Chapter Eight The Generalized GCH—The Bounded Case

CHAPTER EIGHT

The Generalized GCHThe Bozlnded Case

A. Statement of Easton's Theorem Suppose that in the preceding chapter we had taken Z to be

w x on,where n is a positive integer. Then a proof like that of

Theorem 7.1 would have given

l=Img(K,) # K(SC(dj)) for each m < n. Thence by Theorems 3.45-3.47, we conclude l=Img(Kn)I K(SC(6)).

(8.1)

A proof like that of Theorem 7.2 gives tImg(K,+ 1 ) = K(SC(K,N for

c1

(8.2)

2 n. A proof like that of Theorem 7.8 gives

k1mgWn) = K(SC(Img(Kn - 1 ) ) ) -

(8.3)

Finally a proof like that of Theorem 7.9 gives l=Img(Kn)= K(SC(Img(KJ))

(8.4)

for 0 s m < n. 127

128

8. THE GENERALlZED GCH-THE

BOUNDED CASE

Suppose we should take I = w x w,? Analogously to (8.1) we would conclude !=Img(K,) I K(SC(i5)).

(8.5)

Can we conclude analogously to (8.4) CImg(K,) = K(WImg(Kfn))) for 0 5 m < w ? The answer is that we cannot. Indeed, we cannot conclude this no matter what Boolean algebra we employ, since it can be disproved in set theory. This is done by means of Konig's theorem. Formula 3.16 on p. 177 of Bernays and Fraenkel [l] states the generalized form of Konig's theorem, as given by Ph. Jourdain. In words, this takes the following form. Let J be a nonempty index set. Let f and g be functions from J to cardinal numbers such that for each j E J we havef(j) < ~ ( j )Then .

To apply this to the case at hand, we take J to be w ,f ( j ) = K j , and . Then we have

g ( j ) = fc,

That is

Now suppose that 2UO

= K,

.

Then

N, < (N,)No

= (2Na)Uo = 2 N o x N o = 2 N o

- K,,

which is a contradiction. Carrying out the above argument in our model gives !=Img(N,) # N(SC(i5,)).

(8.6)

kImg(K,+,) I K(SC(i5)).

(8.7)

So by (8.5)

A . STATEMENT OF EASTON'S THEOREM

129

However a proof like that of Theorem 7.2 will give (8.2) for a 2 o. Hence we conclude bImg(K,+

1)

=W~(Img(~,)))

(8.8)

for 0 I m I o. Clearly one can push such results considerably farther. However the question immediately arises concerning in what other ways one can generalize GCH consistently with AxC and the set theory axioms. Can one have generally

2N"= N,+Z for instance? Even more generally, for what functions E can one have 2N"= KE(=)

(8-9)

for all ordinals a? The classical GCH says that this can be done when E(a) = a 1. Godel [ S ] has shown that GCH is consistent with AxC and the axioms of set theory. The axioms of set theory impose certain requirements on E. Thus the argument used in the proof of Theorem 7.9 implies the following requirement.

+

Requirement 1. If a and E(a) I E(P).

are ordinals and a I p, then

As we saw above, Konig's theorem implies certain additional requirements. We now formulate these in precise terms.

Definition 8.2. We write Cf(a) for the cofinality of a, namely the least cardinal B for which there is a function f such that

f(V)
if

V


We now state our second requirement. Requirement 2. For each ordinal a, we must have

(8.10)

130

8.

THE GENERALIZED GCH-THE BOUNDED CASE

To show that this is a necessary condition for (8.9),we generalize the proof of (8.6). Suppose that (8.10) is violated. Then there is a cardinal /? I,K, and a function f such that f ( q ) < KE(a) KE(a)

if

q
1

= f(q)* rC B

Then by Konig’s theorem HE(.)

=

C f(q) < n H E ( . )

r
SCB

= WE(a)Y?*

Now suppose (8.9) holds. Then 2% < (2Na)P = 2 K a x B = 2Na. Are there any other requirements on E? The reader will note that Cantor’s theorem imposes the requirement that E(a) > a. However, as Cf(/?) I /? when /? is a cardinal, we see that Requirement 2 implies that E(a) > a. There is a conjecture that (8.9) is consistent with the axioms of set theory plus AxC for each E which satisfies Requirements 1 and 2. The first enunciation of this conjecture has been attributed to Robert Solovay. Easton [4] has used forcing to prove a slightly weaker result. Definition 8.2. We say that a is regular if it is a cardinal and Cf(a) = a. Otherwise, we say it is singular. Easton’s Theorem. Let E satisfy Requirements 1 and 2, and also the further requirement that if K, is singular, then E(a) is the least y such that: (i) y 2 E ( p ) for each /? < a ; and (ii) Cf(K,) > K,. Then it is consistent with the axioms of set theory plus AxC that

2H“= HE(,) holds for each ordinal a.

In the present chapter we shall prove a “ bounded ” version of Easton’s theorem to the effect that Easton’s theorem holds for a restricted class of E’s, characterized mainly by the fact that

B. SPECIFICATION OF THE BOOLEAN ALGEBRA

131

E(a) = a + 1 for all sufficiently large a. In the next chapter we shall prove the “ unbounded ” version of Easton’s theorem for unrestricted E’s. In this chapter and the next, our proofs owe much to those presented by Easton [4]. Easton acknowledges a great deal of assistance from Solovay, and cites Solovay [lo].

B. Specification of the Boolean Algebra To the restrictions on E set forth in the hypothesis of Easton’s theorem, let us add the “ boundedness restriction,” principally to the effect that the set of a’s for which E(a) # a + 1 is bounded. Definition 8.3. We say that E satisfies the boundedness restriction with respect to 8 iff:

(i) 8 is not a limiting ordinal; (ii) E(8) is not a limiting ordinal; (iii) E(a) = E(8) for 8 5 a (iv) E(a) = a

-= E(8);

+ 1 for E(O) S a.

Theorem 8.2. If E satisfies the restrictions of Easton’s theorem and 8 is a nonlimiting ordinal, then there is an E* which satisfies the restrictions of Easton’s theorem and also the boundedness restriction with respect to 8, and for which E(a) = E*(a) for a < 8.

-=

PROOF.We define E*(a) = E(a) for a 8, E*(a) = E(8) + 1 for 8 I a I E(O),and E *(a)= M: + 1 for E(8) < a. Then the various restrictions on E* are seen to be met. This theorem assures us that if we are interested in making a special definition of E only for a’s in a bounded set and do not care what happens outside this set, then we can do so with an E which satisfies the boundedness restriction but in which only the restrictions of Easton’s theorem apply to the a’s of interest.

132

8.

THE GENERALIZED GCH-THE BOUNDED CASE

Accordingly, throughout the rest of this chapter we shall assume that E satisfies the boundedness restriction with respect to some fixed 8. We shall not specify anything further about 8. Our basis sets will be We define X by (2.24) with Z = products of the form

n

{ q j )l j

J}

(8.11)

for each J such that

K(J n K E , , ) ) < K, for each u < E(8) for which K, is regular. It is clear that Axioms 2.1 and 2.2 are satisfied. We define the Boolean algebra by Definition 2.9. Definition 8.4. If P is the product (8.1 I), we write s ( P ) for J.

In accordance with Definition 2.15, we refer to s ( P ) as the support of P. One can rephrase the definition of a basis set by saying that it is a product (8.1 1) with the property that Kc(@) n H E , , , ) < K,

(8.12)

for each u < E(0) for which K, is regular. For the present chapter the choice of B is quite irrelevant. For simplicity, we take B to be the single identity automorphism. We take r to be the filter of all subgroups of 9. We readily verify that Theorems 2.29-2.31 fail, while Theorem 2.32 is no longer applicable. The theorems of Chapter 3 all hold; however, Theorems 3.44-3.47 are no longer applicable. Thus we have kAxC, but must seek new proofs of the theorems on cardinality.

C.Substitutes for the Cohen Combinatorial Lemma Not only do we not have the countable chain condition (unless 8 = 0), but we do not seem to have a strict generalization of the Cohen combinatorial lemma. However, we do have a generalization of Theorem 7.4, and this suffices for our needs.

C. SUBSTITUTES FOR THE COHEN COMBINATORIAL LEMMA

133

Definition 8.5. If Pi (i = 1,2) are products as in (8.1 1) using

gi and J i , respectively, then the agreement set of P I and P , is defined to be the set ofj's in Jl n J , for which gl(j) = g 2 ( j ) .

Definition 8.6. Relative to any a, we specify the a-decomposition of a product P o f the form (8.1 1 ) by p L = (I {B;") 1 j PU =

n {B;") l j

E s(p) n

K,,,,}

s ( ~ -) K,,,,}.

(8.13) (8.14)

Theorem 8.2. For each ci and each P of the form (8.1 l ) , we have P=PL

AP'.

(8.15)

PROOF.Use (2.6). Incidentally, we note that if E(a) 2 E(O), then P L = P and PU=1. We shall use such abridgements of terminology as speaking of " the P relative to a " to indicate that we are speaking of the P L given by (8.13). Theorem 8.3. Suppose that either E(O) I a or ci < E(O) and K, is regular. For k E K , let Qk E A . Let Q be a basis set such that

s ( Q ) n K E ( ~=) A.

(8.16)

Then there is a basis set R and a set H of basis sets B such that: (i) R I Q ; (ii) s(R) n KE(a)= A ; (iii) K(n) IK,;

(iv) if B E n, then s ( B ) E K,,,,; (v) if B E n, then either there is a k E K such that B A R < Qk or else B A R A Qk = 0 for each k E K ; (vi) if P is a basis set then there is a B E ll such that P A B # 0.

134

8.

THE GENERALIZED GCH-THE BOUNDED CASE

PROOF.By induction on v, for v < K,, we shall define R, and ll, so that they satisfy (i) through (v) with R, and ll, in place of R and ll ; moreover, for p 5 d I v R, I R,. (8.17) First let v = 0. Case 1. Q

no= { l } .

A

Qk = 0 for each

k E K. Then take R, = Q and

Case 2. There is a k E K for which Q A Qk # 0. Then by Definition 2.2 there is a basis set B for which B I Q A Qk . Then take RO = B u and no= { B L } relative to a (that is, take B L and B to be the a-decomposition of B).

Now assume that R, and ll, have been defined for q < v so as to satisfy (i)-(v) and also (8.17). Let s, be the sum of the supports of all the members of ll, for q < v. Then s, E KE(,).Take llv*to consist of every basis set S whose support is a subset of s,. We first establish that K(ll,*) I K,. Case 1. a = j? 1 and CL < E(8). Then by (8.12), each member of ll, has fewer than K, members in its support. There are at most K, members of n,. So there are at most

+

K, x K, x v IK, members of s,. Now by (8.12), each possible support of an S in llv*has at most K, members. As each member of the support of S must be a member of s, there are at most K, choices for each member of the support of S. So the possible number of supports for an S of ll,* is at most (N,)NP = (2NP)NP = 2 N P ' U P = 2 N P = Ha.

For each possible support of an S of llv* one can find a t most 2'P S's with that support. So the number of S's cannot exceed 2u~XK,=K,XK,=Ka. Case 2. E(8) I a. Clearly s, has at most KE(o,members, since in (2.24) we took I = K E ( e ) . By (8.12), each possible support of an S in llv*has fewer than KO members, by (i) of Definition 8.3. Then by (ii) of Definition 8.3, there are at most KE(e) possible supports for

C. SUBSTITUTES FOR THE COHEN COMBINATORIAL LEMMA

135

an S of lI,*.For each support, there are at most 2NaS’s with that support. So there are at most HE(,, S’s. That is, there are at most

K, S’S.

Case 3. a < E(0) and u a limiting ordinal. By the hypothesis of the present theorem, K, is regular. So we can still appeal to (8.12). So as in Case 1 there are at most K, members of s,. As K, is regular, we see by (8.12) that each basis set with its support included in s, must have its support comprised entirely within the first K, members of s, for some < u. However, there are at most K,+lpossible basis sets with support comprised entirely within the first K, members of s,. So

K(ll,*) I

c

Kp+1= K,.

@-=a

Choose a well-ordering S,, of llv*.We now define members B,, of ll, and corresponding R,,* by induction on p. We require that (i) and (ii) hold for each R,,* and Ru*5 R,* if T I r~ I p. At the step p we define

Then s(T,,) n = A by (ii). But s(S,,) G HE(,) by the construction of the S,,. So S,,A T,, is a basis set. Case 1. S,,A T,, A Qk = 0 for each k E K. Then take B,, = S, and R,* = T,,. Case 2. There is a k E K for which S,,A T,, A Qk # 0. Take a basis set B with BSS,,AT,,AQ,. Then take B,, = B L and R,* = B U relative to a. Finally, we take ll, to consist of all the B,, and set

Having defined lI, and R, for v < K, , we put

ll,

ll= V
and

R=

n Rv.

V
Clearly R and ll satisfy (i)-(v). So we must prove (vi).

136

8. THE GENERALIZED GCH-THE BOUNDED CASE

Take P a basis set, and suppose if possible that P A B = 0 for each B E ll . By induction on v let us choose B, E ll, in such a way that if 9, denotes the agreement set of P and B,, for u 5 v then for CT
s ( P ) n s(B,,) E 9,.

(8.18)

We take B, E ll, . Now suppose that the B, have been chosen for u < v in such a way as to satisfy (8.18). Take 9 = s(P) n C s(B,).

(8.19)

UCV

It is clear that 9 is a subset of the s, introduced in the process of defining H,. Then P with its support restricted to Y will be an S,, E ll,*.Corresponding to this S,, a B,, was chosen to go into II,. It was chosen so that s(S,,) E s(B,,) and S,, A B,, = B,, . Hence the agreement set of B,, and S,, must include s(S,,), which is just 9. Hence, since S,, is just P with its support restricted to 9,we must have Y included in the agreement set of B,, and P. We pick this B,, from H, to be our B, . Then

9 E 9,

so that by (8.19) we get s ( P ) ns(B,) c 9,

(8.20)

for each u < v. Thus we have extended (8.18) to v. By the definition of the agreement set of P and B,, we have Now else we would have P

9, E s ( P ) n s(B,). 9, z s ( P ) n s(B,,), A

B,, # 0. So by (8.18) 9, E 9,

for u < 1. By (iv), we have for each v.

9, # 9,

9, E s ( P ) n H E ( . )

(8.21)

(8.22)

c . SUBSTITUTES FOR THE COHEN COMBINATORIAL LEMMA

137

Case 1. a < E(0). Then by (8.12), s ( P ) n HE(,)has fewer than K, members. Hence by (8.21) and (8.22) we must come to a v < K, for which

9, = s(P>n K E ( . ) . But then by (iv) we must have contradicting P

A

9, = s (P ) n s(B,), B, = 0.

Case 2. E(8) I a. Then by (8.12), s ( P ) n HE(=)has fewer than

KO members. Again, by (8.21) and (8.22) we must come to a v c K, for which 9, = 4P) n H E ( , ) 7

and again we get a contradiction.

