Jakob Kellner and Saharon Shelah- More on the Pressing Down Game

8/3/2019 Jakob Kellner and Saharon Shelah- More on the Pressing Down Game

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MORE ON THE PRESSING DOWN GAME.

JAKOB KELLNER AND SAHARON SHELAH

Abstract. We investigate the pressing down game and its relation to theBanach Mazur game. In particular we show: Consistently relative to a super-compact, there is a nowhere precipitous normal ideal I on 2 such that playernonempty wins the pressing down game of length 1 on I even if player emptystarts. For the proof, we construct a forcing notion to force the following:There is normal, nowhere precipitous ideal I on a supercompact such thatfor every I-positive A there is a normal ultrafilter containing A and extendingthe dual ofI.

We investigate the pressing down game and its relation to the Banach Mazurgame. Definitions (and some well known or obvious properties) are given in Sec-tion 1. The results are summarized in Section 2. This paper continues (and sim-plifies, see 2.2) the investigation of Pauna and the authors in [14].

1. Definitions

We use the following notation:

For forcing conditions q p, the smaller condition q is the stronger one. Westick to Goldsterns alphabetic convention and use lexicographically biggersymbols for stronger conditions.

E

= { : cf() = }. NS is the nonstationary ideal on . The dual of an ideal I is the filter {A : \ A I} and vice versa. For an ideal I on and a positive set A (i.e., A / I), we set I A to be

the ideal generated by I { \ A}.

We always assume that is a regular uncountable cardinal and that I is a


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2 JAKOB KELLNER AND SAHARON SHELAH

PD(I) is played like PD(I), but empty wins the run if

n Sn = . PDe(I) is played like PD(I), but empty can first choose S1 to be an

arbitrary I-positive set. PDe is defined analogously.

So we have four variants of the pressing down game, depending on two param-eters: whether the winning condition for player nonempty is = or /I, andwhether empty has the first move or not.

We now analogously define four variants of the Banach Mazur game:

Definition 1.2. BM(I) is played as follows: Set S1 = . At stage n,empty chooses an I-positive subset X of Sn1, and nonempty chooses anI-positive subset Sn of X. Empty wins the run if

n Sn I.

The ideal game Id(I) is played just like BM(I), but empty wins the run if

n Sn = . BMne(I) is played just like BM(I), but nonempty has the first move. Idne(I) is defined analogously.

More generally, we can define the Banach Mazur game BM(B) on a Booleanalgebra B: The players choose decreasing (nonzero) elements an B, nonemptywins if there is some (nonzero) b B smaller than all an. Then BM(I) is equivalentto the corresponding game BM(BI) on the Boolean algebra BI = P()/I (since Iis -closed), the same holds for BMne(I) and BMne(BI); we could equivalently usethe completion ro(BI) instead of BI. Also the /I versions of the pressing downgame can be played modulo null sets, i.e., on the Boolean algebra BI, in the obviousway. For the = versions of the games, the version played on BI does not makesense.

In the /I version, the pressing down and Banach Mazur games have naturalgeneralizations to other lengths : At a limit stages , we use


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MORE ON THE PRESSING DOWN GAME. 3

IfI is nowhere -saturated,1 then c&cset(I, ) is equivalent2 to BM(I) (andtherefore to c&c(BI, ), cf. 1.10).

If I is +

-saturated, then c&cset

(I, ), c&c(BI, ), PDe(I) and BM(I) areall equivalent, cf. 3.2 and 3.3.

We also consider the variant of c&cset(I, ) where empty wins if the intersectionis empty (as opposed to null). However, then we have to allow empty to cut intoarbitrary pieces, not just into I-positive ones, while empty still has to choose apositive piece. (Otherwise nonempty always wins, by fixing some and picking thesets containing .)

Definition 1.4. c&cmin(I, 2) is played as follows: First, empty choosessome positive S1. At stage n, empty cuts Sn1 into two arbitrary pieces,and nonempty chooses an I-positive piece Sn.

In c&cmin(I,


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b(BM)

b(Id) a(BM) b(PDe)

a(Id) b(PDe ) a(PDe)

a(PDe )

~~~~

~~~

ddd

dddd

cc

cccc

c

cc

cccc

cc

bb

bbbb

b

Figure 1. The trivial implications (for empty moving first)

Facts 1.8. In the pressing down games, we can assume without loss of gen-erality that nonempty chooses at stage n a set of the form Sn = f1n (n) Sn1 for some n.

4

PD is monotone in the following sense: if J I, then PD(J) is stronger

than PD(I). The same holds for PD, but not for PDe or PDe or for any

of the Banach Mazur games. In particular, PD(I) is stronger than PD(NS) for all normal I.

