i f f : N -, (0, l} is an arbitrary sequence of 0’s and 1’s and S , is its set of finite initial sequences, then (4,Is E S,} is contained in an n-type of T. Since different sequences result in different n-types and since there are c such sequences, it follows that SdT) has cardinality c. W
{$jl
{$jl
($jl
Thus we can confine ourselves to the case in which S,(T) is countable for each n. We introduce a new condensation cE to analyze the models of such a theory.
31 8
13.
THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
DEFINITION 13.108: Let 4(x) be a formula of L. We say that d, is an interval formula of T if T t (3x)+(x)A i(Vx)$(x) and
T I-
(~X)(VY)(VZ)((X
< y < z A 4(x) A 4b))
+
4(Y)).
Thus 4 is an interval formula of T if and only if for every model A of T, { a E A I A k 4 [ u ] } is a proper interval of A. The set of interval formulas of T is denoted i(T).For any model A of T and any a E A we define c E ( a )= (blfor every interval formula #(x) of T , A k 4[b] if and only if A k 4 [ a ] } . Note that cE is a condensation since, as is easily verified, if b E c,(u), then C E ( b ) = C E ( ~ ) . DEFINITION 13.109: An interval formula 4(x) of T is called a terminal formula of T if T t (Vx)(Vy)((x < y A 4 ( x ) )+ & y ) ) . Similarly, #(x) is an initial formula of T if T t (Vx)(Vy)((x < y A 4( y ) ) -+ &x)).
Given an interval formula #(x) of T , we let +'(x) denote (Vy)(&y) -+ y < x) and c$-(x) denote ( V y ) ( + ( y ) --* y > x). Then $'(x) is a terminal formula of T unless +(x) is a terminal formula of T , and &(x) is an initial formula of T unless $(x) is an initial formula of T . If @x) is an interval formula of T , then any model A of T can be written as Cap k
4-CaN +
4C4 +b
( A
4"aIh
with the first (last) term absent if 4 is an initial (terminal) formula of T. Given a model A of T and a E A , we let @a be the set of all interval forthen either 4+(x) E or +-(x) E Oa, mulas of T satisfied by a. If +(x) $ so that 4(x) cannot be consistently added to Thus @ a is a maximal consistent set of interval formulas of T. Conversely, if Q, is a maximal consistent set of interval formulas of T , then @ s P for some 1-type P by Exercise 13.67.4 since @ G F,(T),and hence there is a countable model A k T and an element a E A whose 1-type is P, and therefore @a = @. DEFINITION 13.110: A maximal consistent subset of the set I ( T ) of interval formulas of T is called an interval type of T. The collection of interval types of T is denoted ZT(T).If @ is an interval type of T and A i= T, we let A(@)= { u E A ( @ ,= @}. Note that A(@) is an interval of A if it is non-empty. If A ( @ )# $3,we say that @ is realized in A, and otherwise we say that @ is omitted in A . We let IT(A) denote the set of interval types of T realized in A .
Note that for any model A k T, there is a natural correspondence between c,-condensation classes cE(u)of A and interval types @a realized in
5.
THE NUMBER OF COUNTABLE MODELS OF A COMPLETE THEORY
319
A ; thus b E cE(aJif and only if Qb = @a, so that A(@,) = cE(a).This correspondence induces a linear ordering on IT(A) by @, < I T ( A ) @ b if and only if cE(u)< cE(b);thus I T ( A ) = c E [ A ] . Let and be two interval types of T and assume that $ ( x ) E 0,but + ( x ) $ 0,.We may assume, without loss of generality, that $ J + ( x E) 0,. From this we see that if Ol,Q2 E I T ( A )n I T ( B ) , then < I T ( A ) Q2 if and only if O1 m2. Thus the various linear orderings induced on the subsets IT(A) of I T ( T ) by the correspondences with c E [ A ]are all compatible. Since S,,(T)is assumed to be countable for all n, by Theorem 13.82 there is a countable saturated model S of T . Since each interval type is contained in a l-type and since each l-type is realized in S, it follows that every interval type is realized in S , so that I T ( S )= I T ( T )and I T ( T )2 c E [ S ] ;in particular, I T ( T ) is countable. DEFINITION 13.111: An interval type CD is called a principal interval type if there is a formula $(x) E @ that generates 0;that is T k (Vx)[$(x) -, $(x)] for all 4(x) E 0.Otherwise, @ is called a non-principa1,or limit, interval
type. We now obtain a characterization of the principal interval types. LEMMA 13.112: Suppose that 4(x) is an interval formula of T. Then {@ E I T ( T ) )$(x) E @} has a first mid last element.
Proof: Suppose that it has no last element. Then there is a countable < < a2< . . . of elements of I T ( T ) that is cofinal in sequence {@ E I T ( T ) l 4 ( x )E 0.). For each n, choose a formula 4,,(x) with 4,,(x)E @, and $n'(x) E On+1 . Let C = {$(x)} u {$,,+(x)In < 01). This is a consistent set of interval formulas of T since any finite subset is contained in some @,,. Hence C c P for some 1-type P of T ; but the set of interval formulas of T in P forms an interval type, so that C c Q, for some @ E ZT(T). But each On< CD and 4(x) E Q, contradicting the assumption that the sequence {a,,In < w } is cofinal in ICD E I T (T )1 + ( x ) E a}. H COROLLARY 13.113: Let @ E I T ( T ) . Then @ is a principal interval type fi and only i f @ has an immediute predecessor and an immediate successor
in ZT(T). Proof : If @ has an immediate predecessor
and an immediate successor in I T ( T ) ,then there are formulas Cpl(x)E CDl and 4 2 ( x )E O 2 neither of which is in @. Then 4 1 + ( x )A $ J 2 - ( . x ) E @ and clearly generates
Q2
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Conversely, if $(x) generates @, then, by Lemma 13.112, {@ E
w T ) I $ - ( x ) E @}
has a last element, which is the immediate predecessor of 0,and
{a E ZT(T)($+(X)E @] has a first element, which is the immediate successor of 0 in ZT(T). The proof of Lemma 13.112 can be adapted to yield the following result. THEOREM 13.114:
I T ( T )is complete.
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Proof : Let IT( T ) = A B. We must show that either A has a last element or B has a first element. Otherwise, we can choose a sequence (@,,In < o) cofinal in A and a sequence {'PI,/n < o} coinitial in B. If +,,(x)E @, and +,,'(x) E a,,+ for each n, and $,,(x) E Y,,and $,,-(x) E Y,,+ for each n, then {+,,+(zc))n < w } u {$"-(x)In < o}is consistent and hence is contained in some @ E I T ( T ) which is bigger than each a,, and less than each Y,,. This contradiction implies that IT( T ) is complete.
COROLLARY 13.115: I T ( T )is scattered. Proof: If q 5 ZT(T),then ZT(T)has c cuts, by Exercise 2.33.4. But since IT( T )is countable and complete, this is impossible.
We now obtain critical information about the intervals S(@) of the saturated model S of T. THEOREM 13.116: Let S be a saturated model of T and let @ be an interval type of T. Then either S(@) is a single element of S or S(@) is self-additive. Proof: We first observe that if +(x) is any formula such that S k + [ a ] for some a E S(@), then { b E S(@)JSb (P[b]) is cofinal in S(@). For suppose that C E SPD). If there is a formula $(x)E@ such that S Y +[b] whenever b 2 c and S k $ [ b ] , then the interval formula (3y)(xI y~ +( y) A $(y)) is satisfied by all elements of S(@)up to a, but by no elements of S(@) beyond c. Since CD is an interval type, this is impossible. Hence for every $(x) E @ there is a b 2 c such that S k $[b] and S I=+ [ b ] . Thus X = {$(v2)1$(x)E @} u {&vz), v2 2 u , } is consistent with the 1-type of
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321
c, and hence, since S is saturated, there is an element b E S that, together with satisfies all formulas of C. Thus h 2 c, S I: 4 [ b ] , and b E S(@), as was to be shown. From this it follows that for any 1-type P of T that is realized in S(@), the set of elements of S ( 0 ) whose 1-type is P is cofinal in S(@).For if c E S(@), then P u {c I u } is finitely satisfiable by the argument above, and hence u> is satisfied is satisfied in S, since S is saturated; but since @ G P, P u { c I by some element of S(@). We use this to show that S(@)has no proper definable intervals. For if A is a proper interval of S(@)that is, without loss of generality, not a terminal interval, we choose a E A and, using the conclusion above, we choose b > A so that a and b have the same 1-type in S. Since S is saturated, there is an automorphism of S that maps a to b ; but any automorphism of S induces an automorphism of S(@).Thus a and b have the same 1-type in S(@),so that A cannot be a definable interval of S(@).Thus S(@)is either self-additive or S(@)has exactly one element. c,
EXERCISE 13.117: Show that if S is a saturated model of T , @ is an interval type of T ,and S(@)has a single element,then that element is definable.
