moti gitik, ralf schindler and saharon shelah- pcf theory and woodin cardinals

8/3/2019 Moti Gitik, Ralf Schindler and Saharon Shelah- Pcf theory and Woodin cardinals

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Pcf theory and Woodin cardinals

Moti Gitik, a Ralf Schindler, b and Saharon Shelah c,

a School of Mathematical Sciences, Tel Aviv University, Tel Aviv 69978, Israel b Institut f ur Formale Logik, Universitat Wien, 1090 Wien, Austria

c Institute of Mathematics, The Hebrew University of Jerusalem, Jerusalem, 91904, Israel, and Department of Mathematics, Rutgers University, New Brunswick, NJ 08903, USA

[email protected], [email protected], [email protected]

Mathematics Subject Classication. Primary 03E04, 03E45. Secondary 03E35, 03E55.Keywords: cardinal arithmetic/pcf theory/core models/large cardinals.

Abstract

Theorem 1.1. Let be a limit ordinal. Suppose that 2 | | < and 2 | |+

| |+ . Then for all n < and for all bounded X | |+ ,

M #n (X ) exists.

Theorem 1.4. Let be a singular cardinal of uncountable conality. If { < | 2 = + } is stationary as well as co-stationary then for all n < and for all bounded X , M #n (X ) exists.

Theorem 1.1 answers a question of Gitik and Mitchell (cf. [GiMi96, Question5, p. 315]), and Theorem 1.4 yields a lower bound for an assertion discussedin [Gi ] (cf. [Gi , Problem 4]).

The proofs of these theorems combine pcf theory with core model theory.Along the way we establish some ZFC results in cardinal arithmetic, moti-vated by Silvers theorem [Si74], and we obtain results of core model theory,motivated by the task of building a stable core model. Both sets of resultsare of independent interest.

1 Introduction and statements of results.

In this paper we prove results which were announced in the rst two authors talksat the Logic Colloquium 2002 in Munster. Specically, we shall obtain lower boundsfor the consistency strength of statements of cardinal arithmetic.

The rst and the third authors research was supported by The Israel Science Foundation.This is publication # 805 in the third authors list of publications.

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Cardinal arithmetic deals with possible behaviours of the function ( , ) for innite cardinals , . Easton, inventing a class version of Cohens set forcing(cf. [Ea70]) had shown that if V |= GCH and : Reg Card is monotone and suchthat cf(( )) > for all Reg then there is a forcing extension of V in which() = 2 for all Reg. (Here, Card denotes the class of all innite cardinals, andReg denotes the class of all innite regular cardinals.) However, in any of Eastonsmodels, the so-called Singular Cardinal Hypothesis (abbreviated by SCH ) holds true(cf. [J78, Exercise 20.7]), i.e., cf( ) = + 2cf( ) for all innite cardinals . If SCHholds then cardinal arithmetic is in some sense simple, cf. [J78, Lemma 8.1].

On the other hand, the study of situations in which SCH fails turned out to bean exciting subject. Work of Silver and Prikry showed that SCH may indeed fail(cf. [Pr70] and [Si74]), and Magidor showed that GCH below does not imply 2 =+1 (cf. [Ma77a] and [Ma77b]). Both results had to assume the consistency of a

supercompact cardinal. Jensen showed that large cardinals are indeed necessary: if SCH fails then 0# exists (cf. [DeJe75]). We refer the reader to [J95] for an excellentlywritten account of the history of the investigation of SCH .

The study of SCH in fact inspired pcf theory as well as core model theory. Weknow today that SCH is equiconsistent with the existence of a cardinal witho() = ++ (cf. [Gi89] and [Gi91]). By now we actually have a fairly completepicture of the possible behaviours of ( , ) under the assumption that 0 doesnot exist (cf. for instance [Gi02] and [Gi ]).

In contrast, very little is known if we allow 0 (or more) to exist. (The existence of 0 is equivalent with the existence of indiscernibles for an inner model with a strong

cardinal.) This paper shall be concerned with strong violations of SCH

, where wetake strong to mean that they imply the existence of 0 and much more.It is consistent with the non-existence of 0 that is a strong limit cardinal (in

fact that GCH holds below ) whereas 2 = , where is a countable ordinalat least as big as some arbitrary countable ordinal xed in advance (cf. [Ma77a]).As of today, it is not known, though, if can be a strong limit and 2 > 1 .The only limitation known to exists is the third authors thorem according to which0 20 + 4 (cf. [Sh94]).

Mitchell and the rst author have shown that if 2 0 < and 0 > 1 then 0exists (cf. [GiMi96, Theorem 5.1]). Our rst main theorem strengthens this result.The objects M #n (X ), where n < , are dened in [St95, p. 81] or [FoMaSch01,p. 1841].

Theorem 1.1 Let be a limit ordinal. Suppose that 2| | < and 2| |+

< | |+ ,whereas | | > | |+ . Then for all n < and for all bounded X | |+ , M #n (X )exists.

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Theorem 1.1 gives an affirmative answer to [GiMi96, Question 5, p. 315]. Oneof the key ingredients of its proof is a new technique for building a stable coremodel of height , where will be the | |+ of the statement of Theorem 1.1 andwill therefore be a cardinal which is not countably closed (cf. Theorems 3.7, 3.9,and 3.11 below).

Let be a singular cardinal of uncountable conality. Silvers celebrated theorem[Si74] says that if 2 > + then the set { < | 2 > + } contains a club. Butwhat if 2 = + , should then either { < | 2 > + } or else { < | 2 = + }contain a club? We formulate a natural (from both forcing and pcf points of view)principle which implies an affirmative answer. We let ( ) denote the assertion thatthere is a strictly increasing and continuous sequence ( i | i < cf()) of singularcardinals which is conal in and such that for every limit ordinal i < cf(),+i = max(pcf( {+ j | j < i })). Note that if cf (i) > then this is always the case on

a club by [Sh94, Claim 2.1, p. 55]. We show:

Theorem 1.2 Let be a singular cardinal of uncountable conality. If () holdsthen either { < | 2 = + } contains a club, or else { < | 2 > + } containsa club.

The consistency of () , for a singular cardinal of uncountable conality, isunknown but the next theorem shows that it is quite strong.

Theorem 1.3 Let be a singular cardinal of uncountable conality. If () failsthen for all n < and for all bounded X , M #n (X ) exists.

The second main theorem is an immediate consequence of Theorems 1.2 and 1.3:

Theorem 1.4 Let be a singular cardinal of uncountable conality. If { < | 2 = + } is stationary as well as co-stationary then for all n < and for all bounded X , M #n (X ) exists.

The proofs of Theorems 1.1 and 1.3 will use the rst many steps of Woodinscore model induction. The reader may nd a published version of this part of Woodins induction in [FoMaSch01]. By work of Martin, Steel, and Woodin, the

conclusions of Theorems 1.1 and 1.4 both imply that PD (Projective Determinacy)holds. The respective hypotheses of Theorems 1.1 and 1.4 are thereby the rststatements in cardinal arithmetic which provably yield PD and are not known to beinconsistent.

It is straightforward to verify that both hypotheses of Theorems 1.1 and 1.4imply that SCH fails. The hypothesis of Theorem 1.1 implies that, setting a = { Card( a ). The question if somesuch a can exist is one of the key open problems in pcf theory. At this point neitherof the hypotheses of our main theorems is known to be consistent. We expect futurereasearch to uncover the status of the hypotheses of our main theorems.

Theorems 2.1, 1.2, and 2.5 were originally proven by the rst author; subse-quently, the third author found much simpler proofs for them. Theorems 2.4 and2.7 are due to the third author, and theorems 2.6 and 2.8 are due to the rst au-thor. The results contained in the section on core model theory is due to the secondauthor.

We wish to thank the members of the logic groups of Bonn and Munster, inparticular Professors P. Koepke and W. Pohlers, for their warm hospitality duringthe Munster meeting.

The second author thanks John Steel for xing a gap in an earlier version of the

proof of Lemma 3.5 and for a discussion that led to a proof of Lemma 3.10. Healso thanks R. Jensen, B. Mitchell, E. Schimmerling, J. Steel, and M. Zeman for themany pivotal discussions held at Luminy in Sept. 02.

2 Some pcf theory.

We refer the reader to [Sh94], [AbMa ], [BuMa90], and to [HSW99] for introduc-tions to the third authors pcf theory.

Let be a singular strong limit cardinal of uncountable conality. Set S 1 ={ < | 2 = + } and S 2 = { < | 2 > + }. Silvers famous theorem statesthat if 2 > + then S 2 contains a club (cf. for instance [HSW99, Corollary 2.3.12]).But what if 2 = + ? We would like to show that unless certain large cardinals areconsistent either S 1 or S 2 contains a club.

The third author showed that it is possible to replace the power set opera-tion by pp in Silvers theorem (cf. for instance [HSW99, Theorem 9.1.6]), providingnontrivial information in the case where is not a strong limit cardinal, for exam-ple if < 20 . Thus, if is a singular cardinal of uncountable conality, and if S 1 = { < | pp( ) = + }, S 2 = { < | pp( ) > + }, and pp( ) > + then S 2contains a club.

The following result, or rather its corollary, will be needed for the proof of The-

orem 1.4. The statement ( ) was already introduced in the introduction.

Theorem 2.1 Let be a singular cardinal of uncountable conality. Suppose that () there is a strictly increasing and continuous sequence ( i | i < cf()) of

singular cardinals which is conal in and such that for every limit ordinal i < cf(),+i = max(pcf( {+ j | j < i })) .

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Then either { < | pp( ) = + } contains a club, or else { < | pp() > + }contains a club.

Proof. Let ( i | i < cf()) be a sequence witnessing () . Assume that both S 1and S 2 are stationary, where S 1 { < | pp( ) = + } and S 2 { < | pp( ) >+ }. We may and shall assume that S 1 S 2 {i | i < cf()} and 0 > cf().

Let > be a regular cardinal, and let M H be such that Card( M ) = cf(),M cf(), and ( i | i < cf()), S 1, S 2 M . Set a = ( M Reg) \ (cf() + 1).

We may pick a smooth sequence ( b | a ) of generators for a (cf. [Sh96, Claim6.7], [AbMa , Theorem 6.3]). I.e., if a and b then b b (smooth), andif a then J (a ) = J + . Hence, for every C , pp() = + if and only if 2 = + .So Theorem 2.1 applies and gives the desired conclusion. (Theorem 1.2)

Before proving a generalization of Theorem 2.1 let us formulate a simple com-binatorial fact, Lemma 2.2, which shall be used in the proofs of Theorems 1.1 and1.4. We shall also state a consequence of Lemma 2.2, namely Lemma 2.3, which weshall need in the proof of Theorem 1.4.

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Let be innite cardinals. Then H is the set of all sets which are heredi-tarily smaller than , and [H ] is the set of all subsets of H of size . If H is anyset of size at least then a set S [H ] is stationary in [ H ] if for every modelM = ( H ; ...) with universe H and whose type has cardinality at most there issome (X ; ...) M with X S . Let = ( i | i A) H with i for alli A, and let X [H ] . Then we write char X for the function f iA

+i which

is dened by f (+i ) = sup( X +i ). If f , g iA +i then we write f < g just in

case that f (+i ) < g (+i ) for all i A. Recall that by [BuMa90, Corollary 7.10] if

|A|+ i for all i A then

max(pcf( {+i | i A})) = cf( ({+i | i A}), < ).

If is a regular uncountable cardinal then NS is the non-stationary ideal on .

Lemma 2.2 Let be a singular cardinal with cf() = 1. Let = ( i | i < )be a strictly increasing and continuous sequence of cardinals which is conal in and such that 2 0 < . Let < Let > be regular, and let : [H ] NS .Let S [H ] be stationary in [H ] .

There is then some club C such that for all f iC +i there is some

Y H such that Y S , C (Y ) = , and f < char Y .

Proof . Suppose not. Then for every club C we may pick some f C iC

+i such that if Y H is such that Y S and f C < char

Y , then C (Y ) = .

Dene f iC +i by f (

+i ) = sup C f C (

+i ), and pick some Y H which is such

that Y S and f (+i ) < charY . We must then have that C (Y ) = for every

club C , which means that ( Y ) is stationary. Contradiction! (Lemma 2.2)

The function to which we shall apply Lemma 2.2 will be chosen by inner modeltheory. Lemma 2.2 readily implies the following.

Lemma 2.3 Let be a singular cardinal with cf() = 1. Let = ( i | i < )be a strictly increasing and continuous sequence of cardinals which is conal in and such that 2 0 < . Suppose that () fails. Let > be regular, and let : [H ]2

NS .