Just as in the case of BM, b(PD) cannot hold for = 1 (cf. 5.2). Other than in the case of Id, the property a(PDe) has no consistency

strength (cf. 2.1).

What is the effect of empty moving first?

Facts 1.9. For the Banach-Mazur games, the distinction whether emptyhas the first move or nonempty is a simple density effect: For example,nonempty wins BMne(I) iff there is some S I+ such that nonemptywins BM(I S); similarly simple equivalences hold for empty winning; forcharacterizing BM in terms of BMne; and for the = version.

We will see in Lemma 2.6 that this is not the case for the pressing downgame.

The cut and choose game is related to the pressing down and Banach Mazurgames [12, 4, 18, 19]. As usual, ro(B) denotes the completion of the Booleanalgebra B, and BI = P()/I.

Facts 1.10. c&c(B, ) is equivalent to c&c(ro(B), ); but c&c(B, ) willgenerally not be equivalent to c&c(ro(B), ).

c&c(B, ) is equivalent to the Banach Mazur game on B. So in our case, c&c(BI, ) is equivalent to BM(I). c&c(BI, ) is equivalent to PD(I), cf. 3.2. (However c&c(ro(BI), ) might

be a stronger game.)

The set-partition version of the cut and choose game is also related to the pressingdown and Banach Mazur games: If we assume that I is nowhere -saturated, then

4This of course means: PD is equivalent to the game where nonempty is restricted to movesof this form.


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b(BM) b(PDe) b(PD)

a(BM)

b(Id) b(PDe ) b(PD)

a(Id) precip. -semi precipitous

))

//

//

//

))

//

//

Figure 2. Some properties stronger than -semi precipitous

BM(I) is equivalent to c&c(BI, ), i.e., to the game where nonempty partitions thecurrent set into many positive pieces. In contrast, PDe(I) is equivalent to thegame where nonempty partitions the current set into many positive pieces withthe additional requirement that the partition is maximal in the family ofI-positive

sets (i.e., for every positive set S there is an A in the partition such that S A ispositive).

Winning strategies for games on a Boolean algebra B have close connections tothe properties of B as Boolean algebra and as forcing notions, again see [12, 4, 18,19]:

Facts 1.11. B having a -closed positive subset implies b(BM(B)). b(BM(B)) is also denoted by B is strategically -closed and implies that

B is proper. a(BM(B)) is equivalent to B is -distributive.

It is not surprising that we will get stronger connections if we assume that the Bhas the form BI = P()/I for a normal ideal I. We will mention only one example:

Fact 1.12. If BI is proper, then a(BM(I)) holds.For a proof, see 3.5.

2. The results

2.1. Some Observations. Some of the facts for precipitous ideals can be shown(with similar proofs) for PD, but there are of course strong differences as well:

Lemma 2.1. (1) b(c&cminne

(I, 2)) implies that is measurable in an inner model.

(2) So in particular, b(PD(I)) implies that as well.(3) However, a(PDe(I)) has no consistency strength. In particular, for = 2,

a(PDe(I)) is implied by CH for every I concentrated on E21

.

(4) b(PD(I)) implies > 20 and that I is not concentrated on E0 .

The proofs can be found in 5.5, 5.2 and 5.3.In this paper, we are not interested in the property empty does not win the

pressing down game, since it has no consistency strength. Also, the effect of whomoves first in Banach Mazur games is trivial. The remaining properties are picturedin Figure 2. All these properties are equiconsistent to a measurable (e.g., for I =

NS2 E21

). In fact, they imply that I is -semi precipitous, see Definition 4.1,which in turn implies that is measurable in an inner model. We claim that none ofthe implications can be reversed. In this paper, we will prove some strong instancesof this claim by assuming larger cardinals: We show


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b(PDe) doe not imply precipitous, and

a(BM) does not imply b(PD).

We also claim that (consistently relative to a measurable) b(Id) does not imply b(PD),

but we do not give a proof here. With these claims (for which we assume cardinalslarger than a measurable) it is then easy to check that no implication of Figure 2can be reversed.

In [14], Pauna and the authors showed that, assuming the consistency of a mea-

surable, b(PD(I)) does not imply b(BMne(I)) for I = NS2 E21

. In fact, a slightly

stronger statement holds (with a simpler proof):

Lemma 2.2. It is equiconsistent with a measurable that b(PD(I)) holds (even for

length 1) buta(Idne(I))) fails for I = NS2 E21

.

(For a proof, see 5.8.) Note that a(Idne(I))) fails just means that I is nowhereprecipitous.

