On measurable cardinals violating the continuum hypothesis

Annals of Pure and Applied North-Holland

Logic 63 (1993) 227-240

227

On measurable cardinals violating the continuum hypothesis* Moti Gitik School of Mathematics, Raymond and Beverly Sackler Faculty of Exact Sciences, University, Tel Aviv, Israel

Tel Aviv

Communicated by T. Jech Received 26 November 1991

Abstract Gitik, M., On measurable cardinals Applied Logic 63 (1993) 227-240. It is shown that an extender This together with the Weak is necessary for a measurable

violating

the continuum

hypothesis,

Annals

of Pure

and

used uncountably many times in an iteration is reconstructible. Covering Lemma is used to show that the assumption O(K) = K+~ K with 2” = K+* ((u > 2).

Core models with extenders were constructed by Jensen [7] and Mitchell & Steel [ll]. Suppose that there is no inner model with a strong cardinal. We shall use the following two properties of the core model with maximal sequence of extenders. (1) Any elementary embedding i : X(S)+ M, with M transitive, is an iterated ultra-power of X( 3). (2) If 6 is an ordinal so that there is the maximal K, X2 < K C 6 with o 4(~) > 0, then the following holds: for every a E 6 of cardinality SK, there exists a* c 6, a* E X( 9) )a*)X(flcK and a*za. Let us refer further to (l), (2) as WCL (Weak Covering Lemma). Assuming WCL we are going to show that the assumption O(K) = K+- is necessary for a measurable K with 2”= K+~ (a> 2). And O(K) = K+@+ 1 is necessary for a = /? + 1 where p is a limit with cf p G K. The new point will be to show that it is possible to reconstruct an extender which was used in an iteration Correspondence to: M. Gitik, School of Mathematics, Raymond and Beverly Sackler Faculty of Exact Sciences, Tel Aviv University, Tel Aviv, Israel. * Main results of this paper were presented at UCLA during Fall ‘91 in a short course given by the author. The present version is much influenced by this course. We would like to thank all the participants and especially Tony Martin and John Steel for various remarks and corrections. 0168-0072/93/$06.00

0

1993 -

Elsevier

Science

Publishers

B.V. All rights reserved

228

M. Gitik

more than w,-many times. Notice that it is trivial for measures. The proof will use WCL and will not depend on a special structure of K(S). So it is possible to replace V and X(9) by any two universes W, 1 W, satisfying WCL. Extending methods of [2] it is possible to construct models satisfying “2” = K+ a + K is a measurable” starting with O(K) = K+ Lywhere & = p + 1 and p is a successor ordinal, or 0 is a limit ordinal of cofinality >K. Together with Woodin’s model for limit p’s of cofinality SK from O(K) = K+‘+’ + 1, this will give the exact strength of failure of GCH over a measurable (modulo WCL), except the case when a is a limit ordinal of cofinality >K. 1. Reconstruction

of extenders

Let U E LY’f(6~) be an extender. By [l] there exists an extender I/* G ?P’(‘K), 6, which has the same ultrapower as II and is (K, h)-normal, i.e., for every a<& [h,lua=a, where ~,:‘K+K, &.(a)=~,. Letj:V*M be the canonical embedding of V into the transitive collapse of the ultrapower by U*. If for some Y “M c M then we say that U* is v-closed and if Y = crit(U*) then simply closed. For every sequence a0 < a, < . - . < an-l = supp X and if X’={a bsuppX(aeX} then (yo,...,yn_i)~cX iff ( CX~_~, y,_l)> E X’. Actually only the ultrafilters (U(,, , a < A) ((a”, YOL *. . ) can be used. Just fix some JG:[K]<“* K. Let (Y(~

Theorem

1.1. Let i : V + N be an elementary Then in i 1 X(9)

closed under w1 sequences. w,-many times.

embedding with N transitive and no extender was used more than

Remark. The proof will actually show that closed extenders after o,-many steps.

are recognizable

Proof. Suppose otherwise. By WCL, i 1 X(9): .X(9)-+ X(i(9)) is an iterated ultrapower. Let i* : X( 9) + XC(i(9)) be the corresponding increasing iteration, i.e. with nondecreasing sequence critical points. Notice that if the same extender was used &many times in i 1 X(9) it will be the case also in i*, for any ordinal S. Let i* be the direct limit of embeddings i, : N,-, N, for Y < 8, where No = X(9), NB = .X(i(P)), N,, = X(i,(9)) and i,,,, : N,+ N>/i,(S)(y,, 6,) = N,,,,. Suppose that there are ( vcu1 (Yc o,) and 6 < oivo’“‘(y,,,,) such that for every IX< o,, ivmv,+,:NV*+ N~~*/i,,(F)(yY,, i,,,,,(S)), i.e., i,,,(S)(y,,,, 6) is used in the iteration i*, w1 + l-many times. Our aim will be to show that then i,,,(S)(y,,,, ivoy,,,,(8)) E N.

Measurable cardinals violating the continuum hypothesis

229

By the next lemma, it will be enough in order to derive a contradiction. use the lemma with X(9*) = NV,,, B* = i,,,(9), (Y= y,,,,, /3 = 6 and j = i,,,0. Lemma

1.2. Suppose

X(9*).

Assume

that j : 3T(9*)-+

is an iterated ultrapower

that for some transitive model M of ZFC M k “X(~(JF*))

core model with maximal sequence j(S*) transitive model

5Y(j(S*))

and every embedding

is an iterated ultrapower”.

Just

Suppose

of X(j(.F*)

that for

some

of

is the into a

(a, p) E

dom 9*, 9*( a; /3) E M and crit( j) 2 (Y.Then ( LY,/3) E dom j( 9*) and 9*( (Y, p) = j(S*)(a,

P).

Proof. Since j(9) is a maximal sequence in .%!(j(%)) for M, it is enough to show that Ult(x(j(%*)), S*((Y, /I)) is well-founded. If N* c N then this is trivial. Let us denote X(9*) by N and X( j(S*)) by N*. Note 9((u) rl N = $??(a) fl N* since crit(j) 2 LY.Pick some regular A big enough to include everything used in the iteration process. Inside N we iterate now every extender of 9* sitting on LY or above cy &many times. Let N* be the resulting model. Since Nh c N, Ult(N,, %*(a, p)) is well-founded. It is easy to find an embedding k: N*- Nh. Let us define an embedding k* :Ult(N*, P*((Y, /3))+ Ult(N,, $*(a, /I)). Set

k*(kl.m.d = [k(g) 14w~.o Since 9((u) fl N* = Y((u) fl N = ??‘(a) fl N,, k* will be an elementary witnessing well-foundness of Ult(N*, 9*((u, 6)). 0

embedding

Let us assume for simplification of notation that v. = 0. Denote yVyy, by K~ for every N d wi and let U denote i, ,,,,(%)(K,,, i,, ,,,,(6)). Since the sequence (K, ( cx< q) is in N, it is possible to define the normal ultrafilter which is below U (in the Rudin-Kiesler order). A set X E K~, , X E X(&,,(S)) belongs to it iff all but boundedly many K,‘S are in X. In the same fashion, using different sets of indiscernibles it is possible to reconstruct in N every measure over VK,,,, which is below U. The problem is to put all these measures together in order to catch the hypermeasure U. Note that U (or more precisely its ultrapower) is a limit of a direct system consisting of such measures (their ultrapowers) with elementary embeddings connecting them. We shall define special sequences of indiscernibles and a well-order on them, which will be appropriate for the order of appearing measures over V,,,,, in U. Definition

1.2.1. A sequence rZ= (z, ) a < co,> is called a good sequence iff (a) for every @, K, s rn < K,+ 1; (b) for some a(], for every a 3 Q, r, = iYn,,Yg(r~,,).

For arbitrary sequences t = (r, 1 a< co,) and b= (0, [ a< 0,) Z2 ii iff for all but countably many LY’S,r, 2 0,. Claim 1.2.2. Each good sequence appears in U.

3 defines a measure

U, over

K(,,,

let us write

in

N which

230

M. Gitik

Proof. For X E K~, , X E YC(i,,(s)) set X E U, iff 3aVp 2 (Yrfl E X. Clearly U, E N. Let us check that U, appears in U. Pick CQas in Definition 1.2.1(b). Use r,,, to define the measure U,eo in i,,,,(X(%)) as follows X E Km,, iff rn,, E iV,,,V,,,+,(X). Now i v‘?,,v o,,(UT.,,,>= UF

IJ

Clearly, the order c on [K,,] a1 is a well-order on good sequences. The basic approach is to use good sequences with their order to generate all the measures appearing in U and then to take the direct limit for reaching U. The problem is that inside N it is hard to distinguish between good and not so good sequences. Let us define inside N by induction a sequence of good sequences ( tB I/? < p*>. Case 1: 0 = 0.

Set t”=

(K,

( a<

WI).

Case 2: /3 = p’ + 1.

Set @ = (rB,’ + 1 ) a < ml). Clearly such defined such. Also tB is the least sequence >SB’.

tP is good provided

tB’ is

Case 3: /3 is a limit ordinal and cf /3 > ol.

Let us show that there exists the least upper bound of (@’ ( 6’ < 6) and it is a good sequence. It is enough to show that the least good upper bound % of (@’ 1j3’ < /3) is also the least upper bound. Suppose otherwise. Let ij be an oi sequence above every Tp’ (/3’ < p) and below ?. Actually we should also consider the possibility when ii is below f only unboundedly many times. But then we just shrink ourselves to these particular places and proceed in a similar fashion. W.1.o.g. let ne < r, for every CE< w1 and z, = iov,(zo) for every (Y> 0. Consider good sequences 3” = (ivar,,p(q(y)1 (Ys /3 < o,) for LY< ol. Using an appropriate inductive assumption, we can assume that each such sequence appears in ( fp’ ) p’ < /3). Pick now y < 6 which is above the indices of fin’s (a < w,) in ( iB’ ( p’ < p). Then, there is cue< wi so that for every (Y3 (Y(), rl, > r’,. But it implies &(q,) > &(z@ = ~1; for every /3 3 LY.This means that ijO’> fy which contradicts the choice of y. So f is the least upper bound of ( rp’ I/3’ -=z/3). Set tp = t. Case 4: p is a limit ordinal of cojinality o.

As in the previous case let us pick the least good upper bound Z of ( rc(’ 1/3’ < fi) and show that it is the least upper bound. Suppose otherwise. Let 17be an w,-sequence, fl tfl for every p’ Qa. Since cf rc,, = wl, there exists g< w1 and {&!, 1n < o} EKE+, such that i,,,(P9 = Pn for every n. On the other hand, there exists x, c < x < or, such

Measurable cardinals violating the continuum hypothesis

231

since ij > 1On for every 12 and cf that for every ntF K 01 = co1 > co. But then, for every n < o, x G y < wl, r]$ = iyxyy(qx) > iyxyy(z~) = @ which implies fl x > ton. Contradiction. So t is the least upper bound of ( ZB’) j3’ < p). Set ts = ;t. Case 5: /3 is a limit ordinal of cofinality w,. Then in YC(i,,(.F)), cf /3 < K~, by the Covering Lemma. Let us define now a game with a winning strategy leading to the right choice of ,p. We define a game 99as follows. The first move: Player I picks a subset B of p + 1, B E X(&,,,(S)) of cardinality K@, in .7t(i,,(S)). Player II answers by picking a sequence ? = (7,, 1 y E B) such that (a) for every y E B, vu E flIidw, Ki+l; (b) there is cr* < w1 such that for every my,a* G (Y< o,, for every i, j < K,, Yi<

Yj iff

gy, 1 (WI\~)
1 (OI\~)

where ( yi 1i < K,, ) is an enumeration of B in YC(i,,(%)). Suppose now the first IZmoves in % are defined. I: BO B, ... 11: q” y’ . .

B,_, -n-l Y

.

.

Then the n + lth move of Player I is a subset B, of /3 + 1, B, E X(&,,(9)) of cardinality of K,, in X(i,,(9)) and B, 1 B,_]. Player II answers by picking a sequence P” = (Y”y1 y E B,) satisfying (a), (b) as -t,above and in addition also (c) starting with some LY(IZ - 1, n) < o, Y”y1 (WI\(Y)= V’ny_i1 (w,\e) for every (~3 cu(n - 1, n) and y E B,_l with index
in the X(&,,,(S))-least

Claim 1.2.3. Player II has a winning strategy. Proof.

Suppose otherwise. Then, since the game is open and so determined, the player I has a winning strategy. Let o be such a strategy. Consider the following play, where I will play according 0 but II chooses his moves in V. BO

B, . . . 60

t1

B, ...

f”

... ...

Just let each t,, be a good sequence for measures in B,. Notice that each B, is in X(&,( 9)) which is a direct limit of X(i,(S)) (a < o,), and so there always exists a good sequence t,, which is a legal move in the play. But N is closed under w-sequences of its elements and also for every & < wl, t,(m) can be coded by an

232

M. Gitik

ordinal. So the play (Bo, to, . . . , I?,, tn, . . . ) is in N. Since I used his winning 0 strategy during the play it cannot be infinite. Contradiction. For a winning o(~,, U {p}), i.e. o({P)) = f6, i.e. order to continue

strategy o for Player II, let a( {p}) be the last sequence of the one corresponding to p. Clearly, there is o E N such that the good sequence for /3. So the following will be enough in the process of picking good sequences inside N.

Claim 1.2.4. For every winning strategy o E N for II, a((j3)) 2 T6. Proof. Suppose otherwise. Let u be a winning strategy for II in N and the inequality a( {/I}) > Zp does not hold. Let a({P}) = co and assume for simplicity that for every a! < ol, ~“(a) < rs( (Y). Define a function fo, by setting fi,( (u) = Y”((Y)for every a. Let B; = {i_(v()(a)) ) cu< co,}. Using the Covering Lemma, we pick B1 E p, B, E X( 9) of cardinality K,, there containing BI. We proceed in playing against o by moving B, U {p} next. Let o(B,) = {v;) y E BI U {p}}. Define a function fi on w1 as follows. f,(a) = min{E 1there is y E B, with index

Proof. Pick a* < ~r)~to be big enough to satisfy conditions (b) and (c) of the definition of $9. So above it the sequences (Vi ( y E B,) are in the right order and Then y = iaw,(yO(a)) E BI, y < p respect V”. Let now & 3 a*. Consider Ye’. and the index of y in the X(&,(9))-least enumeration of B1 U {p} is less than K,. The last property holds since w.1.o.g. we can assume that the support of B1 U {p} is below LY*and so at stage K, of the iteration we already get the preimage of this enumeration. So, by (c), V;(o) < Yfj(Ly)= V”(Ly)=&,(a). But then i,,,(Y:(a))

< i,,

(Y”((Y)) = y. Hence, OGf,(a)

S ~:(a).

q

Now we continue and define B; = {ino,(fi( a)) 1 a < w1 and fi( CX)L 0} , its covering B;’ E X(&,,,(S)) of cardinality GK,, and the next move B2 = B1 U B;‘. Let ~(B,U{P})=(Y~~\YEB~U{P}). Define a function f2 on w1 as follows. &(cu) = min{E ) there is y E B2 with index

Measurable cardinals violating the continuum hypothesis

We continue in the same fashion to obtain there will be an infinite decreasing sequence

233

Bg, fi B4, f4 * . . and so on. Finally

&,>f,>.*.>fn>... where

>

means

impossible

since

less for

decreasing

infinite

picking

all but

boundedly

(Y big enough

sequence

many

a’s

below

we will have j,(a)

of ordinals.

u,.

>f,(a) qClaim

Contradiction.

this

is

> . . . , i.e.,

But

a

1.2.4.

Notice that the construction of 3”‘s for j3 of cofinality Cases O-4, can be presented using game 99. 0

different

Let us turn now to the cofinality w-times in an iteration i 1 YC(S)

hypermeasure

o.

Suppose

some

from

o,,

i.e.

is used

where i : V-t N, “‘N c N. Can it be reconstructed inside N? We do not know the answer to this question. It seems to us that it is possible. The following provides an answer under stronger assumptions. Proposition

1.3.

Let i : V+

N be an elementary embedding

with N transitive and

closed under w,-sequences, K = crit(i). Suppose that O(K) < (K+~")~'~~. Let i* the increasing rearrangement of i r 5Y(!9) be the direct limit of embeddings i, : NC,+ N, for Y < 8, where N,, = X(S), N, = X(&(g)) and iv,,+, : N,,+ Np/iy(S)(y,,, 6,) = N v+l. Suppose that there are ( Y,, 1n s CO)and 6 < oivc’(“)(y,,,,) such that for every n s w t%v”+,.N,,,~N~/i,,(S)(y,,,, . i.e., the hypermeasure Then i,,,(Wy,,,, Remark.

i,,,(B)( yy,,, 6) is used in the iteration i* o + l-many times.

j L,,,,(W)

If i,,,,(S)(y,,,

iv,,y,(~))J

E N.

6) is a closed extender,

then w + 1 can be replaced

by o.

Proof. Let us preserve the notations of Lemma 1.2. The basic idea used in the proof will be to consider the game similar to that of Lemma 1.2, but in which the player II should survive o, + 1 moves in order to win. If II has a winning strategy, applied.

then the arguments similar to those in the end of Lemma 1.2 can be The fact that O(K) < (K+~)"'* will be used for construction of a winning strategy for Player II. Define in N by induction a sequence of good sequences ( tfi I/3 < p*). The cases /I = 0, /3 = /3’ + 1 and p is a limit ordinal of cofinality >o are as in Lemma 1.2. Let us consider the cofinality-w case. We split it into two cases according to p < (&J’K(i,,,(:~)) or p > (K<+)“(‘,,z(*)). N otice, that the length of the hypermeasure we are going to reconstruct O(K) < (K+~~)lrc(.~).

is less than

(K:“)~(‘“~(~~)), since

we assumed

that

Case A: /jj < (&J.%&~))_ Notice, that all the measures { U,p. 1p’ < p} are different. Also the Pth measure should be above all of them in the Rudin-Kiesler order. So it should be different from each iY?r (p’ < /3). Let us prove the following.

234

M. Gitik

Claim 1.3.1. Let II, be a measure over Then there exists the least sequence

K,

in K(iru(9))

defining

defined by the sequence

U, and this sequence

ij.

is a good

sequence. Proof. For every n < w let ?j” = (i,,,y,n(qn) ) m 2 n’, n’ = min{l 1K~> qn}). Den-

ote U,. by U,, and II, by U. We claim that U = U,, for some n. Suppose otherwise. By the Covering Lemma there exists a set YE X(&,(9’)) consisting of measures on K, so that U$Y, Yz{Zl,\n<~} and (Y(“lE(ioJ(4))=~w. Let Y={V,l y< K,}. We work for a while inside X(&,,(9)). For every y < K", pick X, E U \ V,. Let X = {r < K, ( Vy (y
For every sequence

E generating

U, ,$a t.

Proof. Suppose otherwise. Let E be a sequence every n there is m > n such that &,, < r,. p = (&,&&;I) [ m 2 n’, n’ = min{f ( K[ > qn} for above shows that all but finitely many of them many m’s 5” < t and each of the Em’s is a good choice of Z. qSubclaim qClaim 1.3.1.

generating U and so that for Consider then the sequences II < o). The argument used generate U. But for infinitely sequence. This contradicts the

Now pick fP to be the least sequence among the sequences generating measures which are different from { Uip. 1j3’ < /?}. Note that by Claim 1.3.1, the least sequence generating a measure is good and so the order ‘G’ is a well-order on the set of such sequences. It completes the definition ( tP ( p C (K~)"('"'(~))). Notice, that the same idea can be used in order to define tP’s up to some p 6 (K;++)%(~~~(~))below the point where the same measures would appear in different places in the sequence of &is. Case B: p >

(K:)~~~",(~)).

We define a game similar to that of Lemma 1.2. On stage 6 Player I picks a subset B6 of j3 + 1 which is in X(&,(S)), cardinality K, there and contains l.J6,<6 Bar. The answer of Player II is a sequence +” = ( 9’;3) y E B6) so that (a) for every Y E Ba, $ E h,, K,+~;

is of

Measurable cardinals violating the continuum hypothesis

(b) there

is n* < w such that for every ~1, n* Q n < o, for every i, j < K,

yj < y, iff ?f, 1 (w\n) where

( y; ( i < K,)

(c) for every

< Yt, 1 (w\n)

is an enumeration

6’ < 6 there

Y; 1 (w\n) where

235

is n(S’,

of B6 in X(&,,(9)); S) < m such that for every n 2 n(6’,

6)

= “y”’ 1 (w\n)

y E B6, and has index

order type K,. The game lasts wi + 1 steps.


well-ordering

II loses if at some stage
legal moves. Let us show how to finish the proof provided

of B,, by

+ 1 he has no

that II has a winning

strategy.

Claim 1.3.2. For every winning strategy CTE N for Prayer II, o( {p}) 2 Z”. Proof. Suppose otherwise. Let o be a strategy witnessing this. We define J’s and B;‘s (i < w, + 1) as in Claim 1.2.4. Let us describe the construction only for a limit stage i. First we pick some Blz {Bjz 1i’ < i} in %“(i,(9)) of cardinality K, there and consisting of a set of .X(&(9)) of cardinality K,. Let Bi = IJ B!. Suppose o(B;) = (I$( instead of w,.

y E B;).

Define

the function

Subclaim 1.3.3. For every j < i d ml, 0 a&(n)


h as in Claim

1.2.4 but on w

for all butfinitely many n’s.

Proof. Let j < i. Pick n * to be big enough such that Yi, 9j satisfy the conditions (b) and (c) of the game above n* and also A(n) 2 0 for every n 2 II*. Let n ,> n*. Then fi(n) = -~$(n) f or some y E B,, y’ < y. By the definition of So y’ E B,. By the choice of n* and n > n*, v:(n) = -~iy(n) =fi(n) Bi+,, Y’ E Bj+,. and v;,(n) < v;(n). Hence, if 5 = ~$(n) then i,,,(E) < &,(~$(n)) =&(6(n)) = y’. so, 0 a$@)
Suppose, finally we obtain B,,, (VF’ ) y E B,,) and (A ) j c co,). For every j < oi let n(j, 0,) be as in the condition (c) of the definition of the game, i.e., aboven(j,o,) (V$‘,IYEB~) and (Vy’ly~B,,) actually agree. Pick a stationary Scw,andn*
of the game

is satisfied

above

II* for

(iii) for every n 2 n* J(n)

3 0.

Let now k, j E S, k > j. Pick some n 2 n *. Since the agreement with w, , k and j agree over It. Then as in Subclaim 1.3.3 fk(n)
236

M. Gitik

Claim 1.3.4. Player II has a winning strategy. Proof. Let t be the least good Pick

n* < w and

simplification P*, f* E X(9).

the

of the Th en

least notations for

all

sequence

function that but

f

of indiscernibles : K:“*

+

/?J

/3 = i,,,,,,,(p*)

finitely

many

in and

m’s,

(T6 ( p’ < p).

above YC(i,(9)).

Suppose

f = &,(f*) fm:

K?*

-

for z(m)

for some where

[K(:~*]~~~], g = i,,,,,(g*) fm = i,,,,,,,(f*). Let g* E Y{(s) be the least enumeration and g, = i,,,,,,,(g*) for every m < CO. The sequences of functions (fn 1n < w),

(g, 1n < CO) belong

to N.

Let us define a winning strategy u for the player II. Suppose that the player I has picked a subset B of p of cardinality SK, in X(&,,(9)). Consider an ordinal P(B) = g-‘(f-“‘(B)) and the sequence of indiscernibles for it Z”@. For every cy of-“‘(B) there exists the canonical projection n, of UTmo, onto Uia. For all but finitely many n’s, .7c,(zflCB’(n)) = r”(n). In order to choose the sequences of indiscernibles for measures in B, let us proceed as follows. Let y E B and let LY=f-l(r). Apply .76, to tlYB). Then r] DI= ( na(zHA’(n)) ( n < o) will be a good sequence of indiscernibles for a which is almost equal to t”. Now let us use (fn 1n < 0). Set p* = (f,(n,(z”“‘(n))) ) n < w). Then, as it is not hard to see, 0” will be a good sequence almost equal to t4 The sequence (0” 1 a E B) will give a good sequence of indiscernibles for all the measures of B. Use it with the answer of II, i.e., a(B). Notice that the sequence (0” 1a E B) is defined completely inside N using Z as a parameter. Continue the definition of o in the same fashion. Notice only that it is always possible to satisfy the condition (5)’ of the definition of the game on every stage E, o, 2 5 2 o, since if (B,, 30, . . . > B,., ZEf, . . .I E’ < 5) B, is the play according to o the sequence of indiscernibles ZE. for BE. (5’ < E) is obtained from the single sequence of indiscernibles for g-‘(f-“‘(BE,)). qiClaim 1.3.4. 171.3. Combining Proposition 1.3 with [3] it is possible to find the precise bound values i(K) for i : V * N, “IN c N, under the assumption +K O(K) = K+"'. We

on do

not know what the precise bounds are without this assumption. The following theorem follows from Theorem 1.1. Theorem

1.4 (WCL). Suppose is a measurable cardinal and 2” = K+~ L(4) a 2. 2 K+ oI or there exists an inner model S’(t)-measurable cardinal, where A is the next measurable above t.

Proof. Suppose that there theorem. Then X(9) exists then some hypermeasure i r X(.9), where i : V+ N contradicts Theorem 1.1.

for an with a

is no inner model L(5) as in the statement of the and the Weak Covering Lemma holds. If O(K) < K+~, was used more than w, + l-times in the iteration is an ultrapower of V by a measure over K, which 0

Measurable

If

2”

=

improved

K+P+’

cardinals

for a singular

violating

the continuum

p of cofinality

237

hypothesis

SK then

O(K) TIC+‘+’

can

be

to O(K) Z=K+~+’ + 1.

Theorem 1.5 (WCL). Suppose K is a measurable cardinal 2” = IC+~+’ for a singular cardinal p of cofnality SK. Then O(K) 3 K +F’ + 1 or there exists an inner model L(9) with a PA(-c)-measurable cardinal, where A is the next measurable above z.

Proof. Suppose that O(K) u OVCX pf(“K), K+“s 6 i : V + N is as in Theorem

Then, as in Theorem 1.4, some hypermeasure was used in the iteration i 1 3X(S) where 1.4. Since there is no measurable in YC(9) cardinal I,

4 K+‘+‘.
between K and K+‘+’ (otherwise we would have a ?@(K)-hypermeasure with A measurable), the cofinality of p and of K+[’ is 5 K in Y{(9). Assume for simplicity that the hypermeasure U was used as the first hypermeasure moving K. Let j: Yl(!F)+ X(j(S)) be the canonical embedding by U. Since K+~ S 6 < K+/‘+‘, j(~) will satisfy the same inequality. Then, in Yt(9) (K+[~)~~@>~(K) but in YC(j(S)) (K’“)“~@<~(K). So there are K+~+’ subsets of K+~ of cardinality K in X(9) which are not in X(j(S)). Since YC(j(9)) 2 (P7’p(~))w“w, there are no measurables between K and K+~ + 1 in YC(j(S)). So, every subset A of K+~, A E N, [A( d K, can be covered by a set A* E X(j(S)) of cardinality SK in X(j(9)). Since “N c N, all the subsets of K+@ of cardinality K are in N; in particular, all the subsets of K+~ of cardinality K which belong to X(9). But the total number of their covers is ((Kfp)K)“/Ll(i(4))= K’? Hence some A* c K+~, o f cardinality K should contain A* E X(j(3)) YC(9), i.e., 2” 2 K+~ in YC(%) which is impossible.

K+~+’

subsets Contradiction.

which 0

belong

to

Remark. The proof above actually shows that an extender with unclosed ultrapower cannot be a part of a closed ultrapower. H. Woodin had constructed models with 2” = K+~ from O(K) = K+~ + 1. By Theorem 1.5 it is the exact strength for (Y= b + 1 with a singular /3 of cofinality SK.

The methods of [2] can be used to show that the assumptions O(K) = K+[‘+’ are sufficient for models with K measurable and 2” = K+~+‘, where p is a successor ordinal or ordinal of cofinality above K. It is unclear if it is possible to weaken the assumptions used by Woodin “O(K) = K+ a + 1” t0 “O(K) = K+n” for limit a’s with cf (Y> K. ’ The ideas of this section will be extended in [6] and applied to problems on the strength of the singular cardinals hypothesis. ’ It

is possible by [S].

23X

M. Gitik

2. On the strength of K-compactness Applying Lemma 1.2, we can answer negatively the following question of Mitchell [lo] : Is O(K) = K++ sufficient for a model where every K-complete filter on K can be extended to a rc-complete ultrafilter? A cardinal K with this property is usually called K-compact cardinal. Theorem 2.1 (WCL).

If there exists a K-compact cardinal then there exists an inner

model with a strong cardinal (i.e., L( 9) with o s(~) = On). Proof. Suppose otherwise. Then ?C(9) exists and by Lemma 1.2 {i(K) 1i : V-+ N, N transitive, “IN E N and crit(i) = K} is a set. Let t be a regular cardinal above

the supremum of all its elements. We shall use K-compactness of K in order to construct an elementary embedding i : V+ N, “N c N and i(K) > t. Denote by gK+(r) the set {a ( a c_ r and Ia1S K}. For every (Y< KC let fE :K * K be a canonical function representing a. We shall define a K-complete ultrafilter I!J~concentrating over the set X, = {(a!,, ( Y
Measurable cardinals violating the continuum hypothesis

239

N be the transitive collapse of this direct limit and i : V + N be the appropriate elementary embedding. Then crit(i) = K and i(K) 2 z, which contradicts the choice of z. q Let us conclude the paper with some questions. Question 1. Suppose some extender is used o-times in an iteration where i : V + N, “N G N. Can it be reconstructed inside N? We think that this should be true.

i 1 .9”(g)

The following are related less technical questions. Question 2. What is the precise bound on values

an elementary

embedding wih the the critical point

Question 3. What is the exact strength of “2” =

(Yis a limit ordinal of cofinality

where i : V -+ N, “N c N is

I K?

K+~

+

K

is a measurable”,

where

>K?~

By Woodin [12], O(K) = K+” + 1 is sufficient; on the other hand O(K) = K+~ is necessary. It is unclear how to use the methods of [2] here, since it may be impossible to add a Rudin-Kiesler increasing sequence of ultrafilters over K of length K+~ without making first 2” = K+? Let us now state two questions concerning the K-compactness of K. Question

4. Is the existence

K-compact cardinal

of a strong cardinal enough for a model with a

K?

In order to construct such a model a lot of filters of the core model should be destroyed thus, for example, NSzng ( t h e nonstationary ideal restricted to singular cardinals) where the set {a < K ) cf mK’“3(~)< (u} should be made nonstationary. If we restrict ourselves only to the filters containing the closed unbounded filter, then the following is unclear. Question 5. What is the strength of the following principles?

(a) Every K-complete filter over K containing all closed unbounded subsets of K can be extended to a K-complete ultrafilter. (b) For every stationary subset S of K there exists a K-complete ultrafilter containing S and all the closed unbounded subsets of K. It is unclear if 3K O(K) = K++ is not enough. The best we can show to exist is an up repeat point [4] for the definition. Concerning (b) our conjecture is that its strength is exactly an up repeat point. ‘It is shown in [S] that O(K) = K+~ is sufficient limit ordinal of cofinality >K.

for “2%= K+~+

K is a measurable”,

where

(Y is a

240

M. Gitik

References [l] S. Baldwin, Between strong and superstrong, J. Symbolic Logic 51 (1986) 547-559. [2] M. Gitik, The negation of SCH from O(K) = K++, Ann. Pure Appl. Logic 43 (1989) 209-234. [3] M. Gitik, On the Mitchell and Rudin-Kiesler ordering of ultrafilters, Ann. Pure Appl. Logic 39 (1988) 175-197. [4] M. Gitik, Some results on nonstationary ideals, to appear. [5] M. Gitik and M. Magidor, Extender based forcing, J. Symbolic Logic, to appear. [6] M. Gitik and W. Mitchell, Indiscernible sequences for extenders and the singular cardinal hypothesis, to appear. [7] R. Jensen, Core model up to strong cardinal, handwritten notes. [S] M. Magidor, Changing cofinalities of cardinals, Fund. Math. 99 (1978) 61-71. [9] W. Mitchell, The core model for sequences of measures I, Math. Proc. Cambridge Phil. Sot. 95 (1984) 229-260. [lo] W. Mitchell, Hypermeasurable cardinals, in: M. Boffa, D. van Dalen and K. McAloon, eds., Logic Colloquium 78 (North-Holland, Amsterdam, 1979) 303-317. [ll] W. Mitchell and .I. Steel, Fine structure and iteration trees, to appear. [ 121 H. Woodin, Private communication.