-=

Theorem 8.4. For t K, and for k E K, let E A . Take a and Q as in Theorem 8.3. Then there are basis sets R, and sets ll, of basis sets such that for z < Nu(i) through (vi) of Theorem 8.3 hold with R , , n,, K,, and Q r , k in place of R, n, K , and Qk.Also, for p I t < K, we have (vii) R, I R , . PROOF.We define R, and II, by induction on z. If z = 0, we take the K of Theorem 8.3 to be KO.Then Theorem 8.3 gives us an R and n which we take to be R , and no.Now suppose we have defined R, and n, for p < z. In Theorem 8.3 we take K to be K, and Q to be R,. Then Theorem 8.3 gives us an R and ll which we take to be R, and n,.

n,<,

Theorem 8.5. Suppose that either E(0) I p or p < E(8) and K, is regular. For T < K, and for k E K, let Q7,k E A . Let Q be a basis set. Then there is a basis set R and sets n, of basis sets such that for z < K,: (i) R I Q ; (ii) K(I7,) I K,; (iii) if B E IT,, then s ( B ) c KE(,);

138

8.

THE GENERALIZED GCH-THE BOUNDED CASE

(iv) if B E Il, then either there is a k E K, such that B A R I Qr,k or else B A R A Qr,k= 0 for each k E K,; (v) if P is a basis set then there is a B E ll,such that P A B # 0; (vi) if we form the 8-decompositions of Q and R, then

QL= RL.

PROOF.We take the 8-decomposition of Q into QL and Qu. In Theorem 8.4 take 8 for a and Qu for Q. Then we get R, and II, with

S(K) n K E ( B ) = A We then take R=QLh

n R,.

r
Now let B E & . Then either there is a k e K, such that B A R, < Q,,k or else B A R, A Qr,k = O for each k E K,. AS R I R,, weget B A R 5 Qr,kinthe first case and B A R A Qr,k = 0 in the second case. Theorem 8.6. Let f be a function from ordinals to nonlimiting ordinals such that if q < 7 thenf(q)
For 7 < K, and f o r k E K, let Q,,k E A. Let Q be a basis set. Then there are basis sets R, and set ns,,, of basis sets such that for n < y and 7 < Nf(,,):

(0 R, 5 Q ; (ii) R,, I R, for (r I q ; (iii) K(n,,,)I Kf(,); (iv) if B E then s ( B ) E NE(/(,)); (v) if B e l l , , " , then either there is a k e K, such that or else B A R, A Q,,k = 0 for each k E K,; B A R, s (vi) if P is a basis set then there is a B E II,,, such that PABZO;

c.

SUBSTITUTES FOR THE COHEN COMBINATORIAL LEMMA

139

+

(vii) if we form the f ( q 1)-decompositions of R, and R q + l ,then (4JL = @,+AL; (viii) If 0 < y and cr is a limiting ordinal, and we formf(a)decompositions, then (RUY =

(n.,).. ,
PROOF.We carry out a series of definitions by induction on q. Let q = 0. Take fl to bef(0) in Theorem 8.5. Then for our Q there is a basis set R and sets n,. We take Ro to be R and

+

K , o = nr

+

To go from q to q 1, we take fl = f ( q 1) in Theorem 8.5 and take Q to be R,. Take the R that we get to be R,+l and take the ll, that we get to be nr,n+l. Let 0 < y and r . ~a limiting ordinal. Then take fl =f(o) in Theorem 8.5 and take Q to be R,,. Take the R that we get to be Ru and take the n, that we get to be n r , u *

nn<,

Theorem 8.7. Let a be an ordinal. For T < N, and for k E K, let Qr,kE A. Let Q be a basis set. Then there is a basis set R and sets n, of basis sets such that for T c K,: (i) (ii) (iii) (iv) BAR

R

< Q;

N(nJI Nu; if B E n,, then s ( B ) E NE(,); if B E n,, then either there

is a k E K, such that Qr,k= 0 for each k E K,;

I QT,kor else B A R A (v) if P is a basis set then there is a B E II, such that P A B # 0; (vi) if B E n,, then N(s(B)) < K, . PROOF. If K, is regular and a < E(8), or if E(8) s a, then we use Theorem 8.5. So let a < E(8), and let K, be singular. Then there is a y < K, and a function g such that 9(9)< K,

if

v <7

140

8. THE GENERALIZED GCH-THE

BOUNDED CASE

Clearly we can take g increasing. For each q, there is a fi such that g(q) = K, Definef(q) = 1 for this fi. Thenfsatisfies the con-

.

+

ditions of Theorem 8.6. We take

and where in the summation we use only those q's for which t < Kf(,)so that IIz,"will be defined.

D. Cardinality Relations As Requirement 2 involves cofinality, we need to show something about cofinality in the model. Our next theorem is an important step in that direction.

Theorem 8.8. If q and p are cardinals and q < Cf(p), then

-(Jm : G(4, P , f )

*

H(4,li,S)

(8.23)

where G(u, u , f ) denotes (x) : X

E U . 3

. (E,y) . y € v . ( x , y ) Ef

(8.24)

and H(u, u , f ) denotes

( z ) : ~ ~ ~ . ~ . ( E x , y ) . x ~ u . y ~ v . ~ ~ (8.25) y.(~,y)~~ PROOF.Suppose not. Therefore if we write S = II(Ef) : G(4, P , f )

. Wq,2i,f)ll

(8.26) then S # 0. Then by Theorem 3.54 there is an f E V such that

s = II w,P , f )

*

H(4, P , f ) II.

From this we have

s II G(4, P , f ) II s IIW%P,f)II.

(8.27) (8.28)

D. CARDINALITY RELATIONS

141

So for each z < q, we obtain by (8.27)

s 5 II(ElY) . Y E P . (5, Then by Theorem 3.13, there is a y

s 5 IIY E P

*

E

Y >Efll.

(8.29)

V such that

(f, Y> E f l l .

(8.30)

We writef(f) for this y , so that

s IIlf(f)< PII.

Then by Theorem 3.19

sI

c Ilf(?)

U
= 811.

(8.31)

Lemma 1. There is a basis set R with RIS

(8.32)

and sets l-Ir of basis sets such that for z < q

K(l-4) Iq ;

(8.33)

and, if P is a basis set with P I R, then there is a B E l-Ir and a 0 < p such that PABZO (8.34)

B

A

R5

Ilf(9= 811.

(8.35)

PROOF.As S # 0, there is a basis set Q with Q I S.Take this Q in Theorem 8.7. Take also K, = q, K, = p, and Qr,u

llf(Q = 811.

=

Consider the R and set of ll, whose existence is asserted by Theorem 8.7. As Q IS, we have (8.32) by (i), while (8.33) follows by (ii). Now let P be a basis set with P I R. Then by (v) there is a B E l-Ir such that (8.34) holds. Moreover, (iv) tells us that either there is a CT < p such that (8.35) holds, or else

B

h

R

A

Ilf(?)

= 811 = 0

(8.36)

for each u < p. So we must show that this cannot be the case. As A B I P I R I Q 5 S, we have by (8.31) and (8.34)

P

'f

A

B

A u
Ilf(f)

= 811 # 0.

142

8. THE GENERALIZEDGCH-THE BOUNDED CASE

So by Theorem 2.27 there is a basis set B1 and an index a < p such that Bl IP A B A Ilf(.r) = 811As P IR , this gives Bi IB

A

R

A

IIf(.r)

= 811,

contradicting (8.36). As P s R,we have

BARZO (8.37) by (8.34). For each basis set B with B A R # 0 and for each t < q there is at most one a for which (8.35)holds. For if (8.35) holds for a1 and a 2 , then 0 z B A R Ilf(.r) = 8111 A Ilf(.r) = 8211 1181 = ~211,

which can happen only if al = a2. We may order the members of Il,as B,,v with v < q. Then for t < q and v < q, we write a,,vfor the unique a < p such that

0 z 4,"A R 5 llf(.r) = 811 if there is one; otherwise take ar,v= 0. Take

(8.38) (8.39)

As in fact q < Cf(p), we must have a < p. Lemma 2. For t < q and'a I y

R

A


Ilf(9= ?I1 = 0.

(8.40)

PROOF.Suppose otherwise. Then there is a basis set P with lJ IR

A

Ilf(8 = 711.

(8.41)

So by Lemma 1 there must be a B,,v E l l r and a 6 < p such that P A B,,v # 0 B,,v A R II l m = $11.

(8.42) (8.43)

D. CARDINALITY RELATIONS

143

By (8.41) and (8.42)

So by (8.43)

0 # 4,"A R

* llf(z) =?I*

0 # llf(z) = $11

A

(8.44)

Ilf(z) =Nl.

Therefore 0 # 118 =7ll, so that 6 = y by Theorem 3.24. But by (8.38), (8.43), and (8.44), we see that 6 = So by (8.39), we have 6 c cr, whereas y 2 cr. Then by Lemma 2 we have

Ilf(9 =?I1 s R' for cr I y c p. However, if y < cr, then 117 2 811 = 0. So for y < p Ilf(z)

= 7 . 7 2 811 5 R'.

Then by Theorem 3.20

IIWr). Y F .f(z)

.

=Y y 2

all IR'.

By a standard transformation of the predicate calculus with equality 118 ~ f ( z < ) $11 IR'.

As this holds for r < v, we obtain by Theorem 3.20 II(Ex) . x

E ij

.8 I f ( x )

< PI1

IR'.

Recalling the meaning off(x), we may write this as II(Ex,y). X E 4 . y < ,tl . 8 Iy . ( x , y )

~ f l l IR'.

But by (8.28) and (8.25)

s III(Ex,y). X E i j . y c p . 8 s y . ( x , y ) Efll. So S I R'. But this contradicts (8.32). If we know that i j and fi are formal cardinals whenever q and p are intuitive cardinals, this theorem will tell us that if r,~ and p are cardinals and q < Cf(p), then C i j c Cf(jl). So we proceed through a succession of theorems which will lead to the conclusion that fi is a formal cardinal whenever p is an intuitive cardinal.

144

8.