Of course, precipitous cannot generally imply a winning strategy for nonemptyin any game, since precipitous is consistent with 20 . However, we can getcounterexamples for > 20 as well: Just adding Cohens destroys any winningstrategy for nonempty (for any ideal on 2), but preserves precipitous. So we get:

Lemma 2.3. It is equiconsistent with a measurable that CH holds, NS2 E21

is

precipitous but b(PD(J)) fails for any normal ideals J on 2.

2.2. Large cardinals. To see that not even a(BM(I)) implies any winning strategy

for nonempty, we assume CH and a 3-saturated ideal I on 2 concentrated on E21

.Saturation is preserved by small forcings, in particular by adding some Cohens, andsaturation (together with CH) implies a(BM(I)). So we get:

Lemma 2.4. The following is consistent with CH plus an 3-saturated ideal on2:CH holds, a(Id(I)) holds for some I on 2, butb(PD

(J)) fails for any normal Jon 2.

See 5.11 and 5.12. (It seems very likely that saturation is not needed for this,but the construction might get considerably more complicated without it.)

As mentioned in Lemma 2.2 it is possible that b(PD(I)) holds for a nowhereprecipitous ideal, i.e., for an ideal such that a(Idne(I)) fails. It seems harder toeven get b(PDe(I)) for a nowhere precipitous ideal. Here we use a supercompact:

Theorem 2.5. It is consistent with a supercompact that for = 2 there is anowhere precipitous I such that b(PDe(I)) holds (even for length 1).

(See 6.1.)Note that (as opposed to 2.3, 2.4) we just make a specific ideal non-precipitous,

and we do not destroy all precipitous ideals. It seems very hard (and maybe im-possible) to do better, and it is not clear whether we can avoid cardinals largerthan a measurable. It is not known how to kill all precipitous ideals5 on, e.g., 1

5Since we are only interested in normal ideals, it would be enough to kill all normal precipitousideals. This doesnt help much, though; it is not known whether the existence of a precipitousideal does imply the existence of a normal precipitous one. Recently Gitik [7, 6, 3] proved someinteresting results in this direction.


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with reasonable forcings.6 And it might be even harder to do so while addition-ally preserving b(PDe(I)) for some ideals: By recent results by Gitik [6] (and later

Ferber and Gitik [3]) a -semi precipitous ideal does imply a normal precipitousideal under in the absence of larger cardinals and under some cardinal arithmeticassumptions.

2.3. Moving first. Let us now investigate the effect of whether empty moves first.If we compare Ge and Hne for any games G and H, then these variants will be dif-

ferent for trivial reasons: For example, b(BMne(NS2)) does not imply b(PDe (NS2)):

Let U be a normal ultrafilter on , Levy-collapse to 2, and let I1 be the idealgenerated by the dual of U (which is concentrated on E21 ). Then nonempty

wins BM(I1) and therefore BMne(I1 + E20

) as well. But nonempty can never

win PDe (I1 + E20

), since nonempty cannot win PD(E20 ).So the games are very different (for trivial reasons) when we change who has the

first move. However, for the Banach Mazur game, the effect of who moves first isa simple density effect, 7 as we have mentioned in 1.9. For example, b(BMne(I))holds iffb(BM(I S)) holds for some positive S.

This is not the case for the pressing down games. Of course we still get:

b(PDe(I)) holds iffb(PD(I S)) holds for all S I+.

The same holds for PDe .

But unlike the Banach Mazur case, we can have the following:

Lemma 2.6. It is equiconsistent with a measurable thatb(PD(I)) holds butb(PDe(I S)) fails for all positive S, e.g., for I = NS2.

(See 5.8.) So in other words, b(PD(I)) can hold but for all positive S there is apositive S S such that b(PD(I S)) fails.

3. Empty not winning

Lemma 3.1. CH implies a(PD(I)) for every I on 2 that is not concen-

trated on E20 .

More generally, if 0 < for all < and 0 = , then empty winsPD(I) iff E> I.

So if I is concentrated on E> (and the same cardinal assumptions hold)thena(PDe(I)) holds.

Proof. Assume that I is concentrated on E0 . Just as in [5], it is easy to see thatempty wins PD(I): For every E0 , let (seq(, n))n be a cofinal sequence in. Let Fn map to seq(, n). If empty plays Fn at stage n, then the intersectioncan contain at most one element.

So assume towards a contradiction that E> / I and that empty has a winning

strategy for PD(I). The strategy assigns sets Xt and regressive function ft to nodest in the tree T =


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Generally, fix t T. We can assume by induction that one of the following caseshold:

t corresponds to a partial run rt with (positive) partial result Xt; then weset ft to be emptys response to rt.

Xt I; then we set ft 0.

In both cases we set Xt = Xt f1t ().

Let b be a branch of T (i.e., b ). We set Xb =

n Xbn.

