cardinal arithmetic and woodin cardinals

7/30/2019 Cardinal arithmetic and Woodin cardinals

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arXiv:math/0203081v1[math.L

O]8Mar2002

Cardinal arithmetic and Woodin cardinals

Ralf Schindler

Institut fur Formale Logik, Universitat Wien, 1090 Wien, Austria

[email protected]

http://www.logic.univie.ac.at/rds/

Abstract

Suppose that there is a measurable cardinal. If 21 < , but 0 > 1, thenthere is an inner model with a Woodin cardinal. This essentially answers aquestion of Gitik and Mitchell (cf. [GiMi96, Question 5, p. 315]).

We refer the reader to [AbMa] for an introduction to cardinal arithmetic and

to Shelahs pcf theory (cf. also [BuMa90]). The perhaps most striking result ofShelahs in cardinal arithmetic is that if is a strong limit cardinal then

2 < min((20 )+, 4).

Magidor was the first one to produce a model of set theory in which the GCH holdsbelow , but 2 = +2 (cf. [Ma77a], [Ma77b]). It is now known how to producemodels in which there are arbitrarily large countable gaps between and 2 , whilethe GCH holds below (cf. for instance [GiMa92]). A strong cardinal is more thanenough for this purpose. In fact, many equiconsistencies are known. We refer thereader to [Gi] and [Mi] and to the references given there.

It is, however, open if it is possible to have that is a strong limit cardinal,but 2 > 1 (cf. [Gi, Section 7, Problem 1]). This problem is just one of thekey open problems of pcf theory in disguise. Gitik and Mitchell have shown that astrong cardinal is not enough for producing such a model:

Theorem 0.1 ([GiMi96, Theorem 5.1]) If 20 < and 0 > 1 then there is asharp for a model with a strong cardinal.

The purpose of this note is to prove the following theorem which in a certainsense improves Theorem 0.1. It will say that you will need at least a Woodin cardinalin order to produce a model in which is a strong limit cardinal, but 2 > 1.

As far as we know this is the first statement in cardinal arithmetic which is knownto practically imply the consistency of a Woodin cardinal.

Theorem 0.2 If 21 < , 0 > 1, and there is a measurable cardinal then thereis an inner model with a Woodin cardinal.

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Our proof of Theorem 0.2 will make use of Shelahs pcf theory. Specifically, wellneed the following theorems which are due to Shelah. Recall that ifa is a set ofregular cardinals then pcf(a) is the set of all possible cofinalities of

a/U where U

is an ultrafilter on a.

Theorem 0.3 ([BuMa90, Theorem 5.1]) Let 20 < . Then pcf({n: n < }) ={ 0 : is regular }.

Theorem 0.4 ([BuMa90, Theorem 6.10]) Let 20 < . Letd pcf({n: n < })and pcf(d). There is then somed d such that Card(d) = 0 and pcf(d).

We shall also need the following simple combinatorial fact, Lemma 0.5. Let and be cardinals with . H is the set of all sets which are hereditarily smaller

than , and [H]

is the set of all subsets of H of size . Recall that S [H]

is called stationary if and only if for all models M = (H; ...) of finite type andwith universe H there is some X S such that X is the universe of an elementarysubmodel ofM, i.e., (X; ...) M. We say that S [H]

is -stationary if and onlyifS {x: x x} is stationary, i.e., if for all models M = (H; ...) of finite type andwith universe H there is some X S such that X X and X is the universe ofan elementary submodel ofM. We let NS1 denote the non-stationary ideal on 1.

Lemma 0.5 Let 21 be regular, and let : [H]21 NS1. There is then

a pair (C, S) such that C is a closed unbounded subset of 1, S is -stationary in[H]

21 , and C (X) = for all X S.

Proof. Suppose that for every club C 1 the set

{X [H]21 : C (X) = }

is not -stationary. This means that for every club C 1 there is a model MC offinite type and with universe H such that for every (X; ...) MC with X Xand Card(X) = 21 we have that C (X) = . As there are only 21 many subsetsof 1, there is a model M of finite type and with universe H such that for everyclub C 1, if (X; ...) M is such that 21 X then (X; ...) MC for everyclub C 1. Pick (X; ...) M with X X, Card(X) = 21, and 21 X. We

shall have that C (X) = for every club C 1, which means that (X) isstationary. Contradiction!

