Chapter II Section 1 Ordinal Numbers

CHAP. I1 SEC. 11

ORDINAL NUMBERS

2101

CHAPTER II

In Chapter I we introduced the fundamental Godelian theory of semisets

TSS and the fundamental Godelian theory of sets TS. Wc also considered

some general models of these theories. Since we have not proved very many theorems in these theories, we have only been able to establish general properties of the models. In this Chapter we shall therefore define a numbcr of set-theoretical concepts and prove a number of theorems in TSS; the models will be dealt with in later chapters. We investigate ordinal numbers and well-orderings in Sect. 1, and equivalence and subvalence of sets in Sect. 2. Sect. 3 is devoted to a weak form of the axiom of choice and its consequences; in particular we consider the consequences for powers of sets. Sect. 4 is devoted to Boolean algebras and Sect. 5 to ordered sets.

SECTION 1

Ordinal numbers Throughout this and following sections we shall work within the theory of semisets TSS or one of its extensions. We first consider the theory of ordinal numbers due to von Neumann. Let us recall here the following remark of Godel:

2101. “The ordinal ct will be the class of all ordinals less than a. For instance, 0 = the null (empty) set, 1 = (O}, 2 = (0, 11, o = the set of all 89

2102

[CHAP. I1 SEC. 1

ORDINAL NUMBERS

integers, etc. In this way, the class of ordinals will be well ordered by the relation, so that a < p corresponds to a E p. Any ordinal will itself be well ordered by the erelation since an ordind i s a class of ordinals. Moreover, any element of an ordinal must be identical with the segment generated by itself, since this segment is the class of all smaller ordinals.”

2102. Reniarli. If R is a relation then x R y means the same as (x, y ) E R . 2103. DEFINITION (TSS). a) Arelation R is an ordering (Ordg(R)) if it has the following properties: (1) (Vx, y, z ) ( x R y & y R z

.

--f

xRz)

(2) (Vx) (1x R x )

(transitivity), (irreflexivity).

b) A relation R is a linear ordering (LOrdg ( R ) ) , if it has the following properties:

( I ) R is an ordering, (2) (VX,

~3

E

C ( R ) ) ( x R y v ~ R vx x

= y)

(trichofom~?)

2104. DEFINII‘ION (TSS). a) X is ordered (linearly ordered) by a relation R if the relation R n X 2 is an ordering (a linear ordering) and X c C ( R ) . b) A one-to-onc mapping F of X onto Y is called an isomorphism between X and Y w.r.t. the relations R and S if (Vx, y

E

X ) (xRy

= (F’x)S(F’y)).

2105. D E r l N l I ION (TSS). X is a n ordinal (Ord ( X ) ) if it has the following

properties:

a) x is complete, b) (VX, J E X ) (X E J’ v y E x v x = I)), c) (Vu c X ) ( u =I= 0 -,( 3 z E u ) ( z n u = 0)). 2106. Remurk. The predicate Ord ( X ) is obviously normal: it is easy to find a PUP-formula equivalent to Ord ( X ) .

2107. DEFINITION (TSS). x E On = Ord (x). On is the class of all ordinal numbers; ordinal numbers are the ordinals which are sets; we use lower-case letters from thc beginning of the Greek alphabet to denote ordinal numbers. 2108. LI:MMA (TSS).(a) Ord (0). 90

CHAP. I1 SEC. I ]

2112

ORDINAL NUMBERS

(b) If Ord ( X ) , Ord (Y), Ord (2) then

x $6 x ;

l ( X E Y& Y E x) ; l ( X E Y& Y E Z &

zEX) .

(c) If Ord ( X ) and y E X then Ord ( y ) . (d) If X , Yare ordinals, then X n Y is an ordinal. Proof. (a) is obvious. (b) If X is an ordinal and X E X then (X} c X ; putting u = {X} we obtain a contradiction with (c) in the definition of an ordinal. Similarly put u = ( X , Y}, u = {X, Z } in the remaining cases. (c) Suppose Ord (X) and y E X. Then y is complete. For suppose u E u E y , then, by (b) in the definition of an ordinal, u E y v u = y v y E u ; the last two cases contradict (b) of the present lemma and so u E y. If u, u E y then u, D E X since X is complete; hence u E u v u = D v u E u. Finally, if 0 ==! u -C y then u E X and therefore by (c) in the definition there is a z E u such that z n ti = 0. (d) is obvious. ,1109. LEMMA (TSS). Let Y be a non-empty real subclass of On. Then there exists z E Y such that z n Y = 0. Proof. Choose y E Yand set u = y n Y . If u = 0 we are finished If u 0 then u E y and thus there exists z E u such that z n ti = 0, i.e. z n y n Y = = 0. Since z E y we have z E y , so L n Y = 0.

+

2110. LEMMA (TSS). If X and Y are real ordinals and if Y is a proper subclass of X then Y E X. Proof. Since X - Y is non-empty there exists z E X - Y such that z n n ( X - Y) = 0; hence z E Y. It suffices to prove z = Y. Suppose z c Y. Then there exists z1 E Y - z such that z1 n (Y - z ) = 0; hence z1 c z . If rl = z then z E Y , a contradiction; thus z1 c z. Since z,z1 E X and X i s an ordinal we have either z1 E z or z E z , . But z E z1 implies z E zl, which contradicts z1 c z,and so z1 E z. Since z1 E Y we have z1 $6 Y - z , a contradict ion. 2111. LEMMA (TSS). If X and Y are different real ordinals then either X E Y o r YEX. Proof. Set Z = X n Y. Z is a real ordinal and 2 E X , Z E Y. If Z = X then X c Y; hence X c Y and so X E Y by Lemma 2110. Similarly if 2 = Y then Y E X . The remaining possibility Z c X & 2 c Y can be excluded; for by Lemma 21.10 it implies Z E X& 2 E Y, i.e. 2 E X n Y = 2 which contradicts Lemma 2108. 2112. LEMMA (TSS). Let Y be a non-empty real subclass of On. Then there is exactly one element z of Ysuch that z n Y = 0. 91

2113

ORDINAL NUMBERS

[CHAP. I1 SEC. I

Prooj”. By 2109 there is at least one such z . Now we prove the uniqueness. Let zl, z2 EY, z1 z 2 . z1 n Y = 0, z2 n Y = 0. Then by the preceding Lemma e.g. z1 E z2 so that z1 E z 2 n Y, a contradiction. Thus the lemma is proved.

+

2113. DEFINITION (TSS). Let Y be a non-empty real subclass of On. An element z of Y is called minimal ( z = Min (Y)) if z n Y = 0. 2114. LEMMA (TSS). The class On is linearly ordered by the erelation E. Proof. The irreflexivify of E on On (a4 a) follows from Lemma 2t08. For the transitivity suppose a E p and p E y ; then fl E y and hence a E y. The trichotomy follows from Lemma 2111. 211 5. LEMMA (TSS). (a) On is a n ordinal; it is a real class and is not a set (consequently On is not a semiset).

(b) If X is an ordinal and X is not a semiset then X

=

On.

Proof. (a) On is an ordinal by Lemma 2108(c), Lemma 2LlL and Lemma 2109. On is a real class by Metatheorem 1438. (For Ord ( X ) is a PUPformula.) On is not a set by Lemma 2108(b). (b) Let X be a n ordinal and suppose X is not a semiset. Then X E On and, for every ordinal number a, X $ a, i.e., for any a there is a p such that a E p E X . By the completeness of X every a is an element of X and so X = = On. 2116. Reinark We have seen that there are ordinals which are sets (the ordinal numbers) and that there is just one ordinal (the class On) which is not a semiset. Nothing is asserted about ordinals which are semisets but not sets; it can be shown by the ultraproduct model that the existence of such ordinals is consistent but we shall not be interested in them in the present book. 21 17. LEMMA (TSS). Let a be a n ordinal number. Then

(a) cx u (a} is a n ordinal number, (b) there is no

such that z

E

p& p E ( a u ( a } ) .

Proof. (a) Obvious. (b) Let z E p, p E cx u ( a } . Then either j? E a or p E (a}, i.e. a E fi it follows in either case that a E a - a contradiction. 92

j? = a. Since

CHAP. I1

SEC. I f

2124

ORDINAL NUMBERS

2118. DEFINITION (TSS). a

+ 1 = CL u (a}. (the successor of a).

2119. METADEFINITION. We define in TSS: l = O + l ,

2 = t + l ,

3=2+1

etc.

(TSS). If X is a class of ordinal numbers then its sum U ( X ) 2120. LEMMA is an ordinal. (Obvious.)

2121. DEFINITION (TSS).For X c On define Sup ( X ) = U ( X ) (the supremum of X )

If Sup (X) E X , we write Max ( X ) instead of Sup ( X ) (the maximum of X). 2122. LEMMA (TSS). The supremum of a set of ordinal numbers is an ordinal number. 2123. THEOREM (TSS). (The principle of the Transfinite Induction.) Let X be a real class satisfying the following conditions: (a) 0 EX, (b) (Va) (a E X

-+

(c) ( v x ) (x E x

(a

-+

+ 1) EX),

s u p (x) EX).

Then X contains all ordinal numbers. Proof. Suppose that not all ordinal numbers are in X. Then On - X is a non-void real class of ordinal numbers; let z = Min (On - X ) . We will prove z E X , a contradiction. Obviously z E On n X. Set z1 = Sup (z). By (c) z1 E X . If z = z1 we are finished. If z E z1 then z G z1 and z1 2 z ; hence z E z , a contradiction. If z1 E z set z 2 = z1 1. By (b) z 2 E X . We claim that z 2 = z. Otherwise we would have z 2 E z. Then z1 E z2 E z and hence z1 E Sup ( z ) = zl, a contradiction.

+

2124. THEOREM (TSS). (Construction by Transfinite Recursion.) If G is a real function (i.e., a function and a real class) then there i s exactly one real function F on the class On with the following property: (va) (F’a = G‘(F”a)) .

Proof. Lct K be the class of all set functionsf such that D (f)is an ordinal number andf’a = G’(f”a) for each U E D(f). It is easy to check that the 93

2125

ORDINAL NUMBERS

[CHAP. 11 SEC. 1

condition defining K can be written as a PUP-formula, hence K is a real class. Now we prove that, for an] f , g E K , a E D ( f ) n D ( 9 ) implies f ’ a = = g’a. Assume f , g E K and f ’ a += g’a for some a E D ( f ) n D (9). Let Y = { a E D ( f ) n D ( g ) ; f ’ a =k g’a}. Y is real by Metatheorem 1438. Hence, there is a least a E D ( f ) n D (9) with f ’ a =Ig’a. = Then f ” a = g”a and so f ’ a = G ( f ” a ) = G’(g”a) = g’a, a contradiction. Set F = u ( K ) ; obviously, F is a function and F’a = G’F”a whenever O L ED (F). We prove that F is not a set. If F is a set then D ( F ) is a set by Axiom (A4) and therefore D(F) is an ordinal number. Set f = F, y = D ( j ) and g = f u ((G’(f”y), y ) ) . Then f , g E K , D (9) = y + 1, hence y E D ( j ) = y which contradicts 2108. Hence F is not a set and consequently K is not a set by 1427. We now prove that D(F) is not a semiset. S e t X = { ( f , a ) ; f ~ K & & D(f) = a). Obviously X is a one-to-one mapping, D(X) = D(F) and W ( X ) = K . If D ( F ) is a semiset then K is a semiset and therefore a set since K is real. But we proved that K isnot a set and so D (F) is not a semiset. Furthermore, F is not a semiset since Sm ( F ) -+ Sm (D (F)). It follows by 1440 that F is real and by 2115 that D (F) = On. The proof of uniqueness is straightforward. Remark. I n the course of the above proof we gave the definition of F in terms of G. Hence we are justified stating the following matemathical version of Construction by Transfinite Recursion:

2125. METATHEOREM. There is a godelian term F ( G ) with one class variable G such that TSS

t Real (G) & Un (G) =

. + . Real (F (G)) & Un (F ( G ) )& D (F ( G ) )=

On& (VLY) ( F (F)’ a = G’(F (G)” a)) .

Demonstration. The formula Un ( f ) & D (f)E On & (Va) ( f ’ a = G’(f”a)) is normal in TSS, i.e. there is a N F q(f,G) equivalent to the former one in TSS. Hence, by Corollary 1122 there is a godelian term K (G) such that T S S I- K (G) = ( f ; Un (f)& D ( f ) E On& (Va E D ( f ) ) (f’a= G’(f”a))}. The formula x E U(K (G)) is equivalent to (If) (q(f,G ) & x ~ f in) T S S and is therefore normal; hence there is a godelian term F (G) such that TSS t F (G) = U(K (G)). In the same way as Theorem 2124 was proved we can prove in TSS: If G is a real function then F(G) is a real function, D ( F (G)) = On and F (G)’ a = G’(F (G)” a) for every a.

2126. DEFINITION (TSS). An ordinal number a is called isolated if + 1 = a.An ordinal number which is not

a = 0 or if there is a p such that

94

CHAP. I1 SEC. 11

3129

ORDINAL NUMBERS

isolated is called a Zimif number. The classes of all isolated and all Iimit numbers are denoted by On, and On,, respectively. 2127. THEOREM (TSS). Both On, and Onll are real classes and proper classes, i.e. they are not sets. Proof. For any x we have x E On, iff Ord (x)& (x = 0 v (3y E x) (x = = y u ( y } ) ) . The last formula is evidently equivalent to a PUP-formula, hence, by Metatheorem 1438 On, is real. Since On,, = On - On,, OnlI is also real. Suppose that the isolated numbers form a set and let y = Sup(OnI). Then y + 1 is again isolated, a contradiction. Suppose that the limit numbers form a set and let y = Sup (On,,). Then for each nonempty set x of ordinals, x E On - y implies Sup (x) E x and consequently x has a maximal element. Let a be a set satisfying the axiom C l of infinity, i.e. suppose that 0 E a and that (x} E a whenever x E a. Let X be the class of all functions f such that D (f)= B - y for some fi and such that f’y = 0 and f ’(a 1) = { f ’ a } for each a such that a + 1 E D ( j ) . It is easy to write the condition defining X as a PUP-formula; hence X is real. We first prove that every f E X is one-one and has its values in a. Let a be the least element of D (f)such that f ’a = f ’p for some P E a.There exist a, and Po such that a = a, + 1 and j? = Po 1. Then f ’a, = f’P,, a contradiction. Hence f is one-one. Suppose that W (f) is not included in a and let a be the least element of D (f)which is not in a. Then f ’ a = {f ’ao}, a = a, + 1, f ’ a , E a and hence f ’ a E a, a contradiction. We can prove in a similar manner that if D(f) c D (9) then f E 9, whenver f and g are in X. Set F = U(X). We prove that F is not a set. Suppose that F is a set and let a, be the maximal element of the domain of F. Set a = a, + 1, g = = F u (({F’a,], a>}. Then we have g E X and D (9) = a 1, a contradiction. Hence F is not a set and consequently X is not a set. Since X is real X is not a semiset. Analogously to the proof of 2124 we prove that neither F nor D (F) are semisets. But then F is real by 1440. F is one-to-one and its range is a set. Hence F is itself a set, contradiction.

+

+

+

2128. DEFINITION (TSS). The least limit number is denoted by o (or

0,).The elements of

m, n etc.

o are natural numbers and are denoted by the variables

2129. THEOREM (TSS). (The induction principle.)

If a real class X contains 0 and with each natural number n also its successor n + 1, then X contains all natural numbers. 95

2130

[CHAP. I1 SEC. 1

ORDINAL NUMBERS

Proof. Suppose that o - X 9 0 and let n be the least natural number not in X . The number n is a successor of a number m E X and hence is itself in X , a contradiction. The following theorem is a variant of Theorem 2124: 2130. THEOREM (TSS). If GI, G , are real mappings and a is a set then there is exactly one real mapping F such that

t 1)

F’O = a ,

(2)

F’(a

(3)

F’M = G;(F”a)

+ 1) = G;(F‘a), for M limit

,

The proof is analogous to the proof of Theorem 2124. We define K as the class of all functions f such that D ( f ) E On and f fulfils (L), (2), (3) for any a for which these equations make sense. K is real (because it is defined by a PUP-formula), u ( K ) is real by Lemma 1440 and F = u ( K )fulfils (I) -(3). 2131. COROLLARY (TSS). (Construction by Recursion.) Let a be a set and let G be a real mapping. Then there is exactly one mappingfsuch that D (f)= o , f ’ O = a andf’(n + 1) = G’(f’n)for each n. (Set G I = G , and let G , be an arbitrary real mapping. (Take f = F o where F fulfils (1)-(3). f is a set because F is a real mapping.) The reader may formulate Metatheorems analogous to Metatheorem 2125 which express the fact that we may “describe” F in terms GI, G 2 ,a (fin terms of G and a) by means of godelian operations. In particular define:

r

2132. DEFINITION (TSS).

Roughly speaking Unv (x) is the “infinite union” x u u(x) u uU(x) u ... It follows from the definition of Unv(x) that T S S t (Unv(x) is a set). (Unv(x) is the union of the domain of values of some set function f x . ) The following definition is a generalization of Definition 2132: 2133. DEFINITION (TSS). If R is a real regular relation then

Unv, (x) = u ( W ( f ) ) where f ’ 0 96

=

EXtR (x), f ’ ( n

+ 1.) = R”(f’n) .

CHAP. I1 SEC. 11

ORDINAL NUMBERS

2137

The definition is justified by the following easy 2134. LEMMA(TSS). If R is a real regular relation then, for every x, ExtR (x) is a set. Proof. Clearly Ext, ( x ) = W (R n (V x {x})); R n (V x {x}) is real and it is a semiset (indeed, ExtR(x) is a semiset and if Ext, ( x ) c a then R n (V x {XI) c a x { x } ) , hence R n (V x {x}) is a set and therefore W ( R n (V x {x})) is a set. Note that T S S k Unv (x) = Unv, (x). 2135. Remark. The first place in this Section where axiom C1. was used is the proof of 2127. Hence if we denote by CLbiSthe assumption On,, 0 ("limit numbers exist") and by TSSbi"the theory T S S with C1 replaced by CIbis,then we can define o in TSSbi"and prove Theorem 2131. Hence in TSSbiSdefine f'0 = 0, f ' ( n + 1) = { f ' n } , a = W (f);it follows in TSSbis that 0 E a and that ( y } E a for any y E a . This means that C1 is provable in TSSbiS.Consequently, C1 can be equivalently replaced by Ctbi" in the axioms of TSS.

+

In the rest of this Section, we shall consider the theory TSS' (i.e. TSS + + Dl). Recall Metatheorem 1453 which enables to prove many classes to

be real. We shall now investigate well orderings.

2136. DEFINITION (TSS'). A relation R is a well-ordering (WOrdg(R)) if it has the following properties:

(0) R is a real class, (L) R is a linear ordering, (2) each nonempty real subclass of C ( R ) has a least element, i.e. (VX) [0

+X

c C (R) & Real ( X ) . -+

(3x E X ) [ X n Ext, (x) = 0 ] ] .

2137. DEFINITION (TSS'). A relation R is called a regular well-ordering (RWOrdg (R)) if it has the following properties: (0) (1) (2) (3)

R is a real class, R is regular, R is a linear ordering, each nonempty subset of C ( R ) has a least element.

A real class P is well-ordered (regularly well-ordered) by a real relation R if P G C (R) and R n P z is a well-ordering (a regular well-ordering). 97

2138

ORDINAL NUMBERS

[CHAP. I1 SEC 1

2138. LEMMA (TSS’). A real relation R is a regular well ordering if and only if it is regular and a well ordering. Proof. A real regular relation which is a well ordering is obviously a regular well ordering; conversely, let R be a regular well ordering. Let X be a nonempty real subclass of its field. To find the least element of X , take an arbitrary z E X and set u = X n Ext, ( 2 ) . If u = 0 we are finished; otherwise, take the least element x of u. We assert that X n Ext, ( x ) = 0. Assume the contrary and take y E X n Ext, ( x ) . We have y R x , x R z whence y R z . Since y E u we have u n Ext, ( x ) =k 0, which contradicts to the minimalify of x in u and our assertion is proved.

2139. Remark. The formula RWOrdg(R) is normal in TSS’. 2140. THEOREM (TSS‘). The class of all ordinal numbers is regularly well ordered by E (the erelation). Proof. The theorem is an immediate consequence of preceding lemmas. 2141. THEOREM (TSS‘). Let P , R be real and suppose that P is regularly well ordered by R . Then there exists a unique real ordinal X and a unique real isomorphism F between P and X w.r.t. R and E. Proof. Let P be regularly well ordered by R and suppose, for simplicity, that 0 4 P. First we shall prove the existence of an ordinal X and a n isomorphism F. Define the function G as follows: G’x = y if y is the least element of P - x, and G’x = 0 if P - x = 0. By Theorem 2124 there is a real function F such that F’a = G’(F”a) for each a. Hence F’a is the least element of P - (F’a). Set X = { a ; F’a + 01. Then F X is the required isomorphism. To prove the uniqueness suppose first that there are two real isomorpbisms F , and F , between P and an ordinal X . Let a be the least ordinal in X such that F ; a Fia. Then F;P = FiP whenever P E E . By trichotomy we have e.g. ( F ; a , F i a ) E R . Then there exists y E X such that F;a = F i y . Obviously y E a and hence F ; y = F i y . It follows that F;a = F;y, a contradiction. Now suppose that there is a real isomorphism F , between P and XI and a real isomorphism F , between P and X , where X , and X , are different ordinals. We have e.g. X i EX^. Define the function G on X , as f(>llcws: G’a = B = F;/I = F i a . G is obviously an isomorphism between X , and X , (w.r.t. E). Since G’X, EX^, we may denote by y the least elen-ent cf X , such that G’y E y. We have G’(G’7)E G’y and so by the minirrality of y, either y E G’y or y = G’y, contradicting G’y E y.

r

=+

98

CHAP. I1 SEC. 11

2145

ORDINAL NUMBERS

We conclude our treatment of ordinal numbers with some metarnathematical results concerning the possibility of definitions by induction in TSS’ even in the case where the arguments are proper classes. Although we cannot deal with ordinal functions which assume proper classes as values, we can employ relations such that the extensions of ordinal numbers (e.g.) are proper classes with certain properties. 2242. METADEFINITION. Let T be a theory stronger than TC. Let X be a variable. A term T(X) is local in T if TWX)=UT(y) YCX

(i.e. if we can prove that u E T ( X ) iff there is a subset y of X such that u E E

T (Y)>*

Note that if T ( X ) is local in T then T t X and Yare variables of the same sort).

c Y -+ T ( X ) c T ( Y ) (where X

2143. METALEMMA. A term T ( X ) is local in T iff the following is provable in T: T(X) =

u T(Y)

YEF

whenever F E P ( X ) contains for each x c X a superset y called a S-coJinaZ system of subsets of X ) .

zx ( F r a y

be

2244. METADEFINITION. Let T be a theory stronger than TSS’. A godelian term T (X) of T is iterable in T if (a) T ( X ) has the set images property (i.e. T t T (x) is a set), (b) T (X) is local. 2145. METATHEOREM. Let T be a stronger than TSS’. Let T(X) be a godelian term of T iterable in T. Then there is a godelian term S ( X ) (called the class iteration of T(X)) such that the following is provable in T: If X =# 0 is real and H = S ( X ) then H is a real relation, D ( H ) is an ordinal, H”{O} = X and H”{a} = T (H”a) for every a > 0.

Demonstration. PutK(X) = { r ; Re1 ( r ) & D(r) E On&(3z c X ) ( r ” { O } = = z) & (Va > O)(a E D(r) --f r”{a} = T(r”a)). The formula defining K(X) is normal (see 1126), hence the godelian term K(X) can be constructed following 1122. Furthermore, by Metatheorem 1454 T !- Real (X) 3 99

2146

[CHAP. I1 SEC. 1

ORDINAL NUMBERS

-+ Real (K ( X ) ) . Put S (X) = U ( K (X)); we show that S ( X ) has the desired properties. We proceed in T. Denote S ( X ) by H . Then H is real relation and D ( H ) is an ordinal. Evidently H”{O) = X. Further, H”(a} = U r”{a}

and H”a

T(ur”a)

=

u

r”a.

rsK(X)

=

We

have

H”{a} =

u

rcK( X)

=

U T (r”a) r

reK(X)

and

T(H”cr). If we prove UT(r”a) = T ( u r ” a ) we have H”{a) r

r

r”{a}

=

r

T(H”a). It suffices to prove that ( F a ; r E K(X)} is a n G-cofinal system in Ur”cr. Let q 5 u r ” a ; we find ro E K(X) such that q c rza. Define =

r

r

( r , u ) E Q = . u E q & r E EC (X) & u E P a . Q is a relation and D ( Q ) is a set; by (Dl), there is a semiset relation e E Q such that D(Q) = D(Q). Put W (e) = u; u is a subsemiset of K(X). Let u s a ; we can suppose that a C E K ( X ) (since K ( X ) is real). U ( a ) is a relation and U(a)” ( 0 ) is a subset xo of X. There exists ro E K ( X ) such that ro”{O} = xo and 0 < /3 5 a -+ r y { p } = = T ( r Y P ) ; it follows that r”a c rycc for any r E a. (Prove r, s E K(X)& & cr E D ( r ) n D ( s ) & r”{O} C s”(O} . -+ r”{a} C s ” { a } by transfinite induction.) Therefore q c ro”a. 2146. METATHEOREM. Let T (X) be iterable in T where T is stronger than TSS‘ and let O(X) be the range of the class iteration of T(X). Then the following is provable in T: For every real X, Z = 0 ( X ) is the least real

Z and T(Z) c Z . class with the properties X Demonstration. We use the denotation from the proof of the preceding Metatheorem. In T we have Z = W(H), T(W(H))

=

U T(r”cr) = U

reK(X)

rsK(X)

r”(a} E W ( H ) ;

0 < acOo

aeon

furthermore, it is obvious that X E 2. If W is real, X C Wand T ( W ) c W, then by induction H ” { M } E W for each a; hence Z -C W. 2147. Example. Define H as follows (for X real):

H”fO}

H”{n and let

=

x,

+ I > = U(H”(n + 1))

for n E o

Unv ( X ) = W ( H ) .

Verify that the term Unv(X) is definable in TSS’ by a normal formula, that in case X is a set this term coincides with the previously defined term Unv (X) (cf. Definition 2132), and that TSS‘ k Comp (Unv ( X ) ) .

CHAP. I1 SEC. 21

EQUIVALENCE AND SUBVALENCE

2201

SECTION 2

Equivalence and Subvalence of Sets. Cardinal Numbers

Throughout the present Section we shall work within the theory TSS’. We shall define equivalence of sets (having the same power) and subvalence (having smaller than or equal power). We will not define “the power of x” for arbitrary sets x, because without assuming the axiom of choice or some stronger axiom of regularity we cannot define an operation assigning to each set its “power”. However we shall prove in TSS‘ a number of useful statements concerning cardinalities which can be applied in those theories where the axiom of choice is not assumed (or even does not hold), namely a number of statements on finite sets. On the other hand, in TSS’ we can define cardinal numbers as powers of well-orderable sets and prove a number of statements about them. The concept of equivalence will be defined for classes in general (not only for sets). Equivalence of two objects (sets or classes) usually means the existence of a one-one mapping of one object onto the other. But observe that - at least as far as sets are concerned - we have two possible ways to define equivalence. Given two sets x and y we may ask whether there is a set f which is a one-one mapping of x onto y or whether there is a class with this property. (It must necessarily be a semiset.) More generally, for real classes X and X we may ask whether there is a real class which is a one-one mapping of X onto Y or whether there is simply a class with this property. This gives the following definitions: 2201. DEFINITION (TSS’). (1) Let X , Y be real classes. X is equivalent to Y (or X has the same power as Y, X E Y ) if there is a real class which is a oneone mapping of X onto Y. X is subualent to Y ( X Y) if there is a real class which is a one-one mapping of X into Y. X is strictly subualent to Y ( X < Y) if X Yand not X w Y.

<

101

2202

EQUIVALENCE AND SUBVALENCE

[CHAP. I1 SEC. 2

(2) Let X , Y be arbitrary classes. X is absolutely equivalent to Y ( X & Y ) if there is a class which is a one-one mapping of X onto Y. X is absolutely suhoalent to Y ( X Y ) if there is a class which is a one-one mapping of X into Y. X is absolutely strictly subvalent to YifX Yand not X & Y.

3

3

2202. Retilurk. Sets x and y are equivalent iff there is a set which is a oneone mapping of x onto J’; x is absolutely equivalent to y iff there is a semiset which i s a one-one mapping of .x onto JJ. Similarly for subvalence. Obviously, X 2 Y -+ X 2 Y and X Y+X Y are provable in TSS‘; further, the reflexivity, symmetry and transitivity of = are provable in TSS’; i.e. X M X , X = Y - + Y = X . X M Y & Y w Z . + X M Z for real X , Y, 2 ; similarly for & instead of = (and arbitrary X , Y , Z ) . Also X X and X Y & Y Z . -+ X Z are provable in TSS‘ and similarly for 4 instead of

<

4

<

<

<

< <.

2203. DIIFIKITION (TSS’). A relation R is an equivalence if the following hold for all x. J’. z E C ( R ) : (s,s) E R ,

(1)

(2)

(I, J) E

R

-+ (y, x) E

(x. J,) E R & (J’. z )

(3)

E

R

.

R

. --+

(x, z }

E

R.

Evidently. the relation Eq defined by (x, y) E Eq E x w y is an equivalence. (Note that the formula x w J’ is normal i n TSS‘.) In the present section we shall be interested in the notions of equivalence and wbvalence (=, 4 ) ;we shall deal with absolute equivalence and absolute wbvalence later. For the moment we shall only show that the Cantor-Bernstein theorem holds for both notions. 2204. TIII-OKEM (TSS‘). (a) X 1‘.

c1a\\e\

(b)

x,

x 3 Y&

Y

3 x . -+ x

< Y& Y < X .

4

X w Y for any real

& Y for any classes X , Y.

Proof. Given one-one mappings F of X into Yand G of Yinto X we construct a one-one mapping H of X onto Y; then we show that H is real provided that F and G are. Obviously it is sufficient to construct a one-one mapping o f X onto G”Y which is real provided that F and G are. Let K bo the class I02

EQUIVALENCE AND SUBVALENCE

CHAP. I1 SEC. 21

2208

of all one-one mappings f such that D ( f ) is a natural number

(1)

(2) (3)

1‘0E X (VH)

(PI

+ 1 E D(f)

.

- G”Y,

-i f’(i1

+ I.)

= G’(F’(f’n)).

For any f , g E K , n + t E D ( f ) n D(g) implies f ’ ( n + 1 ) = g’(n + 1 ) = = f ’ n = g’n. Define X E W = ( 3 f ~ K ) ( 3 n ) ( =x j ’ n ) ; S = X - W ( = = G”Y - W). Note that if F and G are real then K , S and Ware also real. Now define H’x = G’(F’x) for x E W, H’x = x for x E S. We claim that H is the required mapping. First, H is real if F , G are. Further, H f S is a oneone mapping of S onto S and H f W is a one-one mapping of W onto (G”Y - S); hence H = (H W ) u ( H S) is a one-one mapping of X onto G”Y, q.e.d.

r

r

2205. LEMMA (TSS’). Every set x is strictly subvalent to P ( x ) . Proof. The mapping f on x which assigns to each y E X the set {J’} is one-one and into P ( x ) . Hence x P ( x ) . Suppose now that x M P ( x ) . Let f be a one-one mapping of x onto P ( x ) . Consider the set a of all J’ E x which do not belong to their images, i.e. y 4 f ’ y . Since a E P (x) there is a y, E x such that a = f’y,. If y o E a , then yo $f’y,, a contradiction. If y o 6 a , then yo ~ f ’ y ,= a , a contradiction. Hence x x P ( x ) .

<

2206. DEI-INITION (TSS’). X is a cardinal (Card (X)) if it is a real ordinal and is not equivalent to any of its elements. Ordinal numbers which are cardinals are called cardinal numbers. The class of all cardinal numbers is denoted by Cn. 2207. LEMMA (TSS’). The class On is a cardinal. Proof. On cannot be equivalent to any ordinal number a,since it is a proper class. 2208. LEMMA (TSS’). The class Cn of all cardinal numbers is a proper class. Proof. Suppose the contrary and set t = Sup (Cn). Clearly ( is equivalent to each a > 5. Let the function F assign to each a > ( the set of ordering5 of 5 which are isomorphic to the natural ordering of a (by E).The function F is defined for every a > ( and assumes different values at different points, 103

220Y

[CHAP. I1 SEC. 2

EQUIVALENCE AND SUBVALENCE

i.e. is one-one. Since all of its values F’a belong to P (P (c x c)), the range of F is a set. Consequently the domain of F is a set too, a contradiction. 2209. LEMMA(TSS’).If X is a real class which can be regularly wellordered then X is equivalent to a cardinal, and this cardinal is unique. Proof. By Theorem 2141 every real class X which can be regularly wellordered js equivalent to an ordinal. Each ordinal is equivalent to a cardinal. The uniqueness is obvious. 2210. LEMMA(TSS’). The supremum of any real class X of cardinal numbers is a cardinal. Proqf. The assertion holds whenever X is a proper class since then Sup X = = On. Let X be a set and let SupX $ X . Suppose that there is a B E SupX equivalent to Sup X . Then there is a y E X such that p E y E Sup X . We have then p z y by 2204, since y is a cardinal number, this is a contradiction. 2211. LEMMA(TSS’). x c FI + x < n . Proof. By induction. Let n be the least natural number equivalent to a proper subset, let x be such a subset and let f be a one-one mapping of n onto x. Since n 0 let n = no + 1. The mapping f no is one-one and onto x - { ~ ‘ F I , } If . f ’ n , = no or if no is not in the range o f f , then x {f’n,} c no, which contradicts the induction hypothesis. If no = f’i and i E ti,, define f on no as follows: f y =f ’ n , if j = i and f? = f ’ j otherwise. The function f is then a one-one mapping of no onto a proper subset, a contradiction.

+

2212.

THEOREM

r

(TSS’). a) Every natural number is a cardinal number.

b) The class of all infinite cardinal numbers is isomorphic to On. Proof. a) follows from 2211. because m E n 4 m c n and hence m i.e. each n is a cardinal number. b) follows from 2208, 2141, 2115.

< n,

221 3. DEFINITION (TSS’). The unique enumeration of infinite cardinal numbers by ordinal numbers is denoted by N. The a-th infinite cardinal number is denoted by Nu or 0,. 2214. Remark. Both o, and H, denote the same operation on ordinal numbers. The former notation is used if the ordinal number o, is involved, whereas the latter one is used of the cardinality of the number o,is under consideration; however, both symbols may be used interchangeably. 2215. LEMMA (TSS’).Each o,is a limit ordinal number. 104

CHAP. I1 SEC. 21

EQUIVALENCE AND SUBVALENCE

2219

Proof. Suppose w, = p + 1. Define f as follows: f’y = 0 if y = p , f’y = y + 1 if y E w and f’y = y otherwise. Then f is a one-one mapping of o,onto 8, so that w, x j?,a contradiction. Tn what follows we investigate the properties of finite sets. 2216. DEFINITION (TSS’). A set x i s j n i t e (in the sense of Tarski), if every non-void subset of the power-class of x has a maximal element with respect to inclusion, i.e.

A set x is countable

if

x x

No.

2217. LEMMA (TSS’). (a) A singleton is finite. (b) The sum of two finite sets is finite. Proof. (a) Obvious. (b) Let y be a non-void subset of P ( x , u x,). Set u E y, = ( 3 v ~ y(u ) = u n xl). If y, = (0) then y contains only subsets of x, and hence has a maximal element. Tf y, =l= (0) then there exists z1 E y, such that (Vu E y,) ( l z , c u). Set u ~ y -, ( + ~ y ) ( u = u n x 2 & z l = u n x , ) .

If y, = { 0 } ,then z1 is maximal in y, i.e. (Vu E y) (1 z, c u) . If y, =I= (0) take a maximal z2 in y, and set z = z, u z,; then z is maximal in y. (TSS’). (a) If x1 is subvalent to x2 and x, is finite then x, 2218. LEMMA is also finite. (b) A set which is equivalent to a finite set is itself finite. Proof. (a) Letf be a one-one mapping of x1 into x2 and let y be a non-void set of subsets of x,. Set 7 = {f”z; z E y}. Obviously 7 E P (x,) and 7 =k 0 and so J has a maximal element z. Then the inverse image of z under f is maximal in y. (b) follows from (a).

2219. LEMMA (TSS’). (a) Each natural number is finite. (b) No is infinite. Proof. (a) Denote by F the class of all finite sets. F contains the void set and, by 2217, if x E F then x + 1E F . Hence, by the Induction Principle, F contains all natural numbers. 105

2220

EQUIVALENCE A N D SUBVALENCE

[CHAP. 11 SEC. 2

(b) We have ooE P (ao) and oohas no maximal element with respect to inclusion.

2220. THEOREM(TSS’). (a) A set is finite if and only if it is equivalent to some natural number. (b) If x contains a countable subset ( i t . if KO ,Y) then x is infinite. Proof. (a) If s = n then x is finite by preceding lemmas. If x is finite q q x ) s e that Y i \ not equivalent to any natural number II and set u E y = = u c .Y & (31)( 1 1 x n ) . We have y =+ 0 and y G P (x). Let z be a maximal element of y and denote by the natural number equivalent to 2 . Since x -2 0, tahe q E X - z and set z,= z u ( 4 ) . We have z 1 =I z and z,w ,v tio 1- 1. a contradiction. (b) fo!lows from 2219(b) and 2218(a).

<

-+

3321. COROLLARY (TSS‘). x is finite if and only if x

< coo.

2222. Remurk. The converse of 2220(b) is not provable in TSS‘, see

62 17. 2223. LEMMA (TSS’). The power set of a finite set is finite. Proof. It suffices to prove that Y ( n ) is finite for each 11. Let ti be the least number with an infinite power set. Clearly n =+ 0 and thus n = no + I. for some no. We have P ( n ) = P ( n o ) u (P ( n ) - P (no)). By the induction hypothesis, P ( t i , , ) is finite. Definef’by settingJ’x = x u { H ~ }for x E no. ‘The function.f’is a one-one mapping of P ( n o ) onto P ( n ) - P ( n o ) and hence P ( 1 1 ) is a union of two finite sets; thus P ( n ) is finite, a contradiction.