Assume that b corresponds to a run of the game; this is the case iff Xbn isI-positive for all n. Then Xb I, since empty uses the winning strategy. If b doesnot correspond to a run, then Xb I as well. So

(1) Xb I for all branches b

Xb and Xc are disjoint for different branches b, c; and for all there is exactlyone branch b such that X

b . We assume = 0 from now on. By definition,

for all nfbn() = b(n)

Since fbn is regressive, b (n) < for all n . In other words, b .

Fix an injective function : . Since 0 < for < , we can find aclub C such that

for all C E>.

This defines a regressive function g : C E> by g() = (b ). Since Iis normal and does not contain E>, there is a positive set S and a (orequivalently a branch b of T) such that g() = , i.e., b = b for all S. Thisimplies that S Xb is positive, a contradiction to (1).

Lemma 3.2. If I is normal, then PDe is equivalent to c&c(BI, ).

Proof. A regressive function defines a maximal antichain in BI of size at most .On the other hand, let A be a maximal antichain of size . We can choosepairwise disjoint representatives (Si)i for the elements of A, and define

f() =

1 + i if Si and 1 + i < ,

0 otherwise.

f1(0) I. (Otherwise there is an Si in A such that T = Si f1(0) I+, pick T \ (1 + i + 1), contradiction.) So the partition A is equivalent to the regressivefunction f.

Together with 1.10 we get:Corollary 3.3. If I is normal and +-saturated, then BM(I) and PDe(I) areequivalent. The same holds for BMne(I) and PD(I).

IfI is +-saturated, then it is precipitous, i.e., a(Id(I)) holds [9, 22.22]. However,I can be concentrated on E0 (for example, could be 1), which negates a(BM(I)).However, with Lemma 3.1 we get:

Corollary 3.4. If I is +-saturated, 0 = , 0 < for all < , and I isnormal and concentrated on E>0, thena(BM(I)) holds.


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In the rest of the section, we show that properness implies a(BM(I)). This isnot needed for the rest of the paper.

For any Boolean algebra B, b(BM(B)) implies that B is proper (as a forcingnotion), cf. e.g. [11, Thm. 7]. For Boolean algebras of the form BI = P()/I wealso get:

Lemma 3.5. Assume > 20. If BI is proper then a(BM(I)) holds.

Normality of I is not needed, just


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This is of course no surprise: the proof is a simple generalization of the proof [8,Theorem 2] for precipitous; Jech and others have used in fact very similar gener-

alizations. (E.g., in [12] it is shown more or less that pseudo-precipitous ideals are-semi precipitous.)

Proof. We assume that there is a forcing P and a nameD for the V-generic filter.

In particular:

(2)P forces that in V[G] there is an elementary embedding j : V Mfor some transitive class M in V[G].

If we are only interested in consistency strength, we can use Dodd-Jensen coremodel theory as a black-box: (2) is equiconsistent to a measurable cardinal, whichfollows immediately, e.g., from [9, 35.6] and the remark after [9, 35.14]: KV =KV[G], and there is a measurable iff there is an elementary embedding j : K M(which also implies M = K). However, this only tells us that there is some ordinal

which is measurable in an inner model, and not that this ordinal is indeed .To see this, we can either use more elaborate core model theory (as pointed out

by Gitik, cf. [20, 7.4.8,7.4.11]). Alternatively, we can just slightly modify the proofof [8, Theorem 2] (which can also be found in [9, 22.33]). We will do that in thefollowing: Let K be the class of strong limit cardinals such that cf() > . Let(n)n be an increasing sequence in K such that |K n| = n. Set A = {n >n } and = sup(A).

By a result of Kunen, it is enough to show the following:

(3)There is (in V) an iterable, normal, fine L[A]-ultrafilter W suchthat every iterated ultrapower is wellfounded.

We have a nameD for the V-generic filter.

D does not have to be normal, but

there is some p0 P and 0 such that p0 forces that [Id] = 0. We set

J = {x : p0 x /D}, and

U = {x P() L[A] : x / J }.

U is generally not normal, but the normalized version of U will be as required.U is an L[A] ultrafilter: Let x be in L[A]. We have to show: x or \ x are

in J.

There is a formula and a finite E K such that (in L[A]) x iff < and (,E,A).

Assume G is P-generic over V and contains p0. [Id] = 0, so x D[G] iff

0 j(x). By elementarity (in V[G]) 0 j(x) iffj(L[A]) thinks that (0, j(E), j(A)).

But j() = for every K.

So we get x D[G] iff (in L[A]) (0, E , A) holds, independently of G

(provided G contains p0). In other words, if there is some generic G suchthat x

D[G], then x

D[G] for all generic G (containing p0); i.e. p0

forces that x D[G]; i.e. \ x J.

