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Covering games and the Banach-Mazur game:

arXiv:math/9207203v1 [math.LO] 25 Jul 1992Covering games and the Banach-Mazur game:k-tactics.Tomek Bartoszynski∗Department of Mathematics,Boise State University,Boise, Idaho 83725Winfried Just†Department of Mathematics,Ohio University,Athens, Ohio 45701Marion Scheepers‡Department of Mathematics,Boise State University,Boise, Idaho 83725AbstractGiven a free ideal J of subsets of a set X, we consider games whereplayer ONE plays an increasing sequence of elements of the σ-completionof J, and player TWO tries to cover the union of this sequence by playingone set at a time from J. We describe various conditions under whichplayer TWO has a winning strategy that uses only information aboutthe most recent k moves of ONE, and apply some of these results to theBanach-Mazur game.1IntroductionLet J be a free ideal of subsets of a given set.

By ⟨J⟩we denote the σ-ideal generated by J (⟨J⟩could turn out to be the power set of ∪J). Twoconcrete examples of ideals motivated much of our work.

The one is JR,the ideal of nowhere dense subsets of the real line R. In this case ⟨JR⟩is theideal of meager sets of reals. The other is [κ]<λ where ω = cof(λ) ≤λ ≤κare cardinal numbers.We are interested in games of the following type: Player ONE playsa set On ∈⟨J⟩during inning n, to which TWO responds with a set∗Supported by Idaho State Board of Education grant 92-096†Supported by NSF grant DMS-9016021 and Research Challenge Grant RC 89-64 fromOhio University‡Supported by Idaho State Board of Education grant 91-093.1

Tn ∈J. ONE is required to play an increasing sequence of sets; TWO’sobjective is to cover Sn∈ω On with Sn∈ω Tn.

As long as TWO remembersthe complete history of the game, this task is trivial. However, it oftenhappens that TWO needs to know only the last k moves of the opponentin order to win.

A strategy that accomplishes this is called a winningk-tactic.We consider four such games, MG(A, J), MG(J), the “monotonicgame”, SMG(J), the “strongly monotonic game”, and V SG(J), the “verystrong game”. The study of these games was initiated in [S1], and moti-vated by Telgarsk´y’s conjecture that for every k > 0 there exists a topo-logical space (X, τ) such that TWO has a winning k + 1-tactic but nowinning k-tactic in the Banach-Mazur game on (X, τ) (see section 4.4for more information).

However, we find the games considered here ofinterest independent of the original motivation. The game MG(J) wasintroduced in [S1], as was the game SMG(J); the games MG(A, J) andV SG(J) appear here for the first time.In sections 2 and 3, we introduce and discuss pseudo Lusin sets, theirredundancy property and the coherent decomposition property of ideals.These properties, together with the ω-path partition relation, are the maintools for constructing winning k-tactics in our games.

These combinatorialproperties of ideals are very likely of independent interest - they havealready appeared in the literature in various guises.In section 4 we apply the results of sections 2 and 3 to give variousconditions sufficient for the existence of winning k-tactics for TWO inthe games mentioned above. Not surprisingly, as the game becomes morefavorable for TWO, weaker conditions suffice.

Among other things, ourresults show that in the Banach-Mazur game on the space that inspired theinvention of meager-nowhere dense games, TWO has a winning 2-tactic.The appendix is devoted to a proof of an unpublished consistencyresult of Stevo Todorcevic, which we use in section 4.Our notation is mostly standard. One important exception may bethat we use the symbol ⊂exclusively to mean “is a proper subset of”.Where we otherwise deviate from standard notation or terminology weexplicitly alert the reader.

For convenience we also assume the consis-tency of traditional (Zermelo-Fraenkel) set theory.All statements wemake about the consistency of various mathematical assertions must beunderstood as consistency which can be proven by means of that theory.The reader might find having a copy of [S1] and [S2] handy when readingparts of this paper a bit more comfortable than otherwise.We are grateful to Stevo Todorˇcevi´c for sharing with us his insightsabout the matters we study here, and for his kind permission to presentin this paper some of his answers to our questions.2

2The irredundancy property.For a partially ordered set (P, <) which has no maximum element we letadd(P, <)be the least cardinal number, λ, for which there is a collection of cardi-nality λ of elements of P which do not have an upper bound in P. Thiscardinal number is said to be the additivity of (P, <). Note that add(P, <)is either 2, or else it is infinite.

In the latter case (P, <) is said to be di-rected.We attend exclusively to directed partially ordered sets in thispaper. Isbell [I] and some earlier authors also refer to the additivity of apartially ordered set as its lower character; they denote it by ℓ(P, <).A free ideal J on a set S is partially ordered by ⊂.The partiallyordered set (J, ⊂) is directed.When add(J, ⊂) = ℵ0, the symbol ⟨J⟩denotes the σ-completion of J (i.e., the smallest collection which containseach union of countably many sets from J).

We say that J is a σ-completeideal if J = ⟨J⟩.The other important example for our study is the set ωω of sequences ofnonnegative integers; we use c to denote the cardinality of this set. We sayg eventually dominates f and write f ≪g if: limn→∞(g(n) −f(n)) = ∞.It is customary to denote add(ωω, ≪) by b.A well known theorem of Miller ([M], p. 94, Theorem 1.2) states thatadd(⟨JR⟩, ⊂) ≤add(ωω, ≪)(= b).Again, for an arbitrary partially ordered set (P, <) the symbolcof(P, <)denotes the least cardinal number, κ, for which there is a collection Xof cardinality κ of elements of P such that: for each p ∈P there is anx ∈X such that p ≤x.

This cardinal number is said to be the cofinality of(P, <). Some authors (see e.g.

[I], p. 397) also call this cardinal numberthe upper character of (P, <) and denote it by u(P, <). It is customary todenote cof(ωω, ≪) by d.A theorem of Fremlin ([F], Proposition 13(b)) states that(d =)cof(ωω, ≪) ≤cof(⟨JR⟩, ⊂).Let (P, <) be a directed partially ordered set.

The bursting numberof (P, <) ([I], p. 401) is the smallest cardinal number which exceeds thecardinality of each of the bounded subsets of (P, <). This cardinal num-ber is denoted by burst(P, <).

More important is the principal burstingnumber of (P, <), denoted bu(P, <) and define asbu(P, <) = min{burst(Q, <) : Q is a cofinal subset of P}(following [I], p. 409). It is always the case that add(P, <) ≤bu(P, <).3

Definition 1 A directed partially ordered set (P, <) has the irredundancyproperty if:bu(P, <) = add(P, <).The cofinal subfamily A of (P, <) is said to be irredundant if burst(A, <) ≤add(P, <).Not all σ-complete ideals have the irredundancy property. Here is anad hoc example.

Let S1 and S2 be disjoint sets such that Si has cardinalityℵi for each i. Define an ideal J on the union of these sets by admitting aset Y into J if: Y ∩S1 is countable and Y ∩S2 has cardinality less thanℵ2. Then add(J, ⊂) = ℵ1 and cof(J, ⊂) = ℵ2.

No cofinal family of J isirredundant.A refined version of the classical notion of a Lusin set is instrumentalin verifying the presence of the irredundancy property in many directedpartially ordered sets.Since what we’ll define is not exactly the sameas the classical notion, we call our “Lusin sets” pseudo Lusin sets (moreabout this after the definition). Let κ and λ be infinite cardinal numbers.Let (P, <) be a directed partially ordered set.Definition 2 A subset L of P is a (κ, λ) pseudo Lusin set if:1. λ is the cardinality of L and2.

for each x ∈P the cardinality of the set {y ∈L : y ≤x} is less thanκ. (κ, λ) pseudo Lusin sets are interesting only when κ ≤λ.

If a directedpartially ordered set (P, <) has a (κ, λ) pseudo Lusin set, then add(P, <) ≤κ and λ ≤cof(P, <). Moreover, every partially ordered set has an(add(P, <), add(P, <)) pseudo Lusin set.

Thus, if add(P, <) = cof(P, <),then these are the only types of pseudo Lusin sets in (P, <).Let J be a free ideal on a set S. The uniformity number of J, writtenunif(J), is the minimal cardinal κ such that there is a subset of S whichis of cardinality κ, which is not an element of J.Consider the partially ordered set (⟨JR⟩, ⊂). If L ⊂R is a Lusin setin the classical sense (i.e., L is uncountable and every meager set meetsL in only countably many points), then {{x} : x ∈L} is an (ω1, | L |)pseudo Lusin set.

There will be pseudo Lusin sets even when there are no(classical) Lusin sets: If unif(⟨JR⟩) > add(⟨JR⟩, ⊂) then every set of realnumbers of cardinality ℵ1 is meager, whence there is no Lusin set in theclassical sense. Now let {Mα : α < add(⟨JR⟩, ⊂)} be a family of meagersets such that1.

Mα ⊂Mβ whenever α < β < add(⟨JR⟩, ⊂) and2. ∪α

For example, it is consistent that4

the real line is a union of ℵ1 meager sets and that each set of real numbersof cardinality less than ℵ2 is meager (see e.g. [M], §6).The reader should also compare our notion of a (κ, λ) - pseudo Lusinset with Cichon’s notion of a (κ, λ) - Lusin set (see [Ci]).The connection between the irredundancy property and the existenceof certain pseudo Lusin sets is given by the following proposition.

Theargument in its proof is well known in the special case when P is thecollection of countable subsets of some infinite set, ordered by set inclusion(see the proof of 4.4 on p. 409 of [I]).Proposition 1 Let (P, <) be a directed partially ordered set. Then thefollowing statements are equivalent:1.

There is an (add(P, <), cof(P, <)) pseudo Lusin set for (P, <),2. (P, <) has the irredundancy property,3.

There is a cofinal (add(P, <), cof(P, <)) pseudo Lusin set for (P, <),Proof. That 1. implies 2:Let L = {xξ : ξ < cof(P, <)} be such a pseudo Lusin set and let {aξ :ξ < cof(P, <)} be a cofinal subfamily of P. For each ξ < cof(P, <)choose zξ ∈P such that xξ, aξ ≤zξ.

Put A = {zξ : ξ < cof(P, <)}.Then A is an irredundant cofinal family.That 2. implies 3:Let A be an irredundant cofinal family. We may assume that thecardinality of this family is cof(P, <).

Then A is an example of acofinal (add(P, <), cof(P, <)) pseudo Lusin set.It is clear that 3. implies 1.Corollary 2 Let κ > λ ≥ℵ0 be cardinals, λ regular. If cof([κ]<λ, ⊂) =κ, then ([κ]<λ, ⊂) has the irredundancy property.Proof.

Let {Sα : α < κ} be a pairwise disjoint subcollection from [κ]<λ.Then this family is a (λ, κ) pseudo Lusin set for this ideal.Ap-plying the cofinality hypothesis we conclude that this ideal has theirredundancy property.The ideal of finite subsets of an infinite set has the irredundancy prop-erty; the set of one-element subsets of such an infinite set forms an ap-propriate pseudo Lusin set for this ideal.Lemma 3 Let κ > λ be an uncountable cardinal numbers, λ regular.Then the following statements are equivalent:1. The ideal ([κ]<λ, ⊂) has cofinality κ.2.

There is a free ideal J such that:(a) add(J, ⊂) = λ,5

(b) cof(J, ⊂) = κ and(c) (J, ⊂) has the irredundancy property.Proof. The proof of 1⇒2 is trivial.

We show that 2 implies 1. Let Jbe a free ideal on the set S such that cof(J, ⊂) = κ and add(J, ⊂) = λ, and (J, ⊂) has the irredundancy property.

Let L ⊂J be an(λ, κ) pseudo Lusin set for J. Also let C ⊂J be a cofinal family ofcardinality κ.

For each X ∈C define: SX = {Y ∈L : Y ⊆X}.Then the collection B = {SX : X ∈C} is cofinal in ([L]<λ, ⊂).The following examples play an important role in our game-theoreticapplications.Example 1: The ideal of countable subsets of an infinite set.Let κ be an uncountable cardinal number. Then add([κ]≤ℵ0, ⊂) = ℵ1and bu([κ]≤ℵ0, ⊂) ≥ℵ1.