There is then a club C and a limit point of C with cf( ({+i | i < }), < ) >+ and such that for all f iC

+i there is some Y H such that Card( Y ) = 2 ,

Y Y , C (Y ) = , and f < char Y .

Let us now turn towards our generalization of Theorem 2.1. This will not beneeded for the proofs of our main theorems.

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Theorem 2.4 Suppose that the following hold true.(a) is a singular cardinal of uncountable conality , and ( i | i < ) is an

increasing continuous sequence of singular cardinals conal in with 0 > ,(b) S is stationary, and ( i | i S ) and ( i | i S ) are two sequences of

ordinals such that 1 i i < and + ii < i+1 for i < ,

(c) for any S which is a limit point of S , for any A S with sup(A) = and for any sequence ( i | i A) with i < i for all i A we have that

pcf({+ i +1i | i A}) ( , + ] = ,

(d) pcf({+ +1i | < i }) = {+ +1i | < i } for every i S ,

(e) pp( i ) = + ii for every i S , and

(f ) S is the set of all S such that either ( ) > sup(S ), or ( ) cf() > 0 and {i S | i = i } is a stationary subset of , or ( ) pcf({+ +1 j | < j , j < }) {

+ +1 | < }.

Then there is a club C such that one of the two sets S 1 = {i S | i = i }and S 2 = {i S | i < i } contains C S .

It is easy to see that Theorem 2.1 (with some limitations on the size of pp( i )as in (b) and (e) above) can be deduced from Theorem 2.4 by taking S = {i +i }, i = 1, and i = 2. Condition (c) in the statement of Theorem 2.4 plays the role of the assumption ( ) in the statement of Theorem 2.1.

Proof of Theorem 2.4. Let us suppose that the conclusion of the statement of Theorem 2.4 fails.

Let, for i S , a i = {+ +1i | < i }, and set a = {a i | i S }. We may x asmooth and closed sequence ( b | a ) of generators for a (cf. [Sh96, Claim 6.7]).I.e., (b | a ) is smooth and generating, and if a then b = a pcf(b ) (closed).

For each S = S 1 S 2, by [Sh94, I Fact 3.2] and hypothesis (d) in thestatement of Theorem 2.4 we may pick a nite d {+ +1 | < } such that

{b | d} {+ +1 | < }.If S 1 then = and so pp() =

+ by (e) in the statement of Theorem

2.4. By [HSW99, Corollary 5.3.4] we may and shall assume that {b | d}contains a nal segment of {a j | j < }, and we may therefore choose some() < such that

{b | d} {a j | () j < }.

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There is then some and some stationary S 1 S 1 such that () = for every S 1 . Let C be the club set { < | = sup( S 1 )}.

If S then condition (c) in the statement of Theorem 2.4 implies that we mayassume that {b | d} contains a nal segment of the set {+ +1i | < i i then condition (1) in the statement of Theorem2.5 can be replaced by i S pp( i ) ++i , giving the same conclusion.

Theorem 2.6 Let be a singular cardinal of uncountable conality, and let ( i | i cf(). Suppose that there is 0 < such that for every with 0 < < ,pp() < . Let S cf() be such that

(1) i S pp( i ) > +i , and (2) if i S is a limit point of S and X = { j | j i S } with j < j pp( j )

regular then max(pcf( X )) = +i .Then S is not stationary.

Proof. Let us suppose that S is stationary. Assume that 0 = 0 otherwise justwork above it. We can assume that for every i < cf() if < i then also pp( ) < i .

Let > be a regular cardinal, and let M H be such that Card( M ) = cf( ),M cf(), and ( i | i < cf()), S M . Set a = ( M Reg) \ (cf() + 1). If thereis < such that for each i S |(a \ ) i | < cf() then the previous theoremapplies. Suppose otherwise. Without loss of generality we can assume that for everyi S and < i |(a \ ) i | = cf( ).

Let (b | a ) be a smooth and closed (i.e. pcf( b) = b ) sequence of generatorsfor a .

Claim. For every limit point i S max(pcf( a i )) pp( i ).

Proof. Fix an increasing sequence ( j | j < cf(i)) of cardinals of conalitycf() with limit i and so that (a j ) = j . Now max(pcf(a j )) pp( j ) forevery j < i , since |a j | = cf( ) = cf( j ). There is a nite f j pcf(a j )such that a j {b | f j }. Assume that |i | = cf( i). Otherwise just runthe same argument replacing i by a conal sequence of the type cf(i). Consider = max(pcf( {f j | j < i })). Then pp( i ), since | {f j | j < i } | cf(i).

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So, there is a nite g pcf( {f j | j < i }) + 1 pp( i ) + 1 such that{f j | j < i } {b | g}. By smoothness , then a i {b | g}. Since

the generators are closed and g is nite, also pcf(a i ) {b | g}. Hence,max(pcf( a i )) max( g) which is at most pp( i ). (Claim)

For every limit point i of S nd a nite set ci pcf(a i ) such that a i {b | ci } By the claim, max(pcf( a i )) pp( i ). So, ci pp( i ) + 1.

Set di = ci \ i . Then the set a i \ {b | di } is bounded in i , since we just removed a nite number of bs for < i . So, there is (i) < i such that (i ) a i \ {b | di } . Find a stationry S S and such that foreach i S (i) = . Let now i S be a limit of elements of S . Then there is < i such that {d j | j < i } \ b +i . Since otherwise it is easy to constructX = { j | j i S } with j < j pp( j ) regular and max(pcf( X )) > +i . Now ,

by smothness of the generators, b +i should contain a nal segment of a i . Whichis impossible, since pp( i ) > +i . Contraduction. (Theorem 2.6)

The same argument works if we require only pp( ) < for s of conality cf().The consistency of the negation of this (i.e., of: there are unbounded in many swith pp( ) ) is unkown. Shelahs Weak Hypothesis states that this is impossible.

The Claim in the proof of Theorem 2.6 can be deduced from general results like[Sh94, Chap. 8, 1.6].

Let again be a singular cardinal of uncountable conality, and let ( i | i < cf())be a strictly increasing and continuous sequence which is conal in and such that

0> cf(). Let, for n + 1, S

ndenote the set {+ n

i| i < cf()}. By [GiMi96], if

there is no inner model with a strong cardinal and + i = ( + i )K for every i < cf()

then for every n < , if for each i < cf() we have 2 i + ni then there is a clubC cf() such that pcf({+ ni | | i C }) S n . Notice that () from the statementof 2.1 just says that pcf({+i | | i < cf()}) S 1, or equivalently pcf(S 1) = S 1.

The following says that the connection between the + ni s and the S n s cannotbe broken for the rst time at + 1.

Theorem 2.7 Let be a singular cardinal of uncountable conality, and let ( i | i cf(). Let, for n + 1 , S n denote the set {+ ni | i < cf()}. Supposethat for every i < cf(), pp( + i ) = + +1i .

If for every n < there is a club C n cf() such that pcf({+ ni | i C n }) S n then there is a club C +1 cf() such that pcf({+ +1i | i C +1 }) S +1 .

Proof. Set C = n be a regular cardinal, and let M H be such that Card( M ) = cf( ), M cf(), and ( i | i < cf()) {C n | n max( m, n ). By the smoothness of ( b | a ), + n j b + m forevery j (C ) \ ( + 1). But

pcf({+ n j | j (C ) \ ( + 1) }) \ = {+ n }.

So + n b + m and hence b + n b + m . This, however, is impossible, since n > m .Contradiction! (Claim 2)

We now have that + +1i

b for unboundedly many i C +1 . By Claim2, by the smoothness of (b | a ), and by the choice of C +1 , we therefore getthat + +1 j b for unboundedly many j E . Hence again by Claim 2 and bythe closedness of (b | a ), + +1 b . Bo by the smoothness of (b | a ),b + +1 b . This implies that =

+ +1 , and we are done. (Theorem 2.7)

M. Magidor asked the following question. Let be a singular cardinal of un-countable conality, and let ( i | i < cf()) be a strictly increasing and continuoussequence which is conal in . Is it possible to have a stationary and co-stationaryset S cf() such that

tcf(i< cf( )

++i / (Club cf( ) + S )) = ++

andtcf(

i< cf( )

++i / (Club cf( ) + (cf( ) \ S ))) = + ?

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should be at least + ji . (Claim)

Let us return to + ki . By the Claim, k + ki . But + ki {b + l | 1 l < k }.

So b + ki {b + l | 1 l < k }. Let l k 1 be least such that b + l b + ki isunbounded in i . Then l + ki . Hence for some j 1 < j 2 n, j 1 = j 2 .

Let + li = j 1 = j 2 , where l n. Then b + li b + j 1 b + j 2 by the smoothness of (b | a ), since j 1 b + j 1 and j 2 b + j 2 . But now b + j 2 \ 1 j


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Proof. Let U be an iteration tree on (( M T , M ), ). We want to absorb U by an iteration tree U on M . The bookkeeping is simplied if we assume thatwhenever an extender E U is applied to M T to yield U 0+1 then right before thatthere are many steps of padding. I.e., letting P denote the set of all +1 lh( U )with E U = , we want to assume that if crit( E U ) < then + 1 = + 1 + forsome such that + 1 P for all [, ).

Let us now construct U . We shall simultaneously construct embeddings

: M U M U

,

where lh( U ) \ P , such that lh(E U ) = lh(E U ) whenever < lh( U ) \ P and or else (, ] P . The construction of U and of the maps is a standard recursive copying construction as in the proof of [MiSt94, Lemma

p. 54f.], say, except for how to deal with the situation when an extender is appliedto M T .Suppose that we have constructed U + 1 and ( | ( + 1) \ P ), that

+ 1 P for all [, + 1 + 1), and that crit( E U + ) < . We then proceed asfollows. Let + 1 = + 1 + . We rst let

: M (E U ) M U +1 ,

and we let : M E U ult( M , E

U ).

We may dene k: ult( M , E U ) M U

+1

by settingk([a, f ]ME U ) = [ (a), f ]

M (E U )

= (f )( (a))

for appropriate a and f . This works, because P (crit( E U )) M = id. Noticethat k lh(E U ) = lh(E U ).

We now let the models M U +1+ and maps

: M T M U

+1+ ,

for , arise by copying the tree T onto M U +1 , using . We shall also havemodels M T and maps

: M T M T ,

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for , which arise by copying the tree T onto ult( M , E U ), using . Notice thatfor there are also copy maps

k : M

T M U

+1+

with k0 = k. Because lh(E T ) whenever < , lh(E T ) () whenever < ,so that in particular k () = k (), and thus k lh(E U ) = k lh(E U ).

We shall also have that = , so that we may dene

k : M U +1 = ult( M , E U ) M T

by settingk ([a, f ]ME U ) = (f )(a)

for appropriate a and f . Let us now dene+1 : M U +1 = ult( M , E U ) M U

+1

by +1 = k k . We then get that +1 lh(E U ) = k lh(E U ) = k lh(E U ) = lh(E U ). (Lemma 3.1)

Let be an inaccessible cardinal. We say that V is n-suitable if n < andV is closed under M #n , but M

#n +1 does not exist (cf. [St95, p. 81] or [FoMaSch01,

p. 1841]). We say that V is suitable if there is some n such that V is n-suitable. If is measurable and V is n-suitable then the core model K below n + 1 Woodincardinals of height exists (cf. [St ]).

The following lemma is a version of Lemma 3.1 for M = K . It is related to[MiScSt97, Fact 3.19.1].

Lemma 3.2 (Steel) Let be a measurable cardinal, and suppose that V is suit-able. Let K denote the core model of height . Let T be a normal iteration treeon K of length + 1 < . Let , and let be a cardinal of M T such that (E T ) > whenever [ , ). Then the phalanx ((M T , M T ), ) is iterable.

Proof sketch. As K c is a normal iterate of K (cf. [ScSt99, Theorem 2.3]), itsuffices to prove Lemma 3.2 for K c rather than for K .

We argue by contradiction. Let T be a normal iteration tree on K c

of length + 1 < , let , and let be a cardinal of M T such that (E T ) > whenever [ , ). Suppose that U is an ill behaved putative normal iteration tree onthe phalanx (( M T , M T ), ). Let : V V +2 be such that V is countable andtransitive and {K c, T , , , U} ran( ). Set T = 1(T ), = 1(), = 1( ), = 1(), and U = 1( U ).