THE GENERALIZED GCH-THE BOUNDED CASE

Theorem 8.9. If

r]

Wij) K(2i).

and p are ordinals and K(r]) I K(p), then

PROOF.If K(g) 5 K(p), then there is a one-to-one correspondence from r] to a subset of p. Transferring this correspondence into the formal system gives our theorem. Corollary. If !=K(ij) = K(jl).

r]

and p are ordinals and K(g) = K(p), then

Theorem 8.10. If a and a are ordinals and a < K a + , , then

C N I S ) < K(Img(Ka+1))* PROOF.Let fl s a. If we take r] = K, and

p = Ka+l,then g < Cf(p)). So by Theorem 8.8, the formula (8.23) holds. If

K(ij) 2 KU),this would say that there is a g which establishes a one-to-one correspondence between all of p and part of ij. By letting the other members of i j correspond to 0, we could infer the existence of an f of the sort whose existence is denied by (8.23). So a consequence of (8.23) is CK(Img(K,J) < N(Img(Ka+i)). Let CT < K,+l.Then there is a f l < a such that K(o) = K(K,). So by the corollary to Theorem 8.9

CK(8) = K(Img(K,J). Combining the last two formulas gives our theorem. Theorem 8.11. If

r]

and p are ordinals and K(r]) < K(p), then

CK(ij) < K(J).

PROOF. There must be an a such that K(p) = K,. So by the corollary to Theorem 8.9

b W f i ) = Wmg(Ka))*

(8.45)

Case 1. u is not a limiting ordinal. Then our theorem follows by (8.45) and Theorem 8.10.

E. PROOF OF EASTON’S THEOREM

Case 2.

ct

145

is a limiting ordinal. Then there is a p c a such that

< KB+l.So by Theorem 8.10

W?)< K(Img(K.CB+1)). However, by Theorem 8.9 WImg(K,+ 1)) 5 ~(Img(K,)). Then our theorem follows by (8.45). This theorem is the equivalent of Theorem 3.44. From it we prove the equivalents of Theorems 3.45-3.47 by analogous proofs. Then, as noted earlier, Theorem 8.8 becomes a formal theorem about cofinality.

E. Proof of Easton’s Theorem We are finally ready for our key result, which we do in three steps.

Theorem 8.22. If

ct

is an ordinal, then

Fh3(NE(,)) 5 K(SC(K,)). PROOF. First take the case where ct < E(8) and K, is regular. Then as K, x KE(,)= KE(,), we may identify coordinates less than HE(,)by two subscripts, the first < N u and the second < H E ( , ) . Hence we may define q,, with D(qJ = Vuuby the condition that if X E VH, then

(8.46) This is analogous to Definition 6.3. Then a proof like that of Theorem 6.3 gives

bFC Ka

(8.47)

for p < KE(,).A proof like that of Theorem 6.4 gives b n

# qfi

(8.48)

146

8.

THE GENERALIZED

GCH-THE

BOUNDED CASE

for t~ < p < Thus we have at least KE(a) distinct subsets of Kd.To prove our theorem, we define a formal one-to-one cor-

respondence between fi and qP. Write momentarily p = HE(.)+ 3. We define f with D ( f ) = V, by the condition that if x E Vs then

We verify without difficulty that f is extensional, so that by Theorem 3 . 9 f ~V and

c

IIx E Sll

P < Kd.)

IIX

= ( P , 4JIL

(8.49)

Lemma 2. PROOF.

Take t~ < HE(.).Then

c

P < HE(*)

II (5, 4 2 = ( P , 4J II = 1*

But (is, qa) is in D ( f ) . So by the definition ofJ

t ( 6 46) E f . Therefore C(Ez) . (is, z)

As this holds for each t~ < 3.20.

Ef.

our lemma follows by Theorem

147

E. PROOF OF EASTON’S THEOREM

Lemma 3. k(X,y,Z):(X,Z)Ef.(y,Z)Ef.3.X=y.

PROOF.We wish to show first that II(x,

Case 1. p # II(x,

(P, qJ

2) = 0.

( Y , 2) = (8, q,>Il IIIX = Yll.

*

Since

2) =

(P, q r )

( Y , 2) = ( 8 , 4u)Il IIhjt= 4011

*

we conclude by (8.48) that the left side must be 0. Hence the left side is IIIx = yl(. Case 2. p = (T. Then the result to be proved reduces to

II (X,

2) =

( P , 4J

*

( Y , z> = (P, 4J

II

IIX = Yll

which is certainly valid. If we now sum on p and conclude our lemma.

(T

on the left, we can use (8.49) to

Lemma 4. k(x,y, z) :( x , y )

E

f.(x,

z ) E f. 3 . y = 2.

PROOF.Proceed as in the proof of Lemma 3. From these four lemmas, our theorem follows. Next consider the case where a < E(8) and Na is singular. By (i) and (ii) of Easton’s theorem, E(a) has the minimum value allowed by Requirements 1 and 2. As these reflect classical theorems of set theory, we can use these classical theorems of set theory to prove our theorem. Finally, let E(0) Ia. Then a reference to (iv) of Definition 8.3 discloses that we have E(a) = a

+ 1.

So our theorem takes the form

t ImgWa+,) IN(SC(W).

148

8. THE GENERALIZED GCH-THE

BOUNDED CASE

Since a + 1 is the minimum value allowed by Requirements 1 and 2, we could parallel the proof of the previous case. However, Cantor's theorem gives the desired result more simply.

Definition 8.7. We define C to be the set of all pairs such that: (i) (ii)

(n, s)

is a nonempty set of basis sets;

K(n)I K,;

(iii) if B E n, then s ( B ) E KE(,); (iv) s E II x K,;

(v) if B E n, then K(s(B))< K,.