For uncountable cardinal numbers κ it is alwaysthe case that κ ≤cof([κ]≤ℵ0, ⊂). A set of the form {{αξ} : ξ < κ} (wherethis enumeration is bijective and λ ≤κ) is an (ω1, κ) pseudo Lusin set for[κ]≤ℵ0.

The only difficult cases to decide whether or not the irredundancyproperty is present are those where κ < cof([κ]≤ℵ0, ⊂); this occurs forexample when κ has countable cofinality. It turns out that for these theirredundancy property is not decidable by the axioms of traditional settheory:1.

In [To3], Todorˇcevi´c shows that if for each uncountable cardinal λof countable cofinality the assertions(a) cof([λ]≤ℵ0, ⊂) = λ+ and(b) □λare true, then for each uncountable cardinal number κ there is acofinal family K ⊂[κ]ℵ0 such that |{A ∩X : X ∈K}| ≤ℵ0 forany countable subset A of κ.Such a family K is an example ofan (add([κ]≤ℵ0, ⊂), cof([κ]≤ℵ0, ⊂)) pseudo Lusin set for ([κ]≤ℵ0, ⊂).These particular examples of pseudo Lusin sets are called cofinalKurepa families.Thus it is true in the constructible universe, Lthat ([κ]<ℵ0, ⊂) has the irredundancy property for each infinite κ.2. One might ask if any hypotheses beyond ZFC are necessary to ob-tain the conclusion that ([κ]≤ℵ0, ⊂) has the irredundancy property.Todorˇcevi´c has shown in [To2] that for an infinite cardinal numberκ the following statements are equivalent:(a) bu([κ]≤ℵ0, ⊂) = ℵ1.

(b) ([κ]≤ℵ0, ⊂) has the irredundancy property.He also noted (p. 843 of [To4]) that the version(ℵω+1, ℵω) →(ω1, ω)of Chang’s Conjecture implies that ℵ1 < bu([ℵω]≤ℵ0, ⊂) (and thusthis ideal does not have the irredundancy property). Now [L-M-S] es-tablished the consistency of the above version of Chang’s Conjecturemodulo the consistency of the existence of a fairly large cardinal.6

3. This takes care of uncountable cardinals of countable cofinality.

Whatis the situation for those of uncountable cofinality? It is clear that([κ]≤ℵ0, ⊂) has the irredundancy property if κ is ℵn for some fi-nite n or if, for some m < ω, κ is the m-th successor of a singularstrong limit cardinal of uncountable cofinality.

In fact, the axiomaticsystem of traditional set theory has to be strengthened fairly dra-matically before one could create circumstances where there is acardinal number of uncountable cofinality which is strictly less thanthe cofinality of its ideal of countable sets; it follows from Lemma4.10 of [J-M-P-S] that if there is a cardinal number of uncountablecofinality which is smaller than the cofinality of its ideal of countablesets, then there is an inner model with many measurable cardinalnumbers.Information about the ideal of countable subsets of some infinite setcan be used to gain information about some other ideals, using the notionof a locally small family.Definition 3 A family F of subsets of a set S is locally small if:|{Y ∈F : Y ⊆X}| ≤ℵ0for each X in F.If the ideal of countable subsets of an infinite set has an irredundantcofinal family then that cofinal family is ipso facto locally small. If thereis an (ω1, cof(J, ⊂)) pseudo Lusin set for the σ-complete free ideal J onthe set S, then J contains a locally small cofinal family.Example 2: The ideal of meager subsets of the real lineAssume that add(⟨JR⟩, ⊂) = cof(⟨JR⟩, ⊂) (This equation is for exam-ple implied by Martin’s Axiom).

Then ⟨JR⟩has the irredundancy prop-erty. In this case one may insure that the cofinal family which witnessesthe irredundancy is a well-ordered chain of meager sets.

By the resultscited from [M] and [F], the hypothesis implies that b = d. It is well knownthat the reverse implication is not provable.Irredundancy does not require having a well ordered cofinal chain ofmeager sets. For let an initial ordinal be given.

According to a theoremof Kunen ([K], p. 906, Theorem 3.18) it is consistent that the cardinalityof the real line is regular and larger than that initial ordinal, and at thesame time there is an (ω1, c) pseudo Lusin set. It follows that ⟨JR⟩hasa locally small cofinal family of cardinality c. In particular, ⟨JR⟩has theirredundancy property.

If the continuum is larger than ℵ1 it also followsthat this ideal has no cofinal well-ordered chain.Stevo Todorˇcevi´c has informed us that it is also consistent, modulo theconsistency of a form of Chang’s Conjecture that ⟨JR⟩does not have theirredundancy property. Actually, something apparently weaker than thatform of Chang’s Conjecture is used: we present this result of Todorˇcevi´c’sin Theorem 4, which he kindly permitted us to include in this paper.7

Theorem 4 (Todorˇcevi´c) If “ZFC+MAℵ1+ there is no Kurepa familyin [ℵω]ℵ0 of cardinality larger than ℵω” is a consistent theory, then so isthe theory “ZFC + bu(⟨JR⟩, ⊂) > add(⟨JR⟩, ⊂) = ℵ1”.Proof Let P be the set of finite functions with domain a subset of ℵω andrange a subset of ω (in other words, P is the standard set for addingℵω Cohen reals). For p and q in P we write p < q if q ⊂p.

For D acountable subset of ℵω we write P(D) for the set of elements of Pwhose domains are subsets of D.Suppose we have a sequence {Nξ : ξ < θ} (θ > ℵω) of P-namesfor meager sets of reals. Let Dξ ∈[ℵω]ℵ0 be the support of Nξ i.e.,Nξ ∈VP(Dξ).

By the hypothesis of the theorem and by Theorem 1of [To3] there is an uncountable set A ⊂θ such that D = ∪ξ∈ADξis countable. Thus, Nξ ∈VP(D) for each ξ ∈A.

Since P(D) isessentially the poset for adding one Cohen real and since MAℵ1holds, VP(D) |= “ ∪ξ∈A Nξ is meager” (because VP(D) |= “MA(σ −centered)”).The hypothesis of Theorem 4 is consistent modulo the consistencyof the relevant form of Chang’s Conjecture, because that form of theconjecture is preserved by c.c.c. generic extensions.3The coherent decomposition propertyLet J be a free ideal on a set S and let ⟨J⟩be its σ-completion.

Let A bea subcollection of ⟨J⟩.Definition 41. A has a coherent decomposition if there is for eachA ∈A a sequence (An : n < ω) such that:(a) An ∈J for each n,(b) An ⊆Am whenever n < m < ω, and(c) For all A and B in A such that A ⊂B, there is an m such thatAn ⊆Bn whenever n ≥m.The collection {(An : n < ω) : A ∈A} is said to be a coherentdecomposition for A.2.

The ideal J has the coherent decomposition property if some cofinalsubset of ⟨J⟩has a coherent decomposition.It is worth mentioning that if J has the coherent decomposition prop-erty and if ⟨J⟩has a cofinal chain, than the family ⟨J⟩itself has a coherentdecomposition. We now explore the coherent decomposition property forour examples.Example 1: (continued)Theorem 5 Let A be a locally small family of countable sets such that(A, ⊂) is a well-founded partially ordered set.Then A has a coherentdecomposition.8

Proof. Let Φ : A →α be a function to an ordinal α such that Φ(A) <Φ(B) for all A ⊂B in A (i.e., a rank function).

Since A is locallysmall we may assume that α is ω1.For A in A with Φ(A) = 0, choose a sequence (An : n < ω) of finitesubsets of A such that A = ∪n<ωAn and An ⊆An+1 for all n.Let 0 < β < ω1 be given and assume that we have already assignedto each A in A for which Φ(A) < β, a sequence (An : n < ω) incompliance with 1 and 2. Now Let B be an element of A such thatΦ(B) = β.

Write F(B) = {A ∈A : A ⊂B}.To begin, arbitrarily choose a sequence (Sn : n < ω) of finite setssuch that B = ∪n<ωSn. For each A ∈F(B), define gA : ω →ω suchthat for each n < ω,gA(n) = min{k < ω : An ⊆S0 ∪.

. .

∪Sk}.Then {gA : A ∈F(B)} is countable since A is locally small. Let f ∈ωω be a strictly increasing function such that gA ≪f for each A inF(B).

Define:Bn = S0 ∪. .

. ∪Sf(n)for each n. Then (Bn : n < ω) is as required.Corollary 6 Let J be a free ideal on a set S and let A be a locally smallfamily of sets in ⟨J⟩such that (A, ⊂) is a well-founded partially orderedset.

Then A has a coherent decomposition.Proof. For each B in A, let (Sn(B) : n < ω) be a sequence from Jsuch that B = ∪n<ωSn(B).

Also write Γ(B) = {A ∈A : A ⊆B}.Then B = {Γ(A) : A ∈A} is a well-founded, locally small collectionof countable subsets of A. Choose, by Theorem 5, for each A ∈A asequence (Γ(A)n : n < ω) of finite subsets of Γ(A) such that:1.

Γ(A) = ∪n<ωΓ(A)n where Γ(A)n ⊆Γ(A)n+1 for each n, and2. for all A and B in A with A ⊂B there exists an m such that:Γ(A)n ⊆Γ(B)nfor all n ≥m.For each A in A and each n < ω define:An = ∪{Sj(B) : j ≤n and B ∈Γ(A)n}.Then the sequences (An : n < ω) are as required.Corollary 7 If ([κ]≤ℵ0, ⊂) has the irredundancy property, then it has thecoherent decomposition property.Proof An irredundant cofinal family is necessarily locally small.

We maythin out any cofinal family to a well-founded cofinal family. Nowapply Theorem 5.9

Example 2: (continued)We show that the ideal of meager sets of the real line has the coher-ent decomposition property, and also that it has a second combinatorialproperty which plays an important role in our game-theoretic applica-tions. It is convenient, for this section, to work with the set ω2, with theusual Tychonoffproduct topology (2 = {0, 1} is taken to have the discretetopology) in place of R. For a subset S of the domain of a function g,the symbol g⌈S denotes the restriction of g to the set S. For s an ele-ment of <ω2, the symbol [s] denotes the set of all those x in ω2 for whichx⌈length(s)= s. Subsets of ω2 of the form [s] where s ranges over <ω2,form a base for the topology of ω2.

Let f ∈ωω be a strictly increasingsequence and let x be an element of ω2. Define:Bx,f = {z ∈ω2 : ∀∞n (z⌈[f(n),f(n+1))̸= x⌈[f(n),f(n+1)))}.Now also fix an n ∈ω and defineBnx,f = {z ∈ω2 : (∀k ≥n)(z⌈[f(k),f(k+1))̸= x⌈[f(k),f(k+1)))}.Then Bmx,f ⊆Bnx,f whenever m < n < ω; also, Bx,f = ∪n<ωBnx,f.Proposition 8 For x, y ∈ω2 and strictly increasing f, g ∈ωω, the fol-lowing assertions are equivalent:1.

Bx,f ⊂By,g.2. (a) Bx,f ̸= By,g and(b) (∀∞n )(∃k)(g(n) ≤f(k) < f(k+1) ≤g(n+1) and x⌈[f(k),f(k+1))=y⌈[f(k),f(k+1)))Proof.

That 1 implies 2 requires some thought:If 1 holds, then (a) of 2 holds. Assume the negation of 2(b).

It reads:(∃∞n )(∀k)(¬(g(n) ≤f(k) < f(k+1) ≤g(n+1)) or ¬(x⌈[f(k),f(k+1))= y⌈[f(k),f(k+1))))Put S = {n < ω : (∀k)(¬([f(k), f(k+1)] ⊆[g(n), g(n+1)]) or ¬(x⌈[f(k),f(k+1))=y⌈[f(k),f(k+1)))}. Our hypothesis is that S is an infinite set.Consider an n in S. For each k, there are the following possibilities:1.

¬([f(k), f(k + 1)] ⊆[g(n), g(n + 1)]2. [f(k), f(k+1)] ⊆[g(n), g(n+1)], but x⌈[f(k),f(k+1))̸= y⌈[f(k),f(k+1)).Put Sn = {k : 2 holds for k}.

We consider two cases.Case 1: There are infinitely many n for which Sn is nonempty.Choose an infinite sequence (n1, n2, n3, . .

.) from S such that:1.

Snm ̸= ∅,2. nm+1 > g(nm + 1), and3.

(∃k)(g(nm + 1) < f(k) < g(nm+1)), for each m, and10

4. f(1) < g(n1).This is possible because f and g are increasing, and S is infinite.Put T = ∪∞j=1[g(nj), g(nj + 1)). Define z, an element of ω2, so thatz⌈T = y⌈T and z(n) = 1 −x(n) for each n ∈ω\T .

Then z ∈Bx,fwhile z ̸∈By,g. Thus 1 fails in this case.Case 2:There are only finitely many n ∈S for which Sn isnonempty.We may assume that Sn = ∅for each n ∈S.

Consider n ∈S. Wethen have that for each k ∈ω, [f(k), f(k + 1)) ̸⊆[g(n), g(n + 1)).We distinguish between two possibilities:1.

(∃k)(g(n) ≤f(k) < g(n + 1)) or2. (∀k)(f(k) ̸∈[g(n), g(n + 1))Case 2 (A): Possibility 1 occurs for infinitely many n ∈S:Choose n1 < n2 < n3 < .

. .

from S such that• 2 · nj ≤nj+1 for each j,• for each j there is a k such that g(nj + 1) < f(k) < g(nj+1),• for each j there is a k such that f(k) ∈[g(nj), g(nj+1)), and• f(1) < g(n1).Put T = ∪∞j=1[g(nj), g(nj + 1)) and define z so that z⌈T = y⌈T , andz(n) = 1−x(n) for each n ∈ω\T . From the hypothesis of Case 2(A)it follows that z ∈Bx,f, but z ̸∈By,g.

Thus, 1 of the Propositionfails also in this case.Case 2 (B): Possibility 1 occurs for only finitely many n ∈S:We may assume that possibility 2 occurs for each n ∈S. Choosek1 < k2 < k3 < .

. .

such that for each j there is an n ∈S with[g(n), g(n + 1)) ⊂[f(kj), f(kj + 1)).For each j choose nj ∈Ssuch that [g(nj), g(nj + 1)) ⊂[f(kj), f(kj + 1)). As before defineT = ∪∞j=1[g(nj), g(nj + 1)).

Finally, define z so that z⌈T = y⌈T andz(n) = 1 −x(n) for each n ∈ω\T . Then z ∈Bx,f and z ̸∈By,g,showing that 1 of the Proposition fails also in this case.This completes the proof of the Proposition.Lemma 9 Let f and g be strictly increasing elements of ωω for whichthere is some k < ω such that g(n + k) = f(n) for all but finitely many n.If Bx,f ⊆By,g, then Bx,f = By,g.Proof.

Assume that Bx,f ̸= By,g and suppose that By,g ̸⊆Bx,f. Weshow that Bx,f ̸⊆By,g.

Let z be an element of By,g\Bx,f. Fix Nsuch that11

1. z⌈[g(n+k),g(n+k+1))̸= y⌈[g(n+k),g(n+k+1)) and2. f(n) = g(n + k)for each n ≥N.Since z is not an element of Bx,f, there are infinitely many n ≥N for which z⌈[f(n),f(n+1))= x⌈f(n),f(n+1)).Consequently the setS = {n ≥N : x⌈[f(n),f(n+1))̸= y⌈[f(n),f(n+1))} is infinite.Nowdefine t such that t⌈[f(n),f(n+1))= y⌈[f(n),f(n+1)) for each n ∈S, andt(m) = 1 −x(m) for each m ∈ω\(∪n∈S[f(n), f(n + 1))).

Then t isin Bx,f but not in By,g.Under the hypothesis of Lemma 9, x(n) = y(n) for all but finitelymany n.Proposition 10 Let x, y be elements of ω2 and let f, g be increasing el-ements of ωω. Of the following two assertions, 1 implies 2.1.

Bx,f ⊂By,g.2. f ≪g.Proof.

Assume that Bx,f ⊂By,g. Fix, by Proposition 8, an N such that(∀n ≥N)(∃k)([f(k), f(k + 1)] ⊆[g(n), g(n + 1)] and x⌈[f(k),f(k+1))=y⌈[f(k),f(k+1))).For each n ≥N choose kn such that [f(kn), f(kn +1)] ⊆[g(n), g(n+1)].

It follows that kn + 1 ≤kn+1 for each n ≥N (since f and g areincreasing).Claim: [f(kn), f(kn + 1)] ⊂[g(n), g(n + 1)] for infinitely many n.Proof of the claim: For otherwise, fix M ≥N such that [f(kn), f(kn+1] =[g(n), g(n + 1)] for each n ≥M. Then we have kn+1 = kn + 1 foreach n ≥M.

It follows that g(n) = f(n + (kM −M)) for all n ≥M.Then Lemma 9 implies that Bx,f = By,g, contrary to the fact thatBx,f is a proper subset of By,g.This completes the proof of theclaim.Thus, there are infinitely many n for which kn+1 > kn+1. Let m > 1be given, and fix L ≥M such that |{n < L : kn+1 > kn + 1}| ≥k1 + m. Then kn > (n + m) for each n ≥L; we havef(n + 1) < f(n + m) ≤f(kn) < g(n + 1)for each n ≥L.

In particular, m ≤g(n + 1) −f(n + 1) for eachn ≥L. This completes the proof that f ≪g.Proposition 11 Let x and y be elements of ω2 and let f and g be in-creasing elements of ωω.

If Bx,f ⊂By,g, then there is an m < ω suchthat Bnx,f ⊆Bny,g whenever n ≥m.12

Proof From our hypotheses and Proposition 8 there is an m such that foreach n ≥m there is a k such that [f(k), f(k + 1)) ⊆[g(n), g(n + 1))and x⌈[f(k),f(k+1))= y⌈[f(k),f(k+1)). By Proposition 10 there is anM > m such that f(j) ≤g(j) for each j ≥M.We show thatBnx,f ⊆Bny,g for each n ≥M.Let z be an element of Bnx,f.

Then z⌈[f(j),f(j+1))̸= x⌈[f(j),f(j+1)) foreach j ≥n. But consider any j ≥n.

Then there is a k such that[f(k), f(k +1)) ⊂[g(j), g(j +1)); k ≥j for any such k, by the choiceof M. It follows that z⌈[g(j),g(j+1))̸= y⌈[g(j),g(j+1)). Thus, z is alsoan element of Bny,g.Proposition 12 For each X ∈⟨JR⟩there are an x in ω2 and an increas-ing f in ωω such that X ⊂Bx,f.Proof.

Let X be a meager set. We may assume that X = ∪∞n=0Xn whereXn ⊆Xn+1 and Xn is closed, nowhere dense for each n. Fix a well-ordering of <ω2, and define (sn : n < ω) and f in ωω as follows:Take s0 = ∅and f(0) = 0.

Assume that s1, s2, . .

. , sn and f(1), .

. .

, f(n)have been defined so that:1. s1 is the first element of <ω2 such that [s1]∩X1 = ∅and f(1) =length(s1),2. sj+1 is the first element of <ω2 such that [t⌢sj+1] ∩Xj = ∅foreach t in ≤f(j)2, and f(j +1) = Pj+1i=0 length(si) for each j < n.Then let sn+1 be the first element of <ω2 such that [t⌢sn+1]∩Xn = ∅for each t in ≤f(n)2; put f(n + 1) = f(n) + length(sn+1).Finally, set x = s⌢1 s⌢2 s⌢3 . .

..Claim: X ⊆Bx,f.For suppose that z is not an element of Bx,f. Then there are in-finitely many n for which z⌈[f(n),f(n+1))= x⌈[f(n),f(n+1)); in otherwords, there are infinitely many n for which z⌈[f(n),f(n+1))= sn+1.Now fix an m. Choose an n > m such that z⌈[f(n),f(n+1))= sn+1.From the choice of sn+1 it follows that [z⌈f(n+1)] ∩Xm = ∅; inparticular, z ̸∈Xm.

Consequently, z is not an element of X.Proposition 13 Each Bnx,f is in JR.Proof. Consider an s fromω2 for which [s] ∩Bnx,f ̸= ∅.Choose msuch that f(m) > length(s) and m > n.Then choose t from<ω2 such that length(s⌢t) ≥f(m + 1) and s⌢t⌈[f(m),f(m+1))=x⌈[f(m),f(m+1)).Then [s⌢t] ∩Bnx,f = ∅.It follows that Bnx,f isnowhere dense.Consequently, Bx,f is a meager set for each x in ω2 and for eachincreasing f from ωω.Theorem 14 ⟨JR⟩has a cofinal family which embeds in (ωω, ≪) andwhich has the coherent decomposition property.13

Proof. By Propositions 13 and 12 the family of sets of the form Bx,fwhere f is an increasing element of ωω and x is an element of ω2,is a cofinal family of meager sets.

By Proposition 11, this familyhas the coherent decomposition property. Also, the mapping whichassigns f to Bx,f is, according to Proposition 10, an order preservingmapping.Example 3: Cardinals of countable cofinalityHere is a result which is quite analogous to Theorem 5.Theorem 15 Let λ be an uncountable cardinal number which has count-able cofinality.

Let λ0 < λ1 < . .

. be a sequence of infinite regular cardinalnumbers which converges to λ.

Let (A, ⊂) be a well-founded family of sets,each of cardinality λ, such that|{Y ∈A : Y ⊆X}| ≤λfor each X in A. Then A has the coherent decomposition property.

Inparticular:There exists for each A ∈A a sequence (An : n < ω) such that:1. |An| ≤λn for all n,2.

An ⊆An+1 for all n,3. A = ∪∞n=0An and4.

if A ⊂B, then there is an m < ω such that An ⊆Bn for all n ≥m.Proof. Let Φ : A →λ+ be a rank function.

For all A in A with Φ(A) = 0,choose (An : n < ω) arbitrary, subject only to 1, 2 and 3.Let 0 < γ < λ+ be given and assume that (An : n < ω) has beenassigned to each A from A for which Φ(A) < γ, in such a way that1, 2, 3 and 4 are satisfied. Consider B in A with Φ(B) = γ. WriteF(B) for {A ∈A : A ⊆B} and write F(B) = ∪∞n=0Fn(B) where1.

F0(B) ⊆F1(B) ⊆. .

., and2. |Fn(B)| ≤λn for all n.Also let B = ∪∞n=0Xn where X0 ⊆X1 ⊆.

. .

and Xn ≤λn for alln. Finally put Bn = (∪{An : A ∈Fn(B)}) ∪Xn for each n. Then(Bn : n < ω) is as required.Corollary 16 Let λ be a cardinal number of countable cofinality.If([κ]≤λ, ⊂) has the irredundancy property then it has the coherent decom-position property.14

4ApplicationsThe ω−path partition relation is the one other combinatorial ingredient inour technique for constructing winning k-tactics, or for defeating a givenk-tactic for TWO. For a positive integer n, infinite cardinal number λ anda partially ordered set (P, <), the symbol(P, <) →(ω −path)nλ/<ωmeans that for every function F : [P]n →λ there is an increasing ω-sequencep1 < p2 < .

. .

< pm < . .

.such that the set {F({pj+1, . .

. , pj+n}) : j < ω} is finite.

The negation ofthis assertion is denoted by the symbol(P, <) ̸→(ω −path)nλ/<ω.This partition relation has been studied in [S2]. The reader should consultthis reference about the various facts concerning the ω-path relation whichare used in the sequel.4.1The game MG(A, J)For a free ideal J on an infinite set S and for a family A in ⟨J⟩with theproperty that for each X ∈A there is a Y ∈A such that X ⊂Y , thegame MG(A, J) is defined so that an ω-sequence (O1, T1, .

. .

, On, Tn, . .

. )is a play if for each n,1.

On ∈A is player ONE’s move in inning n,2. Tn ∈J is player TWO’s move in inning n, and3.