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By [St96, 9] there are and and maps : M T N and : MT N such

that N and N agree below (), and () = (). (Here N and N denote models from the K c construction.) We may now run the argument of [St96,9] once more to get that in fact U is well behaved. But then also U is wellbehaved after all. (Lemma 3.2)

In the proofs to follow we shall sometimes tacitly use the letter K to denotenot K but rather a a canonical very soundness witness for a segment of K whichis long enough. If M is a premouse then we shall denote by M| the premouseM as being cut off at without a top extender (even if E M = ), and we shalldenote by M|| the premouse M as being cut off at with E M as a top extender(if E M = , otherwise M|| = M| ). If M then + M will either denote thecardinal successor of in M (if there is one) or else + M = M OR.

Lemma 3.3 Let be a measurable cardinal, and suppose that V is suitable. Let K denote the core model of height . Let 2 be a regular cardinal, and let M K || be an iterable premouse. Then the phalanx ((K, M ), ) is iterable.

Proof. We shall exploit the covering argument. Let

: N = X V +2

such that N is transitive, Card( N ) < , {K, M , } X , X , and K = 1(K ) is a normal iterate of K , hence of K || , and hence of M . Such a map

exists by [MiSc95]. Set M = 1(M ) and = 1(). By the relevant versionof [St96, Lemma 2.4] it suffices to verify that ((K, M ), ) is iterable. However, theiterability of (( K, M ), ) readily follows from Lemma 3.1. Using the map , we maythus infer that (( K, M ), ) is iterable as well. (Lemma 3.3)

Lemma 3.4 Let be a measurable cardinal, and suppose that V is suitable. Let K denote the core model of height . Let be a cardinal of K , and let M be a premouse such that M| + M = K |+ M , (M ) , and M is sound above .Suppose further that the phalanx ((K, M ), ) is iterable. Then M K .

Proof. This follows from the proof of [MiScSt97, Lemma 3.10]. This proof shows that (( K, M ), ) cannot move in the comparison with K , and that eitherM K or else, setting = + M , E K = and M is the ultrapower of an initialsegment of K by E K . However, the latter case never occurs, as wed have that = crit( E K ) < so that + K || = + K and hence E K would be a total extenderon K . (Lemma 3.4)

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where j = crit( F P i ) (we have j < i ). Set i = sup( i ) for i .The proof of [MiSc95] now shows that we may and shall assume that the following

hold true, for every i .

Claim 1. If S i is a set premouse then S i K || ( i ).

Proof. This readily follows from the proof of [MiScSt97, Lemma 3.10]. Cf. theproof of Lemma 3.4 above. (Claim 1)

Claim 2. If S i is a weasel then either S i = K or else S i = ult( K, E K ) where i is such that crit( E K ) < and (E K ) ( i ).

Proof. This follows from the proof of [MiScSt97, Lemma 3.11].Fix i, and suppose that S i is a weasel with S i = K . Let

R ik = S ik S ik 1 ... S i 0 = S i

be the decomposition of S i , and let j : S i j S i for j k (cf. [MiScSt97, Lemma3.6]). We also have

T 0 i k : K MT i k

= P ik .

Notice that we must have

= crit( T 0 i k ) < = crit( ),

as otherwise M T i k couldnt be a weasel. Let us write

: P ik ult( P ik , E ( ik )) = R ik .

Notice that we have (R ik ) ( i j ) and c(R ik ) = (cf. [MiScSt97, Lemma 3.6]. Itis fairly easy to see that the proof of [MiScSt97, Lemma 3.11] shows that we mustindeed have E K i k = , crit( E

K i k

) = , (E K i k ) ( ik ), s(E K i k

) = , and

R ik = ult( K, E K i k ).

The argument which gives this very conclusion is actually a simplied version of theargument which is to come.

We are hence already done if k = 0. Let us assume that k > 0 from now on.We now let F be the (, )-extender derived from ik T 0 i k , where

= max( {( i )}c(S i ))+ S i .

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We shall have that (F ) ( i ) and s(F ) \ ( i ) = c(S i ) \ ( i ). Let us writet = s(F ) \ ( i ). We in fact have that

t = j


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However, this holds if and only if a U 01(X ), i.e., if and only if X (E U 0 )a , because,using the hull- and denability properties of K , ik T 0 i k (X ) =

U 0 (X ) =

U 01

(X ).We have indeed shown that G = E U 0 . But we have that G N = M U . This is

a contradiction! (Subclaim)

By the Subclaim, s ran( V 1 ), and we may dene an elementary embedding

: M U 1 S i

by setting M

U 1 [ 1, 2, 3] S i [ 1, (V 1 ) 1( 2), 3],

where is a Skolem term, 1 < ( i ), 2 s, and 3 for some appropriate thick

class . However, t = ( V 1 )

1(s), and S i = H

S i (( i ) t ). Hence is onto,and thus S i = M U 1 = ult( K, E U 0 ).

If we now let be such that E K = E U 0 then is as in the statement of Claim 2.(Claim 2)

Let us abbreviate by S M the phalanx

((S i | i < ) M , (i | i < )) .

Claim 3. S M is a special phalanx which is iterable with respect to specialiteration trees.

Proof. Let V be a putative special iteration tree on the phalanx S M . ByClaims 1 and 2, we may construe V as an iteration of the phalanx

((K, M ), ).

The only wrinkle here is that if crit( E V ) = (i ) for some i < , where S i =ult( K, E K ), then we have to observe that

ult( K, F ult( K || ,E V )) = ult(ult( K, E K ), E V ),

and the resulting ultrapower maps are the same.Lemma 3.3 now tells us that the phalanx (( K, M ), ) is iterable, so that V turns

out to be well behaved. (Claim 3)

By [MiScSt97, Lemma 3.18], Claim 3 gives that

((R i | i < ) M , (i | i < ))

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is a very special phalanx which is iterable with respect to special iteration trees. By[MiScSt97, Lemma 3.17], the phalanx

((P i | i < ) M , ( i | i < )) ,

call it P M , is nally iterable as well.

Claim 4. Either M is an iterate of K , or else M P .

Proof. Because P M is iterable, we may coiterate P M with the phalanx

P = (( P i | i ), ( i | i < )) ,

giving iteration trees V on P M and V on P . An argument exactly as for (b) (a)

in the proof of [St96, Theorem 8.6] shows that the last model MV of V must sitabove M , and that in fact M V = M , i.e., V is trivial. But as (M ) and M

is sound above , the fact that V is trivial readily implies that either V is trivial aswell, or else lh(V ) = 2, and M V = M V 1 = ult( P i , E V 0 ) where crit( E V 0 ) = i < ,(M V 1 ) = (P i ) i , (E V 0 ) , and s(E V 0 ) = .