2223. LEMMA (TSS’). If .Y is a non-void finite set of ordinal numbers then Sup (x) E x (i.e. Sup (x) = Max (x)). Proof. Suppose that 11 0 is the least natural number for which there is a n n-element set x E On which has no maximum. Obviously n =k 1. Set ti = no + I . I f /’ is a one-one mapping of n onto x then by the induction hypothesis the image of under f has a maximum u. The number y = = Max ( x , f ” i i o ) is obviously the maximum of x, a contradiction.

+

2225.

DEI:INITION (TSS’).

Pfi,,( X ) is the class of all finite subsets of X .

2226. DEFINII-ION (TSS’). The relation Sd on the class of all finite sets of ordinals is defined as follows: ( s . .I.)E Sd 106

= .x

=k y & Max ((x

- 4’) u ( y - x)) E y

CHAP. I1 SEC. 21

2230

EQUIVALENCE AND SUBVALENCE

2227. LEMMA (TSS'). The relation Sd is regular and real. Proof. Let y E Pfin(On).If y = 0 then Sd"(y) = 0. If y 0 let a = = Max(y). Since ( x , y ) E Sd implies x E a + 1, we have Sd" { y } E E P ( a + 1) and so Sd is regular. Sd is real by Metatheorem 1453.

+

2228. LEMMA (TSS'). Pfin(On) is linearly ordered by Sd. Proof. The only non-trivial part of the proof is the proof of transitivity. Suppose that (x, y) E Sd and (y, z ) E Sd. Set a1 = Max ( ( x - J') u u (p - x)), u2 = Max ((y - z ) u ( z - 4')). We have a l E y - x , a2 E z - y and a I z2. Suppose that a, E cil. Then y E y = y E z wheneber a2 E y and hence a l E z - x. Let ctl E y and y E X - 2 for some y. Then y E X - y, contradicting the definition of a l . Similarly, if a , E Y and Y E z - x then y E y - x, contradicting the definition of a l . Hence al = Max ((x - z ) u u ( 2 - x)), and since a1 E z we have ( x , z ) E Sd. The case a l E a2 is treated analogously.

+

2 2 3 . THEOREM (TSS'). Pfin(On) is regularly well ordered by Sd. Proof. It suffices to prove that each non-void subset of P,,,(On) has an Sd-least element. Each u E Pfin(On) is a subset of Pfln( a ) for some a ; we take the least CI such that there is a non-void u c P,,, ( a ) having no Sdleast element. Set 5, = Min ((Max (x) ; x E u } ) ; obviously [j E a. Denote by uo the set of all x E u such that b = Max (x). The set u,, has no least element, for otherwise the least element of u o would also be least in u. We have uo c Pfln(/3 + 1) and hence a = b 1 . Let x E u , = x c j?& & x u ( f i ) E u,,. Since u o is non-void, the set u1 is also non-void. Hence, u 1 has a least element (by the induction hypothesis). Denete by z 1 the least element of u 1 and set z = z l u ( p } ; z is the Sd-least elemelit of uo, a contradiction.

+

2230. THLORFM (TSS'). (a) Pfin(On)z On, (b) P t i n (om) x m . Proof. (a) Since Pfin(On) is a proper class regularly well-ordered by Sd, there is a real isomorphism F between On and Pfin(On) with respect to E and Sd. Hence P,,, (On) and On are equivalent. (b) Let F be the above-mentioned isomorphism. We shall prove that F " o , = Pfin(0,) for each 0,. Suppose the contrary and consider the least a, such that F " o , ==! Pfin(oa).Since p E y implies ( ( p } , {y}) E Sd, Pfin(om) contains a subset isomorphic to o,,and Sd"(x} c Pfin(0,)whenever x E Pfin(om). Hence P o , c Pfin(om) and there exists a finite y E o, 107

2231

EQUIVALENCE A N D SUBVALENCE

[CHAP, 11 SEC. 2

such that y $ Po,.Let fl be the maximum of y. Then F ” o , c Prin(/?+ I). Since fl 1 < K,, we have Plin( p + 1) < K,, a contradiction.

+

2231. THEOREM (TSS’). If X is a cardinal such that X /> S o then X x X 2 . Proof. Since X 2 c Prin(Pfi,,( X ) ) , we have X 2 X by the preceding theorem; obviously X X z and so X x X 2 .

<

<

2232. DEFINITION (TSS’). o,is coJinul with op(Conf (o,,op)) if o, 5 - og and if there is a function f such that D(f) = o,,W ( f ) c og I and U(W(f))

=op.

2233. LEMMA (TSS’). If o,is cofinal with wI, and o, 5 o66 ogthen o6 is cofinal with wp. (Obvious.) 2234. DEFINITION (TSS‘). cf(og) is the least cardinal number o, which is cofinal with op. 2235. LEMMA (TSS’). If o, = cf (og)then there is a functionf such that D (,f) = a,,W ( f ) c ob,U(W ( f ) ) = opand (Vl,

K E 0,) (1

<

ti

= f”1


Proof. Let g be a function from o, into opsuch that U(W (9)) = wg. Define the set d by the following induction: O E and, ~ for 0 < 1 < o,, 2 E d iff g ’ l > g’y for all y E (1 n 3.. Then d is a subset of oar g d is a 1 - 1 function and L < K = g’r < g’ti for all 1, ti E d. Clearly d is isomorphic with a n ordinal 6 5 0,; but 6 cannot be less than o, because otherwise some cardinal less than o,would be cofinal with op.Hence there is an isomorphism h of o,onto d ; we setf’i = g’(h’i) for L E 0,.

I‘

2236. LEMMA (TSS‘). cf (cf (cob)) = cf (ap). Proof. Let o, = cf(op), o6= cf(o,). Let f be a monotone mapping of o,into opsuch that U(W (1))= opand let g be a monotone mapping of o6into o, such that U(W ( 9 ) ) = 0,. Define h’i = f ’ ( g ’ i ) for L E 0,; then h is a (monotone) mapping of oainto wp such that U(W (11)) = og;hence o, 5 ma. Evidently w6 I o,,so that o, = oh. 2237. Rerizark (TSS’). There exist infinite cardinal numbers such that cf ( a ) = a (e.g. KO)and infinite cardinal numbers with cf (a) E a (e.g. Nmo). 2238. DEFINITION (TSS’). (a) A cardinal number a is regulur if it is infinite and cf (z) = a. 108

2242

EQUIVALENCE AND SUBVALENCE

CHAP. I1 SEC 21

(b) A cardinal number cz is singular if it is infinite and cf ( a ) E a. 2239. LEMMA (TSS’). The class of all singular cardinal numbers is a proper class. Proof. Suppose that the class of all singular cardinal numbers is a set. Since it is non-void, this set has a supremum y ; we have oo c y. Put f ’ 0 = y, f ’ ( n + 1) = the first cardinal number greater than f ’ n , x = f”o. We have Sup ( x ) # x and y E Sup ( x ) ; furthermore Sup ( x ) is a cardinal number. Sup ( x ) is singular, since it is cofinal with a countable subset; this is a contradiction. 2240. THEOREM (TSS’). For every limit number a, cf(a) is a regular cardinal. To close this Section we define a regular-ordering of On2 which will be useful later. 2241. DEFINITION (TSS’). (a,

p ) Maxlex ( y , S) = . Max (a, p) < Max ( y , 6)

v [(Max (a, p) = Max ( y , S) & ct v [(Max(a,

/?) = Max(y, S)&

v

< y)] v

cz = y &

fi < S)] .

(the maximo-lexicographical ordering). 2242. LEMMA (TSS’). Maxlex is a regular well-ordering of On’; there is an isomorphism Is of On2, On w.r.t. Maxlex, E. Proof. Exercise.

230 I

FIRST AXIOM OF CHOICE

[CHAP. I1 SEC. 3

SECTION 3

The First Axiom of Choice

We shall study three forms of the axiom of choice E l , E2 and E3. E l is weaker than E2 and E2 in turn is weaker than E3. The present section deals with the first of these which is a “set form” of the axiom of choice; it ensures the existence of a selector (choice-function) on each set. In addition, we define the exponentiation of cardinal numbers and prove the recurrence formulas for cardinal exponentiation. It turns out that problems concerning cardinal exponentiation can be reduced to the investigation of a certain unary operaticn 1 on cardinal numbers. 2301. DE121NITION (TSS’). A function f defined on x is called a selector f o r x, if f ’ v E y whenever 0 =I=2’ E x. Sel (x) denotes the set of all selectors for x. 23013. AXIOM ( E l ) . Every set has a selector. In the sequel we shall work in the theory (TSS’, El). 2303. THEOREM (TSS’, El.). Every set can be well ordered. Proof. Let f be a selector for P (x). Define the function F by transfinite recursion: F’O = f’x, F’a = f ’ ( x - F”a) for a < 0. Note that F is a real class by 2124. Since x is not a proper class, there exists a E On such that x = = F”a. If a is the least number with this property, then F ci is a one-one mapping of a onto x. F induces a well ordering T on x, namely (F(/3), F ( y ) ) E ~r = P ~ y f o r f iy, ~ r .

r

2304. LEMMA(TSS’, El). For every set x there is a unique cardinal number equ‘valent to x. ._

2305. DEFINITION (TSS’, E l ) . X is the unique cardinal number equivalent to x. (The cardinality of x.) 110

CHAP. I1 SEC. 31

FIRST AXIOM OF CHOlCE

2310

2306. LEMMA (TSS’, El). Let 5 K, and suppose that y (= K, whenever _. y E x. Then u ( x ) 5 K,. Moreover, if x = K, or if = Kz for some y E x, then UG) = Nu. Proof. For each z E u(x) choose (by EL) some y E x such that z E y and denote it by k’z. For each y E x choose a one-one mapping f,, of y into o, and let g be a one-one mapping of x into 0,. Let 11 be the following function on u(x) : k’z = (g’(k’z),fkPz’z). The function h is obviously one-one and into o, x 0,. This completes the proof by Theorem 223 I . 2307. LEMMA (TSS’, El). For every a, Nu+ is a regular cardinal number. Proof. Take the least a such that Nu+,i s singular. Then there is a subset x of ma+, such that 3 S K, and u(x) = ma+ By 2306, Sup (x) = u(x) is strictly subvalent to X u + , . This is a contradiction. 2308. COROLLARY (TSS‘, El). The class of all regular cardinals js a proper class.

2309. DEFINITION (TSS’, El). The set of all sets which are mappings of the set y into the set x is denoted by exp (x, y). From now on we shall be interested in cardinal exponentiation, i.e. in the cardinalities of the sets exp (x, y ) where x, y are cardinal numbers at least one of which is infinite. From Theorem 2231. we immediately deduce that __

eG(h’,Gj = K, _ for every a and every n + 0. Also, _- - -

as we shall see later, . -exp ( n , K,) = exp (2, K,) for every n 2 2. Hence, we are interested in cardinalit’es of sets exp (K,,K f l ) and exp (2, Nu).These cardinalities have traditionally been denoted by K p , 2’“ respectively. This notation is very common but it i s defective in that the operation symbol for exponentiation is absent. (Let us mention that for this reason it is necessary to write (e.g. n * * m instead of nm in programming languages.) We preserve this traditional notation in the present section for the sake of intelligibility; but we introduce simultaneously another notation which will turn out to be useful in the sequel; ra.nely, we take the Hebrew letter 1for the sign of exponentiation. The reader may rewrite the recurrence formulas given in the sequel in the 1 symbolism.

2310. DEFINITION (TSS‘, El). If x, y are cardinal numbers then x y = - .~ = x 1 y = exp (x, y). We ncw define some important suprema. In each case we give two alternat.ve nntations; one of these is traditional and the other uses the 1 symbolism. ~~

111

231 1

[CHAP. II SEC. 3

FIRST AXIOM OF CHOICE

2311. DEFINITION (TSS', EL).

kp 'PI

2 u

=

E,

=

2 1 gg= sup (2'7; y

1 K, = sup ( K F ;y E

2312. LEMMA (TSS', El). If x, y , s, t E Cn and x

E P}

,

P> . 5 y, s 5 t then xs 5 y'.