LEMMA 13.118: I f @ is a principal interval type, then S(@)is a saturated model of Th(S(@)). Proof : Let P be an n-type of Th(S(@))and Q be an (n such that P c Q. Let @ be generated by O(x).Then
+ 1)-typeof Th(S(0))
Qe(w)= {4(xl, . . . , X , , X , + ~ ) ' ( ~ ) (E~Q ) 2 Pecw) = {4(Xl>..
1
>x,)e(w)]4 E PI.
If ( a l , . . . , a , ) is an n-tuple of elements of S(@)whose n-type is P, then S I: 4 e ( w ) [ a l. ,. . ,a,] for every 4 E P, so that, since S is saturated, there is an element b E S such that S k @('"')[a,,. . . ,a,*,b] for every 4 E Q and such that S 1 O[b]. Hence S(@)is a saturated model of Th(S(@)). We now have all the information we need, so we proceed to prove Rubin's theorem. First consider the possibility that for some isolated cb E I T ( T ) , S(@)is not KO-categorical.By Theorem 13.116, S(@)is self-additive, so that, by Corollary 13.103, Th(S(0))has c countable models; by Lemma 13.118, S(@)is saturated, so that, by Exercise 13.83.1, there are distinct models (A al a< c} such that each A , < S(@). Write S = S, + S(@)+ S,. Then, for each a < c, S, = S, + A , + S , 4 S and { a E SalsaI: O[a]} = A,, where e(x) generates @. Hence these c countable models are all distinct models of T, since any isomorphism from S, to S, must carry A , onto A , .
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Thus we may assume that S(@) is KO-categoricalfor every isolated @ E I T ( T ) .If ZT(T)is finite, then S = S , . . . S,, where each Th(S,) is KO-categorical.By Theorem 13.91, each Si- M ifor some Mi € A l ,so that S = M , . . . M , E A , ; hence T is K,-categorical. Thus we may also assume that ZT(T) is infinite. Since it is scattered, it has an interval of order type w or o*;we assume the former, with no loss of generality. Since I T ( T ) is complete by Theorem 13.114, it has an interval of order type o 1. Thus there are interval types @ 0 , @ 1 , @ 2 r . . . ,@ such that each S(@,) is KO-categoricaland S(@,) S(a1)+ S(@,) . . . S ( @ ) is an interval of S. Let us pause to consider a particular example A = a(l,2) a(2,3) + a(3,4) + . . . of this case. It is easily verified that the c,-condensation classes of A are precisely {a(.,. 1)ln < w } , so that ZT(A) has order type o.Since Z T ( T ) is complete, it must have order type o 1; thus ZT(T) consists of @, < 0,< Q3 < . . . < 0 and S((D,) = a(n,n + 1) for each n < w since a(n,n 1) is KO-categorical.What is S(@)? We observed that cF(a)cannot be finite for any a E S(@) since there is no maximal finite interval of length n in A after any maximal finite interval of length n + 2. Furthermore, given any element a E S(Q there must be a first element b 2 a with no immediate successor and a last element c I a with no immediate predecessor. Thus S ( 0 ) is a sum of intervals each of which has order type w [ . CI o* for some order type a. Moreover, since the elements of A with no immediate predecessor have order type q, we conclude that S(@) = (w w * ) . q. Thus c distinct countable models of T h ( A )can be obtained by replacing one copy of o o*in A (o w * ) . q by each possible w + [ . c t o*. In this example, S(@,) f S(@,) whenever n # m. The procedure we now describe is reminiscent of the example and works whenever (S(@,)) n < w > contains infinitely many non-isomorphic linear orderings. We need the following lemma.
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LEMMA 13.119: Let JV be an infinite collection of C.C,-categorical order n o} of order types such that each types. Then there is a sequence { ~ , \ < zi is the order t y p e of an interval of some z E N and such that each z, is isomorphic to a proper interval of z,+ 1.
Proof : We define the notion of predecessor for KO-categoricalorder types by induction on rank (see Exercise 13.92). Thus if z has rank n + 1 and z = z1 T ~ where , both z1 and z 2 have rank at most n, then the predecessors of z consist of z,, z2, and the predecessors of z1 and 7,; similarly, if t has rank n + 1 and T = a(F), where each z’ E F has rank at most n, then the predecessors of z consists of F together with the predecessors of each T’ E F.
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Now given N , we let A"* consist of Jli together with all predecessors of order types in N.We now define by induction a sequence {z,(n < o) of elements of N * such that each z, has rank n. We will assume, as part of the induction hypothesis, that the set T , of elements of N* that have z, as a predecessor is infinite. We define zo to be 1 and, assuming that z, is defined and T , is infinite, we enumerate the elements ol,.. . , ( T ~of T , that have rank n + 1. This set is finite by Exercise 13.92. Moreover, if T E T,, then, as is easily verified by an induction on rank, some ( T ~is a predecessor of z. Thus if we define T"+ to be the first oithat is a predecessor of infinitely many elements of T,, then the induction hypothesis continues to hold and we obtain a sequence {z,ln < w ) , as desired. Suppose now that { S ( 0 , ) 1 n < o} contains infinitely many nonisomorphic KO-categorical linear orderings. Then, by applying Lemma 13.119, we can find a sequence (t,ln < co} of order types such that each T~ is a predecessor of some S(@,,,J; a subsequence {a,ln < w } can easily be selected so that each (T, is a predecessor of some S((Dq(,)),where the subscripts { q ( n )\ ti < o}form an increasing sequence. Let C, be an interval of S(aq(,,J of order type B, for each n. let g, map C, onto an interval of C,+, for each n, and let (T be the order type of C = U{C,[ n < o},where in this union each C , is identified with g,[C,]. In the example discussed earlier, each C, has order type n and the order type of C is either w, w*, or ( depending on the functions {gnln < ( 1 ) ) chosen. We now show that S(@) has an interval of order type rr. Indeed, if we choose elements {a,lz E Z ) G C so that z1 5 z2 implies aZlI azz and so that { a , , l n N ~ } is cofinal in C and { a , / n ~ Z N ) is coinitial in C, then C = U { [ a _ , , , a , , ] I nN~) ; since each [a-,,a,] is an interval of some S(@"J, each [a_,,,a,] is KO-categorical by Exercise 13.92, and hence is finitely axiomatizable by Theorem 13.106. Let $, be an axiomatization of [a-.,a,] for each n. Add to the language of linear orderings a (-sequence (a,] z E Z } of constant symbols and let C = C ( S ) u ($,'In < wI1 u { g z , I gz2/z1 5 z 2 } u {Cp(%)(zE 2 and Cp E @ } , where t+bn' is the relativization of $, to the interval [a-,, 4.Any finite subset of C can be realized in S by selecting elements of some S(0.,) to correspond to the constant symbols. Hence 'c is consistent and is realized in some countcontains elements { c, I z E Z ) able model s' of T. Since s' < s,we see that s(@) such that each interval [c - ,,c,] Y [ u - ,, a,] ;hence S ( 0 )contains an interval of order type rr. Now B itself cannot be KO-categoricalsince if its rank were k, then every interval of it would have rank at most 2k + 1 by Exercise 13.92; this is impossible since the sequence { rr, 1 YE < o}of intervals of (T all have different
324
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
ranks. Since o is an interval of S(@), it follows that S(@) cannot be KOcategorical. Since by Theorem 13.116, S(@)is self-additive, it follows from Corollary 13.103 that there is a countable model I of T such that I . CI is different from I . fl whenever ct # p. Let S’(@)be a saturated model of Th(S(@))and let S’ be obtained from S by replacing S(@)by S’(@).Since S(@) is saturated, it follows that each I . a < S’(@). If we let S, be the result of replacing S‘(@)by I . c( in S‘, then each S , < S’ and { u E S,I S, k 4 [ u ] for all 4 E @} = I . a. Hence these c countable models are all distinct models of T since any isomorphism from S, to S, must carry I, onto I , . Thus if the sequence {S(@,)ln < a}contains infinitely many distinct KO-categoricallinear orderings, then T has c distinct countable models. We thus need only consider the case in which the sequence (S(@,)ln < 01) contains only finitely many distinct KO-categorical linear orderings. As an example of this case we can take A = A , + 3 + A , + 3 + A , + 3 + . . . , where A, = (q 2 * q ) . n. Here again I T ( A ) has order type w and the condensation classes have order type q, 2 . q, and 3. Thus I T ( T ) consists of @, < Q2 < Q 3 < . . . < 0,and we must again calculate S(@).We leave this calculation to the reader.