Theorem 8.13. K(C) = HE(,).

PROOF.Clearly there are at least KE(,)subsets of HE(,) with fewer than K, members because there are that many with one member. On the other hand, there are certainly no more than (KEc,,)Nasuch subsets. Lemma. (KE(a))Na

= KE(a)

Case 1. E(a) is not limiting. Say E(a) = p E(a) > a by Requirement 2, we have

+ 1. Recalling that

Case 2. E(a) is limiting. Note that since K, < KE(,),we can see that (NE(,JNais just the number of ways of choosing Ha elements out of KE(,).Recall that by Requirement 2, K, < Cf(K,(,,). Hence any subset of HE(,) with K, elements must be bounded. Take a sequence of nonlimiting ordinals whose sum is E(a). Clearly we can take each of them greater than a. As in Case 1, if p is one of these, then the number of subsets of K, with K, members is just K,. Adding these for each fi c E(u), we conclude that there are KE(,)subsets of KE(,)with K, members.

149

E. PROOF OF EASTON'S THEOREM

By our lemma, the number of supports for basis sets satisfying (iii) and (v) is KE(,).For each such support, there are at most K, basis sets with that support. Hence there are KE(,)basis sets B satisfying (iii) and (v). As each ll contains at most K, of these B's, the number o f n's is at most (KE(,))'", which is KE(,) by our lemma. On the other hand, there are KE(,)n's which contain one member each. So there are exactly KE(,)n's.For each such n, there are K, members of II xK,, and hence Ha+, possible s's. But a + 1 I E(cc),so that the number of pairs (n, s) is KE(a)

Ka+l

= KE(a)

*

corollary. k ~ ( i = ) 1mg(KE(,)),

The reader should note the parallelism between the present C and that of Definition 7.1. Also the theorem just proved parallels Theorem 7.5. The lemmas of the next theorem parallel Theorems 7-6-7.8, and their proofs have many similarities. The reader will find it helpful to review the proofs of Theorems 7.6-7.8 before commencing to read the next theorem and its proof. Definition 8.8. Let (n, s) E C and y E V. We define II@((n,s), y)I( to be the sum of all basis sets R such that:

(i) If P is a basis set with P I R, and if y < K,, then there is a B E ll such that PABZO

(8.50)

and either B

A

R I IlT EYll

(8.51)

B

A

R I ll-f~yll.

(8.52)

or

(ii) s = { ( B , y ) I B ~ l l . y < K , .B A R S I l f ~ y l I ) . Definition 8.9. Choose and (n, s) E X , then

large enough so that if y

(v, Img(W, s)))

E

v&4 *

E

D(SC(K8))

8. THE GENERALIZED GCH-THE

150

BOUNDED CASE

Define f with D(f ) = Vs by the condition that if x E V s , then f(x) =

c I&%).

x = ( u , fi) . @(w, u ) * u E K&ll.

WSZ

Theorem 8.14. If u is an ordinal, then CImg(K,(a)) = K(SC(KtJ).

PROOF.By Theorem 8.12, it suffices to prove (8.53)

CK(SC&)) 5 Img(KE(a))* This we do in a series of lemmas.

First we note that clearly f is extensional, so that by Theorem 3.9 f E V and

I ~ x E ~ I II 1 [ ~ ( E Ux) .= (u,*). @(w, u). u E K2c,ll.

(8.54)

W € Z

Lemma 1. w

k(y, z ) : ( y , z) E f . 3 . y E Kz.Z € X PROOF.

For w E E and u E V

II(y, z)

= (u,

*) . @(w, u) . u G K&ll I Ily E K& . z E 511.

Summing on u gives II(Eu) . ( y , z) = ( u , z ) . @(w, u) . u

= K1lI I lly E K, . z E Ell. "

Summing on w E Z and using (8.54) gives II(Y,Z)Efll

Lemma 2. Let y

E

V. Then

c lIw%Y)ll

W € Z

PROOF.

I IIy=K,*zE5II.

= 1.

Suppose not. Then

Choose a basis set Q included in the term on the left. Then (8.55)

151

E. PROOF OF EASTON'S THEOREM

In Theorem 8.7 take this Q. Take also K, = (0) and Q,,, = ~ y l lTake . n = Cn, (8.56) r
and

s = { ( B , y ) I B E n . y < K, . B

R I Ilj E ~ I I } .

(8.57)

B A R A I ( ~ E ~ ~ ( = ~ .

(8.58)

A

By Definition 8.7, (n, s) E Z. Also, (ii) of Definition 8.8 is satisfied. So we wish to verify (i) of Definition 8.8. Let P be a basis set with P I R and let y < K,. Then by (v) of Theorem 8.7, there is a B E lTy such that (8.50) is satisfied. Also by (iv) of Theorem 8.7, either (8.51) holds or else As P I R , we have B A R # 0 by (8.50). So B Then by (8.58) and Definition 2.3 B

A

R

A

A

R is a basis set.

Cl(IljEyJ()= 0.

So by Definitions 2.4 and 2.6

B

A

R I Cc(llj EYII).