2313. LEMMA (TSS', El). ( K ~ ) ' Y = K ~ ( ' p * ' y ) .

Proof. Let f be a mapping of my into exp (a,,a,), i.e. f't is a mapping of al, into a, for each 5 E a,.Define a mapping f of a, x my into o, by f ' ( q , 5 ) = (f't)'q. The function which assigns to eachf E exp (exp (ma, o,), a,)the corresponding f~ exp (ma, m, x my) is one-one and onto. Hence exp(exp(m,, 0,).my) and exp(o,, mPI x my) are equivalent. By 2231. the latter set has cardinality K ~ ( ' ~ * ' y ) . 2314. LEMMA (TSS', El). 2'1 = q w , , > K, for each a. If x E Cn and 2 5 x 5 K, then xu, = 2".. Proof. If f~ exp (2, 0,) (i.e. iff is a function on a, assuming the values 0. I ) we set Flf' = {y, y E ma&f'y = I}; F is obviously a one-one mapping of exp (2, 0,) onto P (ma). This proves the first . -. - assertion; now we prove the . second. Let x 5 Xu. We claim that exp (x, 0,) 2"-. I f f € exp (x, am) then f E x x ma s o that f~ P (x x 0,). Since x x ma = K, the assertion is proved. ~

2315. DEFINITION (TSS', El). P,. (x) is the set of all subsets of x having power at most Ka. 2316. LEMMA (TSS', El). If a 5 P then K F = PN,&,) Proof. Let f be a mapping of a, into mg. Clearly f is a subset of o, x mu .~and has cardinality K,, so that f E P,. (a,x ma). Since a, x mu = N, we have K p 5 PK,
+

CHAP. I1 SEC. 31

2320

FIRST AXIOM OF CHOICE

<= K, < K,. Then N,Urn - (‘Nabcf(Kb) ) .

231 7. LEMMA (TSS’, EL). Let cf (N,)

Proof. Denote cf (a,) by oBo. There is an increasing function g from obo into a cofinal subset of 0,. We claim that PUrn(o,)is subvalent to exp ( PUrn(g’6), oB0). If x E o, is of power at most Nu,set g:y = g’y n x

u

OF PO

for each y E ope. The function associating gx with each x E PKrn (oflo) maps the set P,. (0,) into exp ([ U PRa(g’h)], ope) and is one-one. Since K, <

< K,, there is a 6,

,=#O

E a,,,such

x exp (9’6, a,) whenever 6,

exp ( U PUa(g’d), a,,,)is *SU#O

that o, E 9’6,. By 2316 we have PNa(g’6) x

< 6 < wp0. Hence the cardinality of the set

(G?)’#o

and we have proved K p

(e?)

<= (G?)’#O.

As for the converse inequality is at most K p (since 51 < p) and hence the right-hand side is at most (K?)’,o. Since 3/, 5 tl we have (K?)K,o = = N? by 2313. 2318. LEMMA (TSS’, El). 2’P = (2 ?)d(U#). Proof. We again set oso= cf (0,).Let g be an increasing function with domain wD0and range cofinal with a,. As in the proof of the preceding lemma it can be shown that P (0,) is subvalent to exp ([ U P (g’s)], os0) *-%O

and hence 2Nf’5 (2 ?)‘#o. The converse inequality is obvious; indeed we have (2 = 2Np.

2319. DEFINITION (TSS’, EL). X(K,)

=

% U)~,O

=< (2HB)K,~ =

K$(KB).

2320. THEOREM (TSS’, El). (1) If K, =< K, then K F = 2’.. (2) If K, 5 K, then N> = Max (N,+ Np) (Hausdorff’s recurrence formula). (3) If p is a limit number and N, < cf(K,) then Hp = (Tarski’s recurrence formula). (4) Let K, be singular and let cf(Ks) K, < K,. Denote by K, the cardinal number We distinguish two cases as follows:

Ep

(ep).

(a) if K, = K? for some y < p then

113

2321

FIRST AXIOM OF CHOICE

(b) if K, >

[CHAP. I1 SEC. 3

K F for every y < jl then

Kp

=a

(Gp)= a (K,).

Proof. (1) See Lemma 2314.

RE1,

Gp

(2) and (3): Since K, 5 whenever B is limit, and K p = both cases are included in the formula K p = Max (K,,62) whenever K, < < cf (Kfi).The left-hand side is obviously at least as great as the right-hand side and hence it suffices to prove K p 5 Max (K,,Ep).Here we use 2306 and the fact that exp (a,,a,) = exp (y, a,); for if f E exp (afi,a,),

u

YE-#

then there is a y E assuch that W7(f)c y since K, < cf (Nfi). The cardinality of exp ( y , a,) is just the right-hand side of the equality.

u

YEW#

(4a) Suppose that K, = Rp is equal to K p for some 7 < 8. By 2317 we have = (gp)d(H#) (S\rF)cf(Kg) = KMax(%cf(H#)) = N N U =

~p

~

I

7

$ip.

(4b) Suppose that there is no maximal number among the cardinals

K p (y < B). We claim that cf (a6)= cf (a,). Since cf (a6) cf (ag)i s obvious, it suffices to prove cf(o,) 2 cf(a,). Suppose that cf(a6) <

< cf (a,)and let f be an increasing function of cf (a6) onto a cofinal subset of a,. For every 5 E D ( f ) let g'c be the first y < B such that K F 2 f 'c; set y o = u ( W (9)). Then K,, < K, and Ny; 2 f '5 for all 5, i.e. Ky; = 06. Consequently, for each y between yo and B we have K F = Ky; which contradicts the assumption. Now, we have K p = Kf('#) by 2317 and hence K p = X(K,). 2321. COROLLARY (TSS', El). If /3 is a limit number and 27' every y < B then

< K, for

(i) K> < K, whenener a,y < jl , (ii)

K?

= 3,

(K,) .

Proof. (i) Let 6 (ii) By 2318

=

K;P

Max (a,y ) . Then K> 5 K p = 2" < N,. =

20'

=

H#

(2 ~)c'('~)). Using the assumption we have

# '

2 u = K, and hence K\'a""= 1(K,). 2322. THEOREM (TSS', El.). (1) If K, is regular, then

114

CHAP. I1 SEC. 31

FIRST AXIOM OF CHOICE

2323

(2) If K, is singular, then we again distinguish two cases: H.

(a) if 2 u = 27' for some y < a, then

U.

(b) if 2 u > 2Nyfor every y < a, then N,

2"" = 3, (2 u) . Proof. (1) follows from Theorem 2320 Case (1). Nu

(2a) Let 2 u = 2Hvfor some y < x ; we may suppose that cf (K,) 5 h",. By . N Nu 2318 we have 2" = (2")Cf(Nd = (2%)d(Hd = 2MadN7, Cf(u=)) = 2% = 2 u N.

(2b) It suffices to prove cf (2 u) = cf (8=) and then to use Lemma 2318. The proof is analogous to the proof of Theorem 2320 Case (4b).

Remark. Theorems 2320 and 2322 give the promised recurrence formulas for cardinal exponentiation. 2323. THEOREM (TSS', El). Properties of the function 1.

Prooj. (L) If K, is regular then 3, (K,) = 2'B and the assertion follows from Lemma 2314. Suppose that A", is singular. Let g be a function on cf(a,) whose range is cofinal with a,. Let a be the set of all pairs (6, y ) such that y < cf (0,) and 6 E g'y. Obviously, = K,. We have, of course, K, <= 6 3, (K,), so suppose N, = 3, (K,). Then there is a one-one mapping of a onto exp (a,,cf (ag)). Denote byfdy the image of (6, y ) under this mapping and define a mapping h of cf(a,) into o, as follows: h'y = Min(a, - ($,(y); 6 E g(y)}). The function h differs from all f&;s; for, if h = fcp where (c, y ) E a, then h'y 4 f&(y) whenever 6 E g'y and hence Ii'y =k += f&(y), a contradiction. But the functions f a y were supposed to include all functions in exp (a,,cf (a,));thus we have reached a contradiction. 115

2324

[CHAP. I1 SEC. 3

FIRST AXIOM OF CHOICE

(2) Suppose that cf (1 (His))

=

cf (Kis). Then

a contradiction. 2315. The Continuur~A x i o m :

(Cont) (va) (

F a

=

Nu+

.

23-76. THEOKI 51 (TSS', El). The Continuum Axiom is equivalent to the statement (Vu) (1(xz)= xu+

~3

roof. Since 3 (sJ 5 = 2'~, the Continuum axiom implies (VZ) (1(Xz) = N u , ,). Conversely, suppose that (Vz) (1 ( H a ) = Nu+ and consider the least r with 2'" > N u , If H a is regular then 1( N u ) = 2'I by2322, Ka

Case L. If Xz is singular then by the induction hypothesis 2 u EC,

we have 2'= = 1 ( 2 u)

1 I6

= 1(Xz) =

Xn+ by 2322, Case (2b).

=

K, and

CHAP. I1 SEC. 41

2402

COMPLETE BOOLEAN ALGEBRAS

SECTION 4

Complete Boolean Algebras

This section deals with the basic properties of complete Boolean algebras, which play an important role throughout the book. A complete Boolean algebra is a class which contains the complement of each of its elements and the meet of each of its subsets and which satisfies certain additional conditions. Thus a complete Boolean algebra is determined by a class B, a function C mapping B into B and a function F of P ( B ) into B . It is convenient to consider only real B, C, F. Thus, a Boolean algebra is formed by three classes fulfilling certain conditions. If these classes are sets we speak of a set Boolean algebra. In this case it is more convenient to consider the ordered triple of these sets rather than the three sets separately. To be able to speak of an ordered pair (triple) of classes which are not necessarily sets we introduce the notion of generalized pair (triple) as follows: 2401. DEF~NITION (TSS’).

[x,y, z ] = (x x

(0)) u (Y x (1)) u

(zx

(2)).

2402. DEFINITION (TSS’). B = [ B , C , F ] is a complete Boolean algebra (Cba ( B ) )if B, C , F are real classes, C is a mapping of B into B, F is a mapping of P ( B ) into B and the following conditions hold: (1) F’{x, C’x} = C’F’O for each x E B ,

(2) F’(x, C ’ y } = C’F’O = F’(x, y } = x for each x, y

E

B,

(3) F’z = F‘F’q provided q -C P ( B ) and z = U(q). 117

2403

COMPLETE BOOLEAN ALGEBRAS

[CHAP. I1 SEC. 4

2403. Remark. We introduce the following notation: (the z i t i i t element) , (the zero element), (the meet of x and y) , (the complement of x) .

F’O = 1, c ’ 1 B = OB F’[.Y,y> = .Y C’.V = - B X

If z

5

1’

A

B we write

(the meet or Boolean product of z ) .

F’r = AB3

Iff is a function with values in B we write

The subscript will be omitted if there is no danger of misunderstanding. The conditions (1)-(3) may now be rewritten as follows: 0, ,

(I.)

X A

(I)

S A -J‘=O,=XAj’=.U.

(3)

if z

-

E

N =

B and

z = U(q), then /\z =

A (At). t=l

(This is in fact the associative law.) 2404.

LEMhlA

(TSS’). Tf X. y, z E B then =

X A J’

.Y A

(1. A

Z)

V A X ,

= (X A J’) A Z .

-3405. DEFINITION (TSS’). (1) R is a reflexive ordering of A (Ordg,(R, A)) if for all x, y , z E A we have

(a) x R x , (b) sRy& y R z .

-+ x R z ,

(c) x R y & ~ 7 R x-+. x

= 1’ .

( 3 ) A = [ A , R ] is an ordered class if R is a reflexive ordering of A . If A = [ A , R ] is an ordered class then A is said to be the field of A and the reflexive ordering R is often denoted by

s.

CHAP. I1 SEC. 41

2408

COMPLETE BOOLEAN ALGEBRAS

2406. Remark. If f = {(x, x); x E V } and R is a reflexive ordering of A then R - I is a n (irreflexive) ordering of A. 2407. DEFINITION (TSS‘). Let 5 be a reflexive ordering of A and let z G A , x E A . The element x is the infimum of z (inf.(x, - z)) if

s

(a) (VY E =)(x Y ) (b) (Vw) [(Vy E z ) [w 5 y] 7

-+ w

5

x]

.

2408. THEOREM (TSS’). Let 5 be a complete Boolean algebra. For x, y E B define x y by s A - y = 0, (i.e. x 5 y iff x A y = x). The relation 5 is then a (real) reflexive ordering of B and for each z E B, Az is the infimum of z. If x 5 v then - y 5 - x; - - x = x; x =l= - x unless B contains the single element x. Proof. Clearly x 5 x7 since x A - x = 0, by (1). Suppose x 5 y z, i.e. x A y = x and y A z = y . Then x A z = (x A y ) A z = x A A ( y A z) = s A y = x, i.e. x 5 z. If x 5 y 5 x then x = x A y = y . Hence S is a reflexive ordering. Since x A - x = 0, we have x A x = x. Suppose z E B and let x = A z = F’z; we shall prove that x 5 y for every y E z. We have X A

y = AZ A y = F’{F’Z, y } = F)F”{z, { y } } =

= F’(z u {y}) = F’z = AZ = x

.

Suppose that w 5 y for all y E z; we shall prove w 5 s. IfyEzthenw A y = w ; h e n c e f o r 2 = z u { w } w e h a v e F ’ 1 = ~ z ~ w = = x A w. On the other hand,

Hence w A x = w , i.e. w 5 x, and x = Az is the infimum of z. By (l), - x A - - x = 0,; i.e. - - x 5 x. If x 5 y , then - - x 5 y ; i.e. - - x A - y = 0, and so - y 5 - x. Since - - x x we deduce that - x 6 - - - x; on the other hand, substituting - x for x in - - x 5 I x, we get - - - x - x and hence - - - x = - x. Using x A A - x =0 , we get x A - - - x = O,, i.e. x 5 - - x. Hence x = = - - x. If x = - x then O , = X A - X = X A X = X and 0,=-0,=1,. Since 0, y 5 1, for each y E I?, we have 0, = y = 1, for each y E B.

2409

COMPLETE BOOLEAN ALGEBRAS

[CHAP. I1 SEC. 4

2409. DEFINITION (TSS’). (1) Let B be a complete Boolean algebra. The canonical ordering of the algebra B is the (reflexive) ordering defined by x 5 y X A JJ = X. (2) For x, y E B define X


E X

s y&x * y ,

y = - (-X

X V

A

-y)

(the joirt or the Boolean s u m of x, y), for z G B define

V Z= C’(hC”z) (the j o i n of

2).

2410. DEFINITION (TSS’).Let be a (reflexive) ordering of A and let z E A, x E A. The element x is the supremum of z (sups(x, z)) if

(a) ( V Y E Z ) ( Y 5 x) (b) (Vw) [(Vy E Z ) [ y 5 w ] 9

-,x 5 w ] .

2411. LEMMA(TSS‘). Let B be a complete Boolean algebra and let Vz is the supremum

z E B. Let 5 be the canonical ordering of B. Then of z and we have

C’(/\u) = V(C”u)

and

C’(Vu) = A(C”u)

(De Morgan rules); in particular, for x, y -(X

A V)

= -X

V

-y,

E

-(X

for each u E B

B, V

y)

= -X

A

-y

,

2412. LEMMA (TSS’). Let B be a complete Boolean algebra. For any elements X, y , z E B we have (1)

120

x A y = y A X

x v y = y v x

COMPLETE BOOLEAN ALGEBRAS

CHAP. 11 SEC. 41

X V

(8)

(y

A Z)

= (X

V

y)

A

2415

(X V Z)

Proof. We shall only prove (6) and (7); the proofs of (1)-(5) are easy, while (8) is analogous to (7). - y , i.e. x A y = 0,. Similarly we have (6): If y = -x we have x 5 y, i.e. -x A -y = 4;hence x v y = 1,. On the other hand, assume that x A y = 0, and x v y = 1,. Then we have x A - - y = = Os, i.e. x 5 -y, and -x A - y = O,, i.e. -y 5 x; hence x = -y. -x

(7): We have x A y x and x A z i_ x; hence (x A y) v (x A z ) 5 x. Similarly we have x A y y v z and x A z 5 y v z; i.e. (x A y ) v v (x A z) 5 y v z. Hence (x A y ) v (x A z) 5 x A ( y v z ) . It remains toprove theconverseinequality. Set b = y v z and a = (x A y) v (XAZ). Since x A y 5 a , we have x A y A - a = 0,. Similarly x A z A - a = = 0,. Hence x A - a 5 -y and x A - a 5 - z. Consequently x A - a 5 5 -y A -z = -(y v z) = -b. Thus we have x A - a A b = O , , i s . x A b 5 a ; hence x A ( y v z ) 5 (x A y) v (x A z). 2413. DEFINITION (TSS’). Let B = [ B , C , F ] be a complete Boolean algebra and let x E €3. The partial algebra determined by x is the algebra B I x formed by the following classes:

5

(~1x)’y = -y A x for y ~ ~ ( x , ( F I x ) ’ z = A z ~ x for z ~ B l x .

B l x = {y;y

x),

2414. LEMMA (TSS’). If B is a complete Boolean algebra and if x E B then B I x is a complete Boolean algebra. Proof. Condition (3) is satisfied trivially. We have OBI, = 0,. Since y A -y = 0, we have y A -y A x = 0, hence (1) is satisfied. Finally, if u, u E B 1 x, i.e. u, u 5 x, then u A - u A x = 0, means the same as u A --2) = 0,, i.e. u A u = u. Hence (2) is satisfied. 2415. DEFINITION (TSS’). Let B be a complete Boolean algebra and let Z E B. (a) 2 is aJilter on B if (1)

2

* 0,

121

2416

COMPLETE BOOLEAN ALGEBRAS

(2)

(vx, Y E Z ) ( X

(3)

(vx E z)(Vy E B ) (x 5 y

A

YE

[CHAP. I1 SEC. 4

Z ) , +

y

E

z).

(b) Z is a properjilter if it satisfies (l), (2), (3) above and if (4)

oB#z.

(c) Z is a rornpletejilter if it satisfies (3) above and if

(5)

(vq

c Z ) (A q E Z ) .

(d) Z is a n ultrafilter if it satisfies (l),(2). (4) above and if (6)

( V X E B ) ( X E Zv - X E Z ) .

2416. LEMMA (TSS’). Every ultrafilter is a proper filter; every complete filter is a filter. Proof. Let Z be an ultrafilter; we shall prove (3). Suppose that for some x , y we have X E Z , y # Z and x 6 y . Then - ~ E Zand so x A - - y e Z ; i.e. 0, E Z , a contradiction. Let Z be a complete filter; we shall prove (l),(2). We have 0 E Z and so A 0 = 1 , E Z . ( 2 ) is a particular case of (5). 2417. Rernurks. 1) A complete ultrafilter is an ultrafilter which is a complete filter. Obviously a class Z 5 B is a complete ultrafilter iff it satisfies (4), (5) and (6) above. 2) A proper filter Z is an ultrafilter iff (6’)

(Vx, y

E

B ) (.Y v y E Z

-+

x

E

Z v y

E

Z)

(obvious).

2418. THEOREM (TSS’). Let B be a complete Boolean algebra and let Z be a proper filter on B. Then Z is an ultrafilter iff there is no proper filter Z , on B such that Z c Z,. PROOF.Let 2 be an ultrafilter. If u E Z , - Z then - u E Z and so - u E A - u E Z , , a contradiction. Now suppose that Z is not an ultrafilter. Then there exists u E B such that neither u E Z nor - u E Z . We have z A u =I= 0 , for each z E Z; for, if z A u = 0, for some z E Z then z - t i , so that - u E Z , a contradiction. Now define x E Z , = (32 E Z) (x 2 z A t i ) . Z , is a proper filter and Z , z 2; since u E Z , - Z , we have Z , =IZ= . E

Z1; thus u

122

CHAP. ll SEC. 41

2424

COMPLETE BOOLEAN ALGEBRAS

2419. THEOREM (TSS’). Let 2, be a real proper filter on B and let H be a real one-one mapping of an ordinal onto B. Then there exists a real ultrafilter Z on B such that Z, c 2. Proof. By induction we define H‘a E 2 = (VU) (Fin (u)& u E 2, u (H”a n 2 ) + Au A H’a 0,) . We have 0, # Z and 2 , c Z. If u is a finite subset of Z then Au E 2; hence 2 Is a proper filter. Suppose that there exists an x E B such that neithcr x E Z nor - x E 2. There exist finite subsets u, u of Z such that x A Au = - x A = 0,. Set w = u v U ; then x A AW = --x A AW = 0, SO that A (x v -x) A A w = 0,, i.e. A w = 0,. This contradicts w c 2, since w is finite. Hence Z is an ultrafilter.

+

2420. DEFINITION (TSS‘). Let B be a complete Boolean algebra. An clement u of B is an atom if it is non-zero and if therc is no u E B such that 0, < v < u. An algebra B is atomic if for each non-zero element x E B there exists a n atom y such that y x. B is atomless if it has no atoms. 2421. LEMMA(TSS’). If u is an atom then {x; u 5 x} is a complete ul t rafilter. 2422. DEFINITION (TSS’). Let B be a complete Boolean algebra. A subset

u of B is a partition of B if

Part (B) (briefly Part) is the class of all partitions. If u and u arc partitions, we define u Q ti h

= (vx E u ) ( 3 y E u ) (x 5 = (x

h

y) ,

y, x E u & y E u & x

A

y $. OB} .

2423. LEMMA (TSS’). The relation Q is a reflexive ordering of the class Part; u A u is the infimum of { u , v } in this ordering.

2424. DEFINITION (TSS’). A non-void real class P titionjilter (on B) if (1)

(vx, y E P ) (x h y

(2)

(Vx

E

E

c Part is called a par-

P) ,

P ) (Vy E Part) (x

Q

y

+

y E P) 123

2425

COMPLETE BOOLEAN ALGEBRAS

[CHAP. 11 SEC. 4

2425. DEFINITION (TSS’). Let 2 be a filter on B. Set Pt (2) = {u E Part, 11 n z $:

01

(the partition jilter induced b y a jilter).

2426. LEMMA (TSS’).Pt (2) is a partition filter. ,7427. DEFINITION (TSS’) Let B and B , be coniplete Boolean algebras. (a) B , is a subalgebra of B if

B , c B ,

F , c F ;

C,cC,

(b) A mapping H of B onto B , is a horiiomur.phis/n of B onto B, if H is real and (1)

(vx E B ) ( C i H ’ s = H’C’X) ,

(2)

(vz

c B)(F;H”z

=

H’F’z)

If H is one-one it is called an isomorphism. An isomorphism of B onto itself is called an autornorphism.

-3428. LEMMA(TSS’). Let B , be a subalgebra of B. Then 0, = O,,, 1, = I,, and for z G B , we have V,z = Vz, where V1 is the symbol for the Boolean sum in B , . If x, y E B , then x 5 y = x 5 y , where 5 is the canonical ordering of B , . (Obvious.)

,

,

2429. LEMMA (TSS’). Let B be a complete Boolean algebra; let B , and suppose (VX E B , ) ( - x E B , ) , (VZ

r

c B , ) (F’z

E

cB

B,) .

Then [ B , , C B , , F P ( B , ) ] is a complete Boolean algebra which is a subalgebra of B. (Obvious.) 2430. LEMMA (TSS‘). Let If be a homomorphism of B onto B , . Then (VY, y E B ) (x

5y

-+

H ’ x -I H’y) ,

H’O, = o,, , H’1, = I,, , H’ V u = V H ” u for ZI c B . 124

CHAP. I1 SEC. 41

2436

COMPLETE BOOLEAN ALGEBRAS

The class {x E B, H ’ x = lB,} is a complete filter on B. If H is an isomorphism, then its inverse is an isomorphism of B , onto B. (Obvious.)

2431. THEOREM (TSS‘). Let B be a complete Boolean algebra and let X be a real class such that X E B. Then there exists an algebra B , which is the least subalgebra of B (with respect to inclusion) such that X E B1. Proof. By recursion we define H ” { O ) = X and H “ { a } = {X;(3yEH”sr)(X = - y ) v (3z E H”z)

(X =

F’z))

r

for ct > 0. Set B , = W ( H ) . By 2429, [ B , , C [‘ B , , F P ( B , ) ] is a complete Boolean subalgebra of B; from 2146 it follows that B , is the least subalgebra containing X .

2432. DEFINITION (TSS’). Let B be a complcte Boolean algebra and let X be a real class such that X E B. The class B , of Theorem 2431 will be denoted by Gen, ( X ) (or simply Gen ( X ) ) . If B = Gen, ( X ) then we say that X generates B. 2433. DEFINITION (TSS’). Let B be a complete Boolean algebra; a real class X E B is a base (for B ) if 0, 4 X and if for each non-zero x E B there x. exists JTE X such that y 2434. In the remainder of this section, we restrict our attention to those complete Boolean algebras which are sets. We say that a triple ( b , c,f) is a complete Boolean algebra, if [b, c,f] is a complete Boolean algebra. The set 6 is then called the field of the complete Boolean algebra (6, c,f) etc. 2435. LEMMA (TSS‘). Let b = ( b , c,f) be a complete Boolcan algebra, and let z 5 b be a complete filter. Then there exists u E b such that z = = {x, u 5 x}; z is an ultrafilter iff this u is an atom. Proof. Take u = Az E z. If u is an atom then z is an ultrafilter (see above); if not, then there is some w such that 0, < w < u. Suppose z is an ultrafilter; since w q! z we have - w E z ; hence w 5 - w , a contradiction since w =k 0.

2436. LEMMA (TSS’). If z is a base for a complete Boolean algebra b = = ( b , c,f) then for each u E b there exists some t E z such that u = Vt. Proof. Set t = { u E z , v 5 u } . We have Vt 5 u ; suppose Vt < u. Then u A -Vt =!= 0, and there exists u E z such that v 5 u A -Vt; i.e. v E t and u 5 Vt. Since u 5 -Vt we have v 5 Vt A -Vt = O,, a contradiction.

125

2437

[CHAP. I1 SEC. 4

COMPLETE BOOLEAN ALGEBRAS

2437. COROLLARY. If z is a base for a complete Boolean algebra b = ( b , c, f ) then z generates b.

=

2438. THEOREM (TSS’). Let q be a base for a complete Boolean algebra b = ( b , c, f > and let q , be a base for a complete Boolean algebra 6 , = = ( b , , cl, f , ) . Let Ii be a t - 1 mapping of q onto q , such that

(vu, 2, E q ) (u 5 v = h’u 5 1 h’u)

( s ,is the ordering of ( b , , c , , f , ) ) . Then there is a unique isomorphism g

r

of b onto b, such that g

q = h.

s

Proof. Define g’x = V , { h ’ u , u x} for x E b. Since this must be satisfied by each isomorphism, the uniqueness will follow from the existence.

,

(t) x 5 y = g’x 5 g’y for x, y E b. If x 5 y then we have g’x 5 g‘y immediately from the definition. To prove thc other implication suppose g’x 5 g’y, i.e. V,(h’u, zi 5 x} 5 V l { h ’ v , v 5 v } . It follows that u x implies u 5 y for each u E q . Hence V { u , u 5 x] 5 V { v , v 5 y } . i.e. x 5 J’. This cornplctes the proof of (1); moreover, we have x = y = = g’x = g’y for x, y E b and hence g is one-to-one.

s

(2) g maps b onto b,. For y E b , set x = V { u , h’u 5 y } . Then g’x = y . We also have g’0, = o b , and g’l, = l,,. (3) g ’ h z = A , g ” z for z c b. To prove the inequality S1, consider some w E z;then Az 5 w implies g ’ A z 5 g’w. Hence g’hz 5 A, {g’w, w E z } = A1g”z. To prove the converse, set A,g”z = y. Then w E z implies y & g’w; if y = g’x then x 5 w, and hence x 5 Az. It follows that g’x = y = Alg”z g ’ h z . It can be proved analogously that g ’ v z = = Vlg”z for z c b.

,

s1

(4) g’( -x) = 8’0, = 0,,

= -

, g’x for x E b. We have g’x , g’( -x) A

and g’x v g’(-x) = g’(x v -x) g’( -x) = - (g’x) by Lemma 2412 (6).

= g’(x A -x) = = 9’1, = lb,. Hence

This completes the proof of the Theorem.

2439. DEFINITION (TSS’). We say that an ordering if for each x, y E A (vz

implies x 126

=< y.

5

x)(3t)(t

5 z& t 5

y)

5 on A is separative

2442

COMPLETE BOOLEAN ALGEBRAS

CHAP. I1 SEC. 41

2440. Remurk. The condition can be restated as follows:

2441. LEMMA (TSS’). Let q be a base for a complete Boolean algebra b = ( b , c , f ) . Then the ordering p of b is separative on q.

z

Proof. If x, y E q and x F y, then x A - y < 0;, hence there is some with z 5 x A - y . Now z satisfies the condition

Eq

sl

2442. THEOREM (TSS‘). Let be a reflexive separative ordering of a set q. Then there exist a complete Boolean algebra b = ( b , c,f) and a 1-1 function g which maps q onto some base for b and which satisfies the condition ( v u , u E q) (u

Proof. (1) We say that u hold :

cq

5* u

5

g’u

5 g’u) .

is saturated if the following conditions

5 y and y E u then x E u, (b) if (Vy 5 x) (3z 5 y ) ( z E u) then x E u. (a) if x

Denote by b the set of all saturated subsets of q . (2) If z is a non-void subset of b then n z is saturated. (Exercise.) Define F’O = q , F’z = nz for 0 z E b .

+

(3) For u E b define C’u = {x E q ; 1 ( 3 z 5 x) ( z

E u)]

.

Evidently

u n C‘u = 0.

(4) We prove that C’u is saturated for each u E b. If x 5 y E C’u then 1 (32 5 y) ( z E u); hence 1(3z 5 x) ( z E u ) and so x E C’u. Further, if we

suppose

(vs

5 x) (3t

s s) ( t

E

C’U)

then we may deduce successively the formulas (VS

s x) (3t s s) 1(32 2 t ) ( z

E u)

,

1 (3s 5 x) (vt 5 s) (3z 5 f) ( z E u ) , 127

2443

COMPLETE BOOLEAN ALGEBRAS

l(3s

5 x)(Vt 5

[CHAP. 11 SEC. 4

S)(tEU),

1 (3s ix) (s E u ) , J E

C’U

.

( 5 ) Define b = ( b , C , F ) ; we have Ob = 0 and u A u = u n u for u , u E b. Obviously u n C’U = 0. If z E. b and z = u(g) then = (fit). (Exercise.) tsg

nz n

(6) We prove that u n C’u = 0 iff u n II = u (i.e. u E u). If u c u then u n C’u = 0, since u n C’u = 0 (see (3)). If u $ u, choose x E u such that x $ c. It follows (from saturatedness) that 1 (VJ 5 x) (3: 5 y ) ( z E u), i.e. (3y 5 x) (Vz 5 y ) (z ef u). For any such y we have y E u and y E C’u; hence zi n C‘u =j=0. Thus b is a complete Boolean algebra.

(7) For w E q set g’w = ( x E 4 ; x

5

w] ;

g’w is saturated by the separativity. Evidently, w I

Finally, g”q is a base, which completes the proof.

ii

5 w 2 = g’w, E g’w, .

2443. LEMMA (TSS’ + El). If q is a base for a c. B. a. 6 and if u E b ar?d 0, then there exists u E q such that

+

I) v u = u ,

y = 0, whenever x nad y are distinct elements of a. Proof. Let f be a selector for the power-set of q and let G be defined as follows: 2) x

A

G’O = f ’ { u E q ; u G’ir = f ’ { u

E

q; u 5

U>, zi

- VG”CI) for

CI

> 0.

We may now let u = G”1 where 1 is the least ordinal number such that G’A = 0, i.e. such that u - VG”1 = 0,. ,7444. THEOREM (TSS’, El). If p is a partition of b and if q is a base for b then there exists a partition j Q p such that c q . Proof. Straightforward.

CHAP. I1 SEC. 51

2502

SEPARATIVELY ORDERED SETS

SECTION 5

Ordered and separatively ordered sets

The significance of separatively ordered sets lies in the fact, established in the preceding section, that they are precisely the bases for complete Boolean algebras which are sets. In the present Section we shall study some characteristics of separatively ordered sets and certain methods for constructing such sets. We shall use the axiom of choice (El) in several places. Some statements are formulated not only for separatively ordered sets but for ordered sets in general. 2501. DEFINITION (TSS’). If a = ( a , I , is) an ordered set then a segment of a determined by x is a set of the form Seg,(x) = { y E a ; y S a x } where x

E a.

If it is clear which ordered set is involved we may write 2 instead of Seg, (x). A set z G a is called an exclusive system (in a) (Ex,(=)), if (vx, y

E

z ) (x

*y

+

2 n 9) = 0

An ordered set is K,-multiplicative (v(Ec,, a)) if for any subset u of a which is linearly ordered by 5 and has cardinality at most K, there is an x E a such that (Vy E 2 1 ) (x 5 y). We say that a is K,-partitionable (p,(K,, a)) if there is an exclusive system u E a which has cardinality at least Ec,. 2502. LEMMA(TSS‘). If Ec, 2 K, then for any ordered set a we have

129

2503

SEPARATIVELY ORDERED SETS

[CHAP. I1 SEC. 5

2503. THEOREM (TSS’). If K; is a singular cardinal and if an ordered set a is Xp-multiplicative for every fl < 5 then a is Ks-multiplicative. Proof. Let u E a be linearly ordered by 5 , and suppose u x K,; let f be a one-one mapping of m, onto u. We let g’0 = f ’0 and for any CI > 0 we let E be the least /? such that (Vy < cc) (g’y >,f’p) and g’cc = f’E. Clearly D(g) 5 0,. If D(g) < mc we let S = g; if D(g) = aslet { T , ) , < ~ ( ~ ~be ) a sequence whose supremum is m5and let g’o! = g’z,. It follows that D (8)< < o, and that W (8) is cofinal with u in 2,; i.e., (Vz E u ) (3w E W (y)) (2 2, w). Let Xp be the cardinality of W (9); clearly, a is Kp-multiplicative and hence there is an x E u such that (Vy E W (x 5 , y). By cofinality ($7 E u ) (x 5 , Y ) .

(a))

2504. THEOREM (TSS’, El). Let a be a separatively ordered set and suppose that a has no minimal elements, i.e. that (Vx E a ) (3y E a ) ( y <,, x). If a is K,-multiplicative then it is K,, ,-partitionable. Proof. Using a selector for P ( u ) we construct a function f from ma+ into a such that j ’ f l >,f’y for any p < y < a,+ By separativity we have

y
+ (32

<,

x) (2 n 9 = 0 ) ,

and hence (32 5 , x) (Vt 5 , z ) (t $, y). If follows that for any j? E m,+, we can define a function g’p such that g’j? 5,f’fl and Seg, (g’fl) n Seg, (g’(j? + 1)) = 0. We have g”ma.+ M o,+ and Ex,, (g”ma+ ,).

+

2505. LEMMA (TSS’). Let u be an exclusive system in an ordered set a. If for a11 x E u, ux E 12 is an exclusive system in a then U u, is an exclusive XEU system in a. (Exercise.)

2506. D1:FINITION (TSS’). If u is an ordered set then a n element x of a is called p-saturated (psat, (x)) if for every Naand every y <, x the existence of an exclusive system u E R of cardinality K, implies the existence of an exclusive system of cardinality N, included in 9.

2507. LEMMA (TSS’). If x is p-saturated and y 5 , x then y is p-saturated. ,2508. LEMMA(TSS’). If x is an element of an ordered set a then there is a p-saturated y 5 , x. Proof. For any y 5 , x we let M ( y ) be the least ct such that there is no exclusive system of power K, of elements of 9. We let y o 5 , x be such that M ( y o ) 5 M ( y ) for every y 5 , x. Clearly y o is p-saturated. 130

CHAP. 11 SEC. 51

SEPARATIVELY ORDERED SETS

2510

2509. LEMMA (TSS’, El). If a is an ordered set then there is an exclusive system u of elements of a which contains only p-saturated elemems of a and such that (Vx E a) (3y E u ) (R n 9) =k 0. Proof. Let (xu},<1 be a sequence consisting of all p-saturated elements and let f’0

= xg,

f’u

=

{

x, if (VP < a) (a, n (f’fi) ’) = O , xo otherwise,

for any a > 0. We also let u = f”A. Clearly Ex,(u). If x E a then there is an a such that x, Sax. If xu E u then we let y = xu. If x u # u then thereexists b < a such that x, E u and 12, n R, 0; we then let y = x,. Clearly, y is p-saturated.

+

2510. THEOREM (TSS’, El). Let a be an ordered set, let Kcbe a singular cardinal and suppose that for every b < 5 there is an exclusive system u E a of cardinality K,. Then there is an exclusive system of cardinality K,. Proof. Let u be an exclusive system in a which satisfies the conditions of Lemma 2509. If this system has cardinality ZK, then we are finished; we therefore suppose its cardinality to be
u

XEY,

by = 5 then

u1 is a n exclusive system on 9; otherwise u1 is a subset of u.) It remains to prove that there is no j? < 5 such that, for any x, there is no exclusive system u, c R of cardinailty K,. Suppose there is such a /3 and let 6 < 5 be such that K, > K, and K, > K,,where K, is the cardinality of u. Let u be an exclusive system of cardinality K,. For every X E U we choose a y, E R such that y , & z for some z E u. If we denote by ij the set of all y, such that x E u then for every z E u , 5 n 2 E 2 is an exclusive system and has cardinality less than K,. Hence the cardinality of ij, and therefore also of u, is at most Max (N,, K,) < K,; a contradiction. This completes the proof. 131

2511

[CHAP. I1 SEC. 5

SEPARATIVELY ORDERED SETS

We now define two important indices for complete Boolean algebras. 2 j f l . D I F I N I T(TSS’). I ~ N If b is a complete Boolean algebra then the crrlrbw of b ( p ( b ) ) , i\ the least cardinal K, such that there is no q E b o f cardinality Nzsuch that (VS,

J’ E

9) (X

$. y + X A

y

=

0,)

2512. LEMMA(TSS’. El). If a n ordered set a is a base for a complete Boolean algebra b then p ( b ) is the least cardinal N, such that there is no exclusive system u E CI of cardinality K,. (Exercise.) 2.513. DEFINITION (TSS’). If b is a complete Boolean algebra then the dr-off of b ( v ( h ) ) .is the least cardinal Xg such that there is no K,-multiplicative base for 6. We h a l l now consider- products of ordered sets. 2514. DI~IINITION (TSS’). (I) If s is a nonempty set then 1 C P (s) is called a cut i n s if it has the following properties:

I ( 2 ) If

112

*0,

(vx E

I ) (Vy

G x) ( y E 1 ) .

is a cardinal number then a cut I is nz-additive if (vz E

<

I ) (.

I12 +

uz E I )

( 3 ) A cut is an itleu[ on F if it is ,-additive, i.e. if. (Vx,

(4) The that

I I O U J ~of

)* E

I ) (x u ?‘ E l ) .

a cut I (Norm ( I ) ) is the least cardinal number K, such

1 (3x

E

r) (x /> KJ .

2515. DEf 1 ~ 1 1 1 (TSS’). 0 ~ Let I be a cut in s and let b be a function on s such that, for any x E 5 , b’s is an ordered set; for convenience we write b, = = ( b x . 5,) imtead of b’x. The product of ordered sets b, over 1 is the ordered sct 6 , = ( u , 5 ) where

n‘

I E F

~1

=

{ f’; LJn (f)& D ( f )

f5g =

E

1 & (Vx

E

D (f))( f ’ x

D (f) 2 D ( g ) & (Vx E D (9)) ( f ’ x

E

b,)) ,

5 , g’x) .

2523

SEPARATlVELY ORDERED SETS

CHAP. 11 SEC. 51

2516. Remark. The reader may easily verify that n ' b K is an ordered

set.

XXES

n'

2517. LEMMA(TSS'). 1 ) If all b, are separatively ordered sets without greatest element then 6, is also separatively ordered.

2) Denote

n'bx

XES

by ( a , 5 ) . Iff and g are in a and iff

=g

XES

r D(f) then

g 2 f;if D (f) n D ( 9 ) =: 0 then f u g is the infimum off and g. 3) The empty function 0 is the greatest element of b,.

nz XES

2518. DEFINITION (TSS'). A cut 1 is said to have the singleton property on s if z E 1 and x E s imply z u (x} E 1. 251Y. Remark (TSS'). 1) Clearly if an ideal on s contains all singletons

then it has the singleton property. 2) Let s be infinite and suppose that 1 has the singleton property and s 4 1 (e.g. s = coo and 1 = Pfin(s)). If, for any x E s, the set b, has at least one element, then there are no minimal elements in b,. (Exercise.) Thus

nz XES

the algebra whose base is n Z b xdoes not contain any atoms.

r

XES

2520. DEFINITION (TSS'). For any set x, at (x) is the ordered set (x, I s). (It follows that every two distinct elements of at (x) are incomparable.) 2521. DEFINITION (TSS'). If b,, bl are ordered sets then b, = (b, x b,,

5 ) where

0 6, =

for any u,, L'E,b, and cil, u , E b,. We may prove (in TSS) that if b,, b , are separatively ordered sets (not necessarily without greatest elements) then b, 0 6, is a separatively ordered set. Further, we have the following 2522. LEMMA (TSS'). Let b,, b , be separatively ordered sets with greatest elements lo, 1 , respectively. If a, = ( b i - (li},S i ) (i = 0, 1) and 1 = = P ((0, 1)) then 6, 0 6, and ai are isomorphic.

ni

i=O,l

Proof. Exercise.

n'

2523. THEOREM (TSS', El). If bx is Nu-multiplicative for each x E s and if 1 is N,-additive then bx is K,-multiplicative. XES

133

2524

[CHAP. 11 SEC. 5

SEPARATIVELY ORDERED SETS

Proof. Let ( a , 5 )

=

n*b, and let

u be a linearly ordered subset of a

XES

of power 5 Nu.For every x E s the set { f ' x ; f E u } is linearly ordered by sx. By Nu-additivity we have U D (f) = z E 1. Let f be a function which assof

ciates with each .Y E z a n element f ' x of bx such that f ' x 2 , g'x for any y E u. The element f of u satisfies the condition of N,-multiplicativity. 2524. THEOREM (TSS', EL). If I is an ideal, if Norm ( I ) 5 K,+l and if no b, is K,+l partitionable then n ' b , is not K,-partitionable for any XES cardinal K, > 2". Proof. For any f~ a we shall call the elements of D ( f ) f i x i n g points off. The set of all fixing points o f f belongs to 1 and hence it is of power at most K,. For any f E a we choose some (not necessarily one-to-one) mapping of my onto the set of all fixing points o f f ; for each a < o, we may refer to the a-th fixing point to be denoted by x!. Suppose that u E a is an exclusive system in a ; we shall prove that u 4 2". Let A be a selector for P (u); i.e. we have A'u E u for any nonempty u E u . We shall prove that every element of u can be constructed by iterated application of A. For any f~ u and ci < os we let C(f, a) = = ( g E u; ({f'x:, g'x!})); G(f, a) is the set of all functions in u whose value at the a-th fixing point off determines a segment disjoint from the segment determined by the value o f f a t this point. We let D be the (real) class of all functions h such that D ( h ) E On and W ( h ) c m5. To each It E D we assign a sequence ft of elements of u as follows : f,"= A'u ,

f,"= A'

n C($,

h'p)

(0 <

C(

5 D (h)) .

B
If p < ci the values off," and$ at the (h'p)-th fixing off; are exclusive (in the ordered set corresponding to the (h'B)-th fixing point off;). We let Do be the class of all h E D such that G(f;, h'p) 0 for all a D (h); this

n

+

s

B
means that all f,"for a D(h) are defined as elcments of u. If a < p g 5 D (11) and if h E Do then f: f;, since they differ at the (h'cc)-fixing point off:. We shall prove the following:

+

1) For every h E Do we have D ( h ) < as+ 1. 2) For every f E u there exist /t E Do and a 5 D ( h ) such that f = f:. This will complete the proof since we shall have Do 2°F by 1)and u K, . 2" = 2' by 2 ) .

<

134

<

<

2526

SEPARATIVELY ORDERED SETS

CHAP. I1 SEC. 51

Sub 1): Suppose there exists h E Do such that D (h) = my+ and let f = For each a < my+, we let p’a be the (h’a)-th fixing point of f:. Each p’a is also a fixing point o f f , since f~ G(f,h, h’a). Hence we have p”wt+, < K,+ and so there exists x E s such that the set q = { a < wt+ p’a = x} has power Kc+l. The set w = (f:’x; a E q} is a n exclusive system of power K,+ I in 6,’ a contradiction. This proves 1). Sub 2): Iff

E

u then we construct h

E

Do as follows.

Iff = A’u we are finished sineeft = A’u for all h E Do. Otherwise we let h’O = y, where y is the least number such that f ~ G ( f t ,7); such a number exists since f += and bothf and f; are elements of the exclusive system u. Suppose that we have already constructed h’b for all < a ; hence f j for all j 5 CI. Tff = f: then we are finished. Otherwise we havef E n G($, h’j)

ft

and f $:

ft; we let h’a = y where y is the least number such that f E G(f2, y). B
We have f E

n G(f;,

BSa

h’p) and the induction may be continued. In this

way we construct a function h

E

Do; hence we have

ft = f for a = D (h).

The following theorem can be proved similarly: 2525. THEOREM (TSS’,El). If Norm(Z) 5 wo and if no b, is No-partib, is at most countable. tionable then every exclusive system in UES

Proof. We proceed as above; we construct the fixing points xf forf E

n’6, XES

and n E w, the function G(f, n), the class D of all functions with D (h) E On and W (h) E w and the class Do. For any h E Do the set D (h) is finite; for any f~ u there is some h E Do and some n such that f = f;; hence Do is countable which proved the Theorem.

2526. THEOREM (TSS’,El). Let Norm ( 1 ) 5 Ka where a is a limit number. If NB < K, and if no b, is KB-partitionablethen every exclusive system in b, is of power at most Sup 2”.

n’

b
XES

Proof. We let l ( y ) = {z E 1; z < K Y }for any y < a ; Z(y) is an ideal and 1). We let a(y) = (a(y), 5 ( y ) ) = b, so that a = U a(y

I = U Z(y

+

Y
+

XES

y
+

1). If u S a is an exclusive system in a then u n a(y) is an exclusive 1). system in a(y); we have u = U u n a(y B
+

By the preceding Theorem we have u n a(y follows immediately.

+ 1) < 2’7;

the result now

135

2527

[CHAP. 11 SEC. 5

SEPARATIVELY ORDERED SETS

2527. DEFINITION (TSS’). If t is a nonempty subset of s and if 1 is a cut in s then the restriction of the cut 1 t o t is the set l / t = 1 n P ( t ) . 2528. Reniark (TSS’). 1 ) Thc restriction of 2 to t is a cut in t . 2) If 1 is a cut then 1 = Z/u(l). 2529. THT;oREM(TSS’). Let I be an ideal on s and suppose that, for any x E s, b, is an ordered set. If t _C s, t 0 and s - t =+ 0 then n ’ b , is isomorphic to h, 0 b

n”‘

XE I

l-I2’(‘-‘’

+

XES

X.

XES-t

Proof. Excrcise. We shall now study products of separatively ordered sets as bases for complcte Boolean algebras. l f b is a complete Boolean algebra then the set of all non-zero elemcn~sendowed with the canonical ordering is a base for 6. This lead\ us to the following 2530. DEFINlrroN (TSS’). If b is a complete Boolean algebra with more than two elements then

is called the cnnonicrrl base for b.

6

=

( b - {Ob}. 5 )

is called the canonical base for b w i t h the unit.

As a conseqitence of Lemma 2522 wc obtain the following 2532. LEMMA (TSS’). Let b,, b , be complete Boolean algebras and let 1 be the ideal P((0, I}). Then the separatively ordered sets s i,, b, are xe2 isomorphic.

no

2532. R e i m r k . We shall denote by b, 0b , the complete Boolean algebra with the base 8, c $,. We first prove a theorem on uniqueness. 2533. T H ~ O R E (TSS’, M El). If I is a cut in s and if for any x E s both and d, are bases for a Boolean algebra b, then the complete Boolean algebras c and d which have bases n ’ c , and n ’ d y respectively*) are XES XES isomorphic. c,

*) We say that an ordered set (a, ’) is a base for a complete Boolean algebra 6 if a is a base for b and 5- is the canonical ordering of b restricted to u.

136

CHAP. 11 SEC. 51

2535

SEPARATIVELY ORDERED SETS

Proof. We may suppose w. I. 0.g. that d, = b, for any s. By Theorem 2438 it suffices to prove that nzc, is a base for the algebra d. If u is an XES

n'(6,)such

element of d then there exists an element f of the ordered set

XES

that f 5 u (in the canonical ordering of d). We let g be a function with the same domain as f such that g'x E cx and g'x S x f ' x for all x E D ( 9 ) ; there is such a g, since, for every x, c, is a base for b,. This g belongs to nl c, and we have g 5 f 5 u, so that c, is a base for d. XES

n'bx

2534. TIIEKOEM (TSS'). If and if t 5 s then

n"'b,

is a base for an algebra 8

XES

is a base for some subalgebra

6,

of

=

(6,cT,f)

b.

XES

Proof. We denote n'b, by ( a , 5 ) and n l " b , by ( a / t , S / t ) . Let 6, be the set of all joins Vz such that z E a/f. We shall prove that (6,, C 6, f P (6,)) is a subalgebra. The set 6, is clearly closed under joins, so it is sufficient to show that 6, is closed under complementation. For any f~ n we denote by f / t the unique element g of u / t which assumes in s the same values as f . We have f 5 f / t a n d f , 5 fi -,fl/t 5 f 2 / t . On the othcr hand, for any f E a and g E a / t we have f 5 g -+ f / t 5 g. It follows that

r

/"

f

A

g

=

06

-+

(f/t) A g =

for f

Ob

E u

and g E a/t ,

and

f

A

(Vq) = 0, -+ ( f / t )

A

(Vq) = Or

for f E a

and

q E a/t .

If u E 6, and u = Vz for some z c a / t we let Z be the set of all g E a/f such that g A u = OP We prove - u = VZ. We have Vz v VZ = 1,. Indeed, in the contrary case there exists f~ a such that f A (Vz v VZ) = $, so that ( f / t ) A (Vz v VZ) = h: hence ( f / t ) A Vz = OK and so f / t E 2, a contradiction. Further we have Vz A VZ = 06; for, otherwise there exist g E z and g E Z such that g A S OK, which contradicts the definition of 2. Thus 6, determines a subalgebra which evidently has ( a / t , as a base.

+

2535. THEOREM (TSS', El). Let

s/t)

IT' 6, be a base for a complete Boolean

XES

,

algebra 5 = (6, C, f); suppose that Norm (1) 5 K,+ and that 7 p( b,, K,+ If u E 6 is of power K, then there exists a set t of power

n'

XES

K, such that the algebra determined by the base

a subset.

nr''

b, contains u as

137

2536

SEPARATIVELY ORDERED SETS

n’

[CHAP. I1 SEC. 5

s).

be an exclusive Proof. We denote 6, by ( a , For any f E u let set of elements of a such that Var = f and for each z E a/. let fix ( 2 ) be the set of all fixing points of z. If we set a, = U ar and t = U fi ( z ) then f

r-

z=u

has power 5 Xz, since every set fix ( z ) has power at most Naand a, is the

union of at most Xu sets of power at most Nw It follows that u E where 6, is the algebra whose base is 6,.

n”’

6,

XES

2536. Example. Let b, = at (2) for x E o and let I = P,, (a).Denote by Cant the algebra with the base b = n ’ b , . Then Cant has no atoms. XEO)

Remark. This construction can be generalized; generalized Cantor algebras will be studied in Chapt. 6 Sect. 1.

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