A minicourse on topological games

Topology and its Applications 258 (2019) 305–335

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Topology and its Applications www.elsevier.com/locate/topol

A minicourse on topological games Leandro F. Aurichi a,1 , Rodrigo R. Dias b,∗ a

Instituto de Ciências Matemáticas e de Computação, Universidade de São Paulo, Caixa Postal 668, São Carlos, SP, 13560-970, Brazil b Centro de Matemática, Computação e Cognição, Universidade Federal do ABC, Avenida dos Estados, 5001, Santo André, SP, 09210-580, Brazil

a r t i c l e

i n f o

Article history: Received 22 October 2018 Received in revised form 27 November 2018 Accepted 28 November 2018 Available online 5 March 2019 Dedicated to Marion Scheepers on the occasion of his 60th birthday

a b s t r a c t The purpose of these notes is to introduce the basics of topological games and some of the techniques used in the field to the uninitiated reader. We deal mostly with games involving covering properties and variations of countable tightness, with some emphasis on selective topological games — such as the Rothberger game G1 (O, O) and the Menger game Gfin (O, O). We also illustrate how topological games can be used in contexts that are not game-theoretical in essence, such as preservation of topological properties in products and the D-space problem. © 2019 Elsevier B.V. All rights reserved.

Keywords: Topological game Undeterminacy Selection principle Banach–Mazur game Rothberger space Menger space Luzin set Suslin tree Countable tightness Topological product D-space

1. Introduction These are lecture notes from the minicourse “Introduction to selection principles and topological games” given by the authors at the conference Frontiers of Selection Principles, held in Warsaw in 2017. It also includes some more notes from a second minicourse given by the first named author in the Set Theory & Topology section of the Winter School in Abstract Analysis 2018. As the minicourse was targeted at an audience with no previous knowledge on topological games, we made an effort to start with the very basics and explain the arguments in detail (at least in the beginning). * Corresponding author. 1

E-mail addresses: [email protected] (L.F. Aurichi), [email protected] (R.R. Dias). Supported by FAPESP (2017/09252-3).

https://doi.org/10.1016/j.topol.2019.02.057 0166-8641/© 2019 Elsevier B.V. All rights reserved.

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At the same time, we tried to be concise in order to cover as broad a variety of topics as the time frame permitted. That said, these notes are by no means intended to exhaust the many different aspects of the field, or even to contain “the more important stuff” concerning topological games. Rather, our selection of topics may be somewhat biased, as we ended up choosing properties and games with which we had already worked previously. So we left out a great deal of important concepts and games that are no less relevant to the field than those appearing in these notes. We hope that this text may help newcomers to the field to get a flavour of the main ideas and some of the techniques, and that this will be a first step for them towards exploring the bibliography on topological games and getting in touch with the (very many) topics we have not covered. Here we must add that the quite arduous task of compiling the existing bibliography on topological games has been undertaken by R. Telgársky, who kept an updated online register of the publications on the field up until his passing in 2014; a slightly expanded version of the bibliography he maintained can be found in [11]. These notes are organized as follows. In Section 2, we start with the basic definitions about games and give some examples regarding games involving covering properties. Section 3 is dedicated to the Banach–Mazur game and its relations with the Baire property. In Section 4, we start with an example of a space built from a Suslin tree as a motivation to revisit some of the covering games from Section 2. In Section 5, we give more examples, this time with games involving tightness properties. Finally, in Section 6, we illustrate the use of topological games to obtain results involving properties that are not game-formulated, with an application to van Douwen’s question on D-spaces. 2. Basic notation and some examples In what follows, the set of all natural numbers (including 0) will be denoted by ω. For every set A, the  set of all finite sequences of elements of A (that is, n∈ω An ) will be denoted by A<ω . In all of the games considered in this text: • there will be two players, Alice and Bob, playing against each other; • there will be ω many innings — meaning that the innings will be numbered 0, 1, 2, 3, . . . , and that for each n ∈ ω there will be an n-th inning in the play; • at the end of each complete play of the game, either Alice or Bob will be the winner — there are no draws. 2.1. The point-open game Let us begin with the following game, which is quite simple to introduce: Definition 2.1 (Telgársky [43], Galvin [15]). The point-open game on a set X ⊆ R is played according to the following rules: • In each inning n ∈ ω, Alice picks a point xn ∈ X, and then Bob chooses an open set Un ⊆ R with xn ∈ Un . • At the end of the play x0 , U0 , x1 , U1 , x2 , U2 , . . . , xn , Un , . . . , the winner is Alice if X ⊆

 n∈ω

Un , and Bob otherwise.

Let us illustrate this game by analyzing what happens in some particular cases.

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Example 2.2. Is X is countable — say, X = {an : n ∈ ω} — then Alice can win the play easily: all she has to do to guarantee the victory is to play, in each inning n ∈ ω, the point an . Example 2.3. If X = R, then Bob can be sure to win: all he has to do is play, in each inning n ∈ ω, an  open interval of length 21n . At the end of the play, the open set n∈ω Un will be a union of intervals whose  lengths add up to n∈ω 21n = 2, hence cannot have covered all of the real line. A closer look at the previous example shows that, more generally, we have the following: Example 2.4. If X ⊆ R is not a (Lebesgue) measure zero set, then Bob has a winning strategy in the point-open game in X. We will soon elaborate further on this last example. For the time being, we just wanted to illustrate that, by defining a game played on a set X, one introduces the properties “Alice has a winning strategy in the game played on X” and “Bob has a winning strategy in the game played on X”, which the set X might have or fail to have. So, in case you were thinking “what’s the point in playing a game of infinite length, since you’ll never get to the end of the play?”, the answer is: the game is not defined with the purpose of being played, but rather with the purpose of introducing the properties of Alice or Bob having (or not) a winning strategy in the game — and these properties can lead to some very interesting features of the set X on which the games are played, as we will try to show throughout these notes. In order to do so, we must define more precisely what we mean by a winning strategy in this game. Intuitively, a strategy is a way of playing the game. This means that a fixed strategy for one of the players must inform what decision should be taken for each possible situation that this player might encounter during a play of the game. Thus, we may think of a strategy as a previously defined answer to each possible move of the other player, for each possible sequence of moves preceding the inning at hand. Here we are assuming that the game at hand is a game of perfect information, meaning that, whenever a player must define their next move, it is assumed that they know all the previous moves made so far in the play. This amounts to: Definition 2.5. Let τ = {U ⊆ R : U is open}. • A strategy for Alice in the point-open game on a set X ⊆ R is a function ϕ : τ <ω → X. • A strategy for Bob in the point-open game on a set X ⊆ R is a function ψ : X <ω → τ such that, for all x0 , . . . , xn  ∈ X <ω \ {}, we have xn ∈ ψ(x0 , . . . , xn ).    =Un

So, from the point of view of strategies for Alice and Bob respectively, a play of this game should look like this: Alice x0 := ϕ() x1 := ϕ(U0 ) x2 := ϕ(U0 , U1 ) x3 := ϕ(U0 , U1 , U2 ) .. . xn := ϕ(U0 , . . . , Un−1 ) xn+1 := ϕ(U0 , . . . , Un−1 , Un ) .. .

Bob U0 U1 U2 U3 .. . Un Un+1 .. .

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Alice x0 x1 x2 x3 .. . xn .. .

Bob U0 := ψ(x0 ) U1 := ψ(x0 , x1 ) U2 := ψ(x0 , x1 , x2 ) U3 := ψ(x0 , x1 , x2 , x3 ) .. . Un := ψ(x0 , . . . , xn ) .. .

As you may have noticed, we wrote e.g. ϕ(U0 , U1 , U2 ) instead of ϕ(x0 , U0 , x1 , U1 , x2 , U2 ) — which could seem to be a more natural definition, since the complete history of the play so far is x0 , U0 , x1 , U1 , x2 , U2 . The reason is that we are assuming that Alice plays according to ϕ in all of the innings; this is the meaning of playing according to a strategy fixing in advance — otherwise, it would mean that Alice changed her strategy halfway through the play, which defeats the purpose of choosing a strategy to use in the course of the play. Thus, it suffices to know what Bob’s moves were, and the rest of the play (i.e. Alice’s moves) can be recovered by making use of that assumption — so the complete history of the play up to that point would, in fact, be ϕ( ), U0 , ϕ(U0 ), U1 , ϕ(U0 , U1 ), U2 . This observation allows us to define strategies the way we did, which makes things simpler. Now we want to define a winning strategy as a strategy that leads to a victory — no matter how the other player has chosen their moves. Definition 2.6. Let X be a subset of R. • A strategy ϕ : τ <ω → X for Alice in the point-open game on X is a winning strategy for Alice if, for every sequence Un : n ∈ ω of open subsets of R such that ∀n ∈ ω (ϕ(U0 , . . . , Un−1 ) ∈ Un ), we have    =xn  X ⊆ n∈ω Un . • A strategy ψ : X <ω → τ for Bob in the point-open game on X is a winning strategy for Bob if, for  every sequence xn : n ∈ ω of points of X, we have X  n∈ω ψ(x0 , . . . , xn ).    =Un

This leads to a classification of games of this kind according to their determinacy: Notation 2.7. If Player is a player of a game G, we denote by Player ↑ G the fact that Player has a winning strategy in G, and by Player ↑ G the fact that Player does not have a winning strategy in G. Definition 2.8. A game G is • determined if either Alice ↑ G or Bob ↑ G; • undetermined otherwise — i.e. if Alice ↑ G and Bob ↑ G. Note that Alice ↑ G and Bob ↑ G cannot both hold: if ϕ and ψ were winning strategies in G for Alice and Bob respectively, then a play of G in which Alice plays according to ϕ and Bob plays according to ψ would have to end in a win for both Alice and Bob, which is a contradiction. One might wonder whether it could actually be the case that none of the players have a winning strategy. Indeed, if the game has finite length — i.e. finitely many innings —, then that could not happen:

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Remark 2.9 (Zermelo [46], Kalmár [21]). If G is a game of length N (for some N ∈ ω) that allows no draws, then Alice ↑ G means that ∃a1 ∀b1 ∃a2 ∀b2 . . . ∃aN ∀bN (Alice wins the play a1 , b1 , . . . , aN , bN ), so Alice ↑ G means that ∀a1 ∃b1 ∀a2 ∃b2 . . . ∀aN ∃bN (Bob wins the play a1 , b1 , . . . , aN , bN ), i.e. Bob ↑ G. However, that may well happen for infinite games, as we shall see soon. 2.2. The Rothberger game Let us now discuss another game. Definition 2.10 (Galvin [15]). The Rothberger game on a topological space X is played according to the following rules: • In each inning n ∈ ω, Alice chooses an open cover Un of X, and then Bob picks an open set Un ∈ Un . • At the end of the play U0 , U0 , U1 , U1 , U2 , U2 , . . . , Un , Un , . . . , the winner is Bob if X ⊆

 n∈ω

Un , and Alice otherwise.

In the Rothberger game played on the space X (with topology τ ):  • a strategy for Alice is a function ϕ : τ <ω → O, where O = {U ⊆ τ : X = U}; • a strategy for Bob is a function ψ : O<ω → τ with ψ(U0 , . . . , Un ) ∈ Un for each U0 , . . . , Un  ∈    =Un

O<ω \ {}. We also recall:

Definition 2.11 (Rothberger [32]). A topological space X is Rothberger if, for every sequence Vn : n ∈ ω  of open covers of X, it is possible to choose Un ∈ Vn for each n in such a way that X ⊆ n∈ω Un . Note that the following chain of implications always holds: X is countable ⇓ Bob ↑ Rothberger(X) ⇓ Alice ↑ Rothberger(X) ⇓ X is Rothberger.

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We will now relate the two games we have considered thus far. In order to do that, we need the following concept: Definition 2.12. Two games G and G are dual if • Alice ↑ G ⇔ Bob ↑ G and • Bob ↑ G ⇔ Alice ↑ G . The rather astounding result is then: Theorem 2.13 (Galvin [15]). The point-open game and the Rothberger game are dual on every topological space. Proof. We will start with the implication Alice ↑ point-open(X) ⇒ Bob ↑ Rothberger(X). Let ϕ : τ <ω → X be a winning strategy for Alice in point-open(X). We will use ϕ to describe a winning strategy for Bob in Rothberger(X): • Let U0 be Alice’s initial move in Rothberger(X). Bob will respond by choosing U0 ∈ U0 such that ϕ() ∈ U0 .    =x0

• If Alice’s next move is U1 , let Bob’s move be U1 ∈ U1 with ϕ(U0 ) ∈ U1 .    =x1

• If Alice plays U2 in the next inning, Bob will pick U2 ∈ U2 with ϕ(U0 , U1 ) ∈ U2 , and so forth.    =x2

• In each inning n ∈ ω, Bob will choose some element Un of Alice’s move Un that covers the point ϕ(U0 , . . . , Un−1 ).    =xn

By proceeding in this fashion, the sets U0 , U1 , . . . , Un , . . .  played by Bob in this play of Rothberger(X) are exactly those played by Bob in the following play of point-open(X):  ϕ() , U0 , ϕ(U0 ), U1 , ϕ(U0 , U1 ), U2 , . . . , ϕ(U0 , . . . , Un−1 ), Un , . . .              =x0



=x1

=x2

=xn

But the above is a play of point-open(X) in which Alice has made use of the winning strategy ϕ, hence n∈N Un = X — which means that U0 , U0 , U1 , U1 , U2 , U2 , . . . , Un , Un , . . . 

is a play of Rothberger(X) in which Bob is the winner. We now prove that Bob ↑ Rothberger(X) ⇒ Alice ↑ point-open(X). Let ψ : O<ω → τ be a winning strategy for Bob in Rothberger(X). We will define a winning strategy for Alice in point-open(X). Claim 0. There is x0 ∈ X such that, for every open neighbourhood U of x0 , there is U ∈ O such that U = ψ(U). Proof of Claim 0. Suppose that every x ∈ X has an open neighbourhood Ux that is not of the form ψ(V) for V ∈ O. Then V  = {Ux : x ∈ X} ∈ O, so ψ(V  ) = Ua for some a ∈ X. This contradicts the choice of Ua . 2

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Let then Alice begin point-open(X) by playing the point x0 ∈ X given by Claim 0. Bob will respond by choosing an open set U0 ⊆ X with x0 ∈ U0 . By Claim 0, U0 = ψ(V0 ) for some V0 ∈ O. Claim 1. There is x1 ∈ X such that, for every open neighbourhood U of x1 , there is V ∈ O such that U = ψ(V0 , V). Proof of Claim 1. Pretty much the same thing we did in Claim 0. 2 Now let Alice play x1 ∈ X in the next inning of point-open(X). Bob will respond with an open set U1 ⊆ X such that x1 ∈ U1 . By Claim 1, U0 = ψ(V0 , V1 ) for some V1 ∈ O. Claim 2. There is x2 ∈ X such that, for every open neighbourhood U of x2 , there is V ∈ O such that U = ψ(V0 , V1 , V). Proof of Claim 2. Essentially the same.

2

In general, we have: Big Claim. For each V0 , . . . , Vn−1  ∈ O<ω , there is x ∈ X such that, for every open neighbourhood U of x, there is V ∈ O such that U = ψ(V0 , . . . , Vn−1 , V). Proof of the Big Claim. Analogous to the previous ones. 2 By successively applying the Big Claim, we simultaneously obtain: • a play of point-open(X) x0 , U0 , x1 , U1 , . . . , xn , Un , . . .  • and a play of Rothberger(X) V0 , ψ(V0 ) , V1 , ψ(V0 , V1 ), . . . , Vn , ψ(V0 , . . . , Vn ), . . . .          =U0

=U1

=Un

 Since ψ is a winning strategy for Bob in Rothberger(X), it follows that n∈N Un = X — which means that Alice wins the play of point-open(X) above. The proof of the equivalence Bob ↑ point-open(X) ⇔ Alice ↑ Rothberger(X) is left to the reader. 2 Let us now get back to what we were (implicitly) talking about in the beginning of this section. Question 2.14. Is point-open(X) determined for every X ⊆ R? Or, equivalently: Question 2.15. Is Rothberger(X) determined for every X ⊆ R? Towards an answer to this question, first we quote: Theorem 2.16 (Pawlikowski [31]). Let X be a topological space. Then X is a Rothberger space if and only if Alice ↑ Rothberger(X). A nice proof of Theorem 2.16 was recently obtained in [42]. And then we also quote:

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Theorem 2.17 (Telgársky [43], Galvin [15]). Let X be a topological space in which every point is a Gδ set (that is: for each x ∈ X, the set {x} is the intersection of a countable family of open sets). Then Bob ↑ Rothberger(X) if and only if X is countable. We will write the proof of this theorem in two different ways: first we will provide a more intuitive reasoning, and then we will rewrite the same argument more formally.  Proof, v.1. For every x ∈ X, let Vk (x) : k ∈ ω be a sequence of open sets such that {x} = n∈ω Vk (x). Now let us play the point-open game on X. We will assume that Bob is naive enough to choose one of the sets Vk (x) whenever Alice plays a point x. Now a play against a naive person can be described as the following: • • • • • •

Alice plays x ; Bob chooses k0 (which really means playing Vk0 (x )); Alice plays xk0  ; Bob chooses k1 (which really means playing Vk1 (xk0  )); Alice plays xk0 ,k1  ; and so on.

Thus, when playing against naive people, Alice can play only countably many points — namely, all of the possible points of the form xs for s ∈ ω <ω . And, if Alice has winning strategy in this game, then those points suffice to cover the space: Suppose not. Let y ∈ X be such that y = xs for every s ∈ ω <ω . Note that, in each inning, Bob (even being naive) can play a Vk (xs ) that does not contain y. By doing so, Bob wins — hence the strategy employed by Alice (whatever it might be) was not a winning one. The result then follows from Theorem 2.13. 2  Proof, v.2. For each x ∈ X, let Ax be a countable family of open neighbourhoods of x satisfying Ax = {x}, and write Ax = {Vk (x) : k ∈ ω}. Now suppose that Bob ↑ Rothberger(X). By Theorem 2.13, Alice has a winning strategy in the pointopen game on X; let then ϕ : τ <ω → X be such a winning strategy, where τ is the topology of X. The idea is to look at all of the possible plays of point-open(X) in which Alice makes use of ϕ and Bob answers to a point x only with sets in Ax . We will then gather all of the points played by Alice in all of those plays, indexing them as yt : t ∈ ω <ω . We proceed as follows. First, let Alice begin the play with ϕ() =: y ∈ X. Bob will reply with a set Vk0 (y ) =: U0 ∈ Ay for some k0 ∈ ω. Now Alice’s next move is ϕ(U0 ) = ϕ(Vk0 (y )) =: yk0  ∈ X, to which Bob responds with Vk1 (yk0  ) =: U1 ∈ Ayk0  for some k1 ∈ ω. Alice’s next move will then be ϕ(U0 , U1 ) = ϕ(Vk0 (y ), Vk1 (yk0  ) =: yk0 ,k1  ∈ X, and so forth. Formally, we are recursively defining yt : t ∈ ω <ω  by setting y = ϕ() and, for each s ∈ ω <ω and each k ∈ ω, defining ys k = ϕ(Vs(0) (ys0 ), Vs(1) (ys1 ), . . . , Vs(m−1) (ys(m−1) ), Vk (ys )), where m = dom(s). With this, we construct a countable subset {yt : t ∈ ω <ω } of X. We now claim that {yt : t ∈ ω <ω } = X. Indeed, if there existed some a ∈ X \ {yt : t ∈ ω <ω }, then Bob could play point-open(X) against Alice’s strategy ϕ by always replying to Alice’s move xn with some Un ∈ Axn satisfying a ∈ / Un — this would be <ω possible by the choice of Axn , since a = xn ∈ {yt : t ∈ ω } — thus leading to a play

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x0 , U0 , x1 , U1 , . . . , xn , Un , . . .  of point-open(X) in which Alice follows the strategy ϕ and loses, as a ∈ / the assumption that ϕ is a winning strategy for Alice. 2

 n∈ω

Un . This would contradict

Theorems 2.16 and 2.17 yield: Corollary 2.18 (Pawlikowski [31]). Let X ⊆ R. Then Rothberger(X) is undetermined if and only if X is an uncountable Rothberger space. The original question then becomes whether any such X exists. Consistently, yes: Theorem 2.19 (Mahlo [25], Luzin [24]). The Continuum Hypothesis implies the existence of a Luzin set (i.e. an uncountable X ⊆ R that has countable intersection with every nowhere dense subset of R). Proof. As R has a countable base, it follows that the topology of R has cardinality 2ℵ0 . Thus, assuming the Continuum Hypothesis, we can enumerate the set of all closed nowhere dense subsets of R as {Fξ : ξ < ω1 }.  Now pick recursively, for each α ∈ ω1 , a point xα ∈ R \ ({xξ : ξ < α} ∪ ξ<α Fξ ) — which is possible since R is a Baire space. The set {xα : α < ω1 } thus obtained is a Luzin set. 2 Theorem 2.20 (Rothberger [32]). Every Luzin set is Rothberger. Proof. Let X ⊆ R be a Luzin set, and fix a countable dense subset D = {dn : n ∈ ω} of X. Now let Vn : n ∈ ω be a sequence of covers of X by open subsets of R. For each n ∈ ω, pick U2n ∈ V2n

 with xn ∈ U2n . Note that the set A := X \ n∈ω U2n is nowhere dense (as it is disjoint from R \ X ∪  n∈ω U2n , which is open and dense in R), so only the countably many points in X ∩ A remain to be covered. This can be done by picking sets U2n+1 ∈ V2n+1 for n ∈ ω. 2 Thus, the statement “there exists a subset of R on which the Rothberger game is undetermined” is consistent with ZFC: Corollary 2.21. The Continuum Hypothesis implies that there is X ⊆ R such that Rothberger(X) is undetermined. On the other hand, we also have the following result, obtained by F. Rothberger in Theorem 4 of [33]: Theorem 2.22 (Rothberger [33]). The statement “every Rothberger subset of R is countable” is equivalent to the Borel Conjecture. In order to make sense of the above theorem, we recall: Definition 2.23 (Borel [7]). A set X ⊆ R is said to have strong measure zero if, for every sequence εn : n ∈ ω  of positive real numbers, there is a sequence In : n ∈ ω of open intervals in R satisfying X ⊆ n∈ω In and such that, for each n, the length of the interval In is εn . The Borel Conjecture is the statement “every subset of R that has strong measure zero is countable”. For a proof of Theorem 2.22, we refer the reader to [28, Proposition 8]. Combining Corollary 2.18 and Theorem 2.22, we obtain:

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Corollary 2.24. The statement “Rothberger(X) is determined for every X ⊆ R” is equivalent to the Borel Conjecture. Now consistency follows from R. Laver’s celebrated Theorem 2.25 (Laver [23]). The Borel Conjecture is consistent with ZFC. Thus, Corollary 2.24 and Theorem 2.25 yield: Corollary 2.26. It is consistent with ZFC that Rothberger(X) is determined for every X ⊆ R. Finally, from Corollaries 2.21 and 2.26, we conclude: Corollary 2.27. The statement “Rothberger(X) is determined for every X ⊆ R” is independent of ZFC. 2.3. The Menger game Let us now consider what would happen if Bob could select more than one open set per inning. Definition 2.28 (Telgársky [44] (arguably, Hurewicz [19])). The Menger game on a topological space X is played according to the following rules: • In each inning n ∈ ω, Alice chooses an open cover Vn of X, and then Bob selects a non-empty finite subset Fn ⊆ Vn . • At the end of the play V0 , F0 , V1 , F1 , V2 , F2 , . . . , Vn , Fn , . . . , the winner is Bob if

 n∈ω

Fn is a cover of X, and Alice otherwise.

Exercise 2.29. Define (winning) strategy in the Menger game — both for Alice and Bob. It turns out that this does make a big difference. First, we recall: Definition 2.30 (Hurewicz [19]). A topological space X is Menger if, for every sequence Vn : n ∈ ω of open  covers of X, it is possible to choose non-empty finite sets Fn ⊆ Vn for each n in such a way that n∈ω Fn is an open cover of X.  Note that, if X is σ-compact (i.e. if X = n∈ω Kn with Kn compact for each n ∈ ω), then Bob ↑ Menger(X): All he has to do is pick, in each inning n ∈ ω, a finite subset Fn ⊆ Vn that covers Kn . (In particular, Bob ↑ Menger(R).) Hence, we have: X is σ-compact ⇓ Bob ↑ Menger(X) ⇓ Alice ↑ Menger(X) ⇓ X is Menger.

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But, in fact, a lot more can be said: Theorem 2.31 (Telgársky [44]). Let X be a metrizable topological space. Then Bob ↑ Menger(X) if and only if X is σ-compact. Theorem 2.32 (Hurewicz [19]). Let X be a topological space. Then X is Menger if and only if Alice ↑ Menger(X). We will prove Theorem 2.31 later on. For a proof of Theorem 2.32, we refer the reader to [36, Theorem 13] — we should also mention the notes [42], which include a conceptual proof of this result. So the question Question 2.33. Is there X ⊆ R such that Menger(X) is undetermined? boils down to Question 2.34 (Menger [27], Hurewicz [19]). Is there X ⊆ R that is Menger but not σ-compact? The answer is now provided by: Theorem 2.35 (Fremlin–Miller [14]). There is X ⊆ R that is Menger but not σ-compact. Therefore, by putting Theorems 2.31, 2.32 and 2.35 together, we obtain the following ZFC result: Corollary 2.36. There is X ⊆ R such that Menger(X) is undetermined. 2.4. Selective games Now let us introduce a more general framework: Definition 2.37 (Scheepers [36,37]). Let A and B be non-empty families of non-empty sets. • We denote by S1 (A, B) the statement: for every sequence An : n ∈ ω of elements of A, there is a sequence bn : n ∈ ω such that bn ∈ An for each n ∈ ω and {bn : n ∈ ω} ∈ B. • We denote by G1 (A, B) the game defined as follows: – In each inning n ∈ ω, Alice chooses An ∈ A, and then Bob picks bn ∈ An . – The winner is Bob if {bn : n ∈ ω} ∈ B, and Alice otherwise. Example 2.38. If O is the set of all the open covers of a topological space X, then S1 (O, O) is the statement “X is a Rothberger space”, and G1 (O, O) is the Rothberger game on X. In much the same way, we can define: Definition 2.39 (Scheepers [36,37]). Let A and B be non-empty families of non-empty sets. • We denote by Sfin (A, B) the statement: for every sequence An : n ∈ ω of elements of A, there is a sequence Fn : n ∈ ω such that Fn is a  non-empty finite subset of An for each n ∈ ω and n∈ω Fn ∈ B.

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• We denote by Gfin (A, B) the game defined as follows: – In each inning n ∈ ω, Alice chooses An ∈ A, and then Bob selects a non-empty finite subset Fn ⊆ An .  – The winner is Bob if n∈ω Fn ∈ B, and Alice otherwise. Example 2.40. If O is the set of all the open covers of a topological space X, then Sfin (O, O) is the statement “X is a Menger space”, and Gfin (O, O) is the Menger game on X. Now we have a whole new world of possibilities of games to be considered — which will be explored in the coming sections. Note that the following implications always hold in the general case: Bob ↑ G1 (A, B)

Bob ↑ Gfin (A, B)

Alice ↑ G1 (A, B)

Alice ↑ Gfin (A, B)

S1 (A, B)

Sfin (A, B)

for every A ∈ A there is a countable B ⊆ A with B ∈ B. Let us now turn our attention to another question. In Theorem 2.13, we have seen that: Theorem 2.41 (Galvin [15]). The Rothberger game and the point-open game are dual. In view of this result, it is natural to seek for a game G (that should resemble the point-open game somehow) such that the Menger game and G are dual. The obvious thing to try seems to be the following: Definition 2.42 (Telgársky [43]). The finite-open game on a topological space X is played according to the following rules: • In each inning n ∈ ω, Alice picks a (non-empty) finite set Fn ⊆ X, and then Bob chooses an open set Un ⊆ X with Fn ⊆ Un . • At the end of the play F0 , U0 , F1 , U1 , F2 , U2 , . . . , Fn , Un , . . . , the winner is Alice if X ⊆

 n∈ω

Un , and Bob otherwise.

However: Theorem 2.43 (Telgársky [43]). On every topological space, the finite-open game is equivalent to the pointopen game. The word “equivalent” in the previous result has the following meaning:

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Definition 2.44. Two games G and G are equivalent if • Alice ↑ G ⇔ Alice ↑ G and • Bob ↑ G ⇔ Bob ↑ G . Exercise 2.45. Prove Theorem 2.43. [Hint: Say we are proving that Alice ↑ finite-open(X) implies Alice ↑ point-open(X). We will play these two games in parallel, in the same spirit of Theorem 2.13. Fix a winning strategy σ for Alice in finite-open(X). If Alice opens the play with σ() = F0 = {x01 , . . . , x0k0 }, let Alice play the points x01 , . . . , x0k0 in the first k0 innings of point-open(X) — regardless of what Bob is responding during the play. After those k0 many innings of point-open(X), Bob will have played open sets Vj for j < k0 such that  x0j+1 ∈ Vj for each j < k0 ; thus, U0 = j
 n∈ω

Un , and Bob otherwise.

The compact-open game seems to be a promising candidate, in view of: Theorem 2.47 (Telgársky [44]). On every topological space X: (a) Alice ↑ compact-open(X) ⇒ Bob ↑ Menger(X) (b) Alice ↑ Menger(X) ⇒ Bob ↑ compact-open(X) (c) if X is regular, then Bob ↑ Menger(X) ⇒ Alice ↑ compact-open(X). Exercise 2.48. Prove items (a) and (b) of Theorem 2.47. We will omit the (quite indirect) proof of item (c). But this is a good point to mention Theorem 2.49 (Telgársky [43]). Let X be a regular space in which every compact set is a Gδ . If Bob has a winning strategy in Gfin (O, O) on X, then X is σ-compact.

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Exercise 2.50. Prove Theorem 2.49 by adapting the proof of Theorem 2.17. Note that Theorem 2.31 follows from Theorems 2.47 and 2.49 — but we will see a direct proof for it in Section 4, as Theorem 4.8. Theorem 2.47 immediately suggests the following Question 2.51 (Telgársky [44]). Are the Menger game and the compact-open game dual on every regular space? (Equivalently: Is it true that Bob ↑ compact-open(X) ⇒ Alice ↑ Menger(X) for every regular space X?) In order to deal with Question 2.51, we will consider open covers of a special kind: Definition 2.52. An open cover V of a topological space X is a k-cover if for every compact K ⊆ X there is U ∈ V with K ⊆ U . We write K = {V ∈ O : V is a k-cover of X}. Theorem 2.53 (in essence, Galvin [15]). The compact-open game and the game G1 (K, O) are dual. Proof. Analogous to the proof of Theorem 2.13. (Exercise!)

2

So the question becomes: Question 2.54. Does Alice ↑ G1 (K, O) imply Alice ↑ Menger if X is regular? Or, equivalently: Question 2.55. Does Alice ↑ Menger imply Alice ↑ G1 (K, O) if X is regular? In other words (in view of Theorem 2.32): Question 2.56. Let X be a regular Menger space. Must it be the case that Alice ↑ G1 (K, O)? The answer is: at least consistently, no. More specifically: Example 2.57 (Aurichi–Dias [6]). If there is a Sierpiński set (i.e. an uncountable X ⊆ R whose intersection with every Lebesgue measure zero set is countable), then there is a Menger space that does not satisfy S1 (K, O). The first half of the argument for Example 2.57 is provided by the following fact: Theorem 2.58 (Fremlin–Miller [14]). Every Sierpiński set is Menger. Proof. Let X be a Sierpiński set. By considering a subset of X is necessary, we may assume that X ⊆ [0, 1] and that the outer measure of X is 1 (why?). Now let Vn : n ∈ ω be a sequence of covers of X by open  1 subsets of [0, 1]. For each n, choose a finite Fn ⊆ Vn such that the measure of Fn is at least 1 − n+1 .    The set [0, 1] \ n∈ω Fn has measure zero; hence, as X is a Sierpiński set, we may take an enumeration   X \ n∈ω Fn = {xn : n ∈ ω}. Now, for each n, we cover xn with some Un ∈ Vn and define Fn = Fn ∪{Un };   thus, we obtain X ⊆ n∈ω Fn . 2

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We can now explain Example 2.57 in more detail2 : Proof for Example 2.57. Let X be a Sierpiński set. By the previous theorem, X is Menger. Claim. Every compact subset of X is countable. Indeed, let K ⊆ X be compact. By the Cantor–Bendixson Theorem (see e.g. 1.7.11 in [13]), K may be written as K = C ∪ P , where C is countable, P is perfect (i.e. P is closed and no point of P is isolated in P ) and the union is disjoint. If P were non-empty, we could construct in P a Cantor set of Lebesgue measure zero by means of the classic Čech–Pospíšil argument (see e.g. 3.2.11 in [13]), which would contradict the fact that X is a Sierpiński set. Hence P = ∅ and K = C is countable. Now suppose, by way of contradiction, that X satisfies S1 (K, O). We claim that this implies that X is a Rothberger space — which is a contradiction, since X is not a Lebesgue measure zero subset of R (recall Example 2.4). The argument for showing that X is Rothberger is adapted from [36, Theorem 17]. Let Vn : n ∈ ω be a  sequence of open covers of X, and write ω = j∈ω Aj with the sets Aj being infinite and pairwise disjoint.   For each j ∈ ω, let Wj = { n∈Aj Un : Un : n ∈ ω ∈ n∈Aj Vn }. Now, by the Claim, Wj ∈ K for each  j ∈ ω. Since X satisfies S1 (K, O), we can choose Wj ∈ Wj for j ∈ ω in such a way that X ⊆ j∈ω Wj ,  which in turn leads to a choice of Un ∈ Vn for n ∈ ω such that X ⊆ n∈ω Un . 2 Theorem 2.59 (Sierpiński [39]). The Continuum Hypothesis implies the existence of a Sierpiński set. Proof. Similar to the proof of Theorem 2.19 — exercise! (This time, the starting point is the fact that {A ⊆ R : A is a Gδ } = 2ℵ0 — and one must also recall that every Lebesgue measure zero subset of R is included in some Gδ subset of R that also has Lebesgue measure zero.) 2 Hence, we have: Corollary 2.60. The Continuum Hypothesis implies the existence of a Menger space that does not satisfy S1 (K, O). Therefore: Corollary 2.61 (Aurichi–Dias [6]). It is consistent with ZFC that the compact-open game and the Menger game are not dual. Since this is only a consistent answer to Question 2.51, we are still left with: Problem 2.62. Is the statement “the compact-open game and the Menger game are dual on every regular space” consistent with — hence independent of — ZFC? (Or, reversing the question: Is there a ZFC example of a regular space on which the compact-open game and the Menger game are not dual?) The discussion preceding Problem 2.62 suggests the following (still unanswered) questions, which are also interesting in themselves: Problem 2.63. Is S1 (K, O) equivalent to Alice ↑ G1 (K, O) on every regular space? (Equivalently: is S1 (K, O) equivalent to Bob ↑ compact-open on every regular space?) 2 The original argument used in [6] considered the Sorgenfrey line topology on a Sierpiński set. We thank Piotr Szewczak and Boaz Tsaban for pointing out to us (in a conversation during the Frontiers of Selection Principles conference) that the usual Euclidean topology would work as well.

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Problem 2.64. Is it consistent with ZFC that S1 (K, O) and the Menger property S1 (O, O) are equivalent for regular spaces? (Reversing the question: is there a ZFC example of a regular Menger space that fails to satisfy S1 (K, O)?) Note that Problem 2.62 will be solved if any of Problems 2.63 and 2.64 is answered in the negative, or if both are answered in the affirmative. In light of the proof given for Example 2.57, the following question also emerges: Problem 2.65. Is it consistent with ZFC that every regular Menger space in which every compact subset is countable is a Rothberger space? (Reversing the question: is there a ZFC example of a regular Menger space X such that every compact subset of X is countable yet X fails to be Rothberger?) Note that a negative answer to Problem 2.65 would yield a negative answer to Problem 2.62, by an argument similar to the one presented for Example 2.57. A closer look at that argument suggests in fact the following question, which is interesting in its own right: Problem 2.66. Is it consistent with ZFC that every regular Menger space in which every compact subset is Rothberger is a Rothberger space? (Reversing the question: is there a ZFC example of a regular Menger space X such that every compact subset of X is Rothberger yet X fails to be Rothberger?) We should add that no answer is known to any of the five problems above even in the realm of separable metrizable spaces. In fact, we do not even know the answer to the following question: Problem 2.67. Is there a ZFC example of a non–σ-compact subset of R that satisfies S1 (K, O)? 3. The Banach–Mazur game In this section, we will work with the (arguably) first example of a topological game: Definition 3.1 (see Problem 43 in [26]). The Banach–Mazur game is played on a topological space X between players Alice and Bob as follows. Alice chooses a non-empty open set A0 . Then Bob chooses a non-empty open set B0 ⊆ A0 . Then, in each inning n ∈ ω with n > 0, Alice plays An ⊆ Bn−1 , and then Bob plays  Bn ⊆ An with An and Bn being non-empty open sets. Bob is declared to be the winner if n∈ω An = ∅; otherwise, Alice is the winner. At first glance, this game seems to be related to the Baire property — we are doing some kind of countable intersection of open sets. This intuition is very right: Theorem 3.2 (Banach, Oxtoby [30]). A topological space is Baire if, and only if, Alice does not have a winning strategy for the Banach–Mazur game. Proof. Let X be a topological space with topology τ . Suppose that X is not a Baire space. Then there exist a sequence Dn : n ∈ ω of open dense subsets of X  and a non-empty open set V ⊆ X such that V ∩ n∈ω Dn = ∅. We will define a winning strategy for Alice in the Banach–Mazur game on X as follows. First, Alice begins by playing A0 = V ∩ D0 . For each n ∈ ω, if Bob’s choice in the n-th inning was Bn , then Alice’s move in the next inning will be An+1 = Bn ∩ Dn+1 . Note that this is a legitimate strategy for Alice, since each Dn is open dense. To see that this is a winning   strategy, note that n∈ω An ⊆ V ∩ n∈ω Dn = ∅.

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Now suppose that X is a Baire space. Let σ : τ \{∅} → τ \{∅} be a strategy for Alice in the Banach–Mazur game on X. We will show that σ is not winning. Claim 1. For every sequence V0 , . . . , Vn  of open sets that satisfies Vk ⊆ σ(V0 , . . . , Vk−1 ) for each k ≤ n, there is a family AV0 ,...,Vn  of non-empty open subsets of σ(V0 , . . . , Vn ) such that {σ(V0 , . . . , Vn , V ) : V ∈ AV0 ,...,Vn  } is a maximal pairwise disjoint family of open subsets of σ(V0 , . . . , Vn ); in addition,  {σ(V0 , . . . , Vn , V ) : V ∈ AV0 ,...,Vn  } is dense in σ(V0 , . . . , Vn ). In the proof of the above claim, whose details we leave to the reader, the existence of AV0 ,...,Vn  follows from a straightforward application of the Kuratowski–Zorn Lemma. For the second part, note that, if  W ⊆ σ(V0 , . . . , Vn ) were non-empty and disjoint from {σ(V0 , . . . , Vn , V ) : V ∈ AV0 ,...,Vn  }, then so would be σ(V0 , . . . , Vn , W ); hence, AV0 ,...,Vn  ∪ {W } would contradict the maximality in the first part. Claim 1 allows us to build a tree with sets played by Alice according to σ. If A = σ(V0 , . . . , Vn ) is the set we have placed in a node of this tree, then the set {σ(V0 , . . . , Vn , V ) : V ∈ AV0 ,...,Vn  } of its immediate successors in the tree will be a maximal pairwise disjoint family of non-empty open subsets of A whose union is dense in A. We may think that, when moving from a node of the tree to its successors in the next level of the tree, we have “blown up” the set A into pairwise disjoint smaller open subsets that, in a topological sense, “fill all of the space in A”. Now define, for each n ∈ ω, the set Cn = {σ(V0 , . . . , Vn ) : V0 , . . . , Vn  satisfies Vk ∈ AV0 ,...,Vk−1  for each k ≤ n}. (That is: Cn is the collection of all of the sets from the n-th level of the tree.) Let U = σ() ∈ τ . Note that, since X is Baire, so is also U .  Claim 2. For each n ∈ ω, the set Cn is a family of pairwise disjoint non-empty open sets; moreover, Cn is an open dense subset of U . Exercise 3.3. Prove Claim 2. [Hint: Proceed by induction on n. (Use Claim 1!)]   Now, as U is Baire, there is x ∈ n∈ω ( Cn ). We are finally ready to describe how Bob should play to  win against σ. Since x ∈ C0 , there is exactly one V0 ∈ A such that x ∈ σ(V0 ). So this V0 is Bob’s choice  after Alice plays σ(). Again, since x ∈ C1 , there is exactly one V1 ∈ AV0  such that x ∈ σ(V0 , V1 ), so this will be Bob’s next move; and so forth. Proceeding like this, we obtain the following play of the Banach–Mazur game on X, in which Alice makes use of the strategy σ: Alice σ() σ(V0 ) σ(V0 , V1 ) σ(V0 , V1 , V2 ) .. . σ(V0 , . . . , Vn−1 ) .. .

Bob V0 V1 V2 V3 .. . Vn .. .

 Since x ∈ n∈ω σ(V0 , . . . , Vn−1 ), it follows that Bob wins the play; therefore, σ is not a winning strategy for Alice. 2 Now we are ready to show some kind of common use of games. As we have just seen, the Banach–Mazur game is related to the Baire property. Actually, it is equivalent to Alice not having a winning strategy.

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Since it is possible for the game to be undetermined, asking for Bob to have a winning strategy is stronger. Let us see just how stronger it is. In [9], Cohen showed that if X and Y are both Baire spaces, there is no guarantee that X × Y is Baire — even if we ask that both X and Y are metrizable spaces. On the other hand, the following is a good exercise: Exercise 3.4. If Bob has a winning strategy for the Banach–Mazur games played on X and on Y , then Bob has a winning strategy for the Banach–Mazur game played on X × Y . In particular, X × Y is Baire. For the previous exercise (and for the proof of the next theorem), it is worth mentioning that, in the Banach–Mazur game, we may assume that both players are only allowed to play basic open sets — for any fixed base given in advance. In other words: Exercise 3.5. Let B be a base for the topology of the space X. Then the Banach–Mazur game on X is equivalent to the B–Banach–Mazur game on X, the latter being defined by the same rules as the former, with the extra restriction that both Alice and Bob must necessarily choose elements from B in their moves. [Hint: To prove that e.g. Bob ↑ Banach–Mazur(X) implies Bob ↑ B–Banach–Mazur(X), let ρ be a winning strategy for Bob in Banach–Mazur(X); now, for each sequence A0 , . . . , An  of elements of B, choose an arbitrary BA0 ,...,An  ∈ B with BA0 ,...,An  ⊆ ρ(A0 , . . . , An ), and define ρ (A0 , . . . , An ) = BA0 ,...,An  . We now claim that ρ is a winning strategy for Bob in B–Banach–Mazur(X). Let

A0 , BA0  , A1 , BA0 ,A1  , A2 , BA0 ,A1 ,A2  , . . .

be a play of B–Banach–Mazur(X) in which Bob played according to ρ . Then A0 , ρ(A0 ), A1 , ρ(A0 , A1 ), A2 , ρ(A0 , A1 , A2 ), . . .  is a play of Banach–Mazur(X) (why?) in which Bob played according to ρ. The result then follows from the fact that ρ is a winning strategy for Bob in Banach–Mazur(X) and from the observation that the sets played by Alice in both of the above plays are the same.] Even though Exercise 3.4 seems to be a powerful result, we can prove something more impressive: Theorem 3.6 (White [45]). If X is a space where Bob has a winning strategy for the Banach–Mazur game and Y is a Baire space, then X × Y is a Baire space. Proof. The following are the ingredients for this proof: • σ is a winning strategy for Alice in the Banach–Mazur game played on X × Y ; • ρ is a winning strategy for Bob in the Banach–Mazur game played on X. With these two strategies, we will define a winning strategy ϕ for Alice in the Banach–Mazur game on Y . Note that this is enough by Theorem 3.2. Before we start, note that, by Exercise 3.5, we can assume that both players play basic open sets in X × Y . Let A0 × B0 = σ(). Define ϕ() = B0 , and let V0 = ρ(A0 ). If W0 is Bob’s first move in the Banach–Mazur game on Y , let A1 × B1 = σ(V0 × W0 ). Define ϕ(W0 ) = B1 , and let V1 = ρ(A0 , A1 ). If W1 is Bob’s next move in the Banach–Mazur game on Y , let A2 × B2 = σ(V0 × W0 , V1 × W1 ). Define ϕ(W0 , W1 ) = B2 , and so on.   At the end, since σ and ρ are winning strategies, n∈ω An × Bn = ∅ and n∈ω An = ∅. Therefore,  n∈ω Bn = ∅. Since Bn : n ∈ ω are the choices of Alice by ϕ, we are done. 2

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The previous theorem is even more impressive once you realize that the usual proofs for the classical Baire spaces actually prove the following: Exercise 3.7. Show that Bob has a winning strategy for the Banach–Mazur game played over X if: (1) X is a Hausdorff compact space; or (2) X is a complete metric space. It is worth mentioning that Scheepers obtained some very interesting results connecting selective covering properties (such as the Menger property) and the Banach–Mazur game in [38]. 4. Covering games In this section, we will take a deeper look in some games related to covering properties. The idea is not to show sharp results, but rather to illustrate the usage of some game-theoretic techniques. 4.1. Trees In the next few results, we will see topological spaces obtained from trees; for the basics on trees, we refer the reader to [22, Section II.5]. The main reference for the results in this subsection (where the reader can find the proofs we have omitted) is [3]. Definition 4.1. We say that a tree with height ω1 is a Suslin tree if it has no uncountable chains and no uncountable antichains. Relative to ZFC, the statement “there is a Suslin tree” is undecidable: it was shown to be consistent with ZFC by T. Jech [20], and its negation was proven consistent with ZFC by R. Solovay and S. Tennenbaum [40]. Let us fix some notation and terminology for trees (T, ≤) in general: • Given p ∈ T , we will write Vp = {q ∈ T : p ≤ q}. • We say that q is a successor of p if p < q and there is no r ∈ T such that p < r < q. • We also write Suc(p) = {q ∈ T : q is a successor of p}. On every tree, we will consider the following topology: if p ∈ T then the basic open neighbourhoods of p  are the sets of the form Vp \ q∈F Vq , where F ⊆ Suc(p) is finite. Given a tree with this topology, it is easy to check the following properties: • it is Hausdorff; • it is zero-dimensional; • the union of its first α levels is closed for every α. Furthermore, if the tree is a Suslin tree, then: • every level is countable; • every point has a countable local base. Our next objective is to prove that, with this topology, every Suslin tree S is Rothberger. For our proof, it is better to prove first that S is Lindelöf.

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Proposition 4.2. Every Suslin tree is Lindelöf. Proof. Let C be an open cover for a Suslin tree S. Let X = {x ∈ S : Vx ⊂ C for some C ∈ C}. Let A be a maximal antichain in X. Note that, as C is an open cover of S, every x ∈ S is comparable with some element of X, which in turn is comparable with some element of A. Thus, every x ∈ S is comparable with some element of A, hence A is maximal in S. Since S is Suslin, A is countable. Now, for each a ∈ A, pick Ua ∈ C such that Va ⊆ Ua . Note that C0 = {Ua : a ∈ A} is a countable subset of C that covers all of the points of S, with the possible exception of those points that lie below some element of A. Since there are only countably many such points, we can now cover them with a countable subset C1 of C, and C0 ∪ C1 is the countable subcover we were looking for. 2 Let us now see how to obtain the Rothberger property: Proposition 4.3. Every Suslin tree is Rothberger. Proof. Let S be a Suslin tree and Vn : n ∈ ω be a sequence of open coverings for S. We can assume that every open set here is a basic one. And, since we already know that S is Lindelöf, we can assume that every Vn is countable.  Given a basic open set Vp \ q∈F Vq , we will say that {p} ∪F is the support of this open set. Note that the support is always finite and, since we have countably many open coverings and each one is also countable, the union of all the supports considered here is also countable. Therefore, let α < ω1 be such that every element of a support of one of the basic open sets here has height less than α.  The trick now is the following. If r is in the α-th level and r ∈ Vp \ q∈F Vq , and if p and F are below r,  then Vr ⊆ Vp \ q∈F Vq . Let rn : n ∈ ω be an enumeration of the α-th level. Then we can use the coverings V2k to cover it. By the previous observation, this will cover every element above the α-th level as well. But there are only countably many elements below the α-th level. Therefore, we can use the coverings V2k+1 to cover them. 2 After having seen the last proof, one could wonder whether we could use the same idea to prove that Bob has a winning strategy in the Rothberger game G1 (O, O) played on S. Note that everything seems to work — except how to compute the α. In fact, by Theorem 2.17, Bob has no winning strategy; and, since Alice does not have one either, the game G1 (O, O) is undetermined on S. We will now see a different approach to Theorem 2.17. 4.2. General spaces The following is a fairly general idea. Let σ be a strategy for Bob in G1 (O, O) played on a space X. Let K =



σ(C)

C∈O

Note that |K | ≤ 1. This is true for every T1 space: suppose that there are x, y ∈ K with x = y. Let C = {X\{x}, X\{y}}. Then C is a covering of X. Finally, note that σ(C) cannot contain both x and y. We did this for the 0-th inning. We can do the same thing for the other (possible) innings and define: Ks =



σ(s C)

C∈O

where s is a finite sequence of open coverings. As before, |Ks | ≤ 1.

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Even if we cannot guarantee that a specific Ks is non empty, if σ is a winning strategy, then the set of all of the Ks s is a cover of the space: Proposition 4.4. If σ is a winning strategy, then X =

 s∈O<ω

Ks .

 Proof. Suppose not. Let x ∈ X\ s∈O<ω Ks . Since x ∈ / K , there is an open covering C0 such that x ∈ / σ(C0 ). So let us pretend that Alice played C0 . Now, since x ∈ / KC0  , there is a C1 such that x ∈ / σ(C0 , C1 ). Again, let us pretend that Alice played C1 in the following inning. Proceeding this way, we find a play of the game in which Bob followed σ but none of the choices made by Bob cover the point x, whence σ is not winning. 2 Even if each Ks is small, the previous result alone does not tell us much about X, because we do not have much control over O<ω . What we will do in what follows is: we will reduce the number of the s ∈ O<ω we have to keep, but in a way that the respective Ks s still cover all of X. Suppose, for instance, that X has a countable base. We may assume that Alice will only play coverings made by basic open sets. Therefore, all the possible answers of Bob are basic open sets. So, in this situation, we can prove the following Lemma 4.5. For every s ∈ O<ω , there is a countable Os ⊆ O such that Ks =

 C∈Os

σ(s C).

The key to proving the above lemma is that, for every possible answer, there is a question; since there are few possible answers, we can fix (few) questions that will be enough to get all the possible answers. Exercise 4.6. Prove Lemma 4.5. So let us restrict again what Alice can play. Let O = {Cn : n ∈ ω}. Then, for each n ∈ ω, let OCn  = {Cn,k : k ∈ ω}; and so on. Note that there are only countably many open coverings here and therefore only countably many Ks s. But they still form a covering (it is the same argument as before). So each |Ks | ≤ 1 and there are countably many Ks s that are enough to cover the space. Therefore, the space has to be countable. Thus we have proved: Proposition 4.7. Let X be a topological space with a countable base. If Bob has a winning strategy in G1 (O, O), then X is countable. Of course, this is an immediate consequence of Theorem 2.17. But our focus here is not on the result itself, but rather on the argument that was presented. By the way, the same Theorem 2.17 can be proven by means of a similar argument, without making use of the duality with the point-open game (as we did when proving Theorem 2.17 for the first time). Let us now go back to Theorem 2.31. The relevant half of it is: Theorem 4.8 (Telgársky [44]). Let X be a metrizable topological space such that Bob ↑ Gfin (O, O) on X. Then X is σ-compact. We will now see a direct proof of this theorem due to Scheepers [35], whose argument goes along the same lines of the one we have just described.3 3 Here we should make it explicit that it was Scheepers’s proof that inspired the argument we have just done, and not the other way around.

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We have to change a little bit the definition of each Ks : Ks =

 

σ(s C)

C∈O

The surprising part here is that every Ks is compact: Lemma 4.9 (Scheepers [35]). If X is a regular space and σ is a strategy for Bob in Gfin (O, O), then each Ks defined as above is compact. One can prove this by definition of compactness and using regularity (it is a nice proof). We will do something different: Proof of Lemma 4.9. Let us pretend that we are playing the game inside βX (we can, right?): in each inning, Alice plays a collection of open sets in βX that covers X and Bob selects finitely many of them. At the end, Bob wins if the selected open sets form a cover of X. Define Ks as above, but with the closure taken in βX. Then, Ks is compact by nature. But we have to prove that this new Ks ⊆ X (all we know is that it is a subset of βX). Suppose not. Then there is an y ∈ βX\X such that y ∈ Ks . That means that y is in every answer of Bob. But, as βX is T2 , we can easily produce a covering of X by open subsets of βX such that none of its elements contains y in its closure, which is a contradiction. 2 Then, if we repeat the argument of the “all possible answers are countable” and “for each answer there is a question”, we obtain that X is the union of countably many Ks s. Exercise 4.10. Fill in the details of the proof. Finally, let us mention that regularity plays an important role in these games. Example 4.11. Consider the real line with the usual topology and additionally declare all the countable sets to be closed. On this space, Bob has a winning strategy in Gfin (O, O), but Alice does not have a winning strategy in the compact-open game. Proof. First we show that Alice does not have a winning strategy in the compact-open game. For this, just note that all the compact sets are finite. Therefore, this is the same as playing the point-open game. Thus, if Alice had a winning strategy, the space should be countable by Theorem 2.17. Now we prove that Bob has a winning strategy in Gfin (O, O). We can assume that Alice only plays coverings with basic open sets of the form ]a, b[\C (where C is a countable set). Let Ak : k ∈ ω be a  sequence of pairwise disjoint infinite sets such that k∈ω Ak is the set of the odd naturals. Then, in the even-numbered innings, Bob can pretend that the coverings are usual ones — that is, Bob can ignore the sets C and pretend that each open cover consists only of the corresponding open intervals of form ]a, b[, with no points removed — and answer using a winning strategy for the real line with the usual topology. Note that, in this way, Bob misses countably many points in each even-numbered inning (the points in the corresponding set C). But then the odd-numbered innings can be used: Bob can make use of the innings n ∈ Ak to cover the countably many points that were missed in the 2k-th inning. 2 5. Tightness games This section is similar to the previous one — but now we are mostly dealing with properties that involve closures, instead of coverings. We begin with some classical properties:

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Definition 5.1 (Moore–Mrówka [29]). A topological space X is countably tight at a point p ∈ X if, for every A ⊆ X with p ∈ A, there is a countable B ⊆ A such that p ∈ B. Definition 5.2 (Sakai [34]). A topological space X has countable strong fan tightness at a point p ∈ X if, for every sequence An : n ∈ ω of subsets of X satisfying p ∈ An , we can select xn ∈ An for n ∈ ω so that p ∈ {xn : n ∈ ω}. Note that the statement “X has countable strong fan tightness at p” is equivalent to S1 (Ωp , Ωp ), where Notation 5.3. Ωp = {A ⊆ X : p ∈ A}. Inspired by the above properties, we may now consider the game G1 (Ωp , Ωp ). Clearly, X is first-countable at p ⇓ Bob ↑ G1 (Ωp , Ωp ) ⇓ Alice ↑ G1 (Ωp , Ωp ) ⇓ S1 (Ωp , Ωp ) . We will now prove the following result: Theorem 5.4 (Gruenhage [17]). If X is a separable regular space and p ∈ X is such that Bob ↑ G1 (Ωp , Ωp ), then X is first-countable at p. The key idea here is to find some argument of the kind “for every answer, there is a question” as in the previous section. But there is also an argument about duality hidden in the statement. The following result will help us see this clearly. Lemma 5.5 (arguably, Galvin [15]). If σ : Ωp <ω → X is a strategy for Bob in G1 (Ωp , Ωp ), then for every A0 , . . . , An  ∈ Ωp <ω there is an open neighbourhood U of p such that, for every x ∈ U , there is an A ∈ Ωp such that σ(A0 , . . . , An , A) = x. Proof. Suppose not. Then, for every open neighbourhood U of p, let xU ∈ U be such that xU is not of the form σ(A0 , . . . , An , A) for A ∈ Ωp . Note that B = {xU : U ∈ τp } ∈ Ωp . Therefore σ(A0 , . . . , An , B) ∈ B, which is a contradiction. 2 This formulation prompts us to look at this game: Definition 5.6 (Gruenhage [17]). Fixed p ∈ X, the neighbourhood-point game at p is played as follows: • In each inning n ∈ ω, Alice plays an open neighbourhood Vn of p, and then Bob plays xn ∈ Vn . • The winner is Alice if {xn : n ∈ ω} ∈ Ωp , and Bob otherwise. The previous lemma is the key to proving: Exercise 5.7. If Bob has a winning strategy in G1 (Ωp , Ωp ), then Alice has a winning strategy in the neighbourhood-point game at p.

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The following notation is handy: Notation 5.8. τp = {U ∈ τ : p ∈ U }. Note that, on the other hand, if σ is a strategy for Bob in the neighbourhood-point game and V0 , . . . , Vn  ∈ τp <ω , then {σ(V0 , . . . , Vn , V ) : V ∈ τp } ∈ Ωp With this, we can prove that, if Bob has a winning strategy in the neighbourhood-point game, then Alice has a winning strategy for G1 (Ωp , Ωp ). The other two implications are quite straightforward. Therefore, we obtain: Proposition 5.9 (arguably, Galvin [15]). The games G1 (Ωp , Ωp ) and neighbourhood-point are dual. Exercise 5.10. Fill in the details of the proof. We are now ready to prove Theorem 5.4: Proof of Theorem 5.4. We will work in the neighbourhood-point game — in which, by Proposition 5.9, Alice has a winning strategy σ. Let D be a countable dense subset of X, and suppose that Bob will always pick a point of D. Let V0 = σ() be Alice’s initial move. By our assumption, there are only countably many possibilities for Bob to play in. Let us call them xk , where k ∈ ω. Therefore, for the next inning, all the possibilities for Alice to play are σ(xk ), for k ∈ ω. For each fixed k0 (meaning that Bob played the point xk in the first inning), there are only countably many possibilities for Bob to choose from — say, xk0 ,k , where k ∈ ω. Therefore, there are only countably many possibilities for Alice to play in the following inning — namely, σ(xk0 ,k ), for k ∈ ω —; and so forth. By induction, there are only countably many possibilities for Alice to play in the whole play, for every possible play in which Bob only picks points from D. Let us now prove that the collection B = {σ(xk0 ,k1 ,...,kn  ) : k0 , k1 , . . . , kn  ∈ ω <ω } of all of those possible moves of Alice is a local base for X at p. For this, we will use regularity. Suppose that B is not a local base. Then there is an open set V such that p ∈ V and B  V for all B ∈ B. By regularity, there is an open set W satisfying p ∈ W and W ⊆ V ; note that B  W for all B ∈ B. Now, if Alice plays B, then Bob can always pick an x ∈ (B\W ) ∩ D. And, playing like this, the set {xn : n ∈ ω} of all the points chosen by Bob is not in Ωp , since this set is disjoint from the open neighbourhood W of p. 2 Let us now consider the following variation of the neighbourhood-point game: Definition 5.11 (Gruenhage [17]). Let p ∈ X be fixed. The neighbourhood-point convergence game at p is played as follows: • In each inning n ∈ ω, Alice chooses an open neighbourhood Un of p, and then Bob picks a point xn ∈ Un . n→∞ • The winner is Alice if xn −→ p, and Bob otherwise.

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The following result relates these two games in a rather surprising way: Theorem 5.12 (Gruenhage [17]). Alice has a winning strategy in the neighbourhood-point game at p if and only if Alice has a winning strategy in the neighbourhood-point convergence game at p. Proof. Let ϕ : X <ω → τp be a winning strategy for Alice in the neighbourhood-point game at p. Define ψ : X <ω → τp as follows: • ψ() = ϕ() • ψ(x0 ) = ϕ() ∩ ϕ(x0 ) • ψ(x0 , x1 ) = ϕ() ∩ ϕ(x0 ) ∩ ϕ(x1 ) ∩ ϕ(x0 , x1 ) .. . • ψ(x0 , . . . , xn ) = 

{ϕ(xi0 , xi1 , . . . , xim ) : m ≤ n and ik : k ≤ m is an increasing sequence in {0, . . . , n}}.

In words: ψ(x0 , . . . , xn ) is the intersection of all of the (finitely many) open neighbourhoods of p that Alice plays in every possible play of the neighbourhood-point game in which Bob’s moves thus far constitute a subsequence of x0 , . . . , xn . We claim that this ψ thus defined is a winning strategy for Alice in the neighbourhood-point convergence game at p. Suppose not. Let x0 , x1 , . . . , xn , . . .  be Bob’s moves in a play of this game in which Bob has beaten the strategy ψ. Now let U ∈ τp be such that I = {n ∈ ω : xn ∈ / U } is infinite, and let I = {ik : k ∈ ω} be an increasing enumeration. Then ϕ(), xi0 , ϕ(xi0 ), xi1 , ϕ(xi0 , xi1 ), xi2 , . . . , ϕ(xi0 , xi1 , . . . , xik ), xik+1 , . . .  is a play of the neighbourhood-point game (why?) in which Alice employs the winning strategy ϕ; hence, p ∈ {xik : k ∈ ω}. Contradiction! 2 Are these games in fact equivalent? What about Bob? Example 5.13 (Gruenhage [18]). There is a countable space with only one non-isolated point on which Bob has a winning strategy in the neighbourhood-point convergence game but not on the neighbourhood-point game. (By the way: on this space, the game G1 (Ωp , Ωp ) is undetermined.) This example shows that, even for countable spaces with only one non-isolated point, different versions of tightness games can have very distinct behaviour. We will skip the details of this example to talk about another example of same kind that shows that the differences can be even deeper, in a sense. Example 5.14 (adapted from) Scheepers [37]. Let Y = ω <ω ∪˙ {p} be the space in which • every s ∈ ω <ω is isolated • basic neighbourhoods of p are of the form V (F ) := Y \

 f ∈F

for F ⊆ ω ω finite.

{f  n : n ∈ ω}

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Proposition 5.15 (Scheepers [37]). Alice ↑ G1 (Ωp , Ωp ) on Y . Proof. Alice initiates the game by playing A0 = {k : k ∈ ω} ∈ Ωp . If Bob picks k0  ∈ A0 , Alice’s move in the next inning is A1 = {k0 , k : k ∈ ω} ∈ Ωp . If Bob picks k0 , k1  ∈ A1 , Alice’s move in the next inning is A2 = {k0 , k1 , k : k ∈ ω} ∈ Ωp . By proceeding in this fashion, we obtain f = k0 , k1 , k2 , . . .  ∈ ω ω such that V ({f }) = Y \{f  n : n ∈ ω} is an open neighbourhood of p that does not contain any of the points played by Bob. Thus, Alice wins the play. 2 But (what would seem to be) a minor modification in the game can change the situation completely, as the next result shows. Here, G2 (A, B) is played according to the same rules as Gfin (A, B), but now Bob must pick two elements per inning. Proposition 5.16 (Aurichi–Bella–Dias [5]). Bob ↑ G2 (Ωp , Ωp ) on Y . Proof. We only have to observe that Bob can play in such a way that, at the end of each inning n ∈ ω, the set of all the points he has chosen includes a set {s0 , . . . , sn } such that every branch of ω <ω contains at most one element of this set. (Note that this implies that Bob wins the play.) We can do so by induction. The case n = 0 is trivial. Now suppose this is true for some n ∈ ω. If there is some point a ∈ An+1 that lies in a branch of ω <ω that misses the set {s0 , . . . , sn }, let Bob pick this point together with any other one. If there is no such point, then each point of An+1 lies in a branch that intersects {s0 , . . . , sn }. So there must be two incompatible a, a ∈ An+1 and some sk such that both a and a extend sk . Let Bob pick {a, a } ⊆ An+1 , and we’re done. 2 This is particularly curious in view of Theorem 5.17 (García-Ferreira–Tamariz-Mascarúa [16]). For every space X and p ∈ X, S2 (Ωp , Ωp ) ⇔ S1 (Ωp , Ωp ). Proof. Let An : n ∈ ω be a sequence of sets in Ωp . By S2 (Ωp , Ωp ), we can choose {an0 , an1 } ⊆ An for each  n ∈ ω in such a way that n∈ω {an0 , an1 } ∈ Ωp . So p ∈ {an0 : n ∈ ω} ∪ {an1 : n ∈ ω}, whence p ∈ {an0 : n ∈ ω} or p ∈ {an1 : n ∈ ω}. Either case validates S1 (Ωp , Ωp ). 2 As a consequence of the two previous results, we obtain: Corollary 5.18. S1 (Ωp , Ωp ) and Alice ↑ G1 (Ωp , Ωp ) are not equivalent. We should also mention that — as we learned from N. Hiers at the Frontiers of Selection Principles conference — the difference we have observed between G1 (Ωp , Ωp ) and G2 (Ωp , Ωp ) does not hold in general for the Rothberger game G1 (O, O) and G2 (O, O): Theorem 5.19 (Crone et al. [10]). The games G1 (O, O) and G2 (O, O) are equivalent on every Hausdorff space. 5.1. Productively countably tight spaces If X and Y have countable tightness at p and q respectively, the product X ×Y may fail to have countable tightness at p, q:

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Example 5.20 (Arkhangel’ski˘ı [1]). For an infinite cardinal κ, let Sκ = (κ × ω) ∪˙ {p} be space in which • points in κ × ω are isolated; • basic neighbourhoods of p are of the form Vf = {p} ∪ {α, n : α ∈ κ & n ≥ f (α)} for f : κ → ω. Then Sκ is countably tight for all κ, yet Sℵ0 × Sc is not countably tight. To see the latter, identify c = ω ω ;  then, for each f : ω → ω, let Xf = {n, f (n), f, n : n ∈ ω}. Note that p, p = f ∈ωω Xf but there is no  countable subset X ⊂ f ∈ωω Xf such that p, p ∈ X. (Exercise! (Use the fact that, if A ⊆ ω ω is countable, then there is a function g : ω → ω such that, for each f ∈ A, the set {n ∈ ω : f (n) ≥ g(n)} is finite.)) But it can occur that X, at some point p, has the special property that if Y is countably tight at q, then X × Y is countably tight at p, q. In this case, we say that X is productively countably tight at p. In much the same spirit of Theorem 3.6, here we have another context in which the existence of a winning strategy for some player in a game yields productivity of an associated topological property. Namely: Proposition 5.21 (Gruenhage [17]). If Bob ↑ G1 (Ωp , Ωp ), then X is productively countably tight at p. Proof. Let Y be a space that is countably tight at q ∈ Y . Fix a winning strategy σ : X <ω → τp for Alice in the neighbourhood-point game (this is possible by Theorem 5.9). Let A = {xi , yi  : i ∈ I} ⊆ X × Y be such that p, q ∈ A. We will construct {Us : s ∈ ω <ω } ⊆ τp and {isn : s ∈ ω <ω and n ∈ ω} ⊆ I so that • q ∈ {yisn : n ∈ ω} for each s ∈ ω <ω and • for each f ∈ ω ω , U , xi , Uf (0) , xif (0) , Uf (0),f (1) , xif (0),f (1) , . . . , Uf k , xif k , . . .  f (0)

f (1)

f (k)

f (2)

is a play of the neighbourhood-point game in which Alice makes use of σ. What we are trying to do is: we will build a “tree of moves of the neighbourhood-point game”, where every node is indexed by a finite sequence of natural numbers (which naturally induces the order of the nodes of the tree) and every branch of the tree describes a play of the neighbourhood-point game. Then we will show that the (countably many) points played by Bob corresponding to the nodes of the tree cluster as required — in order to do so, we will fix a basic neighbourhood of p, q in X × Y , and then (by “climbing up the tree along the right branch”) we will be able to build a play of the neighbourhood-point game in which Alice plays according to σ, which will guarantee the required condition. For the construction, we will proceed by recursion. Suppose that k ∈ ω is such that {Us : s ∈ ω
 Ut = σ

xi , xit(0) , xit(0),t(1) , . . . , xit(k−1) t(0)

t(1)

t(2)

t(k−1)

.

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Note that q ∈ {yi : i ∈ I and xi ∈ Ut }; hence, as Y is countably tight at y, there is {itn : n ∈ ω} ⊆ {i ∈ I : xi ∈ Ut } such that q ∈ {yitn : n ∈ ω}. This completes the recursive construction. 

 We now claim that p, q ∈ xisn , yisn : s ∈ ω <ω and n ∈ ω — which will conclude the proof. Let U × V be an arbitrary basic open neighbourhood of p, q in X × Y . Pick nk ∈ ω recursively for k ∈ ω in such a way that y n0 ,...,nk−1  ∈ V . Now in k



 U , xi Un0  , xin0  Un0 ,n1  , xin0 ,n1  , . . . , Un0 ,...,nk−1  , x n0 ,...,nk−1  . . . n 0

n1

n2

ink

is a play of the neighbourhood-point game in which Alice follows the winning strategy σ, so there is k ∈ ω with x n0 ,...,nk−1  ∈ U . Hence in k

  x n0 ,...,nk−1  , y n0 ,...,nk−1  ∈ U × V, ink

ink

as required. 2 The reader may now suspect that the above proof can be adapted to the covering games we have discussed in the first sections. Indeed, the following is true: Theorem 5.22 (Telgársky [43]). If Alice has a winning strategy in the compact-open game on X, then X ×Y is Lindelöf for every Lindelöf space Y . Exercise 5.23. Prove Theorem 5.22. 6. An application to the D-space problem Let us now mention a surprising application of topological games to a problem apparently unrelated to them. Given a topological space X, we say that Vx : x ∈ X is an open neighbourhood assignment (o.n.a.) if each Vx is an open neighbourhood of x. We say that X is a D-space if for every o.n.a. Vx : x ∈ X there is  a closed discrete set D such that X = x∈D Vx . Working with D-spaces is more comfortable if every finite set is discrete, therefore in the next results all the spaces will be assumed to be at least T1 . The most important question about D-spaces is the following: Question 6.1 (van Douwen [12]). Is there a satisfactory example (in van Douwen’s words, “having a covering property at least as strong as metacompactness or subparacompactness”) of a space that is not D? Which, in the following years after the question was asked, turned to be more specific: Question 6.2. Is it true that every regular Lindelöf space is a D-space? Even though these two questions are still unanswered, a collection of results is already known: • • • •

Every Every Every Every

compact space is a D-space. metrizable space is a D-space [8]. second countable space is a D-space. space with a point-countable base is a D space [2].

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• The Sorgenfrey line L is a D-space [12]. • Ln is a D-space for every n ∈ ω [12]. But if we restrict ourselves to the class of the Lindelöf spaces that are not second countable, not that much is known: • Under ♦, there is a Hausdorff hereditarily Lindelöf non-D-space [41]. • Every Menger space is a D-space [4]. Let us show how the Menger property helps in this problem. In a sense, the proof goes very similar to the proof that every σ-compact is D. So let us first show this as a warm-up. Proposition 6.3. If X is σ-compact, then X is a D-space. Proof. Let Vx : x ∈ X be an o.n.a. and let Kn : n ∈ ω be a sequence of compact sets such that X =  n∈ω Kn .   Let D0 ⊆ K0 be a finite set such that x∈D0 Vx ⊃ K0 . Note that K1 \ x∈D0 Vx is compact. Therefore,    we can find a finite set D1 ⊆ K1 \ x∈D0 Vx such that x∈D0 Vx ∪ x∈D1 Vx ⊃ K0 ∪ K1 . We can proceed this way and cover K0 ∪ K1 ∪ K2 , and so on. The important things about this construction are the following: (i) every Dk is finite;   (ii) every Dk+1 ∩ j≤k x∈Dj Vx = ∅;   (iii) k∈ω x∈Dk Vx = X.  With these properties, we can prove that D = k∈ω Dk is the closed discrete set we were looking for.  Indeed, let x ∈ X. Since x ