Construction with opposition: cardinal invariants and games

We consider several game versions of the cardinal invariants t\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathfrak {t}}$$\end{document}, u\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathfrak {u}}$$\end{document} and a\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathfrak {a}}$$\end{document}. We show that the standard proof that parametrized diamond principles prove that the cardinal invariants are small actually shows that their game counterparts are small. On the other hand we show that t<tBuilder\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathfrak {t}}<{\mathfrak {t}}_{Builder}$$\end{document} and u<uBuilder\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathfrak {u}}<{\mathfrak {u}}_{Builder}$$\end{document} are both relatively consistent with ZFC, where tBuilder\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathfrak {t}}_{Builder}$$\end{document} and uBuilder\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathfrak {u}}_{Builder}$$\end{document} are the principal game versions of t\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathfrak {t}}$$\end{document} and u\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathfrak {u}}$$\end{document}, respectively. The corresponding question for a\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathfrak {a}}$$\end{document} remains open.


Introduction
The main purpose of this paper is to propose a measure of robustness of transfinite constructions. The general question is whether a transfinite recursive construction of an object A with a property ϕ can survive outside interference. This is formulated in terms of a transfinite game where two players, the Builder and the Spoiler, take turns in constructing the object A. The Builder tries to make sure the resulting object has property ϕ and the Spoiler wins if the resulting object does not satisfy the property ϕ. The construction envisioned by the Builder is robust if it produces a winning strategy in the game.
Even though the natural scope of such research is much wider, we have restricted ourselves to the case of cardinal invariants of the continuum, and constructions of length ω 1 . For the vast majority of cardinal invariants such considerations are moot as the invariants are super-robust in the sense that the existence of a winning strategy for the Builder is equivalent to the cardinal invariant in question being ℵ 1 . The winning strategy for the Builder would be described by simply taking a witness and playing its elements one by one independently of the moves of the Spoiler. This is the case for instance of all Borel cardinal invariants in the sense of [15]. There are, however, a few cardinal invariants with structure for which such a simplistic strategy fails, e.g. the almost disjointness number a, the tower number t, and the ultrafilter number u. In these games the Builder and the Spoiler agree that they construct an almost disjoint family (resp. decreasing chain) of infinite subsets of ω of size (length) ω 1 , and hence cannot ignore each other's moves, the distinguishing property ϕ being maximality for a and t, and being a reaping 1 family for u.
The starting point for our investigations is the observation that recursive constructions of length ω 1 produced by the (parametrized) ♦ principles tend to be robust in this sense.
The first condition in both definitions defines the structure we mention above, while the second condition is the requirement of maximality. We denote by a the minimal size of an infinite MAD family, and by t the minimal length of a tower. Finally u denotes the minimal character of a non-principal ultrafilter on ω. For more on cardinal invariants of the continuum see e.g. [5].
We have already mentioned that these and similar constructions are robust in the above mentioned sense-the existence of a winning strategy for the Builder in the corresponding game, as we shall see in what follows. Then we shall consider the question of whether the cardinal invariant being ω 1 is sufficient for the existence of a winning strategy for the Builder. We fix the following notation for the rest of the paper: Given an infinite countable ordinal δ, we fix a bijection e δ : ω → δ. We denote by pair(ω 1 ) the countable ordinals of the form β + 2k, with β limit and k ∈ ω, and let odd(ω 1 ) = ω 1 \pair(ω 1 ).
Let us make a simple yet important remark concerning the parametrized ♦principles here: The definition of the function F : 2 <ω 1 → A almost always requires some simple coding. It has to do with the domain of the function. We shall say that the domain consists of pairs (s, X ) where X is a subset of ω (usually infinite) and s is a sequence of subsets of ω of length some countable ordinal α with some structure (e.g. consisting of infinite sets which are almost disjoint or ⊆ * -decreasing) which constitute an approximation to an object we want to construct. The coding can be described as follows: Given such a pair (X , s), where s = s ξ : ξ < α let σ (s,X ) : ω · (1 + α) → 2 defined by σ (s,X ) (n) = 1 if and only if n ∈ X , and σ (s,X ) (ω · (1 + ξ) + n) = 1 if and only if n ∈ s ξ .
For any given α < ω 1 , the set of such σ (s,X ) 's is easily seen to be Borel, and as the values of F outside of this Borel set are irrelevant for our constructions, we can let F outside this set be constant. As guessing happens on a stationary set, we can also ignore the value of F at heights which are not irreducible (i.e. not of the form α = ω · α).