Assume that x is not in J. Then there is some q p forcing that x D.

So \ x J.

U is


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This also shows that (in V) the intersection of 0 many U-elements is nonempty;which implies that every iterated ultrapower is wellfounded (provided iterability).

U is iterable: Let (in L[A]) (x) be a sequence of subsets of . Let G beP-generic over V and contain p0. In V[G], x

D[G] iff 0 j(x). The sequence

(j(x)) is in L[j(A)], and therefore also the set { : 0 jG(x)}. ButL[j(A)] = L[A].

normalizing: Since we now know that U is wellfounded, we know that there issome f : in L[A] representing . Set W = f(U). Then W is as required.

The following follows easily from Kunens method of iterated ultrapowers (see,e.g., [14, 4.3] for a proof):

Lemma 4.3. Assume V = L[U], where U is a normal ultrafilter on . Let V be a forcing extension ofV and D V a normal, wellfounded V-ultrafilter on . ThenD = U.

This implies:

Corollary 4.4. In L[U], the dual of U is the only normal precipitous ideal on ;and every ideal on that is normally -semi precipitous is a subideal of the dualof U.

We will also need the following:

Lemma 4.5. If I is a w2 > . . . is an infinite decreasingsequence in P, then

i wi represents a run of the game, so the result

i A(wi)

has to be nonempty (or even positive in the case of a PD-strategy).


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Lemma 5.2. b(PD) implies > 20.

Actually, we can even restrict nonempty to play functions f : {0, 1}. Inother words, it is enough to assume b(c&cmin(I, 2)), cf. Definition 1.4.

Proof. The proof is the same as [5, 1]: We assume otherwise and identify witha subset X of [0, 1] without a perfect subset. We claim:

(4)For all w P and n there are disjoint open intervals I1and I2 of length 1/n and w1, w2 < w such that A(w1) I1 andA(w2) I2.

Assume that (4) fails for some v0 and n0. Given v < v0 and n > n0, we fix apartition of [0, 1] into n many open intervals of length 1/n and the (finite) set ofendpoints. By splitting A(v) n + 1 many times, empty can guarantee that A(w)has to be subset of one of the intervals for some w < v. Since (4) fails, there has

to be for each n a fixed element I(n) of the partition such that for all v < v0 thereis a w < v with A(w) I(n).

I(n) can contain at most one point x, so the

empty player can continue v0 by first splitting into {x} and A(v0) \ {x}; and thenextending each vn1 to vn such that A(vn) I(n). Then the intersection is empty.This shows (4).

So we can fix an order preserving function from 2 w1 > in P such

that each wi is in N: Using (5) in N, we get a w0 N P such that / A(w0).


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Given wn1, let wn N be the continuation where empty played the regressivefunction

fn() =

0 if < seq(, n)gn() otherwise.

(Note that seq(, n) < is in N for all n.) Assume that

n A(wn). Then seq(, n) for all n, so . On the other hand, gn() N for all n, so .But / A(w0), a contradiction.

Of course this shows the following: b(PD(I)) implies b(PD(I E>0)) (sinceempty can just cut into E0 and E

>0

as a first move).

Recall that b(PD(I)) for any I implies b(PD(NS)) (due to monotonicity). Sothe last lemma gives:

Corollary 5.4. b(PD(I)) is equivalent to b(PD(I E>0)) and implies b(PD(NS))

andb(PD(NS E>0

)).

Lemma 5.5. b(PD(I)) implies that I is normally -semi precipitous.b(c&cmin(I,


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for > 1, (, ) dom(q) and q(, ) = 0 for {0, 1}. Given < , defineQ = {q : dom(q) } and : Q Q by q q ( ). The following is

well known: If q p G, then q p (i.e. is the same as ). Q is -cc and < -closed. In particular, if p forces that C is club, then there is a club C0 V

such that p forces C0 C. The ideal generated by NSV in V[G] is N S

V[G] .

We also need the following simple fact (see, e.g., [14, 6.2] for a proof):

(6)

Let I be a normal ideal concentrated on E, let T be I-positive,p Q and p p for all T. Then there is an I-positive T Tand a q p such that (p) = q for all T

.

So in particular, every q q is compatible with p for all but boundedly many T.

We will also use:Lemma 5.7. Let be inaccessible and T be stationary. The Levy collapsepreserves b(PD(NS T)). The same holds for PD.

Proof. Assume towards a contradiction that q forces that nonempty does have awinning strategy in V[G]. We describe a winning strategy in V: Assume emptyplays f0 (in [V]). Let q0 q decide that in V[G] nonempty chooses 0 as responseto f0 according to the winning strategy in V[G]. So q0 forces that f

10 (0) T

is stationary, therefore f10 (0) T is stationary in V. Generally, let qn qn1decide that nonempty plays n as response to fn. Since Q is -closed, there is aq < qn for all n. So q forces that

f1n (n) T is stationary.

Starting with L[U] and using a Levy collapse we get:

Corollary 5.8. Consistently relative to a measurable, b(PD(NS2)) holds (even for

length 1) butb(PDe

(NS2 S)) fails for every stationary S, and NS2 is nowhereprecipitous.

Proof. Assume V = L[U] and let Q = Levy(1,


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Otherwise some q1 q forces that S is nonstationary. Then there is in V a club

C and a q2 q1 forcing that S C = . Pick T C such that p and q2

are compatible. Then q3 p, q2 forces that T

C S, a contradiction. Thisshows (7).

By our assumption, p forces that nonempty wins PD(NS S). But b(PD(NS

T)) fails in V (since T T and T / U), therefore b(PD(NS2 T)) fails in V[G]

according to 5.7, and by monotonicity b(PD(NS2 S)) fails as well, a contradic-

tion.

We will now force nonempty not to win PD. For simplicity we will assume CHand look at = 2. It turns out that it is enough to add 1 many Cohen reals(actually, many similar forcings also work). First we need another variant of (4)or (5):

Lemma 5.9. Assume CH andb(PD(NS2)). For eachv P there are F(v) v

and F(v) v such that A(F(v)) and A(F(v)) are disjoint.

(We can choose F(v) and F(v) to be immediate successors of v, i.e. we justhave to choose two regressive functions f and f as emptys moves.)

Proof. We fix an injection : [2]0 2. Let S = C E

21

(for some clubset C)

consist of ordinals such that []0 . For each S, pick a normal cofinalsequence : 1 . For i 1 set gi() = ({

(j) : j i}) for S; andset gi() = 0 for / S. So for all i 1, gi is a regressive function. If = thengi() = gi() for some i; and gi() = gi() implies gj() = gj () for all j > i.

Let x(i) be the strategys response to vgi, We can identify x(i) with the se-quence 1x(i) = (i,k)ki. So for all with gi() = x(i) we get (k) = i,k fork i.

Case A: There are k < i < j < 1 such that i,k = j,k. Then set F = gi andF = gj. If g

1i (x(i)) and g

1j (x(j)), then

(k) = (k), so in particular = .

Case B: Otherwise, all the sequences (i,k)ik cohere for all i 1, so let (k)k1be the union of these sequences, with supremum < 2. So for all = in S thereis some k() 1 such that

(k()) = k()). Set h() = 0 for 1{}2\S.So h is a regressive function. Let l be the strategys response to vh. Set F = hand F = gl. If h

1(l) and g1l x(l), then (l) = l which is different to

(l).

Lemma 5.10. Assume CH. Let P1 be the forcing notion adding 1 many Cohen

reals. Then P1 forces b(PD(NS2)).

(The same holds for any other CH preserving 1-iteration of absolute ccc forcing

notions.) Note that since PD is monotone, b(PD(I)) fails for all ideals I on 2.

Proof. Assume that p P forces that is a winning strategy for nonempty for the

game PD(I).Let P be the complete subforcing of the first Cohen reals. P1 forces that

Lemma 5.9 holds. We fix the according P1-namesF and

F. Let N H()

contain p,,

F and

F. Set = N 1. If G1 is P1-generic over V, then

G = G1 P is P1-generic over N (and P-generic over V).


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So in N = N[G] = N[G1 ], we can evaluate the correct values of,

F and

F

for all valid sequences v in N (i.e., the resulting values are the same as the ones

calculated in V1 = V[G1 ]).In V1, pick any real r / V. Using r, we now define by induction a run b of the

game such that each initial segment is in N: Assume we already have the validsequence u N. Extend u with

F(u) if r(n) = 0, and to F(u) otherwise.

So b V1 is a run of the game according to ; nonempty wins the run; sothere is some

n A(b n). But we can in V use this to reconstruct (by

induction) the run b and therefore the real r: Assume we already know r n andthe corresponding valid sequence u = b n. Then is element of exactly one ofA(F(u)) or A(F(u)), which determines r(n) as well as the sequence correspondingto b (n + 1).

On the other hand, adding Cohens, as any -cc forcing, preserves precipitousness(and non-precipitousness) of an ideal, cf. 4.5. So we get:

Corollary 5.11. a(Id(I)) does not imply b(PD(NS)).

If we assume CH and an 3-saturated normal ideal on 2 saturated on E21

, weget the following:

Corollary 5.12. (Saturated ideal.) a(BM(I)) does not imply b(PD(NS)).

Proof. Since P1 has size 1 < 2, cl(I) remains 3-saturated. So in V[G], we canuse 3.4 to see that a(BM(cl(I))) holds.

6. A supercompact

We have seen that b(PD(I)) can hold for a nowhere precipitous ideal I. Itseems harder to show that there could be a nowhere precipitous ideal I satisfyingb(PD

e(I)):

Theorem 6.1. The following is consistent relative to supercompact: I0 is nowhereprecipitous, i.e., a(Idne(I0)) fails, and for every I0-positive set S the dual to I0 Scan be extended to a normal ultrafilter.

Note that this implies b(PDe(I0)), even for the game of length .We will split the proof into several lemmas: First we define the forcing S()

as limit of P. We also define dense subsets P

of the P. Then we define theforcing notion R+1, by doing the usual Silver-style preparation with reverse Eastonsupport. This forcing notion is as required: In the extension, we define 6.5 the idealI0 and show that the theorem holds (6.8 and 6.9).

6.1. The basic forcing. So let us assume that is an inaccessible cardinal, anddefine S() as the limit of the


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If a = 0, we set ga() = for all , and Ba = . Otherwise, we use some bookkeeping8 to find a B0a Aprec(a), and we set:

(8) Ba = B0a \ b 0.

We define the order on Qa by q p if fq fp. We set fa to be the canonical Qa-generic, i.e.,

qG f

q.

We set Aa = { : f() > 0}. (So A0 = , and Aa Ba B0a.)

Note that to write the diagonal union in (8), we have to identify the index setwith . Different identifications lead to the same result modulo club. In particular,

we get:

(9) If b < a and prec(b) = prec(a) then Ba \ Ab is nonstationary.

Obviously Qa is 0 for all < .

We can interpret p Pa to be a condition in Pa in the obvious way; in particularwe can define the order on Pa to be the one inherited from Pa.

Lemma 6.2. Pa is a dense subset of Pa. The order on Pa (as inherited from Pa) is the extension relation. Pa is


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6.2. The Silver style iteration. We now use the basic forcing S() in a reverseEaston iteration, the first part acting as preparation to allow the preservation of

measurability. This method was developed by Silver to violate GCH at a measur-able, and has since been established as one of the basic tools in forcing with largecardinals. We do not repeat all the details here, a more detailed account can befound in [9, 21.4]. Note that here we do not just need to preserve measurability orsupercompactness (for this, we could just use Lavers general result [15]), we needspecific properties of the Silver iteration.

Fix a j : V M such that

(10) M is closed under ++-sequences.

In particular, cf(j()) > +.We will use the reverse Easton iteration (Ra, S(a))a, for S(a) defined as above.

R is the preparation that allows us to preserve measurability (and we will notneed it for anything else); we will look at R Pa for a +, and in particular atR+1 = R P+ (recall that S() = P+). We claim that R+1 forces what wewant. We will also use j(R Pa) M. We get the usual properties:

The definition of R is sufficiently absolute. In particular, we can (in M)factorize j(R+1) = Rj()+1 as R+1 R

, where R is the quotient forcing

R+1j()+1

. Note that R is ).

Otherwise, a is called a null-index.

Here we interpret Ba and ga as R Pa-names in the canonical way, so the j-images are Rj() Pj(a)-names. In particular we can assume that q is in Rj() Pj(a)(or just in Pj(a), since we start from V[G] anyway).


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Lemma 6.4. If a is null and b >T a, then b is null as well. Also, 0 is a positiveindex.

Proof. Pick < j () such that every master condition forcesj(ga) < or / j(Ba).But the empty condition forces j(gb)() j(ga) and j(Bb) j(Ba).

Definition 6.5. In V[G], we define the ideal I0 by A I0 iff there is an X +

of size consisting of null-indices such that

(12) A iX Bi modulo a club set.10

Lemma 6.6. I0 is a normal ideal on

Proof. Assume that Ai Ci lXiBl for all i . Then (

Ai) Ci l

SXiBl modulo a club set.

By elementarity, if q is a G-master condition and if (c, B[G], g[G]) holds in

V[G] for some c V, then for all H containing q we get in M[G][H]

(13) (j(c), j(B)[G][H], j(g)[G][H]).

Lemma 6.7. In V[G] the following holds: If a is a positive index, then Aa isI0-positive.

Note that this implies: a is a positive index iff Ba is a I0-positive set; and I0 isnontrivial (since 0 is a positive index).

Proof. Assume otherwise, and fix an appropriate X and a club set C, i.e.,

(14) A C iX Bi.

Since X consists of null-indices, there is for each b X a b < j() such that everymaster condition forces / j(Bb) or j(gb)() < b. Since cf(j()) > , we can findan upper bound for all b. So every master condition forces

(15) j(Bb) implies j(gb)() < for all b X.

Since a is a positive index, we can find a master condition q forcing Ba andj(ga)() > + 1. Without loss of generality, q is in Pj(a). So we can extend q to q

such that

(16) j(A) and j(fi)() >

Since C is club, is forced to be in j(C). So q forces j(Aa) j(C). Accordingto (14), AaC bX Bv holds in V[G], so q

forces j(Aa)j(C) j(bX Bb).Let Z be the sequence (Bb)bX . Recall that we fixed (in V[G]) some bijection i

of to X, to make Z well defined. So j(Z) uses j(i), a bijection from j() toj(X); and j(Z) means: There is an < such that j(Z)j(i)(). Notethat j(Z)j(i)() = j(Bi()) and set b = i() X. So

(17) j(Bb) for some b X (in particular, b is null-index).

10As mentioned above, iXBi is only defined modulo a club set, since X is not canonicallyisomorphic to (it is just a subset of+ of size ). To avoid ambiguity, we just fix from now onfor each such X a bijection to and make iXBi well defined; still we use subset modulo clubset in the definition on I0.


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We further extend q to some q such that q(j(c), ) = 0 for all c + that are (asordinals) bigger than . We further extend q to some q deciding the b of (17).

So q

forces(18) j(Aa Bb) for the null-index b.

We will get a contradiction by case distinction on the position of b relative to a inthe tree T:

b > b, contradicting (15) and (16). aT c and c according to (16) whichcontradicts (15).

Lemma 6.8. In V[G], empty has a winning strategy for Idne(I0).

Proof. Assume that we have a partial run of the game of length n, correspondingto the node a in T, and empty has played Xn as last move, which is a subset ofAa. Assume that nonempty plays the I0-positive set B0 Aa. Let bT a be suchthat Xn+1 := Ab B0 is I0-positive, and let Xn+1 be emptys answer (and b bethe new T-node corresponding to the new partial run). This is a winning strategy

since fi() decreases along every branch of T. It remains to be shown that we canfind a b T a as above: B0 itself is enumerated as B0c by the bookkeeping at somestage c T a. Recall that Bc = B0c \ d


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We claim that Case 1 has to occur for some a. Otherwise, X is a subtree of T suchthat every node has at most many successors, i.e., there are only many branches

through X. By induction on n, X is covered (module I0) by cX, T-height(c)=nAc.But for any branch b, the set

n Ab(n) is empty (witnessed by the decreasing

sequence fb(n)), a contradiction.So we can pick a T-minimal b such that Case 1 holds. Note that j+ < cf(j()).

For every null-index c there is a witness c < j (), so there is a universal bound .Since b is a positive index, we can find a master condition q forcing j(gb)() > and j(Bb). Recall that Bb Y (mod I0), so q forces that j(Y). We now extendq to q so that it forces / j(Ac) for all c > a. Then q

is as required: / j(Bc)for any null-index c, by a similar case distinction as in the proof of Lemma 6.7.

This ends the proof of Theorem 6.1.

6.3. The Levy collapse. As usual, we can use a Levy collapse to reflect these

properties to 2:Lemma 6.10. Start with the universe V as in Theorem 6.1. After collapsing to2, we get: cl(I) is nowhere precipitous and satisfies b(PDe(cl(I))) (even for thegame of length 1).

Proof. Nowhere precipitous follows from 4.5. Let S be a P-name for a cl(I)-positiveset and p P. Will show:

(20)In V there is a normal ultrafilter U and a q p forcing that S iscl(U)-positive.

Then according to the usual argument, the cl(U)-positive sets have a -closeddense subset, so nonempty wins BM(cl(U) S), and since cl(U) extends cl(I) nonempty wins PD(I S) (even for length 1).

To prove (20), set T = { E1 : p / S}. T is I-positive. For each T pick a witness p p. Let q, T be as in 6 and pick a normal ultrafilterU containing T. We have to show that q forces S to be cl(U)-positive. Assumeotherwise, and pick q q and A U such that q forces A S = . Then q Qfor some < . Pick TA\. Then p and q are compatible, a contradictionto p S.

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Kurt Godel Research Center for Mathematical Logic, Universitat Wien, Wahringer

Strae 25, 1090 Wien, Austria

E-mail address: [email protected]: http://www.logic.univie.ac.at/kellner

Einstein Institute of Mathematics, Edmond J. Safra Campus, Givat Ram, The Hebrew

University of Jerusalem, Jerusalem, 91904, Israel, and Department of Mathematics,

Rutgers University, New Brunswick, NJ 08854, USA

E-mail address: [email protected]: http://shelah.logic.at/

Jakob Kellner and Saharon Shelah- More on the Pressing Down Game

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