On ⊂On+1.Player TWO wins this play if ∪∞n=1On ⊆∪∞n=1Tn.Theorem 17 Let J be a free ideal on a set S. If A is a family of sets in⟨J⟩such that:1. for each X ∈A there is a Y ∈A such that X ⊂Y ,2. (A, ⊂) ̸→(ω −path)kω/<ω for some k ≥2, and3.

A has a coherent decompositionthen TWO has a winning k-tactic in MG(A, J).Proof. Choose a function F : [A]k →ω which witnesses hypothesis 2.Also associate with each A in A a sequence (An : n < ω) such thathypothesis 3 is satisfied.Define a k-tactic, Υ for TWO as follows.

Let (X1, . .

. , Xj) be givensuch that j ≤k, X1 ⊂.

. .

⊂Xj and Xi ∈A for i ≤j.1. If j < k: Then put Υ(X1, .

. .

, Xj) = X11 ∪. .

. ∪X1j .2.

If j = k: Let m be such that15

• m ≥F({X1, . .

. , Xk}) and• Xn1 ⊆.

. .

⊆Xnk for all n ≥m.Put Υ(X1, . .

. , Xk) = Xm1 ∪.

. .

∪Xmk .Then Υ is a winning k-tactic for TWO. For let (O1, T1, .

. .

, On, Tn, . .

. )be a play of MG(A, J) where:• Tj = Υ(O1, .

. .

, Oj) for each j ≤k• Tn+k = Υ(On+1, . .

. , On+k) for each n < ω.For each t ≥1 let mt be the number associated with (Ot, .

. .

, Ot+k−1)in part 2 of the definition of Υ. By the properties of F, the set{mt : t = 1, 2, 3, .

. .} is infinite.

Thus choose t1 < t2 < . .

. such thatmj < mtr for all j < tr.

It follows from the criteria used in thechoices of the numbers mt thatOmtr1⊆. .

. ⊆Omtrmtrfor all r. But Omtrmtr ⊆Tmtr for all r, according to the definition ofΥ.

It follows that ∪∞n=1On ⊆∪∞n=1Tn.Corollary 18 There is a cofinal family A ⊂⟨JR⟩such that TWO has awinning 2-tactic in MG(A, JR).Proof. Let A be the family of meager sets provided by Theorem 14.Thus, there is an order preserving function from (A, ⊂) to (ωω, ≪).But then (A, ⊂) ̸→(ω −path)2ω/<ω holds, since (ωω, ≪) ̸→(ω −path)2ω/<ω holds.By Theorem 14 the family A also satisfies thethird hypothesis of Theorem 17.Corollary 19 Let J be a free ideal on an infinite set.

If A is a family ofsets in ⟨J⟩such that:1. A is locally small,2.

for each X ∈A there is a Y ∈A such that X ⊂Y , and3. (A, ⊂) is well-founded,then TWO has a winning 2-tactic in MG(A, J).Proof.

The proof is analogous to that of Corollary 18; now we refer to theproof of Theorem 5, we observe that ω1 ≤b, and invoke Theorem17.Corollary 20 Let λ ≤κ be infinite cardinal numbers such that:1. λ has countable cofinality,2. λ+ ̸→(ω −path)2ω/<ω, and3.

[κ]≤λ has the irredundancy property.16

Then there is a cofinal family A ⊂[κ]λ such that TWO has a winning2-tactic in MG(A, [κ]<λ).Proof. Let A be a well-founded cofinal family in [κ]λ which is irredun-dant.

Since there is a rank-function from A to λ+ it follows fromhypothesis 2 that (A, ⊂) ̸→(ω −path)2ω/<ω.From Corollary 16it follows that A also satisfies the third hypothesis of Theorem 17.By that theorem TWO then has a winning 2-tactic in the gameMG(A, [κ]<λ).Theorem 21 shows that under certain circumstances there is for eachn a free ideal Jn and a cofinal family An ⊂⟨Jn⟩such that TWO doesnot have a winning n-tactic, but does have a winning n + 1-tactic inMG(An, Jn). We think that Theorem 21 indicates some relevance of thegames as considered here for Telgarsky’s Conjecture (see 3.4).Theorem 21 Let λ be an infinite cardinal number.

If there is a linearlyordered set (L, <) such that:1. cof(L, <) > ω,2. (L, <) →(ω −path)2λ/<ω, but3.

(L, <) ̸→(ω −path)3λ/<ω,then there is for each n a free ideal Jn and a cofinal family An ⊂⟨Jn⟩such that TWO does not have a winning n-tactic, but does have a winningn + 1-tactic in MG(An, Jn).Proof. Let λ and (L, <) be as in the hypotheses.

It follows from Propo-sitions 3 and 4 of [S2] that there is for each integer m > 1 a linearlyordered set (Ln, 1 and (Ln,

Let T be {Xα : α ∈λ}. Put a subset X ofSn in Jn if:X ∩[λ]<ℵ0 is a subset of a union of finitely many elementsof T , and X ∩Ln is bounded above.Then the cofinality of ⟨Jn⟩is cof(Ln,

Define An so that X ∈An if:X ∩Ln = {t ∈Ln : t < z}for some z ∈Ln.Then An is cofinal in ⟨Jn⟩.Claim 1: TWO does not have a winning n-tactic in MG(An, Jn).17

For let Φ be an n-tactic of TWO. For x ∈Ln put Vx = [λ]<ℵ0 ∪{y ∈Ln : y

Define a partition Ψ : [Ln]n →[λ]<ℵ0 so that(Φ(Vx1) ∪Φ(Vx1, Vx2) ∪. .

. ∪Φ(Vx1, .

. .

, Vxn)) ∩[λ]<ℵ0is a subset of ∪{Xα : α ∈Ψ({x1, . .

. , xn})}.By (1) we obtain an ω-path x1

. .

. and afinite set F ⊂λ such that Ψ(xj+1, .

. .

, xj+n) ⊆F for all j. For eachm we define: Om = [λ]<ℵ0 ∪Vxm.

Letting (O1, T1, . .

. , Ok, Tk, .

. .

)be the corresponding Φ-play, we find that TWO has lost this playsince [λ]<ℵ0 ∩(∪∞m=1Tm) ⊆∪α∈F Xα ̸= [λ]<ℵ0.It follows that TWO does not have a winning n-tactic.Claim 2: TWO has a winning n + 1-tactic in MG(An, Jn).First observe that ∪α∈F Xα = [λ]<ℵ0 whenever F is an infinite subsetof λ.Here is a definition of an n + 1-tactic for TWO in this game: Let{tα : α < λ} enumerate [λ]<ℵ0 bijectively. Let Φ : [Ln]n+1 →λ be acoloring which witnesses that (Ln,

For eachX in An let φX be that element of Ln for which X ∩Ln = {t ∈Ln :t < φX}.For U1 ⊂. .

. ⊂Un+1 elements of An, observe that φU1 ≤.

. .

≤φUn+1. For X ⊂Y sets in An such that X ∩[λ]<ℵ0 ̸= Y ∩[λ]<ℵ0 weset Ψ(X, Y ) = min{α : tα ∈Y \X}.Let U1 ⊂.

. .

⊂Un+1 ∈An be given. We define:1.

G(U1, . .

. , Uj) = ∅when j < n + 1,2.

G(U1, . .

. , Un+1) = Xα ∪(Ln ∩Un+1) when φU1 < .

. .

< φUn+1,and Φ({φU1, . .

. , φUn+1}) = α,3.

G(U1, . .

. , Un+1) = Xα ∪(Ln ∩Un+1) where α is minimal suchthat tα ∈Ui+1\Ui for some i ≤n, otherwise.We show that G is a winning n + 1-tactic for TWO.

Thus, let(O1, T1, . .

. , Om, Tm, .

. .

)be a G-play of the game. For typographical convenience we define:1. xi = φOi for each i, and2.

αi = Ψ(Oi, Oi+1) for each i for which this is defined.There are two cases to consider.CASE 1: {i : xi = xi+1} is finite.Choose m such that xi < xi+1 for all i ≥m. Then the set{Φ({xm+k+1, .

. .

, xm+k+n+1}) : k = 1, 2, . .

. }is an infinite subset of λ and it follows from 2. in the definition ofG that this play is won by TWO.18

CASE 2: {i : xi = xi+1} is infinite.Then the set {i : Ψ(Oi, Oi+1) is defined} is infinite.But then itfollows from 3. in the definition of G that TWO wins this play.The hypotheses of Theorem 21 are realized under any of the followingaxiomatic circumstances (one uses Corollary 27 and Proposition 29 of[S2]):1. 2

c < 2ℵ1, i.e., the negation of LH (Lusin’s second Continuum Hy-pothesis)3. There is an infinite regular cardinal number κ such that 2κ = κ+.For the case when λ = ω, the example constructed in the proof ofTheorem 21 shows that hypothesis 2 of Theorem 17 is to some extentnecessary.

This is because:1. A\ has the coherent decomposition property: For choose α1 < α2 <.

. .

< αn < . .

. from ω, and set Tm = Xα1 ∪.

. .

∪Xαm for each m.Then [ω]<ℵ0 = ∪∞m=1Xαm, and Xαj ⊆Xαi for j < i. For A ∈A\we put Am = (A ∩Tm) ∪(A ∩Ln).2.

(Am, ⊂) →(ω −path)mω/<ω, but3. (Am, ⊂) ̸→(ω −path)m+1ω/<ω.At this point it is an open problem whether the hypotheses (and forthat matter the conclusion) of Theorem 21 are satisfied simply in thetheory ZFC (see Problem 9 of [S2]).4.2The game MG(J)MG(J) denotes the version of MG(A, J) where ⟨J⟩= A.

In Problem1 of [S1] it was asked whether there is for each k a free ideal Jk suchthat TWO does not have a winning k-tactic in MG(Jk), but does havea winning k + 1-tactic in MG(Jk). This problem is still open.

In [S1],Corollary 10, it was proven that TWO does not have a winning 2-tactic inthe game MG(JR), but that TWO has a winning 3-tactic in MG(JR) iffor example the Continuum Hypothesis is assumed. We now extend theseresults in two directions.1.

In Problem 3 of that paper it was asked if player TWO has a winning3-tactic if instead of the Continuum Hypothesis one uses the theoryZFC + MA + EH + ¬CH, which is explained below. We now showthat the answer is affirmative.2.

We identify circumstances under which TWO does not have a win-ning k-tactic in MG(JR) for any k; combining this with a consistencyresult of Todorcevic (given in the appendix), it follows that it is alsoconsistent that there is no k for which TWO has a winning k-tacticin MG(JR).19

It follows that the existence of a winning k-tactic for TWO in MG(JR)is not decided by the axioms of traditional set theory. One might nowwonder if it is consistent that for example TWO does not have a winning3-tactic in MG(JR), but does have a winning 4-tactic?

This is not possiblesince a theorem of [S3] implies that either TWO has a winning 3-tactic,or else there is no k such that TWO has a winning k-tactic in MG(JR).Let EH (which abbreviates Embedding Hypothesis) denote the state-ment:every linearly ordered set of cardinality ≤c embeds in (ωω, ≪).The hypothesis EH is a consequence of the Continuum Hypothesis. Laverhas proven ([L]) that the theory ZFC + EH + ¬CH is consistent, andWoodin ([W], pp.

31-47), extending this, has proven the consistency of thetheory ZFC +MA+EH +¬CH. This theory implies that 2

Thus we have:Proposition 22 The theory “ZFC+¬CH+ TWO has a winning 3-tacticin MG(JR)” is consistent.Proof. Consider any model of ZFC+EH+¬CH+2

Let C denote this cofinal chain. By Theorem 14we may assume that this cofinal chain has a coherent decompositionand that it satisfies the partition relation (C, ⊂) ̸→(ω −path)2ω/<ω.Since we also have (P(c), ⊂) ̸→(ω −path)3ω/<ω it follows that:1.

(⟨JR⟩, ⊂) ̸→(ω −path)3ω/<ω, and2. The family ⟨JR⟩has a coherent decomposition.Theorem 17 implies that TWO has a winning 3-tactic in MG(⟨JR⟩, JR).This completes the proof of the proposition.Indeed, our proof of Proposition shows more generally that if J is a freeideal on a set of cardinality at most c, and if ⟨J⟩has a cofinal chain and thecoherent decomposition property, then the theory ZFC + EH + 2

This generalizesTheorem 8(a) of [S1].Next we give hypotheses under which there is no k for which TWOhas a winning k-tactic in MG(JR). In the appendix we give a proof thatthese hypotheses are consistent with ZFC.

This consistency result is dueto Todorcevic.Theorem 23 Assume that cof(JR, ⊂) = λ and that the partition relation(P(c), ⊂) →(ω −path)3λ/<ω holds. Then there is no k for which TWOhas a winning k-tactic in MG(JR).20

Proof. Let k as well as a k-tactic F for T W O be given.

Let X be anowhere dense subset of cardinality c of R\Q. Let A = {Aα : α < λ}be a bijectively enumerated cofinal subfamily of JR.Define a partition Φ : [P(X)]k →λ so thatΦ({X1, · · · , Xk}) = βwhere β is minimal such thatF(Q ∪X1) ∪· · · ∪F(Q ∪X1, · · · , Q ∪Xk) ⊂Aβ.Since (P(c), ⊂) →(ω −path)3λ/<ω, it follows that (P(c), ⊂) →(ω −path)kλ/<ω ( see [S2], Proposition 36).

Accordingly, choose a finiteset G ⊂λ and an increasing ω-sequence X1 ⊂X2 ⊂· · · of subsets ofX such that Φ({Xj+1, · · · , Xj+k}) ∈G for all j. Put On = Xn ∪Qfor all n. Let B be the nowhere dense set ∪{Aα : α ∈G}.

Alsodefine Tj = F(O1 · · · , Oj) for j ≤k, and Tj+k = F(Oj+1, · · · , Oj+k)for all j. Then(O1, T1, O2, T2, · · ·)is an F-play of MG(JR) for which Q ⊂∪∞n=1On and ∪∞n=1Tn ⊆B.Since B is nowhere dense, Q\B ̸= ∅.

It follows that TWO has lostthis play.We now consider games of the form MG([κ]<λ). In Proposition 15 of[S1] it was shown that if TWO has a winning k-tactic in this game forsome k, then TWO in fact has a winning 3-tactic.

It is not known if “3” isoptimal (this is Problem 7 of [S1]). It also follows from [S1], Proposition 5,that if λ →(ω −path)2ω/<ω, then TWO does not have a winning k-tacticin this game for any k. We now present slightly sharper results.Theorem 24 Let λ be an uncountable cardinal number of countable cofi-nality.

Let k > 1 be an integer. The following statements are equivalent:1.

Player TWO has a winning k-tactic in the game MG([λ+]<λ).2. ([λ+]≤λ, ⊂) ̸→(ω −path)kω/<ω.3.

λ+ ̸→(ω −path)2ω/<ω and (P(λ), ⊂) ̸→(ω −path)kω/<ω.Proof. By Theorem 1 and Proposition 15 of [S1] we may assume thatk ∈{2, 3}.

Let λ1 < . .

. < λn < .

. .

be a sequence of cardinal num-bers converging to λ.1. ⇒2.Let F be a winning k-tactic for TWO in MG([λ+]<λ).

Put S =λ+\λ. Define a coloring Φ : [[S]≤λ]k →ω so thatΦ(X1, .

. .

, Xk) = min{n : |F(λ ∪X1, . .

. , λ ∪Xk| ≤λn}.Since F is a winning k-tactic for TWO, Φ is a coloring which wit-nesses the partition relation in 2.21

2. ⇒1.According to Corollary 16, ([λ+]<λ, ⊂) has the coherent decomposi-tion property.

Since [λ+]≤λ has a cofinal chain it follows that thisfamily of sets itself has a coherent decomposition.The partitionproperty in 2 implies that the family [λ+]≤λ satisfies the hypothesesof Theorem 17; thus TWO has a winning k-tactic in MG([λ+]<λ).The equivalence of 2. and 3. is also easy to establish.Corollary 25 Let λ be an uncountable cardinal number of countable cofi-nality. Assume ZFC + EH + λ < c+c = 2

Then TWO has a winning2-tactic in MG([λ+]<λ).Proof. The hypothesis EH +c = 2

. .

, On, Tn, . .

. )is a play if for each n,1.

On ∈⟨J⟩is player ONE’s move in inning n,2. Tn ∈J is player TWO’s move in inning n, and3.

On ∪Tn ⊆On+1.Player TWO wins this play if ∪∞n=1On = ∪∞n=1Tn.Throughout this section we assume that ⟨J⟩is a proper ideal on S.Theorem 26 Let J ⊂P(S) be a free ideal and let A be a cofinal subfamilyof ⟨J⟩such that:1. TWO has a winning k-tactic in MG(A, J),2. there are functions Φ1 : ⟨J⟩→J and Φ2 : ⟨J⟩→A such that:(a) A ⊂Φ2(A) for each A ∈⟨J⟩, and(b) Φ2(A) ⊂Φ2(B) whenever A ∪Φ1(A) ⊆B ∈⟨J⟩.Then TWO has a winning 2-tactic in SMG(J).Proof Let A, Φ1 and Φ2 be as in the hypotheses.

For each A in ⟨J⟩define(A1, . .

. , Ak) so that A1 = Φ2(A) and Aj+1 = Φ2(Aj) for each j < k.Also define: Ψ(A) = Φ1(A) ∪Φ1(A1) ∪.

. .

∪Φ1(Ak).Let F be a winning k-tactic for TWO in MG(A, J). Define a k-tactic, G, for TWO as follows.

Let A ⊂B be given.22

CASE 1: G(A) = F(A1) ∪. .

. ∪F(A1, .

. .

, Ak) ∪Ψ(A).CASE 2: If Ak ⊂B1, we let G(A, B) be the setF(A2, . .

. , Ak, B1)∪F(A3, .

. .

, Ak, B1, B2)∪. .

.∪F(B1, . .

. , Bk)∪Ψ1(B).CASE 3: Otherwise we put G(A, B) = G(B).Then G is a winning 2-tactic for TWO in SMG(J).

For let(O1, T1, . .

. , On, Tn, .

. .

)be a play of SMG(J) during which TWO followed the 2-tactic G.For each j we put M 1j = Φ2(Oj), . .

. , M kj = Φ2(M k−1j).

An inductivecomputation shows that• (M 11 , M 21 , . .

. , M k1 , M 12 , M 22 , .

. .

, M k2 , . .

.) is a sequence of legalmoves for ONE in the game MG(A, J), and that•1.

F(M 11 ) ∪. .

. ∪F(M 11 , .

. .

, M k1 ) ⊆T1, and2. F(M 1j , .

. .

, M kj )∪F(M 2j , . .

. , M kj , M 1j+1)∪.

. .∪F(M kj , M 1j+1, .

. .

, M k−1j+1 )⊆Tj+1 for each j.Since F is a winning k-tactic for TWO in the game MG(A, J), andsince ∪∞n=1On ⊆∪∞n=1M 1n, TWO won the given play of SMG(J).The next corollary solves Problems 10 and 11 of [S1].Corollary 27 Player T W O has a winning 2-tactic in the game SMG(JR).Proof. Fix, by Corollary 18, a cofinal family A ⊂⟨JR⟩such that TWOhas a winning 2-tactic in MG(A, JR).We define Φ1 : ⟨JR⟩→JR and Φ2 : ⟨JR⟩→A as follows:Fix X ∈⟨JR⟩, and choose a sequence (X0, X1, .

. .

, Xn, . .

.) such that:1.

X0 = X,2. Xn+1 ∈A and N · Xn ⊆Xn+1for each n. Put Φ2(X) = ∪∞n=1Xn.Fix X ∈⟨JR⟩and let Φ1(X) be a nowhere dense set for whichΦ2(X) ⊂N · Φ1(X).Then A, Φ1 and Φ2 are as required by Theorem 26.Corollary 28 For each of the ideals Jn constructed in the proof of The-orem 21, TWO has a winning 2-tactic in SMG(Jn).Proof.

Let An be as in the proof of Theorem 21. For each X ∈⟨Jn⟩we let Φ2(X) be an element of An which contains it, and we letΦ1(X) = {aX} where aX ∈Ln\Φ2(X).

Then An, Φ1 and Φ2 are asrequired by Theorem 26.23

Before giving another application of Theorem 26 we give an exampleof free ideals J which show that TWO does not always have a winning k-tactic in the game SMG(J) for some k. These examples are also relevantto the material of the next section.The symbol M(ω, 2) denotes thesmallest ordinal α for which the partition relation α →(ω −path)2ω/<ωholds. M(ω, 2) is a regular cardinal less than or equal to c+.

It in factsatisfies the partition relation M(ω, 2) →(ω −path)nω/<ω for all n. Let κbe an initial ordinal number. It is consistent that M(ω, 2) is equal to ℵ2while c is larger than κ (this is yet another result of Todorcevic).Theorem 29 Let λ be a cardinal number of countable cofinality and letκ be a cardinal number larger than λ.

If M(ω, 2) ≤λ+, then there is nok such that player TWO has a winning k-tactic in SMG([κ]<λ).Proof. Let F be a k-tactic for TWO.Player ONE’s counter-strategy will be to play judiciously chosensubsets from κ.

We first single out those sets from which ONE willmake moves.Choose sets S0 ⊂S1 ⊂. .

. ⊂Sα ⊂.

. .

∈[κ]λ for α < λ+ such that:1. λ ⊂S0,2. ∪{F(Si1, .

. .

, Sij) : j ≤k, i1 < . .

. < ij < α} ⊂Sα for each0 < α < λ+.Now let λ1 < λ2 < .

. .

< λ be an increasing sequence of regularcardinal numbers converging to λ. Define a function Γ : [λ+]k →ωso thatΓ(ξ1, . .

. , ξk) = min{m : |F(Sξ1, .

. .

, Sξk)| ≤λm}.Then, on account of the relation M(ω, 2) ≤λ+, choose an m <ω and a sequence αk+1 < . .

. < αk+m < .

. .

from λ+ such thatΓ(αj+1, . .

. , αj+k) ≤m for all j.Consider the sequence(Sα1, F(Sα1), .

. .

, Sαk, F(Sα1, . .

. , Sαk), .

. .

, Sαk+m, F(Sα1+m, . .

. , Sαk+m), .

. .

).It is a play of the game SMG([κ]<λ) during which TWO used the k-tactic F. To see that TWO lost this play, let Tk denote TWO’s k-thmove. The choice of the sequence αk+1 < .

. .

implies that ∪∞n=1Tnhas cardinality less than λ. The union of the sets played by ONEhas cardinality λ; TWO didn’t catch up with ONE.Corollary 30 For ω = cof(λ) ≤λ < κ cardinal numbers with cof([κ]≤λ, ⊂) = κ, the following statements are equivalent:1.

TWO has a winning 2-tactic in SMG([κ]<λ).2. λ+ ̸→(ω −path)2ω/<ω.24

Proof. It follows from Theorem 29 that 1. implies 2.That 2. implies 1.:By the cofinality hypothesis and by 2. we find, according to Corollary20, a well-founded cofinal family A such that TWO has a winning2-tactic in MG(A, [κ]<λ).

We may assume that there is an enumer-ation {Aα : α < κ} of A for which α ∈Aα for each α. Define Φ1and Φ2 as follows:For X ∈[κ]≤λ define a sequence (X0, . .

. , Xm, .

. .) such that:1.

X0 = X, and2. Xn+1 = ∪α∈XnAαfor each n.Choose Φ2(X) ∈A such that ∪n<ωXn ⊆Φ2(X).Pick zX ∈(κ\Φ2(X)) and pick ρX minimal such that ρX ̸∈Φ2(X),and Φ2(X) ⊂AρX.

Put Φ1(X) = {zX, ρX}.Then A, Φ1 and Φ2 are as required by Theorem 26.This result will be discussed at greater length after Theorem 34.We finally mention that it is still unknown whether there is for eachm a free ideal Jm such that TWO does not have a winning m-tactic,but does have a winning m + 1-tactic in SMG(Jm). This is Problem 9of [S1].

In this connection it is worth noting the following relationshipbetween winning k-tactics in MG(J) and winning m-tactics in SMG(J).The proof uses ideas as in the proof of Theorem 26.Theorem 31 If TWO has a winning k-tactic in MG(J), then TWO hasa winning 2-tactic in SMG(J).4.4The game VSG(J)For a free ideal J on an infinite set S, the game V SG(J) (read “ very stronggame on J”) is defined so that an ω-sequence (O1, (T1, S1), . .

. , On, (Tn, Sn), .

. .

)is a play if for each n,1. On ∈⟨J⟩is player ONE’s move in inning n,2.

(Tn, Sn) ∈J × ⟨J⟩is player TWO’s move in inning n, and3. On ∪Tn ∪Sn ⊆On+1.Player TWO wins this play if ∪∞n=1On = ∪∞n=1Tn.We assume for this section that ⟨J⟩is also a proper ideal on S. Given acofinal family A ⊂⟨J⟩, we may assume whenever convenient that ONE isplaying from A in the game V SG(J).

It is clear that if TWO has a winningk-tactic in SMG(J), then TWO has a winning k-tactic is V SG(J). Theconverse is not so clear.25

Problem 1 Let J be a free ideal on a set S and let k be a positive integer.Is it true that if TWO has a winning k-tactic in V SG(J), then TWO hasa winning k-tactic in SMG(J)?In the next theorem we find a partial converse.Theorem 32 Let J be a free ideal on a set S and let k be a positiveinteger. If add(⟨J⟩, ⊂) = cof(⟨J⟩, ⊂), then the following statements areequivalent:1.

TWO has a winning 2-tactic in SMG(J).2. TWO has a winning k-tactic in SMG(J).3.

TWO has a winning k-tactic in V SG(J).Proof That 1. and 2. are equivalent: This is Theorem 19 of [S1].That 2. implies 3.: Let F be a winning k-tactic for TWO in SMG(J).Define G so thatG(A1, . .

. , Aj) = (F(A1, .

. .

, Aj), Aj ∪F(A1, . .

. , Aj))for j ≤k.

Then G is a winning k-tactic for TWO in V SG(J).That 3. implies 2.: Let G be a winning k-tactic for TWO in V SG(J).Then choose a sequence (Mξ : ξ < cof(⟨J⟩, ⊂)) such that:1. Mξ ⊂Mν for ξ < ν < cof(⟨J⟩, ⊂) and2.

{Mξ : ξ < cof(⟨J⟩, ⊂)} is cofinal in ⟨J⟩.Now cof(⟨J⟩, ⊂) is a regular uncountable cardinal number. We maythus further assume that the sequence (Mξ : ξ < cof(⟨J⟩, ⊂)) hasbeen chosen such that if (U, T ) = G(Mξ1, .

. .

, Mξj ), then U∪T ⊂Mηfor all ξj < η < cof(⟨J⟩, ⊂).For each X ∈⟨J⟩define α(X) = min{ξ : X ⊂Mξ}.For eachξ choose zξ ∈S\Mξ.We now define a k-tactic, F, for TWO inSMG(J).Let X1 ⊂. .

. ⊂Xj ∈⟨J⟩for a j ≤k be given.CASE 1:α(X1) < .

. .

< α(Xj). Let (U, T ) = G(Mα(X1), .

. .

, Mα(Xj))and define F(X1, . .

. , Xj) = U ∪{zα(Xj)+1}.CASE 2: Otherwise, set F(X1, .

. .

, Xj) = {zα(Xj )+1}. Then F is awinning k-tactic for TWO in SMG(J).There is the following analogue of Theorem 26 for the very stronggame:Proposition 33 Let J be a free ideal on a set S. If there is a cofinalfamily A ⊂⟨J⟩such that TWO has a winning k-tactic in MG(A, J),then TWO has a winning 2-tactic in V SG(J).Proof.

Let A ⊂⟨J⟩be a cofinal family such that TWO has a winningk-tactic in MG(A, J). We will define a winning 2-tactic for TWOfor the game V SG(J).

To this end, choose a winning k-tactic, F,for TWO for the game MG(A, J). For each X ∈⟨J⟩choose a set26

A1(X) ⊂. .

. ⊂Ak(X) from A such that X ⊂A1(X), and chooseΨ(X) from A such that Ak(X) ⊂Ψ(X).Let X ⊂Y be sets from ⟨J⟩.CASE 1: G(X) = (F(A1(X)) ∪.

. .

∪F(A1(X), . .

. , Ak(X)), Ψ(X)).CASE 2: Define G(X, Y ) so that:1.

G(X, Y ) = (F(A2(X), . .

. , Ak(X), A1(Y )∪.

. .∪F(A1(Y ), .

. .

, Ak(Y )), Ψ(Y ))if Ψ(X) ⊂Y , and2. G(X, Y ) = G(Y ) otherwise.Then G is a winning 2-tactic for TWO in V SG(J).

For let (O1, (T1, S1), O2, (T2, S2), . .

. )be a play of V SG(J) such that (T1, S1) = G(O1) and (Tn+1, Sn+1) =G(On, On+1) for all n. Then Sn = Ψ(On) and Ak(On) ⊂A1(On+1)for each n. An inductive computation, using this information, showsthat TWO won this play of V SG(J).Combining Theorem 32 and Theorem 29 we see that TWO does notalways have a winning k-tactic in games of the form V SG(J).

CombiningTheorem 32 and Corollary 30 we obtain another game-theoretic charac-terization of the partition relation λ+ →(ω −path)2ω/<ω when λ is anuncountable cardinal of countable cofinality.Analogous to the case of the ideal of countable subsets of an infiniteset, there is for each uncountable cardinal number λ which is of countablecofinality, a proper class of cardinals κ for which the ideal [κ]≤λ has theirredundancy property. It is also a consequence of MA + c > λ that thepartition relation λ+ ̸→(ω−path)2ω/<ω holds.

Accordingly it is consistentthat there is a proper class of cardinals κ such that TWO has a winning2-tactic in the game V SG([κ]≤λ). The following problem (to be comparedwith the upcoming Conjecture 1) is open.Problem 2 Let λ be an uncountable cardinal of countable cofinality.

Is ittrue that if TWO has a winning 2-tactic in the game V SG([λ+]<λ), thenTWO has a winning 2-tactic in V SG([κ]<λ) for all κ > λ?Our next theorem (Theorem 34) applies to abstract free ideals whoseσ-completions have small principal bursting number. It is not clear to uswhether “3” occurring in Theorem 34 is optimal.

One of its applications isthat ZFC+GCH implies that TWO has a winning 3-tactic in V SG([κ]<ℵ0)for all κ. It is very likely that the “3” appearing in this application is notoptimal, as will be discussed later.Theorem 34 Let J be a free ideal such that:1. bu(⟨J⟩, ⊂) = ℵ2,2.

add(⟨J⟩, ⊂) = ℵ1,3. cof(⟨J⟩, ⊂) = λ and4.

[λ]<ℵ0 has the coherent decomposition property.Then player TWO has a winning 3-tactic in V SG(J).27

Proof. Let J be a free ideal (on a set S) as in the hypotheses.

Let A bea well-founded cofinal family of cardinality λ, such that |{B ∈A :B ⊆A}| ≤ℵ1 for each A ∈A.For each A ∈A fix νA ≤ω1 and a bijective enumeration {Jξ(A) :ξ < νA} of the set {X ∈A : X ⊆A}.Choose a sequence (Cξ : ξ < ω1) from ⟨J⟩such that:1. Cξ ⊂Cν for ξ < ν and2.

∪ξ<ω1Cξ ̸∈⟨J⟩.For A ∈A define ξA = min{ξ < ω1 : Cξ ̸⊆A}.For A ⊂B elements from A, define a set τ(A,B) such that (S1, . .

. , Sn)is in τ(A, B) if:1.

2 ≤n < ω,2. S1 = B and S2 = A,3.

Sj+1 ∈{Jξ(Sj) : ξ < νSj and Cξ ⊂Sj−1} for 2 ≤j < n.For (S1, . .

. , Sn) and (T1, .

. .

, Tm) in τ(A, B) define (S1, . .

. , Sn) <(T1, .

. .

, Tm) if n < m and (S1, . .

. , Sn) = (T1, .

. .

, Tn). Then (τ(A, B), <) is a tree.

Each branch of this tree is finite since (A, ⊂) is well-founded. Indeed, τ(A, B) is a countable set.Define F(A, B) to be the set of X ∈A such that X ∈{S1, .

. .

, Sm}for some (S1, . .

. , Sm) ∈τ(A, B).

Then F(A, B) is a countable set.Notice that if C ⊂A ⊂B are elements of A such that C ∈{Jξ(A) :ξ ≤νA and Cξ ⊂B}, then F(C, A) ⊂F(A, B).Let B ⊂[A]ℵ0 be cofinal, well-founded and with the coherent de-composition property.For each B ∈B choose a decompositionB = ∪∞n=1Bn where each Bn is finite, and these decompositionssatisfy the coherent decomposition requirement. By Proposition 15of [S2] we also fix a functionK : [B]2 →ωwhich witnesses that (B, ⊂) ̸→(ω −path)2ω/<ω.Define Φ1 : [A]2 →B such that(∪{F(X, Y ) : (∃(S1, .

. .

, Sn) ∈τ(A, B))(X ⊂Y and X, Y ∈{S1, . .

. , Sn})})is a subset of Φ1(A, B).

Also define Φ2 : [A]2 →A such thatCξ ∪CξB ∪(∪Φ1(A, B)) ⊂Φ2(A, B)where now A = Jξ(B).Note that if A, B, and C are elements of A such that A ⊂B ⊂Φ2(A, B) ⊂C, then Φ1(A, B) ⊂Φ1(B, C).Finally, choose for each A ∈A a Φ3(A) ∈A such that A ∪CξA ⊆Φ3(A).28

Choose for each A ∈A a sequence of sets A0 ⊆. .

. , An ⊆.

. .

suchthat each Ai is in J and A = ∪∞n=0An.We now define a 3-tactic for TWO: First note that for the verystrong game we may make the harmless assumption that playerONE’s moves are all from the cofinal family A. Let A ⊂B ⊂Cbe sets from A.

Here are player TWO’s responses F(A), F(A, B)and F(A, B, C):Case 1: F(A) = (∅, Φ3(A))Case 2: F(A, B) = (∅, Φ2(A, B))Case 3: F(A, B, C) = (D, Φ2(B, C))if Φ2(A, B) ⊆C, where D = Cm1 ∪. .

. ∪Cmris given by: m ≥K({Φ1(A, B), Φ1(B, C)}) is minimal such that (Φ1(A, B))n ⊆(Φ1(B, C))nfor all n ≥m, and (Φ1(B, C))m = {C1, .

. .

, Cr}.Case 4: In all other cases define F(A, B, C) = F(B, C).To see that F is a winning 3-tactic for TWO, consider a play(O1, (T1, S1), O2, (T2, S2), . .

. )of V SG(J) for which1.

(T1, S1) = F(O1),2. (T2, S2) = F(O1, O2) and3.

(Tn+3, Sn+3) = F(On+1, On+2, On+3)for all n.Then T1 = T2 = ∅, S1 = Φ3(O1), S2 = Φ2(O1, O2) and Sn+1 =Φ2(On, On+1) for all n ≥2. From the fact that On ⊇Sn−1 for alln ≥2 it follows thatO1 ⊂O2 ⊂Φ2(O1, O2) ⊆O3 ⊂Φ2(O2, O3) ⊆O4 ⊂.

. .

,whence Φ1(O1, O2) ⊂Φ1(O2, O3) ⊂Φ1(O3, O4) ⊂. .

.. For each klet mk denote the minimal integer such that1. K({Φ1(Ok, Ok+1), Φ1(Ok+1, Ok+2)}) ≤mk and2.

(Φ1(Ok, Ok+1))n ⊆(Φ(Ok+1, Ok+2))n for all n ≥mk.From the properties of K it follows that there are infinitely manyk such that mj < mk for each j < k. Fix i, and fix the smallestj ≥i such that Oi ∈Φ1(Oj, Oj+1). Then let t be minimal such thatOi ∈(Φ1(Oj, Oj+1))t. Then for each k such that mℓ< mk for eachℓ< k, and t < mk, Omki⊆Tk.

It follows that Oi ⊆∪∞n=1Tn. Fromthis it follows that TWO won this F-play of V SG(J).29

Corollary 35 (GCH) For every infinite cardinal number κ, TWO hasa winning 3-tactic in V SG([κ]<ℵ0).The results of Corollaries 30 and 35 should be compared with thoseof Koszmider [Ko] for the game MG([κ]<ℵ0). In Corollary 30 we showthat there is a proper class of κ such that TWO has a winning 2-tacticin SMG([κ]<ℵ0), and thus in V SG([κ]<ℵ0).

This class includes ℵn forall n < ω.In [Ko] it is proven that TWO has a winning 2-tactic inMG([ℵn]<ℵ0) for all n ∈ω ([Ko], Theorem 18). Under the additional settheoretic assumption that both λ holds and λℵ0 = λ+ for all uncountablecardinal numbers λ which are of countable cofinality, Koszmider furtherproves that player TWO has a winning 2-tactic in MG([κ]<ℵ0) for all κ([Ko], Theorem 19).

In light of these results it is consistent that TWOhas a winning 2-tactic in the game SMG([κ]<ℵ0) and thus in the gameV SG([κ]<ℵ0) for all κ.All this evidence leads us to believe that one could prove (withoutrecourse to additional set theoretic hypotheses) that player TWO has awinning 2-tactic in the game V SG([κ]<ℵ0) for all infinite κ. We suspecteven more: that TWO has a winning 2-tactic in SMG([κ]<ℵ0) for all κ.We state this formally as a conjecture:Conjecture 1 Player TWO has a winning 2-tactic in the game SMG([κ]<ℵ0)for each infinite cardinal number κ.One can modify the proof of Theorem 34 to obtain the following result:Theorem 36 Let J be a free ideal on a set S such that1.

bu(⟨J⟩, ⊂) = ℵn for some finite n,2. there is an (ωk, ωk)-pseudo Lusin set in (⟨J⟩, ⊂) for each k ∈{1, .

. .

, n},3. cof(⟨J⟩, ⊂) = λ), and4. ([λ]<ℵ0, ⊂) has the coherent decomposition property.Then player TWO has a winning n + 1-tactic in V SG(J).We now give an example which shows, assuming the Continuum Hy-pothesis, that the hypothesis that add(⟨J⟩, ⊂) = ℵ1 of Theorem 34 isnecessary (see Corollary 38).Theorem 37 Let ωα be the initial ordinal corresponding with c. Thenthere is a free ideal J ⊂P(ωα+1) such that cof(⟨J⟩, ⊂) = ℵα+1 and thereis no positive integer k for which TWO has a winning k-tactic in V SG(J).Proof Define J ⊂P(ωα+1) such that X ∈J if, and only if, |X| ≤ℵαand X ∩ω is finite.

Then cof(⟨J⟩, ⊂) = add(⟨J⟩, ⊂) = ωα+1. ByTheorem 32 it suffices to show that TWO doesn’t have a winning2-tactic in SMG(J).Let F be a 2-tactic for TWO in SMG(J).

For ω < η < ωα+1 putφ(η) = sup(η ∪F(η)). Let C ⊆ωα+1\(ω + 1) be a closed unboundedset such that φ(γ) < β whenever γ < β are in C.30

For each η ∈C define φη : C\(η + 1) →ωα+1 so that φη(β) =sup(β ∪F(η, β)) for all β.Then choose a closed, unbounded setCη ⊆C\(α + 1) such that φη(β) < γ whenever β < γ are in Cη.Let D be the diagonal intersection of (Cη : η ∈C); i.e., D = {ξ ∈C : ξ ∈∩{Cη : η < ξ and η ∈C}. Then D is an unbounded subsetof ωα+1.

Now observe that if η1 < η2 < η3 are elements of D, then1. η2 ∈Cη1,2.

η3 ∈Cη1 ∩Cη2, and thus3. F(η1) ⊆η2 and F(η1, η2) ⊆η3.Define Φ : [D]2 →ω so thatΦ(η, β) = max(ω ∩(F(η) ∪F(η, β))).By the Erd¨os-Rado theorem we obtain an n < ω and an uncountableX ⊂D such that Φ(η, β) = n for all η < β ∈X.

Pick η1 < η2 <. .

. < ηm < .

. .

from X and put On = ηn for each n. Put T1 = F(O1)and Tn+1 = F(On, On+1) for each n.Then (O1, T1, . .

. , On, Tn, .

. .) is an F-play of SMG(J) which is lostby TWO.Corollary 38 Assume the Continuum Hypothesis.

Then there is a freeideal J ⊂P(ω2) such that cof(⟨J⟩, ⊂) = ℵ2, and there is no positiveinteger k for which T W O has a winning k-tactic in V SG(J).We don’t know if there is for each m a free ideal Jm such that TWOdoes not have a winning m-tactic, but does have a winning m + 1-tacticin V SG(Jm).Problem 3 Is there for each m a free ideal Jm such that TWO doesnot have a winning m-tactic, but does have a winning m + 1-tactic inV SG(Jm)?4.5The Banach-Mazur game and an example ofDebsThe Banach-Mazur game is defined as follows for a topological space(X, τ). Players ONE and TWO alternately choose nonempty open sub-sets from X; in the n-th inning player ONE first chooses On and TWOresponds with Tn.

An inning is played for each positive integer. The setschosen by the players must satisfy the ruleOn+1 ⊆Tn ⊆Onfor all n. Player TWO wins the play(O1, T1, .

. .

, On, Tn, . .

. )if the intersection of these sets is nonempty; otherwise player ONE wins.Following Galvin and Telgarsky [G-T], we denote this game by BM(X, τ).31

In the early 1980’s Debs [D] solved Problem 3 of [F-K] by giving examplesof topological spaces (X, τ) for which player TWO has a winning strategyin the game BM(X, τ), but no winning 1-tactic. In all but one of Debs’examples it was known (in ZFC) that TWO has a winning 2-tactic.

Weshow here that also for the remaining example player TWO has a winning2-tactic (Corollary 43). This was previously known under the assumptionof some additional hypotheses.This result eliminates this example as a candidate for providing evi-dence (consistent, modulo ZFC) towards the following conjecture of Tel-garsky:Conjecture 2 (Telgarsky, [T], p. 236) For each positive integer k thereis a topological space (Xk, τk) such that TWO does not have a winning k-tactic, but does have a winning k + 1-tactic in the game BM(Xk, τk).The following unpublished result of Galvin is the only theorem knownto us which gives general conditions under which TWO has a winning2-tactic if TWO has a winning strategy in the Banach-Mazur game:Theorem 39 (Galvin, unpublished) Let (X, τ) be a topological spacefor which TWO has a winning strategy in the Banach-Mazur game.

If thisspace has a pseudo base P with the property that• |{V ∈P : B ⊆V }| < s(B) for each B in P,then TWO has a winning 2-tactic.Here the cardinal number s(B) is defined to be the minimal κ suchthat B does not contain a collection of κ pairwise disjoint nonempty opensubsets; it is said to be the Souslin number of B.This subsection is organised as follows. We first prove a theorem con-cerning k-tactics in the Banach-Mazur game which is analogous to The-orem 5 of [G-T].It provides an equivalent formulation of Telgarsky’sconjecture which allows player TWO slightly more information: TWOmay also remember the inning number.

After this we give our result onDebs’ example.4.5.1Markov k-tactics.Whereas a k-tactic for player TWO remembers at most the latest k movesof the opponent, a strategy for TWO which remembers in addition to thisinformation also the number of the inning in progress is called a Markovk-tactic. This choice of terminology is by analogy with the terminology“tactic” (used by Choquet [C], p. 116, Definition 7.11 for what we call a1-tactic) and “Markov strategy” (used by Galvin and Telgarsky [G-T], p.52 for what we call a Markov 1-tactic).

A k-tactic is the special case of aMarkov k-tactic where the inning number is ignored by the player.Note that if (X, τ) has a dense set of isolated points then player TWOhas a winning 1-tactic in BM(X, τ). Thus we may assume that if at allpossible, player ONE will avoid playing an open set which contains an iso-lated point.

From the point of view of k-tactics for TWO we may therefore32

restrict our attention to topological spaces without isolated points. By thefollowing proposition we may further restrict our attention to topologicalspaces in which each nonempty open set contains infinitely many pairwisedisjoint open subsets.Proposition 40 Let (X, τ) be a topological space with no infinite set ofpairwise disjoint open subsets.Then there is a positive integer n suchthat:τ\{∅} = τ1 ∪.

. .

∪τnwhere each τi has the finite intersection property.Proof. Claim 1: There is a positive integer n such that every collectionof pairwise disjoint nonempty open subsets is of cardinality ≤n.Proof of Claim 1: This is a well known fact: see e.g.

[C-N], Lemma2.10, p. 31.Now let n be the minimal positive integer satisfying Claim 1. LetU = {U1, .

. .

, Un} be a collection of pairwise disjoint nonempty opensubsets of the space. Then U is a maximal pairwise disjoint family.For 1 ≤i ≤n, let τi be a maximal family of nonempty open setssuch that:1.

Ui ∈τi,2. any two elements of τi have nonempty intersection.Claim 2: τ\{∅} = τ1 ∪.

. .

∪τnProof of Claim 2: Assume the contrary and let Y be a nonemptyopen set which is in none of the τi.Then we find for each i anXi in τi which is disjoint from Y (by maximality of each τi). Wemay assume that Xi ⊆Ui for each i.

But then {X1, . .

. , Xn, Y } is acollection of n+1 pairwise disjoint nonempty open subsets of (X, τ),contradicting the choice of n.Each τi has the finite intersection property.Proposition 41 Let (X, τ) be a topological space for which:1.

Player TWO has a winning strategy in the game BM(X, τ) and2. every collection of pairwise disjoint open subsets is finite.Then TWO has a winning 1-tactic in BM(X, τ).Proof.

Write, by Proposition 40,τ\{∅} = τ1 ∪. .

. ∪τnwhere each τi has the finite intersection property, and n is minimal.Choose a pairwise disjoint collection {U1, .

. .

, Un} such that Uj ∈τjfor each j.Claim 3: For each j, if S1 ⊇S2 ⊇. .

. is a denumerable chain fromτj, then ∩∞n=1Sn ̸= ∅.33

Proof of Claim 3: Assume the contrary, and fix j and a chain S1 ⊇S2 ⊇. .

. in τj such that ∩∞n=1Sn = ∅.

We may assume that Sn+1 ⊂Sn ⊂Uj for all n.Let F be a winning perfect information strategy for TWO in BM(X, τ).Consider the play(O1, T1, . .

. , Om, Tm, .

. .

)which is defined so that:1. O1 = S1,2.

Tm = F(O1, . .

. , Om) for all m and3.

Om+1 = Tm ∩Sm+1.Note that since each Sm is a subset of Uj, each response by playerTWO using F is a member of τj, whence each Om is a legal move byONE. We now get the contradiction that TWO lost this play despitethe fact that TWO was playing according to a winning strategy.

Thiscompletes the proof of Claim 3.We now define a winning 1-tactic, G, for TWO. Let U be a nonemptyopen subset of X.

Choose the minimal j such that Uj ∩U ̸= ∅andput G(U) = Uj ∩U. Claim 3 implies that this is a winning 1-tacticfor TWO.Theorem 42 Let k be a positive integer.

If player TWO has a winningMarkov k-tactic in the Banach Mazur game on some topological space,then TWO has a winning k-tactic in the Banach-Mazur game on thatspace.Proof. Let k be a positive integer and let (X, τ) be a topological spacesuch that TWO has a winning Markov k-tactic in the game BM(X, τ).We may assume that this space has no isolated points.