The tower number game
Consider the tower game G t of length ω 1 played as follows: Players Builder and Spoiler take turns playing a ⊆ * -decreasing transfinite sequence Y α : α < ω 1 of infinite sets of ω, the Builder playing at even stages pair(ω 1 ), and the Spoiler playing at odd stages odd(ω 1 ).
The Builder wins the match if Y α : α < ω 1 is a tower; otherwise, the Spoiler wins.
The first instance of the phenomenon discussed in the introduction is the following: , the Builder has a winning strategy in the game G t .
Proof Given an infinite ⊆ * -decreasing sequence s = {Y s ξ : ξ < δ(s)} with δ(s) limit, we will define a strictly increasing sequence {l s i : i ∈ ω} of natural numbers. Fix an increasing sequence {δ i : i ∈ ω} ⊆ δ(s) converging to δ(s). Let We will define F : 2 <ω 1 → 2 using the above described coding. For a decreasing ⊆ * -sequence s = {Y s ξ : ξ < δ(s)} of length an infinite limit ordinal and X ⊆ ω infinite, define F(s, X ) as follows: As the function is defined only using countable intersections and complements using only the fixed sequence {δ i : i ∈ ω} ⊆ δ(s) as a parameter, and since its domain is Borel, it is Borel.
Let g : ω 1 → 2 be a ♦(2, =)-sequence for F. We are going to use g to define a winning strategy for the Builder.
Suppose s = {Y s ξ : ξ < δ(s)} is a partial play of the game with δ(s) an infinite limit ordinal. The Builder is going to choose Y δ(s) as follows: Let s = {Y s ξ : ξ < ω 1 } be a complete match played by the Builder according to the strategy described above. Let X ⊆ ω. Then if δ is an infinite limit ordinal such that F(s δ , X ) = g(δ), it is straightforward to see that X * Y δ = Y s δ (note that δ(s δ ) = δ).
The previous Proposition has non-trivial content as we shall see next.
Theorem 2.1 It is consistent with ZFC that t = ω 1 and the Builder does not have a winning strategy in G t .
Before embarking on the proof, let us do some preparation. Let F be a filter on ω. The Laver-Prikry forcing associated with F, denoted by L F consists of subtrees T ⊆ ω <ω which have a stem σ ∈ T , denoted by stem(T ), such that for every τ ∈ T , either τ ⊆ σ or τ ⊇ σ . Besides, for every τ ∈ T extending σ , the set {n ∈ ω : τ n ∈ T } belongs to F. The order on L F is given by inclusion.
The following lemma is based on ideas of Baumgartner and Dordal [4]. A similar argument appears elsewhere, see e.g. [6, Lemma 41].

Lemma 2.1 L F preserves Y as a tower.
Proof LetẊ be a name for an infinite subset of ω. Without loss of generality, we may assume its increasing enumeration (also denoted byẊ ) dominates the generic Laver real. Fix n ∈ ω. Say that σ ∈ ω <ω favoursẊ (n) = k if given any Define the rank rk n by recursion as follows: Proof Suppose rk n (σ ) is undefined. Build a tree T ∈ L F with stem(T ) = σ such that rk n (τ ) is undefined for all τ ∈ T with τ ⊇ σ. Let S ≤ T be such that S decidesẊ (n), say S Ẋ (n) = k. Let τ = stem(S). Then rk n (τ ) = 0 because τ favoursẊ (n) = k, a contradiction. Fix a pair n, σ such that rk n (σ ) = 1. So σ does not favourẊ (n) = k for any k but {i : σ i favoursẊ (n) = k for some k} belongs to F + . Define a partial function f : ω → ω as follows: dom( f ) = {i : σ i favoursẊ (n) = k for some k} and, for i ∈ dom( f ), let f (i) be some k such that σ i favoursẊ (n) = k. Note that since rk n (σ ) = 0, f −1 ({k}) / ∈ F + for all k ∈ ω. There is α = α(n, σ ) such that f = f α . Let β be larger than all the β α(n,σ ) .
Proof Fix m ∈ ω and T ∈ L F . It suffices to find k > m, k / ∈ Y β , and S ≤ T such that S k ∈Ẋ . Let σ = stem(T ) and n > max{m, |σ |}. In particular, rk n (σ ) > 0. By extending σ if necessary, we may assume rk n (σ ) = 1. By construction, there is Clearly k ≥ n > m, and we are done.
This finishes the proof of the lemma.
Recall yet another version of ♦. For a given uncountable regular cardinal κ and a stationary set E ⊆ κ, we say that the principle ♦ E holds if there is a sequence d γ : γ ∈ E such that for every X ⊆ κ, the set {γ ∈ E : X ∩ γ = d γ } is stationary. Now we are ready to prove the theorem: and CH. Fix a tower Y = (Y α : α < ω 1 ) as above.
Construct a finite support iteration P γ ,Q γ : γ < ω 2 , where the initial segments of the iteration have size at most ℵ 1 . Use the diamond to guess (initial segments of) names of strategies for the Builder. This is a standard argument which has been used a lot, e.g. in [16]; for the reader's convenience we present an outline, following roughly the one in [7, proof of Theorem 8].
Think of the diamond sequence as acting on the product This can be done, because the initial segments P γ of the iteration have size ω 1 , by building up a bijection F : ω 2 → P ω 2 recursively along the definition of the iteration, such that F γ : γ → γ × P γ is a bijection for all γ ∈ E ω 2 ω 1 . Next, again along the definition of the iteration, we produce a namė f ∈ V P ω 2 for a bijection between ω 2 and At stage γ ∈ E ω 2 ω 1 such that S γ is a P γ -name for a subset of γ andḟ [S γ ] ∈ V P γ is a strategy for Builder, we force withQ γ = LḞ whereḞ is constructed fromȦ α andḂ α as above and theḂ α are obtained from theȦ β ,Ḃ β , β < α, using the strategẏ f [S γ ]. In all other cases we letQ γ be Cohen forcing. Force with P ω 2 .
Since towers are preserved in limit steps of finite support iterations (see e.g. [3,4,9]), the lemma implies that Y is still a tower in V P ω 2 . In particular t = ω 1 .

