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Combinatorics on Ideals and Axiom A

arXiv:math/9302203v1 [math.LO] 26 Feb 1993Combinatorics on Ideals and Axiom AJames D. Sharp1/28/19931IntroductionThe following notion of forcing was introduced by Grigorieff[2]: Let I ⊂ω bean ideal, then P is the set of all functions p : ω →2 such that dom(p) ∈I. Theusual Cohen forcing corresponds to the case when I is the ideal of finite subsetsof ω.

In [2] Grigorieffproves that if I is the dual of a p-point ultrafilter, then ω1is preserved in the generic extension. Later, when Shelah introduced the notionof proper forcing, many people observed that Grigorieffforcing was proper.

Oneway of proving this is to show that player II has a winning strategy in the gameGω for P (see [3], page 91. )The notion of Axiom A forcing was introduced by Baumgartner [1].

If Psatisfies Axiom A, then player II has a winning strategy in the game Gω andthus is proper. Indeed, most of the naturally occurring proper notions of forcingare Axiom A (e.g.

Mathias or Laver forcing). Thus it is natural to ask whetheror not Grigorieffforcing satisfies Axiom A.

The main result of this paper is anegative answer to this question. We will prove this by introducing another gameGU and showing that if P were Axiom A then player II would have a winningstrategy in this game.

We will then prove that the game GU is undetermined.2Definitions and PreliminariesThroughout this paper U will denote a p-point ultrafilter and I will denote thedual ideal. We let [X]<ω = {s ⊆X | |s| < ω}.

And we will denote ordinalnames by ˙α. We let Seq(X) denote the set of finite sequences of elements of X.If s = ⟨x0, .

. .

, xn⟩∈Seq(X) and y ∈X, then s∗⟨y⟩= ⟨x0, . .

. , xn, y⟩∈Seq(X).0This paper forms part of the author’s dissertation, written under the direction of SimonThomas1

DEFINITION. The game GU is for two players playing alternatively.Player I plays a partition of ω, {In | n ∈ω}, such that for all n, In ∈I andplayer II plays finite subsets Fn ⊂In.

Player II wins iff[n∈ωFn ∈U. (1)Note that this definition makes sense for an arbitrary non-principal ultrafilterU.

However, if U is not a p-point, then player I clearly has a winning strategy.DEFINITION. A partial order P is said to satisfy Axiom A if there is a collectionof partial orders {≤n| n ∈ω} of P satisfying:i) p ≤0 q implies p ≤qii) p ≤n+1 q implies p ≤n qiii) if ⟨pn | n ∈ω⟩is a sequence such thatp0 ≥0 p1 ≥1 p2 ≥2 .

. .

≥n−1 pn ≥n . .

. (called a fusion sequence),then there is a q ∈P such that for all n, q ≤n piv) for all p ∈P, for all n ∈ω, and for all ordinal names ˙α, there exists q ≤n pand a countable set B such that q ⊩˙α ∈B.DEFINITION.

GrigorieffForcingP(U) = {p : ω →2 | dom(p) ∈I} where q ≤p iffq ⊇p.The main results in this paper are the following two theorems.Theorem 1 Gregorieffforcing does not satisfy Axiom A.Theorem 2 The game GU is not determined.3GrigorieffForcing and GUIn this section we will prove the following Lemma:Lemma 1 If P(U) satisfies Axiom A, then player II has a winning strategy inthe game GU .PROOF. Suppose P(U) satisfies Axiom A.Claim 1 Let p ∈P(U), n ∈ω, and I ∈I , then there exists q ∈P(U) such thatq ≤n p and | I \ dom(q) |< ω.

(2)2

PROOF OF CLAIM 1: Let p, n, and I be given. We may assume I is infinite.Let {Yα | α < 2ω} be an enumeration of ℘(I) and set ˙α = {⟨α, pα⟩| α < 2ω}where pα : I →2 is the characteristic function of Yα.Since P(U) satisfiesAxiom A, by (iv) there’s q ∈P(U) and a countable set B, such that q ≤n p andq ⊩˙α ∈B.

But then | I ∖dom(q) |< ω as required. □Now we describe a winning strategy for player II in the game GU .

Sup-pose player I plays I0 at the oth move, then player II sets p0 = χI0 andplays F0 = ∅. After the n −1st turn, player I has played I0, .

. .

, In−1 andplayer II has played F0, . .

. , Fn−1 and chosen p0≥0p1≥1 .

. .

≥n−2pn−1. At thenth move player I plays In.

Then by the claim there exists pn ∈P(U) suchthat pn≤n−1pn−1 and | In \ dom(pn) |< ω.Thus player II can play Fn =In \ dom(pn). At the end of the game ⟨pn | n ∈ω⟩forms a fusion sequence,and it follows from Axiom A (iii), that ∪dom(pn) ∈I.

And thus player II wins.■Notice that lemma 1 and theorem 2 imply theorem 1.4The Game GU is undeterminedWe shall prove the theorem as two lemmas.Lemma 2 Player I does not have a winning strategy in the game GU .PROOF. Suppose, by way of contradiction, that player I has a winning strategy.Let σ : Seq([ω]<ω) →I be player I’s winning strategy.

Then ran(σ) is countable.Let ⟨Zn | n ∈ω⟩be an enumeration of ran(σ) such that σ(∅) = Z0, let J0 = Z0and for n ≥1 let Jn = Zn \ ∪j

Thus there exists Y ∈U such that | Y ∩Jn |< ω andhence | Zn ∩Y |< ω for all n ∈ω. Now consider the following game where playerI plays by σ and player II playsFn = Y ∩σ(∅, F0, .

. .

, Fn−1)(3)Notice that for all n, there exists some m such thatFn = Y ∩Zm. (4)Clearly ∪Fn = Y ∈U, so player II wins.

■To prove lemma 3 we shall need thefollowing definitions and results from Grigorieff[2].DEFINITION.i) A ⊂Seq([ω]<ω) is a p-tree if it is non-empty and closed under takinginitial segments.ii) If s ∈A, the ramification of A at s, denoted RA(s), is the set of alla ∈[ω]<ω such that s ∗⟨a⟩∈A.3

iii) A is an I-p-tree if for any s ∈A, there exists X ∈U such that [X]<ω ⊆RA(s).iv) H is an I-p-branch of A if it is a branch such that ∪{H(n) | n ∈ω} ∈U.Theorem 3 (Grigorieff[2]) Every I-p-tree has an I-p-branch.Lemma 3 Player II does not have a winning strategy in the game GU.PROOF. Suppose, by way of contradiction, that player II has a winning strategyµ.

Player I will construct a “tree of games” such that along each branch player IIplays by µ, but there will be a branch such that player I wins the correspondinggame. First we need the following technical result.Claim 2 For all n ∈ω and for any sequence I0, I1, .

. .

, In of pairwise disjointelements of I , there exists Y ∈U, Y ⊆(ω \ ∪{In | i ≤n}) satisfying:(∀t ∈[Y ]<ω)(∃I ∈I)[(t ⊂I ⊂(ω \[i≤nIi)) ∧(µ(I0, . .

. In, I) ∩t = ∅)].

(5)PROOF OF CLAIM. Let n, I0, .

. .

, In be given. Suppose the claim is false.Then we can define, by induction on m ∈ω, a sequence of pairs ⟨Xm, sm⟩suchthati) X0 = ω ∖Si≤n Iiii) for all m ≥1, Xm = Xm−1 \ sm−1iii) (∀I ∈I)(sm ⊂I ⊂X0) →µ(I0, .

. .

, In, I) ∩sm ̸= ∅If we set J0 = Si∈ω s2i and J1 = Si∈ω s2i+1 then one of these must be in I , sayJ0 ∈I. But then for all i ≤ω, µ(I0, .

. .

, In, J0)∩s2i ̸= ∅, so | µ(I0, . .

. , In, J0) |=ω, contradicting the assuption that µ is a winning strategy for player II.□Player I constructs a tree A ⊂Seq({⟨I, F, s⟩| I ∈I, F, s ∈[I]<ω}) byinduction on the height n of a node such that, if ⟨⟨I0, F0, s0⟩, .

. .

, ⟨In, Fn, sn⟩⟩∈A theni) Ii ∩Ij = ∅if i ̸= jii) (1 ≤i ≤n)(Fi = µ(I1, . .

. , Ii))iii) (0 ≤i ≤n)(si ∩Fi = ∅)iv) I0 = F0 = s0 = ∅.v) The projection map π : ⟨⟨I0, F0, s0⟩, .

. .

, ⟨In, Fn, sn⟩⟩→⟨s0, . .

. , sn⟩is aninjection.4

Suppose we’ve decided which B = ⟨⟨I0, F0, s0⟩, . .

. , ⟨In, Fn, sn⟩⟩∈A.

Fixsuch a B. By the claim, there exists Y ∈U satisfying property 5.

For eacht ∈[Y ]<ω, choose It satisfying property 5.Let RA(B) = {⟨I, F, s⟩| s ∈[Y ]<ω, I = Is, F = µ(I0, . .

. , In, Is)}.Let τ ⊂Seq([ω]<ω) be the tree obtained from A via the projection map πThen, by the construction, τ is an I-p-tree, and hence has an I-p-branch H.Let ⟨⟨In, Fn, H(n)⟩| n < ω⟩be the corresponding branch in A.

Then, since∪{H(n) | n ∈ω} ∈U, this is a play of GU which player I wins. But player IIhas used the strategy µ in this play, which contradicts the assuption that µ is awinnig strategy for player II.■References[1] Baumgartner, J.

(1983). Iterated Forcing.

In Surveys in Set Theory (Math-ias, A.R.D., ed. ), pp.1-59, Cambridge University Press.

[2] Grigorieff, S. (1971). Combinatorics on ideals and forcing, Ann.

Math. Logic3, 363-94.

[3] Jech, T. (1986). Multiple Forcing, Cambridge University Press.CURRENT ADDRESS:James D. SharpDepartment of MathematicsRutgers UniversityNew Brunswick, NJ 08903email: jsharp@math.rutgers.edu5

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