We now have that M is an iterate of K if either V is trivial and M P or elseif V is non-trivial. On the other hand, if M is not an iterate of K then we musthave that M P . (Claim 4)

Let us verify that M P is impossible. Otherwise P is a set premouse with(P ) , and we may pick some a P () ( (P ) \ P ). As M P , a M .However, by our inductive assumption on M (and by elementarity of ) we musthave that P () M P () K P . Therefore wed get that a P after all.Contradiction!

By Claim 4 we therefore must have that M is an iterate of K . I.e., K and Mare hence both iterates of K , and we may apply Lemma 3.2 and deduce that thephalanx (( K, M ), ) is iterable. This, however, is a contradiction as we chose sothat (( K, M ), ) is not iterable. (Lemma 3.5)

Jensen has shown that Lemma 3.5 is false if in its statement we remove theassumption that 2. He showed that if K has a measurable cardinal (but 0

may not exist) then there can be arbitrary large K -cardinals < 2 such that thereis an iterable premouse M K || with (M ) , M is sound above , but M isnot an initial segment of K . In fact, the forcing presented in [R aSch ] can be usedfor constructing such examples.

To see that there can be arbitrary large K -cardinals < 1 such that there isan iterable premouse M K || with (M ) , M is sound above , but M

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is not an initial segment of K , one can also argue as follows. K HC need notprojective (cf. [HaSch00]). If there is some < 1 such that Lemma 3.5 holds forall K -cardinals in [,1) then K HC is certainly projective (in fact 14).

By a coarse premouse we mean an amenable model of the form P = ( P ;, U )where P is transitive, ( P ;) |= ZFC (i.e., ZFC without the power set axiom), P hasa largest cardinal, = P , and P |= U is a normal measure on . We shall saythat the coarse premouse P = ( P ;, U ) is n-suitable if (P ;) |= V P is n-suitable,and P is suitable if P is n-suitable for some n. If P is n-suitable then K P , the coremodel below n + 1 Woodin cardinals inside P exists (cf. [St ]).

Denition 3.6 Let be an innite cardinal. Suppose that for each x H =


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Let be an inaccessible cardinal, and let X V . We say that V is (n, X )-suitable if n < and V is closed under M #n , but M

#n +1 (X ) does not exist (cf. [St95,

p. 81] or [FoMaSch01, p. 1841]). We say that V is X -suitable if there is some nsuch that V is (n, X )-suitable. If is measurable and V is (n, X )-suitable thenthe core model K (X ) over X below n + 1 Woodin cardinals of height exists(cf. [St ]).

We shall say that the coarse premouse P = ( P ;, U ) is (n, X )-suitable if (P ;) |=V P is (n, X )-suitable, and P is X -suitable if P is (n, X )-suitable for some n. If P is (n, X )-suitable then K (X )P , the core model over X below n + 1 Woodincardinals inside P exists (cf. [St ]).

Denition 3.8 Let be an innite cardinal, and let X H =


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M K (X )|| be such that (M ) , M is sound above , M is not an initialsegment of K (X ), but if K (X )|| N M is such that ( N ) and N is soundabove then N is an initial segment of K (X ). In order to derive a contradiction itsuffices to prove that the phalanx (( K (X ), M ), ) is not iterable.

Let us now imitate the proof of Lemma 3.5. Let

: N V +2

be such that N is transitive, Card( N ) = 1, {K (X ), M , , } ran( ), and N 2 2. Let X = 1(X ), = 1(), K (X ) = 1(K (X )), M = 1(M ), = 1(), = 1(), and = 1(2) = crit( ). We may and shall assume that((K (X ), M ), ) is not iterable. Furthermore, by the method of [MiSc95], we mayand shall assume that the phalanxes occuring in the proof to follow are all iterable.

Let

be largest such that

K (

X )|

does not move in the coiteration withM #n +1 (X ). Let T be the canonical normal iteration tree on M

#n +1 (X ) of length + 1

such that K (X )| M T . Let ( i | i ) be the strictly monotone enumeration of the set of cardinals of K (X )| , including , which are . For each i , let theobjects P i , R i , and S i be dened exactly as in the proof of Lemma 3.5. For i < ,let i = i+1 .

Because (M #n +1 (X )) , and as K (X ) is n-small, whereas M #n +1 (X ) is not,

we have that for each i , P i is a set-sized premouse with (P i ) i such thatP i is sound above i . Therefore, for each i , S i is a set-sized premouse with(S i ) ( i ) such that S i is sound above ( i ).

Let us verify that the phalanx

P M = (( P i | i < ) M , ( i | i < ))

is coiterable with the phalanx

P = (( P i | i ), ( i | i < )) .

In fact, by our inductive hypothesis, we shall now have that for each i < , S i K (X )and hence S i M . Setting i = sup( i ) for i < , we thus have that

((S i | i < ) M , (i | i < ))

is a special phalanx which is iterable with respect to special iteration trees. As inthe proof of [MiScSt97], we therefore rst get that

((R i | i < ) M , (i | i < ))

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is a very special phalanx which is iterable with respect to special iteration trees, andthen that the phalanx

((P i | i < ) M , (i | i < ))

is iterable.We may therefore coiterate P M with P . Standard arguments then show that this

implies that cannot be the index of an extender which is used in the comparisonof M #n +1 (X ) with K (X ). We may conclude that = , i.e., K (X ) doesnt movein the comparison with M #n +1 (X ). In other words, K (X ) = M T .

However, we may now nish the argument exactly as in the proof of Lemma 3.5.The coiteration of P M with P gives that either M is an iterate of M

#n +1 (X ), or else

that M P . By our assumptions on M , we cannot have that P M . Therefore, M

is an iterate of M #n +1 (X ). However, the proof of Lemma 3.2 implies that the phalanx((M T , M ), ) is iterable. This is because the existence of M

#n +1 (X ) means that the

K c construction, when relativized to X , is not n-small and reaches M #n +1 (X ). Butnow ((K (X ), M ), ) is iterable, which is a contradiction! (Lemma 3.10)

Theorem 3.11 Let be a cardinal, and let X H =


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be such that ( Y ; ...) M , ( i | i < ) Y , and that all the objects occuring in theproof of [MiSc95] are iterable. Let K = 1(K ), = 1(), and i = 1( i ) fori < .

We dene P i , R i , and S i in exactly the same way as P i , R i , and S i were denedin the proof of Lemma 3.5. Let T be the normal iteration tree on K arising fromthe coiteration with K . Set +1 = lh( T ). Let ( i | i ) be the strictly monotoneenumeration of Card K ( +1), and set i = ( i )+ K for i . Let, for i , i < be the least such that i < (E T ), if there is some such ; otherwise let i = .Let, for i , P i be the longest initial segment of M T i such that P ( i ) P i K .Let

R i = ult( P i , E ( i )) ,

where i . We recursively dene (S i | i ) as follows. If R i is a premouse thenwe set S

i= R

i. If R

iis not a premouse then we set

S i = ult( S j , F R i ),

where j = crit( F P i ) (we have j < i ). Set i = sup( i) for i .We also want to dene P i , R i , and S i . For i < , we simply pick i < such

that i = i , and then set P i = P i , R i = R i , and S i = S i ; we also set i = i .Notice that well have that

i + R i = i + S i = sup( N + V i ),

because + P ii =

+ K i =

+ N i (the latter equality holds by [MiSc95]).Let (A) denote the assertion (which might be true or false) that

{ (E T ) | + 1 }

is unbounded in . Let us dene some C .If (A ) fails then M T i = M

T for all but boundedly many i < , which readily

implies that there is some < such that P i = P j whenever i, j \ . In thiscase, we simply set C = \ .

Suppose now that ( A) holds. Then cf( ) = cf( ) = > 0, and both [0, )T aswell as {i | i < } are closed unbounded subsets of . Moreover, the set of all i < such that

+ 1 (0, ]T (crit( E T ) < i (E T ) < i )

is club in . There is hence some club C such that whenever i C theni [0, )T , + 1 (0, ]T (crit( E T ) < i (E T ) < i ), and [i , ]T does notcontain drops of any kind. By ( A), is a cardinal in M T , and it is thus easy to see

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that in fact for i C , P i = M T i . Moreover, if i j C then T i j : P i P j is such

that T i j i = id.Let us now continue or discussion regardless of whether ( A) holds true or not.If i j C then we may dene a map ij : R i R j by setting

[a, f ]P iE i [a, T i j (f )]

P jE j ,

where a [ i ]


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where i C is large enough so that i . Due to the existence of the maps ij ,f () is independent from the the particular choice of i, and thus f is well-dened.Obviously, f K for all < . Moreover, f ( i ) = + R ii = sup( Y +i ) for alli C , as desired.

It is easy to verify that in fact f K . Let R be the premouse given by thedirect limit of the system

(R i ,ij | i j C ).

As > 0 this system does indeed have a well-founded direct limit which we canthen take to be transitive; for the same reason, R will be iterable. We may then useLemma 3.5 to deduce that actually R K . However, we shall have that, for < ,f () = + in the sense of the transitive collapse of

H Rn +1 ( { pR ,n +1 }).

Case 1.2. Clause 3 or 4 holds.

In this case, [MiScSt97, Lemma 2.5.2] gives information on how P i has to looklike, for i C . In particular, P i will have a top extender, F P i . By [MiScSt97,Corollary 3.4], well have that (F P i ) i .

Let = crit( F P i ) = crit( F P j ) for i, j C . Of course, () = crit( F R i ) =crit( F R j ) for i, j C . Let = k , where k < . Setting S = S k , we have that

S i = ult( S , F R i )

for all i C .By [MiScSt97, Corollary 3.4], P i is also Dodd-solid above i , for i C . By

[MiScSt97, Lemma 2.1.4], T i j (s(F P i )) = s(F P j ) for i j C . Due to the existence

of the maps ij , it is then straightforward to verify that

F R j ( i s(F R j )) = F R i

whenever i j C .Let us now dene f : as follows. We set f () = + in the sense of

ult( S , F R i ( s(F R i ))) ,

where i C is large enough so that i . f () is then independent from theparticular choice of i, and therefore f is well-dened. Moreover, f ( i ) = + S ii =sup(Y +i ).

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Case 1.2.1. Clause 3 holds.

By [MiScSt97, Lemma 3.10], S i K for all i C . Also, S K .In order to see that f K for all < it suffices to verify that F R i K

for all i C . Fix i C . Let m < be such that m +1 (S ) () < m (S ), and let

: S F R i S i = K || i ,

some i . It is then straightforward to verify that

ran( ) = H K || im +1 (() ( pS ,m +1 ) s(F Ri )) .

This implies that K . But then F R i K as well.But now letting R be as in Case 1.1 we may actually conclude that f K .

Case 1.2.2. Clause 4 holds.

We know that S = K , as is discontinuous at + K . We also know that forall i C , S i = K , as is discontinuous at 1(+i ). We now have Claim 2 fromthe proof of Lemma 3.5 at our disposal, which gives the following. There is some such that S = ult( K, E K ), where crit( E K ) < crit( ), sup(+i Y ), and (E K ) (). Also, for every i C , there is some i such that S i = ult( K, E K i ),where crit( E K i ) = crit( E

K ) < crit( ), i sup( +i Y ), and (E K ) i .

In order to see that f K for all < it now again suffices to verify thatF R i K for all i C . Fix i C . Let

: S F R i S i .

Let us also write: K E K S ,

andi : K E K i S i .

Standard arguments, using hull- and denability properties, show that in fact

i = .Therefore,

((f )(a)) = i (f )((a))

for the appropriate a, f . As (S ) () = crit( F R i ), we may hence compute F R iinside K .

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By letting R be as in Case 1.1 we may again conclude that actually f K .

Case 2. Clause 2 holds.

Let i C . Then R i is a weasel with (R i ) = i and c(R i ) = . This, combinedwith the proof of [MiScSt97, Lemma 3.11], readily implies that

R i = ult( K, E K i ),

where (E K i ) i and s(E K i ) = . Moreover, by the proof of [MiScSt97, Lemma

3.11, Claim 2], crit( E K i ) = crit( T 0i ) < crit( ). Let us write

i : K E K i R i ,

and let us writei : P i R i

for the canonical long ultrapower map. Notice that we must have

i = i T 0i .

Furthermore, if i j C , then well have that

j T 0j = ij i T 0i .

Let us dene f : as follows. We set f () = + in the sense of

ult (K, E K i ),

where i C is large enough so that i . If f () is independent from the choiceof i then f is well-dened, f K for all < , and f ( i ) = sup( Y +i ) for alli C .

Now let i j C . We aim to verify that

E K i = E K j i ,

which will prove that f () is independent from the choice of i.Well, we know that crit( E K i ) = crit( 0i ) = crit( 0j ) = crit( E

K j ); call it . Fix

a [ i ]


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We may now nally let R be the weasel given by the direct limit of the system

(R i ,ij | i j C ).

The above arguments can then easily be adopted to show that f K . (Lemma 3.12)

We have separated the arguments that f K for all < from thearguments that f K , as the former ones also work for a stable K up to , forwhich the latter ones dont make much sense.

The following is a version of Lemma 3.12 for the stable K (X ) up to .

Lemma 3.13 Let > 20 be a limit cardinal with cf() = > 0, and let =( i | i < ) be a strictly increasing continuous sequence of singular cardinals below which is conal in with 0. Let X H =


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This proves the rst part of Lemma 3.13. The moreover part of Lemma 3.13follows from the method by which [MiScSt97, Lemma 3.13] is proven. (Lemma 3.13)

4 The proofs of the main results.Proof of Theorem 1.1. Let be as in the statement of Theorem 1.1. Set

a = { Reg | | |+ < }.

As 2| | < , [BuMa90, Theorem 5.1] yields that max(pcf( a )) = | a |. However,| a | = | | (cf. [J78, Lemma 6.4]). Because | | > | |+ , we therefore have that

max(pcf( a )) > | |+ .

This in turn implies that

{ +1 | < | |+ } pcf(a )

by [BuMa90, Corollary 2.2]. Set

H =< | | +

H + .

We aim to prove that for each n < , H is closed under X M #n (X ).

To commence, let X H , and suppose that X # = M #0 (X ) does not exist. By

[BuMa90, Theorem 6.10] there is some

d { +1 | < < | |+ }

with min( d ) > TC( X ), |d | | |, and | |+ +1 pcf(d ). By [DeJe75], however, wehave that

{f d | f = f d , some f L[X ]}

is conal in d . As GCH holds in L[X ] above X , this yields max(pcf( d )) sup( d )+ .Contradiction!

Hence H is closed under X X #

= M #0 (X ).Now let n < and assume inductively that H is closed under X M #n (X ). Fix

X , a bounded subset of | |+ . Let us assume towards a contradiction that M #n +1 (X )does not exist.

Without loss of generality, = sup( X ) is a cardinal of V . We may and shallassume inductively that if 2 and if X is bounded then M #n +1 (X ) exists.

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We may use the above argument which gave that H is closed under Y Y #together with [ScWo01, Theorem 5.3] (rather than [DeJe75]) and deduce that forevery x H there is some (n, X )-suitable coarse premouse containing x. We claimthat K (X ) stabilizes in H . Well, if 1 then this follows from Theorem 3.9. Onthe other hand, if 2 then this follows from Theorem 3.11 together with ourinductive hypothesis according to which M #n +1 (X ) exists for all bounded X .Let K (X ) denote the stable K (X ) up to | |+ .

Set = | |+ < | |+ . We aim to dene a function

: [H ] NS | |+ .

Let us rst denote by S the set of all Y [H ] such that Y H , ( | | |+ such that pcf(d ). (In fact, | |+ +1

pcf(d ).) By [BuMa90, Theorem 6.10] there is then some d d with |d | | | and pcf(d ). Set = sup( d ). In particular (cf. [BuMa90, Corollary 7.10]),

cf( d ) > + .

However, we claim that

F = {f d | f : f K (X )}

is conal in d . As GCH holds in K (X ) above , |F | | |+ , which gives acontradiction!

To show that F is conal, let g d . Let Y S be such that D (Y ) = and for all d , g(+1 ) < sup(Y +1 ). As D (Y ) = , we have that D C Y .Therefore, f Y (+1 ) = sup( Y +1 ) for all d , and hence g(+1 ) < f Y (+1 )for all d . Thus, if we dene f : by f (+ K (X )) = f Y () for < thenf F and g < f d . (Theorem 1.1)

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Proof of Theorem 1.3. Fix as in the statement of Theorem 1.3. Set

H = + 2cf( ) .We have shown that

{ < | cf( ) > + 2cf( )}is stationary in , i.e. that SCH fails stationarily often below .

This fact immediately implies by [DeJe75] that H is closed under X M #0 (X ).Now let n < , and let us assume that H is closed under X M #

n(X ). Let us

suppose that there is some X H such that M #n +1 (X ) does not exist. We are leftwith having to derive a contradiction.

As SCH fails stationarily often below , we may use [ScWo01, Theorem 5.3]and deduce that for every x H there is some (n, X )-suitable coarse premousecontaining x. By Theorem 3.9, K (X ) stabilizes in H . Let K (X ) denote the stableK (X ) up to .

Let us x a strictly increasing and continuous sequence = ( i | i < ) of ordinals below which is conal in . Let us dene a function

: [H ]2

NS .

Let Y [H ]2 be such that Y H , ( i | i < ) Y , Y Y , and TC( {X }) Y .By Lemma 3.13, there is a pair ( C, f ) such that C is club, f : , f K (X ) for all < , and f ( i ) = char Y (i ) as well as

+i =

+ K i for all i C . We

let (C Y , f Y ) be some such pair (C, f ), and we set (Y ) = \ C Y . If Y is not as justdescribed then we let ( C Y , f Y ) be undened, and we set (X ) = .

By Lemma 2.3, there is then some club D and some limit ordinal i < of D such that

cf( {+ j | j i D}) > +i ,

and for all f iD +i there is some Y H such that Card( Y ) = 2 , Y Y ,

D (Y ) = , and f < charY . Let us write

d = {+ j | j i D}.

We now claim that

F = {f d | f : i i f K (X )}

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is conal in d . As GCH holds in K (X ) above TC( X ), |F| | i |+ , which gives acontradiction!

To show that F is conal, let g d . There is some Y H such thatCard( Y ) = 2 , Y Y , D (Y ) = , and g(+ j ) < sup(Y + j ) for all j D i.As D (X ) = , we have that D C Y . But now if + j d then g(+ j )

moti gitik, ralf schindler and saharon shelah- pcf theory and woodin cardinals

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