On the other hand, for each strategy =ḟ [T ] of the Builder in
is a strategy in V P γ and was used to construct the filterḞ. Hence there is a game according to which the Builder looses, as witnessed by the LḞ -generic set added in V P γ +1 .
Consider the longer version of the tower game G δ t of length δ played as follows: Players Builder and Spoiler take turns playing a ⊆ * -decreasing transfinite sequence Y α : α < δ of infinite subsets of ω, the Builder playing at even stages pair(δ), and the Spoiler playing at odd stages odd(δ).
The Spoiler wins the match if Y α : α < δ is not a tower; otherwise, the Builder wins.
Given the previous theorem, it is natural to define t Builder as the least ordinal δ such that the Builder has a winning strategy in the game G δ t . The previous result then says t < t Builder is consistent. The following is a special case of a result of Vojtáš [17,Theorem 7]; we include the short proof for the sake of completeness.

Lemma 2.2 t Builder is a regular cardinal.
Proof Let α be minimal such that the Builder has a strategy that makes her win in at most α moves. Let {γ ξ : ξ < c f (α)} be a club subset of α such that for even ξ , γ ξ is also even and γ ξ +1 = γ ξ + 1. We construct a strategy for the Builder that makes her win in at most c f (α) steps such that for each runĀ = {A η : η < ξ} according to of length an even ξ , there is a runB = {B γ : γ < δ ξ } according to of length δ ξ such that B γ η = A η for all η < ξ and Suppose we are at step ξ . If ξ is a limit ordinal, then eitherĀ has no pseudointersection and the Builder already won or, sinceĀ is a cofinal subsequence ofB, we can let (Ā) be (B). If ξ is an even successor, say ζ + 1, consider the corresponding gameB whose final move is B γ ζ = A ζ . Notice that since there is no strategy which makes the Builder win in less than α steps, cannot make the Builder win below the set A ζ in less than α steps. In particular, there must be a game according to and extendingB which still has a move, with the Builder following , at stage γ ξ . LetB be this extension of length γ ξ and let (Ā) be this move (B ).
This describes the strategy . It is clear that the Builder must win after at most c f (α) steps.
We may also define t Spoiler as the supremum of all ordinals δ such that the Spoiler has a winning strategy in the game G δ t . It is easy to see that the Spoiler has no winning strategy in G δ t for δ = t Spoiler moves (for otherwise the game could be continued one further move and would still be winning for the Spoiler). Hence, t Spoiler can be characterized as the least δ such that the Spoiler has no winning strategy in the game G δ t . Again we see (this is a special case of [17,Theorem 6]):

Lemma 2.3 t Spoiler is a regular cardinal.
Proof Suppose t Spoiler = α is minimal such that no strategy of the Spoiler of the game with α moves is winning. Let be a strategy of the Spoiler of the game with c f (α) moves. We need to see that is not winning. As in the previous proof, let {γ ξ : ξ < c f (α)} be a club subset of α such that for even ξ , γ ξ is also even and γ ξ +1 = γ ξ + 1. We shall build a strategy of the Spoiler with α moves such that for is not winning, one such runB is won by the Builder. But then the Builder also wins the corresponding runĀ according to , as required.
As in the previous proof, let Now suppose ξ is even and has been constructed for a runB = {B γ : γ < δ ξ }. LetĀ = {A η : η < ξ} be the corresponding run according to . If ξ is limit, consider the move B γ ξ of the Builder. Let A ξ = B γ ξ be the corresponding move of the Builder in the other game. Then let what tells her to play in the other game. If ξ = ζ + 1 is successor, the last move of the Spoiler was B γ ζ . Note that δ ξ ≤ γ ξ are both even ordinals. So, let ξ be such that δ ξ + ξ = γ ξ . Since ξ < α, the Spoiler has a winning strategy of length ξ below the set B γ ζ (i.e with B γ ζ as the first move). Let in the interval [δ ξ , γ ξ ) be this strategy. LetB = {B γ : γ < γ ξ } be an extension ofB following this strategy. Now continue as in the limit case: let B γ ξ be the next move of the Builder (such a move exists because the strategy of the Spoiler was winning so far); let By modifying the proof of Theorem 2.1 a little we see: Proof We first observe: Lemma 2.4 Assume CH and let be a strategy of the Builder (of length ω 1 ). Also assume there are towers (Y β : β < ω 1 ). Then there is a filter F containing a run of the game according to such that L F preserves all Y β .
Proof To see this simply redo the construction before Lemma 2.1 by diagonalizing against ω 1 towers instead of just one. More explicitly, let Y β = (Y β α : α < ω 1 ) for β < ω 1 . Let (S β : β < ω 1 ) partition ω 1 into sets of size ω 1 . Let {γ β α : α < ω 1 } be the enumeration of S β . Let ( f α : α < ω 1 ) list all partial functions from ω to ω with infinite range. Then construct (A α : α < ω 1 ) and (B α : α < ω 1 ) recursively such that for all α, β < ω 1 , Clearly this can be done. Let F be the filter generated by the sequence and CH. Use the diamond to guess (initial segments of) names of strategies for both the Builder and the Spoiler. Simultaneously construct a finite support iteration P γ ,Q γ : γ < ω 2 and a sequence of (names of) towers Ẏ β : β < ω 2 such that Ẏ β : β ≤ γ ∈ V P γ . At stage γ first consider the (name of the) strategy of the Spoiler handed down by ♦ E ω 2 ω 1 . Since CH still holds while there are 2 ω 1 many games following the strategy, one of these games must be winning for the Builder, that is, there is a towerẎ γ ∈ V P γ that is a run according to the strategy. Now, as in the proof of Theorem 2.1, use the lemma to get a filterḞ containing a run of the game according to the (name of the) Builder's strategy handed down by ♦ E ω 2 ω 1 such thatQ γ = LḞ preserves allẎ β , β ≤ γ . By the argument of Theorem 2.1, a strategy of the Builder of length ω 1 cannot be winning. Similarly, if is a strategy of the Spoiler of length ω 1 , there is γ < ω 2 such that V Pγ is a strategy in V P γ and was guessed by the diamond. This means that the tower Y γ is preserved as a run of the game according to which is won by the Builder.
However we do not know:

Open question 2.1 Is t < t Spoiler consistent?
On the other hand, t Builder ≤ h, where h = min{height(T ) : T ⊆ [ω] ω , * ⊇ is a base tree} 4 is the distributivity number of P(ω)/fin. 5 To see this note that the Builder can simply make sure to play along a branch of the base tree T which, of course, produces a winning strategy. In particular, h = ω 1 is sufficient for the existence of a winning strategy for the Builder in the game G t (of length ω 1 ).
This proof actually gives a little more. Note that in general the Builder has a distinct advantage over the Spoiler in that her moves appear on a closed unbounded subset 4 A base tree is a set T ⊆ [ω] ω which is a tree when ordered by ⊇ * and is such that every element of [ω] ω contains an element of T . The existence of such a tree was proved by Balcar, Pelant and Simon in [2], see also [1]. 5 Remember that [ω] ω , ⊆ * is a preorder. Therefore, the set of its classes of equivalence, P(ω)/fin, defined by X ≡ fin Y if and only if X ⊆ * Y and Y ⊆ * X , defines a partial order P(ω)/fin, ≤ fin , where [X ] fin ≤ fin [Y ] fin if and only if X ⊆ * Y . Given a partial order P, ≤ , we say that a set D ⊆ P is dense if for every p ∈ P, there is q ∈ D such that q ≤ p. A subset set D ⊆ P is open if whenever p ∈ D and q ≤ p, then q ∈ D. As usual, we refer only to P as the partial order if the order is clear from the context. For a partial order P, we define its distributivity number h(P) as the minimum α such that for some collection {D ξ : ξ < α} of open dense sets, its intersection ξ<α D ξ is not dense.
of ω 1 (pair(ω 1 ) ∈ Club(ω 1 ), while odd(ω 1 ) is not stationary). Let G * t be the game in which the players switch places, that is, the Builder plays at odd steps while the Spoiler plays at even steps. It is obvious that a winning strategy of the Builder in G * t gives her a winning strategy in G t as well, while the implication goes the other way round for the Spoiler. Furthermore, the winning strategy described here from h = ω 1 is robust in the sense that it is irrelevant in which order the players play; that is, the latter hypothesis implies a winning strategy for the Builder even in G * t . We shall see below (Corollary 2.2) that ♦(2, =) is not sufficient for this.
Define t * Builder and t * Spoiler similarly as the unstarred versions. The four cardinal numbers t Builder , t Spoiler , t * Builder and t * Spoiler are due to Vojtáš [17] in a more general context, where he also showed they are regular cardinals [17, Theorem 6 and Theorem 7]. Also is obvious. A straightforward modification of the proof of Theorem 2.2 actually shows the consistency of t Builder > t * Spoiler . As in Question 2.1, we do not know whether t * Spoiler > t is consistent.
The following lemma is a special case of a result by Foreman [10].
Lemma 2.5 t * Builder = h. Proof This is immediate using [10, Theorem on page 718] and realizing that given a cardinal λ, the Builder has a winning strategy in λ steps in the game G * t if and only if I has a winning strategy in the game G I I λ + played in P(ω)/fin described in [10]. By the above discussion, both ♦(2, =) and h = ω 1 imply the existence of a winning strategy for the Builder in the game G t in ω 1 many steps. Both are consequences of CH. The two statements are independent, however: in the Mathias model, ♦(2, =) holds [15, Theorem 6.6] and h > ω 1 , while in a model of Judah and Shelah [13], h = ω 1 and ♦(2, =) fails 6 . In particular we have:

Corollary 2.2 It is consistent that ♦(2, =) holds and the Builder has no winning strategy in G *
t . In particular it is consistent that t * Builder > t Builder . Proof As remarked ♦(2, =) and h = c = ω 2 hold in the Mathias model. By Proposition 2.1 and Lemma 2.5, t Builder = ω 1 and t * Builder = ω 2 follow. Another classical upper bound of t is the additivity add(M) of the meager ideal M, that is, the least κ such that there is a family of κ many meager sets whose union is not meager. Since, as observed in the Introduction, cardinals like add(M) are equal to their game versions, one might conjecture that t Builder ≤ add(M) holds in ZFC. However, this is not what the proof of t ≤ add(M) gives for the latter uses towers of dense sets of rationals and not just of arbitrary sets of natural numbers. And, in fact, we show the following: Before starting with the proof we review some notions and some facts. Recall that a non-principal ultrafilter U on ω is Ramsey if for every partition {A n : n ∈ ω} of ω such that A n / ∈ U for all n ∈ ω, there is X ∈ U such that X ∩ A n has one element for all n ∈ ω. Say a function ϕ : ω → [ω] <ω is a slalom if |ϕ(n)| ≤ n + 1 for all n ∈ ω. A forcing notion P has the Laver property if given any condition p ∈ P, any function h ∈ ω ω and any P-nameḟ for a function bounded by h, there are q ≤ p and a slalom ϕ such that q ∀n (ḟ (n) ∈ ϕ(n)). A forcing with the Laver property does not add Cohen reals (see Lemma 7.2.3 in [3]) and thus in particular preserves the additivity of the meager ideal, that is, if add(M) = ω 1 holds in the ground model, it still holds in the generic extension. Like standard Mathias forcing (Lemma 7.2.2 and Corollary 7.4.7 in [3]) used in the proof of the previous theorem, Mathias forcing with a Ramsey ultrafilter U, which is forcing equivalent to Laver forcing L U with U (see e.g. Theorem 1.20 in [12]), has the Laver property. Furthermore, the Laver property is preserved in countable support iterations (Theorem 6.3.34 in [3]).

Proof of Theorem 2.3
As in the proof of Theorem 2.1 we assume ♦ E ω 2 ω 1 and CH. Construct a countable support iteration P γ ,Q γ : γ < ω 2 . We use the diamond again to guess (initial segments of) names of strategies for the Builder. Again this can be done, exactly like in the proof of Theorem 2.1, because the initial steps P γ of the iteration have (a dense subset of) size ω 1 . At stage γ consider (the name of) Builder's strategẏ handed down by ♦ E ω 2 ω 1 . As in the argument before Lemma 2.1, we can construct, in V P γ , a run of the game according to˙ such that the ω 1 -sequence of the sets played generates a Ramsey ultrafilterU. Now letQ γ = LU . Force with P ω 2 . By the discussion in the paragraph preceding the proof, the whole iteration has the Laver property, and add(M) = ω 1 thus follows.
To see t Builder = ω 2 , assume is a strategy of Builder for a game of length ω 1 . By ♦ E ω 2 ω 1 there is γ < ω 2 such that V Pγ is a strategy and was used to construct the ultrafilter U. Hence there is a game following which the Builder looses, as witnessed by the L U -generic set added in V P γ +1 .
Note that this gives an alternative proof of Theorem 2.1. However, the original argument is more direct in that it uses less black-boxed forcing theory. Also, in Theorem 2.1, we additionally have the consistency of t < t Builder = add(M).
The order relationship between the cardinals we considered in this section can be summarized in the following diagram.

The ultrafilter number game
Recall that a filter F on ω is a P-filter if for each countable collection {Y n : n ∈ ω} ⊆ F there is a Y ∈ F such that Y ⊆ * Y n for every n ∈ ω. A non-principal ultrafilter F on ω is called a P-point if it is a P-filter.
The ultrafilter game G u is played as before, the Builder and the Spoiler taking turns constructing a ⊆ * -decreasing sequence U α : α < ω 1 (the Builder playing at pair(ω 1 )-stages, while the Spoiler plays at odd(ω 1 )-stages).
The difference is in how we declare a winner. The Builder now has a harder task as she wins the match if the filter generated by {U α : α < ω 1 } is an ultrafilter; otherwise, the Spoiler wins.
Again, the proof of the following result mimicks closely the proof of Theorem 7.8 in [15]. We include it for the benefit of the reader.