But then (8.52) holds by (2.8) and (3.1). Hence by Definition 8.8

Therefore

R I II@(W, s>, Y>II.

As R 5 Q by (i) of Theorem 8.7, we get a contradiction by (8.55).

Lemma 3. k(x) : x G

Ks . 3 . (Ez) . (x,

z ) Ef.

PROOF.By Lemma 2 and Theorem 2.22 (8.59)

However

8.

152

THE GENERALIZED GCH-THE BOUNDED CASE

so that for w E C

Summing over w E C and using (8.59) gives

II(Eu) - ( Y , a ) = ( 0 , *) . @(w, V ) . v E KiII. IIY C k I I IWCE Z aC oE

By Theorem 2.21, we can interchange the order of summation in a double sum. Therefore

IIY

E Kill

Take y off

E

5

C C II(Eu). (.v,a> = ( 0 , *) . @(w, 0 ) . u c Hill.

asZ WEE

D(SC(&)). Then (y, a) is in D(f). So by the definition

Hence

Lemma 4. If w

E X and x, y E V, thf.n

II@(w, x ) . x

c K2 . q w , y ) . y

E Kill

I IIX = yll.

PROOF.If w E Z, then w = (U,s) for some II and s satisfying the conditions of Definition 8.7. Take y < H,, and suppose, if possible, that Il@(w, x ) . @(w, y ) . r* E X . -r* EYll # 0.

153

E. PROOF OF EASTON’S THEOREM

Then by Definition 8.8 and Theorems 2.22 and 2.25 there must be an R, and an R2 such that R1

A

R2

1 1 7 ~ N ~~ E . Y ~ I

A

ZO,

where (i) and (ii) of Definition 8.8 hold for R, together with x and for R, together with y. Choose a basis set P such that P IRl

R,

A

A

117 E x . -y

E yll.

(8.60)

Take this P in (i) of Definition 8.8 and conclude that there is a B E ll such that (8.50) holds and either

R, II I ~ E x I I

(8.61)

R, II I - ~ E x I I .

(8.62)

B A P I B A R,

(8.63)

B

A

or B

A

By (8.60), we have and B

A

P I Ily~xll.

(8.64)

If (8.62) could hold, then by (8.62) and (8.63) we would have B A P I Il-y~xlI,

contradicting (8.64). So (8.61) must hold. Then by (ii) of Definition 8.8, we have ( B yy ) E s. But then by (ii) of Definition 8.8 we have B

A

R, Ill7EYll.

(8.65)

B

A

P I II-r’~yll.

(8.66)

But by (8.60) AsP~R,,weget B A P I Ilr’EYll

by (8.65). With (8.50) and (8.66), this gives a contradiction. Thus our supposition is false. So for each y < Ec,, we have l l ~ ( ~ Y ~ ) * ~ ( W Y Y ) II llY E X = ) 7 E Y l l *

THE GENERALIZED GCH-THE BOUNDED CASE

8.

154

By a similar argument, we conclude

IlW% X I WJ, Y)ll 5 Ilr’ E Y 1 7 E XI1 for each y < K, . So for each y < K, , we have I I W W , Y ) . @(w, Y)ll 5 Ilr’ E x = r’ E Yll. Multiplying over y < K, gives *

II@(w, x) . @(w, y)II 5 II(t) : t

E

K, . 3 . t E x = t E yll

I

Then by Theorem 3.6, we conclude the proof of the lemma.

Lemma 5. b ( x , y , z ) : (x,z) Ef. ( y , z)

PROOF.We first show II(x,

z> = (01, $1) Q(w2

Y

4

*

* v2

W W l , 01) * v1

E f . 3 .x

= K,

*

=y.

( Y , z> = ( 0 2

Y

$2)

*

(8.67)

c Kdll 5 IIx = Yll

if wl E Z and w2 E Z. Case 1. wl # w2 . But

II(x, z> = (01, $1)

( Y , z> = ( 0 2

*

9

$,>l

11%

= $211.

So by Theorem 3.24 the left side of (8.67) is zero, and so (8.67) holds. Case 2. wl = w 2 . Then by Lemma 4 IP(w1,

u1)

*

01

= K,

*

WW2Y

4- 02

Kzll

IIv1 = u2II.

Therefore A5

Ilk z> = (u1,

$1)

*

( Y , z> = ( 0 2

Y

$2)

*

01

= u2Il

where A is the left side of (8.67). So in this case also, (8.67) holds. We now sum the left side of (8.67) on v1 and u2 and on wl for w1 E Z and on w2 for w2 E Z. Then by (8.54) we get II(X9

z> E f . ( Y , z> E f l l

Ilx = Yll.

F. A NOTE ON THE PROOF

155

Finally, by Lemmas 1, 3, and 5, we conclude

kK(SC(K,)) I K(2). Then our theorem follows by means of the corollary to Theorem 8.13.

F. A Note on the Proof We have left one question unresolved, namely whether for a fixed l7 and s the formula O((II, s), y ) is extensional as far as y is concerned. Perhaps it is, but no proof presented itself readily. As it turned out that we could evade the question, we simply did so, although this required that we proceed with some care, particularly in the proof of Lemma 5 in the proof of Theorem 8.14.