By Proposi-tion 41 we may also assume that every nonempty open subset of Xcontains infinitely many pairwise disjoint open subsets (player ONEmay safely avoid playing open subsets not having this property). ByTheorem 5 of [G-T] we may assume that k > 1.Let F be a winning Markov k-tactic for TWO.

For each nonemptyopen set U, let {Jm(U) : 0 < m < ω} bijectively enumerate acollection of infinitely many pairwise disjoint nonempty open subsetsof U.Define a k-tactic G for TWO as follows.Let U1 ⊇. .

. ⊇Uj benonempty open sets, where 1 ≤j ≤k.Case 1: j = 1: Put G(U1) = F(J2(U1), 1).Case 2: j > 1 and Ui+1 ⊆Jl+i+1(Ui) for 1 ≤i < j, for some l. PutG(U1, .

. .

, Uj) = F(Jl+2(U1), . .

. , Jl+j+1(Uj), l + j).Case 3: In all other cases define G(U1, .

. .

, Uj) = G(Uj).34

To see that G is a winning k-tactic for TWO, consider a play(O1, T1, . .

. , Om, Tm, .

. .

)such that• Tj = G(O1, . .

. , Oj) for j ≤k and• Tn+k = G(On+1, .

. .

, On+k) for all n.From the definition of G and the rules of the Banach-Mazur gameit follows that T1 is defined by Case 1 and Tm for m > 1 by Case 2.In particular, writing Sn for Jn+1(On) we find that:1. Tj = F(S1, .

. .

, Sj, j) for j ≤k and2. Tn+k = F(Sn+1, .

. .

, Sn+k, n + k)for all n. Indeed,O1 ⊇S1 ⊇T1 ⊇O2 ⊇S2 ⊇. .

.Since F is a winning Markov k-tactic, it follows that ∩∞n=1On ̸= ∅.4.5.2Debs’ exampleLet σ be the topology of the real line whose elements are of the form U\Mwhere U is open and M is meager in the usual topology. The symbolBM(R, σ) denotes the Banach-Mazur game, played on the topologicalspace (R, σ).

It is known that TWO has a winning strategy but does nothave a winning 1-tactic in BM(R, σ).Corollary 43 Player TWO has a winning 2-tactic in the game BM(R, σ).Proof. Theorem 22 of [S1] and Corollary 27.5Appendix: Consistency of the hypothe-ses of Theorem 23.We start with a ground model V and let P ∈V be a forcing notion ofcardinality ≤c.

For a cardinal κ, denote by Pκ the product of κ copiesof P taken side-by-side with countable supports.Lemma 44 Let λ be an uncountable cardinal. Suppose:1. κ ≥κ1 ≥κ2 ≥κ3 ≥ω2 are cardinal numbers such that• κ is a regular cardinal,• κ →(κ1)2λ,• κ1 →(κ2)3c,• κ2 →(κ3)2λ and35

• κ3 →(ω2)3c.2. Forcing with P adds a real to the ground model.Then c →(ω −path)2λ/<ω holds in the forcing extension V Pκ.Proof.

Let λ, κ, κ1, . .

. , κ3, P be as in the assumptions.

Our argumentclosely follows section 2 of [To1].For sets A, B the symbol A/B denotes {{α, β} : α ∈A, β ∈B, α <β}.Note that V Pκ satisfies c = κ; we prove that κ →(ω −path)2λ/<ωholds in V Pκ.Let [κ]2 = Si<λ ˙Ki be a given partition in V Pκ. Let ˙U be a Pκ-name for a member of [κ]κ.Pick A ∈[κ]κ and for each α ∈A,a qα ∈Pκ such that qα ∥−α ∈˙U and such that the qα’s form a∆-system.

Define H : [A]2 →(λ + 1) so that H({α, β}) = i if i isthe minimal j such that p ∥−{α, β} ∈˙Kj for some p ≤qα, qβ if suchj exists (i.e., if qα and qβ are compatible), and H({α, β}) = λ if qαis incompatible with qβ.By our choice of κ, the partition relation κ →(κ1)2λ holds. Therefore,choose A1 ⊂[A]κ1 and i ≤λ such that H′′[A1]2 = {i}.

Since Pκsatisfies the c+-c.c., we have i < λ.Let ⟨pα,β : {α, β} ∈[A1]2⟩be a fixed sequence of conditions suchthat pα,β ≤qα, qβ and pα,β ∥−{α, β} ∈˙Ki. For α < β < γ in A1 wedefine H0({α, β, γ}) to be a pair (c, d), where c codes pα,β and pα,γ asstructures as well as relations between the ordinals of dom(pα,β) anddom(pα,γ), and d does the same for pα,γ and pβ,γ.

Since there areonly c such pairs, and since κ1 →(κ2)3c holds, choose A2 ∈[A1]κ2and (c, d) such that H′′0 [A2]3 = {(c, d)}. For convenience, assumethat A2 has order type κ2.It follows that for each α ∈A2 thesequence ⟨pα,β : β ∈A2\(α + 1)⟩forms a ∆-system with root p0α(≤qα), and that for each γ ∈A2 the sequence ⟨pβ,γ : β ∈A2 ∩γ⟩forms a ∆-system with root p1γ (≤qγ).

Moreover, the p0α’s and p1γ’sform ∆-systems with roots p0 and p1 respectively. To see the latter,note that we may shrink A2 to a cofinal subset A3 so that the relevantp0α’s and p1α’s do in fact form a ∆-system.

Now consider α, β, γ ∈A3, and α′, β′ ∈A2. Comparing H0({α, β, γ}), H0({α, β′, γ}) andH0({α′, β′, γ}), one sees that the sequence ⟨p0α : α ∈A2⟩forms a∆-system.

A similar argument works for the p1γ’s.Also, p0 is compatible with p1. We call ⟨pα,β :{α, β} ∈B/B⟩adouble ∆-system with root p0 ∪p1.There is no reason why for a given α the conditions p0α and p1α shouldbe compatible: if these were always compatible, our argument wouldyield a consistency proof of c →(ω)2λ, which is false in ZFC.We now save as much of the compatibility between p0α and p1α as isneeded for the consistency proof of c →(ω −path)2λ/<ω.

Thin outA2 to a cofinal subset A3 such that dom(p0α ∪p1α) ∩dom(p0β ∪p1β) =36

dom(p0 ∪p1) for all {α, β} ∈A3/A3. Then in particular p1α and p0βare compatible for {α, β} ∈A3/A3.Now repeat the reasoning above with A2 in place of A, κ2 in placeof κ, κ3 in place of κ1, and ω2 in place of κ2.

Also, p1α will nowplay the role of qα, and p0β the role of qβ for {α, β} ∈A3/A3. Weget a set A4 ⊂A3 of order type ω2 and some j < λ (which maybe different from i), conditions ¯pα,β for {α, β} ∈A4/A4 that form adouble ∆-system with root ¯p0 ∪¯p1, and we get roots ¯p0α and ¯p1γ asbefore.

Now ¯pα,β ∥−{α, β} ∈˙Kj for {α, β} ∈A4/A4.Our choice of ¯pα,β at the beginning of the second run of the argumentinsures that ¯p0α ≤p1α and ¯p1γ ≤p0γ, and hence ¯p0 ≤p1 and ¯p1 ≤p0.Now let G be a generic subset of Pκ. Define:˙X = {α ∈A4 : p0α ∈G},˙Y = {α ∈A4 : p1α ∈G},˙W = {α ∈A4 : ¯p0α ∈G},˙Z = {α ∈A4 : ¯p1α ∈G}.Then ˙Z ⊂˙X and ˙W ⊂˙Y , and all four sets are cofinal in A4.Now ¯p0 ∪¯p1 forces the following facts:(1) ∃δ1 ∈ω2∀α ∈˙X\δ1 {β ∈˙W : {α, β} ∈˙Ki} is cofinal in A4, and(2) ∃δ2 ∈ω2∀α ∈˙Y \δ2 {β ∈˙Z : {α, β} ∈˙Kj} is cofinal in A4.The combination of (1) and (2) suffices to construct in V Pκ an ω-path of the given partition that uses only colors i and j:Let δ = max{δ1, δ2}.

Inductively define an increasing sequence ⟨xn :n ∈ω⟩of ordinals such that x2k ∈Z (and hence in X), x2k+1 ∈W ,and {x2k, x2k+1} ∈˙Ki (by (1)); {x2k+1, x2k+2} ∈˙Kj (by (2)).It remains to prove (1) and (2). We shall prove (1) only; the proofof (2) is similar, and is a special case of [To1], section 2, property(1)].Assume that ¯p0∪¯p1 does not force (1).

Then we can find a condition¯p2 ≤¯p0 ∪¯p1 and a Pκ-name˙D ∈[ ˙X]ω2 and for each β ∈˙D aγβ ∈A4\(β + 1) such that ¯p2 ∥−{β,δ} /∈˙Ki whenever δ ∈˙W\γβ.Working in V , we pick B ∈[A4]ω2 such that for each β ∈B wefind rβ ≤p0β ∪¯p2 such that rβ ∥−β ∈˙D, and rβ decides the valueof γβ. We may assume that the rβ’s form a ∆-system with root≤¯p2 ≤¯p0 ∪¯p1, and that γβ < δ for all {β, δ} ∈B/B.Since⟨pα,β : β ∈B\(α + 1)⟩forms a ∆-system, we may also assume thatdom(rβ) ∩dom(pβ,δ\p0β) = ∅for all δ > γβ in A4.Pick δ ∈A2 such that B∩δ is uncountable and dom(¯p0δ)∩dom(¯p2) =dom(¯p0).

Since ⟨pβ,δ : β ∈B ∩δ⟩forms a ∆-system with root p1δand since dom(¯p0δ) is countable, we have dom(pβ,δ\p1δ)∩dom(¯p0δ) ̸= ∅for only countably many β ∈B ∩δ. So pick a β ∈B ∩δ such thatdom(pβ,δ\p1δ) ∩dom(¯p0δ) = ∅.Define r ∈Pκ as follows:37

dom(r) = dom(rβ) ∪dom(¯p0δ) ∪dom(pβ,δ\p1δ),r|dom(rβ ∪¯p0δ) = rβ ∪¯p0δ,andr(ξ) = pβ,δ(ξ) for ξ ∈dom(pβ,δ\dom(rβ ∪¯p0δ)).Then r is a well-defined condition with the property that r ≤rβ, ¯p0δand pβ,δ. So r forces that {β, δ} ∈X/W and that {β, δ} ∈Ki, whichis a contradiction.If Pκ is as in the assumptions of Lemma 44, then Pκ is a c+-c.c.

poset.If GCH holds in the ground model and λ = ω1, then our proof works ifκ ≥ℵ8. One can obtain the consistency of c →(ω −path)2ω1/<ω with asmaller size of the continuum, but this is not essential for our purposes.Todorcevic has for example shown that, adjoining at least ω2 Cohen re-als to a model of the Continuum Hypothesis, produces a model in whichω2 →(ω-path)2ω/<3.We have actually proved something apparently stronger than c →(ω −path)2λ/<ω in V Pκ, namely a relation denoted by c →(ω −path)2λ/<3.We do not know an answer to the following two problems concerningthe ω-path partition relation:Problem 4 Is it for each integer k > 2 consistent, for some infinite car-dinal numbers κ and λ, that κ ̸→(ω-path)2λ/

Sacks orPrikry-Silver forcing, then (b) and (c) of the lemma hold for everyκ.It is also known that adding any number of Sacks or Prikry-Silver reals side-by-side with countable supports to a model of CH,one obtains a model where the collection of meager sets whose Borelcodes are from the ground model, is a cofinal subfamily of JR (see[M]). Since |ωω ∩V | = ℵ1, we get cof(⟨JR⟩, ⊂) = ℵ1 in the forcingextension.References[C] G. Choquet, Lectures in Analysis, Vol.

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[F-K] W.G. Fleissner and K. Kunen, Barely Baire Spaces, FundamentaMathematicae 101 (1978), 229-240.

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[J] T. Jech, Multiple Forcing, Cambridge University Press (1986). [J-M-P-S] W. Just, A.R.D.

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