Proposition 3.1 ♦ (r) implies the Builder has a winning strategy in the game G u .
Proof For a ⊆ * -decreasing infinite sequence s = {U s ξ : ξ < δ(s)}, we define the strictly increasing sequence {k s i : i ∈ ω} ⊆ ξ<δ(s) U s ξ as follows: Remember that we have fixed a bijective function e δ : ω → δ for every infinite ordinal δ < ω. Let Given C ⊆ ω and an infinite ⊆ * -decreasing sequence s, we define a Borel map F as follows: F(s, C) = {i ∈ ω : k s i ∈ C} if {i ∈ ω : k s i ∈ C} is infinite, and F(s, C) = {i ∈ ω : k s i / ∈ C} otherwise. The function F uses the coding described at the end of the introduction, and is Borel since its domain is Borel and it is defined by countable boolean operations using e δ as a parameter.
Let g be the respective ♦(r)-guessing function for F. We will show that g defines a winning strategy for the Builder as follows: If s = {U s ξ : ξ < δ(s)} is a partial match with δ(s) even, let U δ(s) = {k s i : i ∈ g(δ(s))}. It is not difficult to see that any complete match s = {U s ξ : ξ < ω 1 } according to the strategy defined by g is a ⊆ * -decreasing sequence. It is also straightforward to show that the set F s = {X ∈ [ω] ω : ∃δ < ω 1 (U s δ ⊆ * X )} is a filter. We are done if F s is an ultrafilter. Let C ⊆ ω. Since g is a ♦(r)-sequence, we can find δ < ω 1 such that either |g(δ) ∩ F(s δ , C)| < ℵ 0 or |g(δ)\F(s δ , C)| < ℵ 0 .
Note that it was enough that the set of guesses of the diamond sequence was just non-empty. It is a simple exercise left to the reader to show that Lemma 3.1 CH implies that the Builder has a winning strategy in G u .
In fact, the stronger statement that the Builder has a winning strategy also in the game G * u where the Builder and the Spoiler switch places easily follows from CH. Since CH does not imply ♦(r) by Proposition 8.2 and Theorem 8.3 in [15], we have the following:

Corollary 3.1 The Builder having a winning strategy in G u does not imply ♦(r).
Again, we will show that all of this is not gratuitous. Theorem 3.1 u = ω 1 does not imply that the Builder has a winning strategy in the game G u .
Rather than constructing an ad hoc forcing model for this, we show that this holds in a model constructed by Shelah in [16, Chapter XVIII, Section 4]. We shall review some standard facts about ultrafilters first. Given two ultrafilters U, V on ω, we recall the Rudin-Keisler order ≤ RK defined as follows: U ≤ RK V if and only if there is a function f : ω → ω such that U = {X ∈ ω : ∃Y ∈ V( f [Y ] ⊆ X )}, and they are RK -equivalent, denoted by U ≡ RK V if such f exists which is, moreover, bijective. We recall the following fact, which shows that Ramsey ultrafilters are ≤ RK -minimal: Proof of Theorem 3.1 Let V | CH + 2 ω 1 = ω 2 , let P ω 2 be the countable support iteration used by Shelah to construct a model with a unique P-point ([16, Chapter XVIII, Theorem 4.1]), and let G be P ω 2 -generic.
We shall show that V [G] is the model we need. We will be able to deduce this from the following three facts which hold there: Now as V [G α ] | CH, we may list all strictly increasing functions in ω ω of V [G α ] as {F ξ : ξ < ω 1 }. Next, for ξ < ω 1 , define a function G ξ ∈ ω ω such that whenever F ξ (m) ≤ n < F ξ (m + 1), then G ξ (n) = m. Note that we have G ξ (n) ≤ n ≤ F ξ (n) for every n. Since U 0 generates a Ramsey ultrafilter in V [G α ] by (1), for each ξ there is A ξ ∈ U 0 such that the intervals [G ξ (n), F ξ (n)) for n ∈ A ξ are pairwise disjoint. (Indeed, define an interval partition (I k : k ∈ ω) of ω such that whenever G ξ (n) ∈ I k then F ξ (n) ∈ I k ∪ I k+1 . Since U 0 is Ramsey, there is A ∈ U 0 such that |A ∩ I k | ≤ 1 for all k ∈ ω. By further pruning A to A ξ , we may assume it intersects either only intervals I k such that k ≡ 0 mod 3 or k ≡ 1 mod 3 or k ≡ 2 mod 3. It is easy to see A ξ is as required.) Now, still in V [G α ], play a game (U ξ : ξ < ω 1 ) in which Builder follows the strategy 0 while Spoiler plays sets U 2·ξ +1 such that To see this is possible fix ξ < ω 1 . If U 2·ξ meets only finitely many of the intervals is still infinite. We then let This completes the construction. As the strategy 0 is winning in V [G α ], the sequence of U ξ 's produces a P-point V in V [G α ]. It suffices to show that V does no longer generate a P-point in V [G]. 7 For if this is the case, the sequence of U ξ 's will remain a -legal play in V [G], but Spoiler will win the game, a contradiction.
By (2), all P-points are RK-equivalent to U 0 in V [G]. Hence it suffices to show that V cannot be RK-equivalent to U 0 in V [G]. Suppose it were, and f : ω → ω is the bijection witnessing this. Since, by (3), the extension V [G] is ω ω -bounding over V [G α ], there is ξ such that both f and f −1 are everywhere dominated by F ξ , more explicitly, f (n) < F ξ (n) and f −1 (n) < F ξ (n) for all n. Note that the latter implies that f (n) ≥ G ξ (n) for all n, that is, f (n) ∈ [G ξ (n), F ξ (n)) for all n. It then follows by ( ) that f −1 [U 2·ξ +1 ] = {n ∈ ω : f (n) ∈ U 2·ξ +1 } does not belong to U 0 , and f cannot witness RK-equivalence, the final contradiction.
Let us state the following here explicitly: Open question 3.1 Does the Builder have a winning strategy in the game G u if and only if she has a winning strategy in the game G * u ? It would be tempting to define now cardinals u Builder and u Spoiler as we did in Section 2 for the generalized tower game. This, however, is problematic, for the following reason. Consider the Cohen model, that is, the model obtained by adding at least ω 2 Cohen reals over a model of CH. In this model, all ⊆ * -decreasing sequences have length some ordinal γ < ω 2 while on the other hand u = c ≥ ω 2 . This means that the game G u is always won by the Spoiler, no matter what its length is. The reason for this problem is that a win of the Builder in G u produces a P-point generated by a decreasing chain and not just an arbitrary ultrafilter.
So let us consider the modified ultrafilter game G u in which the Builder and the Spoiler take turns in building a filter base {U α : α < ω 1 }, with the Builder playing at even steps. The Builder wins again if the filter generated by {U α : α < ω 1 } is an ultrafilter; otherwise the Spoiler wins. G * u is defined similarly, with the players switching places. It turns out that for plays of length ω 1 these games are equivalent to the original ones, in the following sense. Proof 1. First assume is a winning strategy of the Builder in G u . We construct a strategy of the Builder in G u by associating with each gameĀ = {A ξ : ξ < ω 1 } according to a gameC = {C ξ : ξ < ω 1 } according to with A ξ = C ξ for even ξ . This means that if the Builder winsC then she also winsĀ and, thus, is a winning strategy. If ξ = ζ + 1 is odd, we let C ξ := A ξ ∩ C ζ and note that this set must be infinite because C ζ = A ζ and the players build a filter base in G u . Also C ξ is a legal move of the Spoiler in G u . For even ξ , simply let A ξ = (Ā) := (C) = C ξ . Again this is clearly a legal move of the Builder in G u . Now assume is a winning strategy of the Builder in G u . Construct a strategy of the Builder in G u by associating with each runC = {C ξ : ξ < ω 1 } according to a runĀ = {A ξ : ξ < ω 1 } according to with A ξ = C ξ for odd ξ . If ξ is odd, let A ξ := C ξ and note this is a legal move for the Spoiler in G u . For even ξ let C ξ = (C) be a pseudointersection of the C ζ for ζ < ξ and A ξ = (Ā). Such a pseudointersection exists because these sets form a countable filter base. Clearly, if the Builder winsĀ, she also winsC.

Similar.
This lemma should be thought of as saying that producing an ω 1 -generated ultrafilter by a game is equally difficult as producing a P-point generated by a ⊆ * -decreasing ω 1 -chain. It is unknown, however, whether u = ω 1 implies the existence of an ω 1generated P-point. 8 Now consider the game G u of arbitrary length and define u Builder and u Spoiler as in the previous section: the former is the least ordinal α such that the Builder has a strategy 8 One may also consider the same games in the context of the previous section: declare the Builder the winner if the sequence {U α : α < ω 1 } has no pseudointersection. Since these games are naturally related to the pseudointersection number p, denote them by G p and G * p . The analogue of Lemma 3.2 obviously holds: the Builder has a winning strategy in G p iff she has a winning strategy in G t , and similarly for the starred games. This can be seen as the game-theoretic version of the classical result stating that p = ω 1 iff t = ω 1 (see e.g. Theorem 6.25 in [5]). The much deeper p = t was proved by Malliaris and Shelah [14]. that makes her win in G u in at most α many steps, while the latter is the supremum of all ordinals α such that the Spoiler has a winning strategy in the game G u with α moves. Clearly u ≤ u Spoiler ≤ u Builder and Theorem 3.1 says that u < u Builder is consistent. Apart from that we know little: Open question 3.2 1. Is u < u Spoiler consistent? Is u Spoiler < u Builder consistent? 2. Are u Builder and u Spoiler cardinals?
Finally note that if we also consider G * u of arbitrary length and the corresponding ordinals, we still have: Proof To see for example the former, let be a winning strategy of the Builder of length α = u * Builder in G * u . We produce a winning strategy of the same length in G u such that whenever {A γ : γ < α} is a run according to then {B γ : γ < α} is a run according to with B γ +2 = A γ +1 for all γ and B γ +1 ∩ B γ = A γ for limit γ . Clearly this works.

The maximal almost disjoint number game
The last example we consider here is the maximal almost disjoint game G a , which is played as follows. To avoid trivialities, it starts by fixing a partition {A n : n ∈ ω} of ω into infinite pieces, and then the Builder and the Spoiler take turns extending it to an AD family {A α : α ≤ β} (the Builder playing at stages in pair(ω 1 ), while the Spoiler plays at ordinals in odd(ω 1 )).
The Builder wins the match if the family {A α : α < ω 1 } is a maximal almost disjoint family; otherwise, the Spoiler wins.
We could also consider the game G * a played according to the same rules but the Spoiler playing at pair(ω 1 ), while the Builder plays at odd(ω 1 ). However, in this case it is easy to see that the two games are equivalent: Proof First assume is a winning strategy of the Builder in G a . We construct a strategy of the Builder in G * a by associating with each gameĀ = {A ξ : ξ < ω 1 } according to a gameB = {B ξ : ξ < ω 1 } according to such that A ξ ∪ A ξ +1 = B ξ ∪ B ξ +1 for all even ordinals ξ . Thus, since the Builder winsB, she must also winĀ, and the strategy is winning. At even ξ , let B ξ = (B ξ ) be the move of the Builder according to . Let A ξ be an arbitrary move of the Spoiler in G * a . Next choose A ξ +1 almost disjoint fromĀ ξ +1 such that B ξ ⊆ A ξ ∪ A ξ +1 and (A ξ ∪ A ξ +1 ) \ B ξ is infinite. This is clearly possible by the inductive assumption on the sequencesĀ ξ andB ξ . Let (Ā ξ +1 ) = A ξ +1 and put B ξ +1 = (A ξ ∪ A ξ +1 ) \ B ξ . Note that this is a legal move of the Spoiler in G a . This clearly works. Now assume is winning for the Builder in G * a and construct winning for her in G a . This is almost the same except that this time, when associating with the -gameĀ = {A ξ : ξ < ω 1 } the -gameB = {B ξ : ξ < ω 1 }, we guarantee that A ξ ∪ A ξ +1 = B ξ ∪ B ξ +1 for all odd ordinals ξ and A ξ = B ξ for all limit ordinals ξ . Details are left to the reader. : k ∈ ω} and define F as follows: Let g : ω 1 → ω ω be a ♦(b)-sequence for F. Without loss of generality, g(δ) is a strictly increasing function for every δ < ω 1 .
We show that g allows us to construct a winning strategy for the Builder as follows: Let s = {A s ξ : ξ < δ(s)} be a partial match of the game G a with δ = δ(s) ∈ pair(ω 1 We show that this is a winning strategy. Let s = {A s ξ : ξ < ω 1 } be a complete match where the Builder played according to the strategy defined by g. We show that s is maximal. Let B ∈ [ω] ω . Consider f ∈ 2 ω 1 coding (B, s), i.e. f (n) = 1 iff n ∈ B, and f (ω · (1 + ξ) + n) = 1 iff n ∈ A s ξ . We should find δ < ω 1 such that B ∩ A s δ is infinite. Aiming towards a contradiction assume that it is not the case, that is {B}∪{A s ξ : ξ < ω 1 } is an AD family, and for every indecomposable ordinal δ (1)-(3) are satisfied. Let δ be an indecomposable ordinal where g(δ) guesses f , so in particular, F(s, B) * g(δ).
Let {i k = i s δ ,B k : k ∈ ω} be the increasing enumeration of I (s, B). For k ∈ ω, let l k = F(s, B)(k), i.e.
Observe that the family {A s e δ (i) \ j<i A s e δ ( j) : i ∈ ω} is disjoint, so the application k → l k is injective. Since we have F(s, B) * g(δ), the set is infinite. It is enough to show X ⊆ A s δ . Indeed let l k ∈ X . Then l k < g(δ)(k) ≤ g(δ)(i k ) and so Since g(δ) is increasing we see that for all i ≥ i k , This implies that l k ∈ A. In particular, A is infinite and A s δ = A. Hence X ⊆ A s δ follows.
An even simpler task is to show that Lemma 4.2 If CH holds, then the Builder has a winning strategy in G a .

Fact 4.1 Any infinite countable almost disjoint sequence can be extended.
If A ξ : ξ < α is a partial match for α an infinite limit ordinal, using Fact 4.1 let the Builder play any infinite set A α extending the sequence.
Let A ξ : ξ ≤ α be a partial match of infinite length, where the Spoiler has played A α with α ∈ odd(ω 1 ). If there is ξ ≤ α such that A ξ ∩ X α is infinite, then let the Builder play any A α+1 almost disjoint from the previous ones using again Fact 4.1. Otherwise, let A α+1 = X α . It is clear now that any complete match A ξ : ξ < ω 1 defines a maximal almost disjoint family.
Since CH does not imply ♦(b) by Proposition 8.2 and Theorem 8.3 in [15], we have the following: We have still the following open question: Open question 4.1 Does a = ω 1 imply the Builder has a winning strategy in G a ?
As in the preceding sections, we may now consider longer games and the corresponding ordinals a Builder and a Spoiler . Obviously a ≤ a Spoiler ≤ a Builder , and a more general version of the preceding question asks whether these three numbers are equal. As for u, we even do not know whether a Builder and a Spoiler necessarily are cardinals.
Also, if we define t N oSpoiler as the minimum ordinal where the Spoiler does not have a winning strategy in the game G t of length α, we have mentioned in Section 2 that t N oSpoiler = t Spoiler . With similar definitions, we do not know whether u N oSpoiler = u Spoiler or a N oSpoiler = a Spoiler hold.