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EXERCISE 13.120: Verify that if A = A , + 3 A , + 3 + . . . , where A , = ( q 2 . q) . n, then S ( @ ) = (3 ( q 2 . q)(w o*)) . 5 * q.
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The phenomenon that occurs in this example, namely, the existence of c . . . , occurs in all cases in which ( S ( @ , ) l n < o)contains only finitely many different models. That is, we claim that there is an increasing sequence { i(n)ln < u}of natural numbers such that i(n 1) 2 i(n) n and such that S(Qicn,) N S(QiCO,) for all n 2 0, S(@i(n)+l) ‘v S(Qi(,)+ 1) for all n 2 1, and, more generally, S(Qi(,)+;)= S(@i(,l+,)for all n 2 j and all j < a. If we now define A , = S(@,(,,)+ S(Qi(,,)+ + . . . S(@‘i(n)+n+ ,), then each A, is isomorphic to an initial interval of A,+ 1. To prove the existence of {i(n)ln< o},we define it inductively together with a contracting sequence {Z(n)ln < w ) of infinite subsets of N . Let B,, B,, . . . ,B,-, be the finitely many distinct models in the sequence { S ( @ , ) l n < w ) , and let Ij(0)= { nI S(@,)N B j } for each j < k. Since { Ij(0)I j < p } is a partition of N , some Ij(0)must be infinite; call it I(0)and let i(0)be its smallest element. Proceeding inductively, we assume that i(0) < i(1) < . . . < i(k) have been defined, i(t + 1) 2 i(r) + t for every t < k, and for every t 5 k and every n E I(t), s(Qn) S(@i(O),S(Qn+ 1) L s ( Q i ( i ) + 11, . . - S ( Q n + t ) S(@i(i)+t). We now partition I(k) - (0, 1 , 2 , . . . , i(k) + k - 1) into { I j @ + 1)1j < p ) , where I j ( k + 1) = { n 2 i ( k ) + klS(@,+k+ ,) N Biand n E I ( k ) } .Then some Ij(k + 1) must be infinite, so we define that one to be I ( k + 1) and we define its smallest A E A, EA,
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FINITELY AXIOMATIZABLE LINEAR ORDERINGS
325
element to be i(k + I). The induction hypotheses continue to hold; hence the existence of (i,,1 n < w } is substantiated. We now show that the sequence {A,ln < w } of KO-categorical linear orderings is pairwise non-isomorphic. For suppose that n < m and that A, 'v A,. Now A , is isomorphic to the initial interval C of A,. Since C is a finite sum of S ( m j ) ,each of which is definable, it follows that C is definable, and, by Theorem 13.38, C is definable in A,. It follows that more 1-types are realized in A , than in C, so that C. and hence A,, cannot be isomorphic to A,. Hence { A , In < w } is pairwise non-isomorphic. Let cr be the order type of A = U{A,I n < 01, where in this union each A , is identified with the corresponding initial interval of A,, Then, as in the preceding case, S(@)has an interval of order type cr and, since cr cannot be KO-categorical,S(@)is not KO-categorical.Then, as in the preceding case, since S(@) is self-additive, T has c distinct countable models. This completes the proof of the main result of this section. THEOREM 13.121: (Rubin) Let T be a complete theory of linear urderings. Then T is either N,-cutegorical or has exactly c distinct countable models.
Using Corollary 13.103, the proof we have given of Theorem 13.121 can be modified to obtain the following result. COROLLARY 13.122: (Rubin) Let T be a complete theory of linear orderings. Then either T is KO-categorical,S,(T) is uncountable for some n, or there is a countable model A of T that has a self-additive interval I such that if A = A , + I + A,, then, whenever CI # p, the linear orderings A , + I ' C I + A , and A , + 1 . /3 + A , are not isomorphic.
56. FINITELY AXIOMATIZABLE LINEAR ORDERINGS
Our goal in this section is to prove that a statement that is true in some linear ordering is true in some finitely axiomatizable linear ordering. This was mentioned earlier as Theorem 13.23: Given any linear ordering A and M ; i f A is any n, there is a finitely axiomatizable M E A? such that A scattered, then M can be chosen from d oThis . was conjectured by Lauchli and Leonard [8] and was finally proved by Myers [12], Shelah, and Schmerl (independently); our proof is patterned on that of Amit [l].
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We restrict our attention for now to scattered linear orderings. The general case will be considered subsequently. Since we know, by Corollary 7.10, that any countable scattered linear ordering is G,-equivalent to an element of Ao,it suffices to prove, by the transitivity of -,, that every element of ,Aois G,,-equivalent to some finitary linear ordering. Thus, proceeding inductively, we could assume that A -,F, where F is finitary, conclude by Lemma 6.5.2 that A . w -,F . w, and then try to show that F . w is finitary. The difficulty with this approach is that F . w need not be finitary even when F is finitary; moreover, conditions under which F . w must be finitary are elusive. We digress for a moment from our goal and elaborate on the sentence above. An example of a finitary F for which F . w is not finitary is provided by F = w* + 5 + w (shown finitary in Exercise 13.18.3). To verify that
F.
= O*
+ 5 + (O + 0") + 5 + (w + o*)+ . . .
is not finitary, we observe that F . w (and even F o + w* in the following technical sense.
+ F) contains a copy of
DEFINITION 13.123: Let I be an interval of a linear ordering A. We say that 1 is a copy of w w* if 1 has order type w + 5 . CI + w* for some order type u, and if the first element of I has no immediate predecessor and the last element of I has no immediate successor. If A has an interval I that is a copy of w + w*, then we say that A contains a copy of w + w * .
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Note that 5 5 does not contain a copy of w + a*,although F F does, where F = m* 5 + w. We will show later, in Corollary 13.142, that any A that contains a copy of w + w* is not finitary. For now, we will show that F . w is not finitary; a similar argument works for F + F. It follows from Exercise 6.1 1 and Lemma 6.5.2 that
+
F .
-,,
W*
+ (5 + 2").0
for any n; so it suffices to show that for every n. PLAYER I chooses, in his first two turns, the endpoints of a copy of 2" in w* + (5 + 2"). w ; PLAYER II must then choose a point of F . w with no immediate predecessor and a point of F . w with no immediate successor (or lose in two more turns). The interval of F . w thus determined must have order type w F . m w* for some m 2 0. But, by Exercise 6.10.2, PLAYER I has the winning strategy in C,(2" - 2, w + F . m + w * ) and hence in G , , 2(w* (5 Z n ) . w, F . 0).Thus F . w is not finitary. This completes the digression.
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The approach in [l] is to use the appropriate condensation map. Thus we will define a condensation c of each linear ordering A E .Aoso that each c(a)is finitary and so that c[A] has lower rank than A , and hence is G,-equivalent to a finitary F . We will then replace the sum A
=
c ( c ( a ) c[A]} E
by a sum that is indexed by F , is (3,-equivalent to A , and is finitary. In order to replace A by a sum indexed by F , we need to think of c[A] as not just a linear ordering, but as a labeled ordering-where each element c(a) E c [ A ] is labeled by the G,-equivalence class of c(a). If it turns out that c [ A ] -,F not only as linear orderings but as labeled orderings, then we can easily take a sum indexed by F that is G,-equivalent to A. As we will see, each c(a) can be taken to be G,,-equivalent to one of a j n i t e set of linear orderings, so that the labeling of c[A] determines a finite partition of c[A]. (We have been speaking of “labelings’’because of similarities to the discussion in 58.6, but it will be more convenient to talk about partitions.) Because of these considerations we need to enlarge our scope so that we consider partitioned orderings, as well as ordinary linear orderings, as fair game for the Ehrenfeucht analysis. DEFINITION 13.124: Let A be a linear ordering and let A , , A l , . . . , A,be an ordered sequence of k subsets of A that partition A , where k is a fixed natural number. The structure ( A ; & , A , , . . . ,A k - 1 ) is called a partitioned ordering. In some contexts, we will use A to denote this structure. If there is a possibility for confusing the linear ordering A with the partitioned ordering A, then we will use the notation A k for the latter. If we wish to emphasize the number k, then we will speak of A k as a k-partitioned ordering. (We will not introduce the notion of partitioned order types, although that too can be easily defined.)
Given two k-partitioned orderings A k = ( A ; A , , A , , . . . , Ak-l)and Bk= ( B ; B , , B , , . . . ,Bk-,),thedefinitionoftheEhrenfeuchtgameG,(Ak,Bk) is included in Definition 13.1. In order for PLAYER II to win, he must so select his moves that, if a, and b, denote the elements of A and B chosen at the tth turn, not only must a, < A a,
if and only if
b, < B b,
if and only if
b, E B,
for all s, t 5 n, but also a, E A j
for all s I n and all j < k. Thus PLAYER 11 must match PLAYER 1’s moves both with respect to the linear orderings and with respect to the partitions.
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If he can do so, then PLAYER 11 has the winning strategy in G,(Ak,B k ) ;if he cannot, then PLAYER I has the winning strategy. For k = 1 these notions coincide with the definitions given for ordinary linear orderings. The exercise below, involving the case k = 2, gives some indication of what happens at the next level. (Assume that G,-equivalence has already been discussed for k-partitioned orderings-it will be soon, and this exercise will thus serve as motivation.) EXERCISE 13.125 : Corollary 6.9 can be interpreted, for n = 2, as saying that every finite linear ordering with more than 3 elements is G,-equivalent to a finite linear ordering with no more than 3 elements. This is the situation for k = 1; the statements below concern the case where k = 2. (1) Show that there is a finite ( A ; A , , A , ) where A has 8 elements which is not G,-equivalent to any finite ( B ; B 1 , B 2 ) ,where B has fewer than 8 elements. (2) Show that every finite ( A ;A 1, A , ) where A has more than 8 elements is G,-equivalent to a finite ( B ; B1, B , ) , where B has no more than 8 elements.
We now proceed by reviewing the analysis of Ehrenfeucht games on linear orderings presented in Chapter 6 and determining to what extent it applies to Ehrenfeucht games on k-partitioned orderings. (We will be referring to this discussion when, later on, we speak of the "adapted" versions of the statements of Chapter 6.) If we define Ak Bk to mean that there is an isomorphism f : A -+ B such that f [ A j ] = B j for j < k, then Lemma 6.3 still holds. If we define Ak + Bk (for A and B disjoint) to be (A
+ B ; A0 U B o , . . . , Ak-1
U Bk-1)
and similarly for generalized sums (of partitioned orderings over an index set that is an ordinary linear ordering) and shuffles, then Lemma 6.5 and Exercise 7.18 still hold. If we define I k to be ( I ; A , n I , . . . ,Ak-1 n Z)
for any interval I of A, then the following form of Theorem 6.6 is correct: G , , ,(Ak,B k )E I1
if and only if (i) for every j < k and every a E A j there is a b e Bj such that C,((A'")k,(B'b)k)E I1 and G,((A<")k,(B
6.
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3 29
Corollary 6.7 and Definition 6.8 can be given as before, so that we can speak of G,-equivalence and G-equivalence of k-partitioned orderings. The careful calculations of Corollary 6.9 and Corollary 6.12 are completely upset, but the proof of Theorem 6.13 is correct without any changes except that f(1) = 2k instead of 2. Thus, for each k and each n, there are only a finite number of G,-equivalence classes of k-partitioned orderings. Consequently, the following version of Corollary 6.9 is correct: For each k and each n, there is a natural number g(k, n) such that every finite k-partitioned ordering with more than g(k,n) elements is G,-equivalent to one with no more than g ( k , n) elements. Corollary 6.9 thus says that g( 1, n) = 2” - 1; the difficulty in calculating explicit values of g(k,n) for k > 1 will be apparent to the reader who tries to show that g(2,2) = 8 in Exercise 13.125. Corollary 6.12 holds if the partitions are “periodic”; that is to say, if F is a finite kpartitioned ordering and a is any order type, then F.o-,F.(o+[.a) for every n. The proof of this adapted version of Corollary 6.12 uses an adapted version of Exercise 6.1 1, where the function g(1,n) = 2” - 1 has to be replaced by a function h(k, n) defined inductively by h(k, 1) = g(k, 1) and h(k, n + 1) = max{2h(k,n) 1, h(k,n) g ( k , n) l}. (Note that in products A . B, as in generalized sums, the index set B is taken to be an ordinary linear ordering, so that A . B can be thought of as B copies of the partitioned ordering A.) We define a k-partitioned ordering to be finitary as before; that is, Ak is finitary if, for some n, whenever Bk is a countable k-partitioned ordering and Bk -” Ak, then Bk Ak. The following result is fundamental for the subsequent analysis. (We noted earlier that for arbitrary finitary linear orderings F, F . o need not be finitary.)
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LEMMA 13.126: Zf F is a jinite k-partitioned ordering, then F . o is finitary.
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Proof : Suppose that F has m elements. We claim that f ( F . u)2 2m 1. To prove this we must show that if B is any countable k-partitioned ordering and if B N~,,,+ F * o,then B F . o.But since 2m + 1 2 3, we know that the order type of B, ignoring its partition structure, is o 4 . a for some order type a (since the order type of F . o is 0).It thus suffices to show that the first w elements of B are actually F . o and that each (-sequence in B is actually F * 5, for then B 2: F . (o+ 5 * a), which by Corollary 6.12 (adapted) is G-equivalent to F . w .
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If PLAYER I chooses the first m elements of F . w, then he would win in the game G,, l(B,F . w ) unless B begins with an F. Proceeding inductively, assume that B begins with n copies of F. If PLAYER I chooses the m elements of the nth copy of F in B and then chooses the next m elements of B, then PLAYER 11 must pick 2m successive elements of F . w, lest PLAYER I win on his (2m 1)st turn. By the periodicity of F . w, the second group of m elements of F . w must be exactly like the first group, so the same must be true of the 2m elements of B ; hence B begins with n 1 copies of F, so, by induction, B begins with F . w. Similarly, for any copy of [ in B, PLAYER I can choose 2m - 1 consecutive elements in that [, and, since PLAYER 11’s selections in F * w must contain a copy of F , we conclude that PLAYER I’S selection from B also contains a copy of F. This copy of F propagates copies of F in both directions, as in the preceding paragraph. Thus each copy of 5 in B is actually F .[, so that B 2 F . (w 5 . a), as required.
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COROLLARY 13.127: If F , , F , , and F , are j n i t e k-partitioned orderings, than F , F , . w, F , . w* F,, and F , . w* F , F , . w arealljinitary.
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DEFINITION 13.128: A basic (finitary) k-partitioned ordering is a kpartitioned ordering that is in one of the following forms, where F , , F,, and F 3 are finite: F , , F , + F , . o , F 1 . w * + F , , F ; o * + F , + F , . w .
We use these notions in the following way.
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DEFINITION 13.129 : Define a condensation c, of a k-partitioned ordering A by stipulating that b c if every subinterval [x, y) of [b, c] is G,#-equivaIent to a finite k-partitioned ordering. (Recall that when condensations are under discussion, the notation [b, c] is to be interpreted as [c, b] if c < b.)
This is clearly a condensation map for each n. Furthermore, assuming that n 2 3, as we will do throughout this section, the order type of each subinterval [x, y) of an interval [b,c] within c,(a) must be either finite or of the form w + 6 . a + a*for some a. (If A F, where F is finite, then A has a first and last element, and all elements have immediate successors and predecessors.) Thus the order type of c,(a) is either finite or is of one of the following forms: w + 4 . a + w * , c . a + w * , w + 5 . a, 5 . a. The use of Ramsey’s Theorem in the proof below is reminiscent of the proof of Proposition 7.8.