Our proof of Theorem 0.2 will use the core model theory of [St96]. The basicidea for its proof will be the following. We shall first use Theorems 0.3 and 0.4 as

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well as Lemma 0.5 for isolating a nice countable set d {+1: < 1} with1+1 pcf(d

). We shall then use a covering argument to prove that pcf(d) 1yielding the desired contradiction. However, the covering argument plays the key

role in choosing the nice d

we start with.

Proof of Theorem 0.2. Suppose not. Let be a measurable cardinal, and let Kdenote Steels core model of height (cf. [St96]). Let > 1 be a regular cardinal.

Fix X H for a while, where X X. Let : HX = X H be such thatH = HX is transitive. Let K = KX =

1(K||1). We know by [MiScSt97] thatthere is an -maximal normal iteration tree T on K (of successor length) such thatMT

K. Let T = TX be the shortest such tree.IfE is an extender then we shall write (E) for the natural length ofE(cf. [MiSt94,

p. 6]). We shall let

(X) = { < 1: + 1 (0, ]T (crit(E

T

) <

H

(E

T

))}.We aim to apply Lemma 0.5 to .

Claim 1. (X) is a non-stationary subset of 1.

Proof. Suppose that S0 = (X) is stationary. Let S be the set of all limitordinals of S0. S is stationary, too. Let F: S OR be defined by letting F()be the least < such that crit(ET ) <

H , where + 1 (0, ]T is unique

such that crit(ET ) < H (E

T ). Let S S be stationary such that F S is

constant. For S let () be the unique such that crit(ET ) < H (E

T ).

Using the initial segment condition [MiSt94, Definition 1.0.4 (5)] (cf. also [SchStZe,

Definition 2.4]) it is easy to see that we must have crit(ET()) = crit(E

T()) whenever

{, } S. Hence ET() = ET() whenever {,

} S. Let us write E for this

unique extender. Well have to have (E) H1 , so that E cannot have been usedin T. Contradiction!

We shall now define : [H]21 NS1 as follows. Let X [H]

21 . If X His such that X X then we let (X) be defined as above. Otherwise we set(X) = . Let (C, S) be as given by Lemma 0.5.a We let

d = {+1: C}.

We know that 1+1 pcf(d

), by [BuMa90, Remark 1.8]. By Theorems 0.3 and 0.4there is a countable d d with 1+1 pcf(d). We shall now derive a contradiction

by showing that pcf(d) sup(d)+ + 1.aThe lift-up arguments which are to follow will be simplified by the assumption, which we may

make without loss of generality, that every element ofC is a limit ordinal.

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Let = sup(d).

Main Claim. For every f

d there is some g K, g: such that

g() > f() for all d

.

Suppose that the Main Claim holds. Then certainly

{[g d]U: g K g: }

is cofinal ind/U. But there are only + many g K, g: , so that we

must cartainly have cf(d/U) +. This contradiction proves Theorem 0.2.

It therefore suffices to prove the Main Claim. Fix f

d for the rest of thisproof. By the choice of (C, S) there is some X H with Card(X) = 21 , X X,C (X) = , and f X. Let : H = HX = X H, and let T = TX be

as defined above. Then if + 1 (0, ]T and +1 d

we do not have thatcrit(ET ) < H (E

T ).

Let (DT(0, ]T){lh(T)} = {0+1 < ... < N+1}, where 0 N < . Let nbe the T-predecessor ofn+1 for 0 n N. Let us pretend that N > 0 and 0 = 0,i.e., that we immediately drop on (0, ]T.b Notice that crit() 1((21)+). Letus assume without loss of generality that C > (21)+.

For each Cthere is a least () [0, ]T such that MT()||H = K||

H . Let

n() be the unique n < N such that n1 < () n. Then D

T((), n()]T =

and (MTn()

) H . Let () be the least such that (MTn()

||()) H

(if() < n then () = MTn()

OR). Let m() be the unique m < such that

m+1(MTn() ||())

H < m(M

Tn() ||()). The following is easy to verify.

Claim 2. We may partition C into finitely many sets C0, ..., Ck, 0 k < , suchthat sup(Cl) min(Cl+1) whenever l < k and such that for all l k, if{, } Clthen n() = n(), () = (), and m() = m().

In order to finish the proof of the Main Claim it therefore now suffices to find,for an arbitrary l k, some g K, g: such that g(+1) > f(+1) for all Cl.

Let us fix l k. Let us write n, , and m for n(), (), and m(), where isany member of Cl. Let

M = MTn||.bI.e., in this note we simply ignore the possibility that we might have

0> 0. In any event, it

can be shown that if we were to use Friedman-Jensen premice rather than Mitchell-Steel premicethen the case that 0 > 0 would not come up by an argument of [Sch02].

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Let l = sup({H : Cl}). We may define

: M M = Ultm(M; l).

By the argument of [MiScSt97], we shall have that M K. In particular, M K.Let us write for the cardinal predecessor of if is a successor cardinal

(otherwise we let = ). Let l = sup(l). Let us define g: l l as follows.

Let < l, and let us write for ()M, i.e., in the sense of M. Let

: M = hMm (

{pM

}) m M,

where M is transitive. We let

g() = ()+M.

Notice that g K. We are left with having to verify that g(+1) > f(+1) for all Cl.

Fix Cl. Let us assume without loss of generality that () < n, the othercase being easier. Then = MTn OR, i.e., M = M

Tn

. Consider

T()n : MT() M

Tn

.

It is easy to verify that

M+1 = Ultm(MT();

1()),

and that there is a map: M+1 M

which is defined by

M

+1(, pM+1) M

(, pM),

where < and is an appropriate term, and which is just the inverse of thecollapsing map obtained from taking hMm ( {pM}). I.e., = +1.

We now use the fact that C, i.e., that if + 1 (0, ]T then we do nothave that crit(ET ) <

H (E

T ). This implies that

T()n

is the inverse of the

collapsing map obtained from taking hMm (H {pM}), and that

H+1 M

T

(). In

fact, H+1 = +K , by [MiScSt97], =

+MT()

. We therefore have that

f(+1) < sup(H+1) =

+M+1 .

We have shown that g(+1) > f(+1).

We would like to prove that if 21 < , but 0 > 1, then Projective Determi-nacy (or even ADL(R)) holds, using Woodins core model induction. This, however,would require a solution of problem #5 of the list [SchSt].

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References

[AbMa] Abraham, U., and Magidor, M., Cardinal arithmetic, in: Handbook of

set theory, Foreman, Kanamori, Magidor (eds.), to appear.[BuMa90] Burke, M., and Magidor, M., Shelahs pcf theory and its applications,

Ann. Pure Appl. Logic 50 (1990), pp. 207-254.

[Gi] Gitik, M., Introduction to Prikry type forcing notions, in: Handbook ofset theory, Foreman, Kanamori, Magidor (eds.), to appear.

[GiMa92] Gitik, M., and Magidor, M., The singular cardinals problem revisited, in:Set theory of the continuum, Judah, Just, Woodin (eds.), Berlin 1992.

[GiMi96] Gitik, M., Mitchell, W., Indiscernible sequences for extenders and the

singular cardinal hypothesis, Ann. Pure Appl. Logic82

(1996), pp. 273-316.

[Ma77a] Magidor, M., On the singular cardinals problem I, Israel Journal of Math-ematics 28 (1977), pp. 1-31.

[Ma77b] Magidor, M., On the singular cardinals problem II, Annals of Mathemat-ics 106 (1977), pp. 517-548.

[Mi] Mitchell, W., The covering lemma, in: Handbook of set theory, Foreman,Kanamori, Magidor (eds.), to appear.

[MiScSt97] Mitchell, W., Schimmerling, E., and Steel, J., The covering lemma upto a Woodin cardinal, Ann Pure Appl.Logic 84 (1997), pp. 219-255.

[MiSt94] Mitchell, W., and Steel, J., Fine structure and iteration trees, LectureNotes in Logic # 3.

[Sch02] Schindler, R., The core model for almost linear iterations, Annals of Pureand Appl. Logic 116 (2002), pp. 207 - 274.

[SchSt] Schindler, R., and Steel, J., List of open problems in inner model theory,available at http://www.logic.univie.ac.at/rds/list.html .

[SchStZe] Schindler, R., Steel, J., and Zeman, M., Deconstructing inner modeltheory, J. Symbolic Logic, to appear.

[St96] Steel, J., The core model iterability problem, Lecture Notes in Logic #8.

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cardinal arithmetic and woodin cardinals

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