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33 1
PROPOSITION 13.130 : Let Ak be a k-partitioned ordering. Then for each a E A, c,(a) is G,-equivalent to a basic (finitary)k-partitioned ordering. Proof: Suppose that c,(a) has no last element. Choose an o-sequence a = a, < a, < a, < . . . that is cofinal in c,(a). Then each interval [ a i , a j ) is G,-equivalent to a finite k-partitioned ordering. By Corollary 6.9 (adapted), we may assume that there are a finite number of finite k-partitioned orderings such that each [ai,aj)is G,-equivalent to precisely one of them. Thus we can partition [ N I 2 into a finite number of sets in such a way that {i,j} and {k,l) are in the same set if and only if i < j , k < 1, and [ a i , a j )w n [ak,a,). Applying Ramsey’s Theorem (Theorem 7.4), we can choose a,, < a,, < ai2< . . . so that, for each r, [u,,,u,,+~)is G,-equivalent to a fixed finite k-partitioned ordering F. Thus {b E c,(a)(b 2 a } -,[a, a,,,) + F . o. But [a, a,,) is G,equivalent to some finite G, so that { b E c,(a)lb 2 a} -,G F o. If c,(a) has P last element c, then { b E c,(a)l b 2 a } = [a, c], which is G,-equivalent to a finite k-partitioned ordering. Similarly, {b E c,(a) b < a } is G,-equivalent to either a finite k-partitioned ordering or to F , . o* G I for some finite F , and G,. Thus in every case c,(a) is G,-equivalent to a basic k-partitioned ordering. H
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Note that c,(a)is naturally G,-equivalent to a basic k-partitioned ordering in the following sense. DEFINITION 13.131 : We say that A is naturally G,-equivalent to F , F , . o,where F , and F , are finite, if A can be written in the form A = B c { B i l i < a}, where B -,F , and B, -,F , for all i. (The definition is analogous for other basic finitary k-partitioned orderings.)
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EXERCISE 13.132: Show that if A m 4 D, where D is a basic finitary k-partitioned ordering and q 2 f ( D ) . then A is naturally G,-equivalent to D.
Now, by Corollary 6.9 (adapted), there is a finite set of finite k-partitioned orderings such that each finite k-partitioned ordering is G,equivalent to one of these. Hence there is a finite set of basic k-partitioned orderings such that each basic k-partitioned ordering is naturally G,-equivalent to one of these. Thus we conclude that there is a fixed finite ordered sequence D o , D,, . . . ,D, - of distinct basic finitary k-partitioned orderings such that, for every Ak and every a E A, c,(u) is naturally G,-equivalent to some D,. [Note that it is possible that c,(a) is naturally G,-equivalent to more than one D,; to arrange for uniqueness seems troublesome and is unnecessary
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since, in the definition below, we insist on a uniform and optimal method of selecting a Di corresponding to each possible c,(a).] DEFINITION 13.133: Given a k-partitioned ordering Ak, we define, for each c,(a), a basic ordering D(c,(a))from the sequence Do, D,, . . . , D,by the following three-step procedure: First, select those Di such that c,(a) is naturally G,-equivalent to D i ; second, eliminate Dj if c,(a) w q D j but c,(a) -q Di for some Di and some q ; and third, let D(c,(a)) be the first remaining Di. We then define ( C , [ A ~ ]to) ~be the s-partitioned ordering ( c , [ A k ] ;Po,!‘,, . . . ,Pspl),where Pi = {c,,(a))~D(c,,(a)) = Di).
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Note that the third step in the procedure for defining D(c,(a))guarantees uniformity. That is, if c,(a) rr c,(b), then D(c,(a))= D(c,(b));consequently, if Ak Y Bk, then (c,[Ak])”‘v (c,[Bk]l,”.The second step is a condition of optimality; no Di is more similar to c,(a) than D(c,(a)).The first step, of course, guarantees that D(c,(a)) will be naturally G,-equivalent to c,(a). Following the program we outlined at the beginning of this section, we now want to show that for A E Ao,c,[A] is also in A. and has lower rank. Then, in trying to show that A is G,-equivalent to something finitary, we ] } argue inductively that A is G,can resolve A into ~ { c , ( a ) ~ c , [ Aand equivalent to a finitary sum of pieces each of which, by Proposition 13.1 30, is also finitary. Before we do so, we must of course define the set of k-partitioned orderings corresponding to .A?’, since that is where A really lives. At the same time, we must present the notion of rank that is most useful for our purposes. (As observed earlier, we have not introduced partitioned consisted of order types, d o will k consist order types, so that, although &lo of partitioned orderings.) DEFINITION 13.134: We will define, by induction on n, a class .MOk(n) of scattered k-partitioned orderings of rank I n . For n = 0, A o k ( 0 consists ) of all finite k-partitioned orderings. Assuming that A o k ( n )has been defined, we define A o k ( n + 1) to consist of all finite sums A I k+ A,k + . . . + A / , where each A: is either in .Aok(n)or is w or w* copies of an element of -&‘:(n). We define -A?‘: = U,A’,k(n). If Ak E A:, we define the rank of Ak to be the smallest n such that Ak E .Mok(n).
LEMMA 13.135: Let Ak E d oand k let 1 be an interval ofA. Then l kE dtok and the rank of’I k is no more than the rank of Ak. then Ak can be written as A I k+ A Z k+ . . + Amk,where (2) l f Ak E dok, each Aik is either jinite or is w or a* copies of a k-partitioned ordering of lower rank than Ak.
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Proof : By induction on the rank of Ak.
A
PROPOSITION 13.136: For each Ak E d o k and each n, ( c n [ A k ] )E” dos and has lower rank than Ak. Proof : We will first show that A k = A l k + . . . + Amkcan be rewritten as Ak = M I k . . . M,k, where for each i, either
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(i) c,(a) = M t for all a E M t , or (ii) M: consists of o or o*copies of something of lower rank than A , each of which contains c,(a) whenever it contains a.
We may assume, by Lemma 13.135.2, that each A,k is either finite or is o or o*copies of something of lower rank. So assume that AIk= Bi I i < o}, where each Bi v B and B has lower rank than A. Suppose that for some a, E B , and b , E B , we have c,(a,) = c,(b,). If a , is the element of B, corresponding (under the assumed isomorphism) to a, and if c,(a,) = c,(a,), then c,(a,) contains all of Atk. Otherwise b , < a,, so that if we set T o = ( a E Bola E c,(ao)} and I , = (a E B , l a E c,(a,)}, then c,?(a0)= T o + I , . Using the isomorphisms between the B i , we can write
I{
+ N o + T o )+ (I, + N , + T,) + (I, + N , + T 2 )+ . . . , where c,(a) = T i + I i + , for each a E Ti + Ii+, and for each i. Setting Ci = N i + T i+ I i + , for each i, we get A,k = I , + c { C i l i < a}.Note that AIk= ( I ,
each Nj, Ti, and I i is in &lok, so that each Cj E Aok. Also the rank of C iis no more than the maximum of the ranks of Ni,T i , and I i , each of which, by Lemma 13.135.1, is no more than the rank of B i , which is less than the rank of A . Finally, for a E I,, c,(a) 2 I,. Examining the various cases of the preceding paragraph, we see that A,k can be decomposed into at most two pieces, one of which is completely included in c,(a) for any of its elements a, and the other satisfies (ii). By decomposing each A,k in this way, we obtain a decomposition of Ak into at most 2rn pieces; by combining adjacent pieces in the same condensation class, we arrive at a decomposition Ak = M I k+ . . . + M,k that has the required properties. We now consider c,[M?] for each i. If c,(a) = M t for a E M i ,then ( C , [ M ; ] ) E~ ,Ao‘and has rank 0. Otherwise M t = x { M i j lj < o},where each M i j = M and is a union of condensation classes. But then ( c , [ M t ] ) ”= ~ ( ( c , [ M , , ] ) “< / jo}; by the induction hypothesis, each ( c , [ M i j ] f Shas rank smaller than M i j and is in A,“. By uniformity, (c,[Mij])” (cn[Mij.])sfor and its rank is smaller than everyj, j’ < (0.This implies that (c,[M,“])”E A?’,‘ that of MF. Thus, in any case, c,[M?] E A’,‘ and has rank smaller than M/‘.
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But c,[Ak] = c , [ M , ~ ] c , [ M ~ ~ ]. . . c,[M,k], so that c,[Ak] E dos and has rank equal to the maximum of the ranks of the c n [ M t ] ,which is smaller than the maximum of the ranks of the M b ; hence the rank of c n [ A k ] is smaller than the rank of Ak. We have thus defined, for each n, a condensation c, of each k-partitioned and we have shown that each c,(a) is finitary and that ordering Ak E AOk, cn[Ak]has lower rank than Ak.To prove that Ak is G,-equivalent to a finitary k-partitioned ordering, we will assume that c , [ A ] is G,-equivalent to a finitary s-partitioned ordering F and replace A = ~ { c , ( aE)c , [ A ] ) by a sum of D;s indexed by F.
DEFINITION 13.137 : Given an s-partitioned ordering F S = ( F ; F o , F , ,...,F 5 - l ) ,
we define the sum FDk to be the k-partitioned ordering ~ ( D “ ( a )E ~Fa} , where Dk(a)2: Di if a E Fi. The following lemma allows us to make the replacement described above. LEMMA 13.138: Assume that ( C ” [ A ~-, ] ) F“. ~ Then Ak -,FDk.
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Now all of the pieces are in place and we can proceed to the proof of the main theorem. THEOREM 13.139: (Amit, Schmerl, Shelah) Every Ak E .,dokis G,equivalent to a finitary k-partitioned ordering. Proof : We proceed by induction on the rank of Ak.If Ak has rank 0, then Ak is finite and there is nothing further to say. If Ak has positive rank, we apply c, to Ak and write Ak = ~ { c , ( aE)c , [ A k ] } . By Proposition 13.136. (c,[Ak])”E .A/ and has lower rank than Ak.Hence by the induction hypoth-
esis, (c,[Ak])”-,F S for some finitary s-partitioned ordering F’. Therefore, by Lemma 13.138, Ak -,FDk. Also, by Proposition 13.130, we can assume that each c,(a) is naturally G,-equivalent to D(c,(a)),which is on the list D o , D,, . . . ,D,- of basic finitary k-partitioned orderings. We thus need only show that FDk is finitary. We first show that, for each element d E FDk, c d d ) is precisely the copy of D j to which d belongs. Since each copy of D j has order type it is clear that each c,(d) includes the copy of D j to which it belongs. On the other hand, those D j that are included
<[,
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in c,(d) determine an interval of F”;if that interval is non-trivial, it must contain some element together with its successor since F” is scattered. Thus F” contains two successive elements, one in Fi and its successor in F j , and c,(d) 2 D , + D,, where D, and D, are consecutive copies of Di and D j in FDk. Since Fs -,( C , [ A ] ) ~we, conclude that c , [ A ] has successive elements c,(b,) and c,(b,), where c,(b,) and c,(b,) are naturally G,-equivalent to Di and D j , respectively. (To make this conclusion, we must assume that n 2 3, but the theorem for n 2 3 implies the same result for n < 3.) Now either c,(b,) has no last element or c,(b,) has no first element since otherwise c,(b,) = c,(b,). Hence either D, has no last element or D, has no first element. There are now three cases, which, of course, are treated similarly. Suppose that D, has no last element and that D , has no first element. Then D , ends in F , . w and D , begins with F , . w * . Since c,(b,) and c,(b,) are naturally G,-equivalent to D, and D,, respectively, we may assume that c,(b,) ends in [ c t , c, ,), where each [c,, c, + ,) -,F , , and that c,(b2) begins with x - r [ d - , - l , L I ) , where each [ L - , , d - J -,F,. But if x ~ c , ( b , )and Y E c,(b,), then
1,
[x, Y ) = [Ix,ci) +
-,,
1[ci+r c, 7
I
F
+ r t 1)
+ 1[ d j - t --f
+ F , . + F,
. O*
1, d j - t )
+ [dj, y )
+ G.
But F , . w + F , . w* is an interval of D, + D, and therefore is G,-equivalent to a finite ordering. Hence [ X , J J ) is G,-equivalent to something finite for every x E c,(b,) and y E c,(b2),so that c,(b,) = c,(b,), which is a clear contradiction. Hence for each d E FDk,c,(d) is precisely the copy of D j to which d belongs. Now by assumption, Fs is finitary; let p = f ( F s ) ) .Each Di is finitary; let q = max{{f(Di)(i< s} u { i i + 1 ) ) . We will show that ,f(FDk)5 Y, where r = p + q + n + 2. Assume that Bk -,FDk. We will show that if we write Bk = x { c , ( b ) E c , ~ [ B ~ ]then } , c,[Bk] - p F“ (and hence cfl[Bk] F”)), and that if D(c,(b))= Di [so that c,(b) O i l , then c,(b) m 4 Di [and hence c,(b) O i l . This will imply that Bk FDk. showing that FDk is finitary, and completing the proof of the theorem. If PLAYER I has a winning strategy in Gp(c,[Bk],FS), then PLAYER I will also have the following winning strategy in G,(Bk,FDk),contrary to assumption. For his first p moves, PLAYER I pretends he is pursuing his winning strategy in the game G,(c,[Bk], F’). choosing, at each of his turns, elements from different condensation classes. At some point, to avoid losing in p turns, PLAYER 11 must either choose a second element from some condensation class or choose, in response to PLAYER 1’s choice of x, an element y such that D(c,,(x))# D(c,(y)).In the first case, PLAYER I can win in n + 2 further moves-let x, < x2 be the two elements in the same condensation
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class, and let y 1 < y 2 be the corresponding elements in different classes. first chooses the end points z1 and z2 of a bounded interval within [y,,y,] that is not G,-equivalent to any finite k-partitioned ordering-which PLAYER 11 cannot duplicate within [xl, x,]-and demonstrates that difference in n more moves. In the second case, PLAYER I can win in q + n + 2 further moves-for if PLAYER 11’s choice x did not satisfy D(c,,(x))= D(c,( y)), where y was PLAYER 1’s choice, then PLAYER I, in his next q moves, demonstrates that fact and wins unless PLAYER 11 goes outside of c,(x) or c,(y), in which case PLAYER I wins in at most n + 2 further moves, as in the preceding case. Similarly, suppose that PLAYER I has the winning strategy in G,(c,(b),Di), where D(c,(b))= Di, so that c,(b) -,D ibut c,(b) N, Di.We claim that it follows that PLAYER I also has winning strategies in G,(c,(b),c,(d) ) for all d E FDk. For suppose, to the contrary, that c,(b) c,(d) for some d E FDk. Then, by Exercise 13.132, since c , ( d ) = D j for somej # i and since q 2 .f(Dj), it follows that c,(b) is naturally G,-equivalent to c,(d); since q 2 n + 1, this implies that D(c,(b)) could not have been Di. Hence PLAYER I has the winning strategy in G,(c,(b),c,(d))for all d E FDk. But then PLAYER I has the following winning strategy in G q + , +3(Bk,FDk), contrary to assumption. PLAYER I first picks an element b E B such that for no ti E FDk is c,(b) c,,(d).Assuming PLAYER 11 picks a particular element d E FDk, PLAYER I uses his winning strategy in G,(c,(b),c,(d)) to achieve victory in q more moves, unless PLAYER 11 picks, on one of his turns, an element outside of c,(b) or c,(d). Then, as in the argument above, PLAYER I wins in ra + 2 additional moves, so that he has the winning strategy in Gq+n+3(Bkr F D ~ ) contrary , to assumption, since q + n + 3 I r. PLAYER I
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The reader who is so inclined can replace the second half of the proof of Theorem 13.139 by an appeal to Corollary 13.12 and thereby obtain an axiomatization of FDk using the given axiomatizations of Fs and D o . D1, . . . ,DsCOROLLARY 13.140 : Every scattered linear ordering is G,-equivalent to a jinitary linear ordering, for every n. Proof : If A is scattered, then by Corollary 7.10 there is an M E A0such that A -,, M. By Theorem 13.139, M is G,-equivalent to a finitary linear ordering.
If # is a statement true in a scattered linear ordering and # has quantifier depth n, then 4 is also true in a finitary linear ordering that, by Proposition 13.19, is finitely axiomatizable.
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COROLLARY 13.141 : Any statenterit trite in a scattered linear ordering is true in a jinitely axiomatizable scuttered linear ordering. W
Before turning to the non-scattered case, we make several additional remarks concerning scattered linear orderings. COROLLARY 13.142: Let A be (1 scattered linear ordering that contains u copy ofo o*.Then A is not ,finitary.
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Proof: Assume that A is finitary and that f ( A ) = n. By Corollary 7.10, there is a B E A0such that A -,, B ; B is then finitary, f ( B ) I n and A B.
By Exercise 7.7.2, there is a number k that is the size of the largest maximal finite interval of B, k = max { Ic,(b)((b E B and (cF(b)lis finite}. Since A B, A has no larger maximal finite intervals. Also since A B,o o* occurs in B ; otherwise, PLAYER I could win the game G,+ 3(A,B), where p 2 n, k, by choosing the first and last elements of a copy of o o*in A. Now replace each copy of o o* in B by 2p to get C. Then, by Exercise 6.11, B - p C, so that B C . But since 2p occurs in C but not in B, B * p + 2 C . This contradicts the assumption that A is finitary.
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The proof of Corollary 13.142depends only on the statement of Corollary 7.10 and nowhere else uses the assumption that A is scattered. Thus, using Corollary 7.21, we can conclude immediately that no linear ordering containing a copy of w o*is finitary. After verifying the corollary and the above observations, Fellner [7] conjectured that the following converse is correct: Let A be a linear ordering of A. that does not contain o + o* copies of any linear ordering, except possibly <; then A is finitary. (To say that A contains o o* copies of B is to say that there is an interval I of A that can be written
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z=C{B,JiEW+[.a+o*},
where each Bi is G-equivalent to B, and that no interval of A containing I has this property.) Fellner [7] verified this conjecture for certain subclasses of .ko.A related conjecture [7] is that for finitary F , F . o is finitary if and only if F F is finitary. We return now to the general case, which, as we will see, can be handled in much the same manner. Our goal is to show that every linear ordering is G,-equivalent to some finitary linear ordering. As before, because of Corollary 7.21, we need only prove this for order types in A. We will define a class A!* of partitioned orderings (whose 1-partitioned orderings will have precisely the order types in A') and show, by induction on rank, that
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every k-partitioned ordering in A’* is G,-equivalent to a finitary k-partitioned ordering. The desired conclusion will then be a corollary. To carry out the main argument we will define a condensation c,* so that, as in the scattered case, (a) If A E A*,then c,*[A] E .,ti!* and has smaller rank than A. (b) Each c,*(a) is G,-equivalent to something finitary. Then we can resolve A into x(c,*(a) E c,*[A]], argue inductively that A is G,-equivalent to a finitary sum of pieces each of which is finitary, and then conclude that, because of the special character of these pieces, their sum is finitary. Because of the similarities to the scattered case, we will be brief. DEFINITION 13.143: We define, by induction on n, a class J k ( n ) of k-partitioned orderings. For n = 0, A k ( 0 )consists of all finite k-partitioned orderings. Assuming that A k ( n ) has been defined, we define A k ( n 1 ) to consist of all finite sums A I k Azk . . * Amk,where each A: is either in .Lk(n),or is o or w* copies of an element of A k ( n ) ,or is a finite shuffle c{B , , . . . , B r ; of elements of A k ( n ) . We define Ak= u , J k ( n ) , and we define A’* = U k A k . If Ak E Ak, we define the rank of Ak to be the smallest n such that Ak E , A k ( n ) .
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LEMMA 13.144: (1) Let Ak E = d kand let I be an interval of A . Then I kE .Ak and the rank of I kis no more than the rank of A‘. ( 2 ) i f A k E J k ,then Ak can be written as A l k A z k . . . Amk, where each A: is eitherjnite, or is w or w* copies of a k-partitioned ordering of lower rank than Ak, or is a j n i t e shufle of k-partitioned orderings of lower rank than Ak. A
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We will define the condensation c,* so that each c,*(a) is either precisely c,(a), and hence is G,-equivalent to a basic k-partitioned ordering, or else is G,-equivalent to a finite shuffle of basic k-partitioned orderings. The following lemma will thus imply that each c,*(a) is G,-equivalent to a finitary k-partitioned ordering. DEFINITION 13.145: An elementary k-partitioned ordering is either a basic k-partitioned ordering or a finite shuffle of basic k-partitioned orderings. LEMMA 13.146 : Let G = u { G G, , . . . ,G,} be an elementary k-partitioned ordering. Then G is @fury. A
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EXERCISE 13.147: Lemma 13.126.1
Prove Lemma 13.146. [Hint: Review the proof of
Referring back to Definition 13.129 and Proposition 13.130, we see that the condensation map c,, is well defined for arbitrary k-partitioned orderings and that for any element of an arbitrary k-partitioned ordering Ak,c,(u) is G,-equivalent to a basic (finitary) k-partitioned ordering, whether or not Ak is scattered. Thus we may assume that for every Ak and every a E A, c,(a) is G,-equivalent to some member of the list Do, D,, . . . ,Dsp1 of basic k-partitioned orderings. Define D(c,,(a)) for each a E Ak and then define (c,[Ak])” exactly as in Definition 13.133. The definitions below, which lead up to the definition of c,*(a), are reminiscent of Definitions 8.38 and 8.39. DEFINITION 13.148: An interval I of cn[A] of order type q or 1 is said to be homogeneous if there is a subset 9 c {Do,D1,.. . ,Ds,} such that if c,(a) E I , then D(c,(a)) E 9,and if, whenever c,(a) << c,(b) are elements of I , then, for each D j E $3, there is an element dj E A such that c,(a) << c,(dj) << c,(b) and D(c,(dj)) = Dj.Given a E A, there is a maximal homogeneous interval I, of c,,[A] that contains c,(a). We define c,*(a) = x{c,(b) E I,}. PROPOSITION 13.149: Let Ak be a k-partitioned ordering. Then for each a E A , c,*(a) is G,-equivalent to a n elementary jinitary k-partitioned ordering. Proof: By Proposition 13.130. c,(a) -,G j for some basic G j . If I, = {c,(u)), then c,*(a) = c,(a). If I , has order type q, then, by Exercise 7.18 (adapted), c,,*(a) is G,-equivalent to a finite shuffle of basic k-partitioned orderings, which, by Lemma 13.146, is an elementary finitary k-partitioned ordering.
We observe that if c,*(a) = Clc,,(b) E l a } ,then c,*(a) is naturally G,equivalent to an elementary k-partitioned ordering in the sense that, in the elementary ordering, each interval c,(b) of c,*(a) is replaced by a basic ordering to which it is naturally (2,-equivalent. We let D o , D,,. . . , D t - be a list of elementary k-partitioned orderings such that D o , D , , . . . ,D,- are basic (as before) and D,,D,+ .. .,Dtare all the finite shuffles of subsets of (Do,D1,. . . ,D,-l>.
,,
DEFINITION 13.150 : Given A k and having defined (c,[Ak])S, we associate with each c,*(a) an element D*(c,,*(a)) from the list Do,D1, . . . , D t - , in
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the following way. If I, is a singleton, then D*(c,*(a))= D(c,(a)). If I , is not a singleton and the subset of ( D o ,D , , . . . ,D,- associated with I , is 9, then D*(c,*(a))is a(9), which is among D,, D S c 1 , . . , D,- We define (c,*[Ak])*to be the t-partitioned ordering ( c , * [ A k ] ;B,,B,, . . . ,B,- ,), where B j = {c,*(a)lD*(c,*(a))= Dj).
,.
Then, as in Definition 13.133, this definition is uniform. That is, ifc,*(a) 2 c,*(b), then D*(c,*(a))= D*(c,*(b)); and consequently, if Ak rr Bk, then (C,*[Ak])f= (C,*[Bk])'. PROPOSITION 13.151 : For each Ak E ,Ak and each n, (c,*[Ak])'E At
and has lower rank than Ak.
Proof: As in the proof of Proposition 13.136, we need to show that A k = A + . . . + Amk can be rewritten Ak = M , + . . . + M,!, where, for each i, either c,*(a) = M f for all a E Mf,or M f is w or a*copies of something of lower rank than A (each copy of which contains c,(a) whenever it contains a), or M f is a finite shuffle of orderings of lower rank than Ak (each copy of which contains c,(a) whenever it contains a). Treating each A,k separately, and proceeding as in the proof of Proposition 13.136, we see that if Ajk is o or w* copies of something of lower rank, then the analysis is still correct. If, on the other hand, A; is a shuffle, then it is clear (by the remarks preceding Definition 13.150) that if b E c,*(a) and b is not in the same copy as a, then c,*(a) 2 A:, whereas otherwise Ajk already satisfies the desired condition. Having represented Ak properly, we now, as in Proposition 13.136, need only analyze c,*[Mjk] for each j , and in particular for those cases where Mjk is a finite shuffle (and c,*[Mjk] is not trivial). So suppose that Mjk = a ( B , , . . . ,B4) and that each copy of Bicontains c,*(a) whenever it contains a. Then cn*[M;] = t ~ ( c , * [ B ~. ]. ,. , c,*[B,]), and since each cn*[Bi] has lower rank than B i , it follows that c,,*[M:] has lower rank than Mjk. DEFINITION 13.152 : Given a t-partitioned ordering
F' = ( F ; F o , F , , . . . ,F,-,), we define the sum F*Dk to be the k-partitioned ordering z { D k ( a ) l aE F } , where Dk(a)= D iif a E Fi. LEMMA 13.153: Assume that (c,*[Ak])'-,F'. Then Ak -,F*Dk.
A
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THEOREM 13.154 : (Amit, Schmerl, Shelah) Every Ak E Mkis G,-equivalent to a finitary k-partitioned ordering. Proof: As in the proof of Theorem 13.139, the only fact needing to be verified is that F*Dk is finitary for a suitably chosen finitary F'. The first step, again, is to show that for each element d E F*Dk, c,*(d) is precisely the copy of D j to which d belongs. To do this, we first rewrite F*Dk as a sum of Di where i < s and, condensing these Di, obtain an s-partitioned ordering that we denote FS. We first show that for each d E F*Dk, c,(d) is precisely the copy of Di to which d belongs, where i < s. Otherwise, those D,included in c,(d) determine a non-trivial interval of FS. If that interval contains some element and its successor, we proceed exactly as in Theorem 13.139. Otherwise, it is dense, so by Theorem 7.17, there is a subinterval of that interval of FS to which there corresponds in F*Dk a shuffle of some subset of { D o , D , , . . . , D,-l}. If that shuffle, or some non-trivial part of it, is contained within some Dk(a) for some a E I', then, as in Theorem 13.139, we could argue that for certain distinct c,(a) and c,(b) in cn[Ak] we should have c,(a) = c,(b); otherwise we could argue that the homogeneous interval I,, for some a E A , was not really maximal. Thus, in any case, c,(d) is precisely the copy of Di to which d belongs, where i < s. It then follows that c,*(d) is precisely the copy of D j to which d belongs, where j < t. By assumption, F' is finitary; let p = f ( F ' ) . Each D j is finitary; let q = max((f(D,)li < t } u { n 1)). We will show that f ( F * D k )5 r, where r = maxip 2n 5, q n 3}. So assume that Bk .vr F*Dk. We will show that if we write Bk = x{c,,*(b)E c,*[Bk]}, then c,*[Bk] F' (and hence c,*[Bk] F r ) , and that if D(c,(b)) = D i [so that c,(b) -,Oil, then c,(b) Dj [and hence c,(b) Oil. This latter implies that if D*(c,*(b)) = D j , then c,*(b) D j , which, together with the above, yields Bk F*Dk, showing that F*Dk is finitary and completing the proof of the theorem. If PLAYER I has a winning strategy in Gp(c,*[Bk],F'), then PLAYER I will also have a winning strategy in G,, 2n+5(Bk, F*Dk), contrary to assumption. For his first p moves, PLAYER I pretends he is pursuing his winning strategy in the game G,(c,*[Bk], F')%choosing, at each of his turns, elements from different c,*-condensation classes. At some point, to avoid losing in p turns, PLAYER 11 must either choose a second element from some c,*condensation class or choose an element of a c,*-condensation class that is not G,-equivalent to the c,*-condensation class of PLAYER 1's last move. In the first case, there are two possibilities. Either the two elements are also in the same c,-condensation class, in which case PLAYER I wins in n 2 further moves, as in Theorem 13.139, or the c,*-condensation class of the two elements must be a shuffle whereas the c,*-condensation classes of the other two elements, since they are distinct, are not part of a similar shuffle.
+ +
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+ + +
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THE FIRST-ORDER THEORY OF LINEAR ORDERINGS
Then, in three more moves, PLAYER I can choose an element whose c,condensation class cannot be duplicated by PLAYER 11’s choice. Hence, as in Theorem 13.139, PLAYER I wins in at most 2n + 1 further moves. In the second case, where PLAYER 11 has chosen an element of a c,*-condensation class that is not G,-equivalent to the c,*-condensation class of PLAYER 1’s last move, there are again two possibilities. Either both of these c,*-equivalence classes are also c,-equivalence classes, in which case PLAYER I wins in 2n + 1 more moves, or at least one is a shuffle, in which case, by choosing an appropriate element of the shuffle, PLAYER I can force PLAYER 11 into the first-case situation and so win in at most 2n + 4 more moves. In this latter case, PLAYER I thus wins in p 2n + 5 moves. The proof that if D(c,(b)) = D i , then c,(b) -4 Di proceeds exactly as in Theorem 13.139, by showing that otherwise PLAYER I has a winning strategy in G,+,+3(Bk,F*Dk).This completes the proof.
+
=
COROLLARY 13.155: (Amit, Myers, Schmerl, Shelah) Every linear ordering is G,-equivalent to a jnitary linear ordering, for each n.
The corollary below can be interpreted as saying that the Boolean algebra of statements of T h ( 9 ) is atomic. COROLLARY 13.156: Any statement true in a linear ordering is also [rue in ajnitely axiomatizable linear ordering. REFERENCES Amit, R., and Shelah, S., The complete finitely axiomatized theories of order are dense, Israel J . Math. 23 (1976). 200-208.[MR 58, #5162] [2] Baldwin, J. T., Blass. A. R.. Glass. A. M. W., and Kueker, D W., A “natural” theory without a prime model. Algebra Ui~irersalis3 (1973). 152-155. [ 3 ] Doner, J., Mostowski, A,, and Tarski. A,, The elementary theory of well-ordering-a metamathematical study, Loyic Colloquium 77 (Proc. Cons. Wroclaw, 1977). I -54. Studies in Logic and Foundations of Mathematics 96, Amsterdam: North-Holland Publ.. 1978. [ M R 8Od: 03027a1 [4] Ehrenfeucht, A,, An application of games to the completeness problem for formalized theories, Fund. Math. 49 (1961), 129-141. [ M R 23, #3666] [5] Ehrenfeucht, A., Decidability of the theory of the linear ordering relation. Notices Amer. Math. Soc. 6 (19591. 556-38. [6] Engeler, E., A characterization of theories with isomorphic denumerable models, Notices Amer. Math. SOC.6 (1959), 161. [7] Fellner, S., Recursiveness and finite axiomatizability of linear orderings. Thesis, Rutgers University, 1976. [XI Lauchli, H., and Leonard, J., On the elementary theory of linear order, Fund. Moth. 59 (1966), 109-1 16. [ M R 33. #7258] [l]
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Morley, M., Categoricity in power., rrans. Amer. Math. Soc. 114 (1965), 514-538. [ M R 31, # 5 8 ] Morley, M., The number of countable models, J . Symbolic Logic 35 (1970), 14-18. [ M R 44,#5213] Mostowski, A,. and Tarski, A,, Arithmetically definable classes and types of well-ordered systems. Bull. Amer. Marh. Soc. 55 (1949), 65, 1192. Myers, D., The Boolean algebra of the theory of linear orders, Israel J . Math. 35 (1980), 234-256. [ M R 81i: 030381 r131 Robinson, A,, Introduction to Model Theory and to the Metamathematics of Algebra, Amsterdam: North-Holland, 1963. Rosenstein. J. G.. KO-categoricity of linear orderings, Fund. Math. 64 (1969), 1-5. [ M R 39. ~ 3 9 8 2 1 Rosenstein. J . G., Theories which are not KO-categorical,in Proceedings Summer School in Logicat b e d s f967. pp. 273 278. Berlin: Springer, 1968. Rosenthal, J. W . . Models of T ~ ( ( w "<)), , Nutre Dame J . Formal Logic 15 (1974), 122132. [ M R 49, 4'88531 Rubin, M., Theories of linear order, IsraelJ. Math. 17 (1974). 392-443. [ M R 50, #I8711 Ryll-Nardzewski. C., On the categoricity in power IN,, Bull. Acad. Polon. Sci. Ser. Sci. Math. Asrro. Phys. 7 (1959),545-548, Svenonius, L., KO-categoricityin first-order predicate calculus, Theoria 25 (1959), 82 -94. Tarski, A,. Grundziige des Systemenkalkiils, Fund. Math. 26 (1936), 301. Tarski, A,. and Vaught, R., Arithmetical extensions of relational systems, Compositio Math. 13(1957), 81-102. Vaught, R. L., Denumerable models of complete theories, in InJnitistic Methods (Symposium on Foundutions ofMathmiutic.i. Warsaw, 1959), 1961, 303-321. [ M R 32, #4011]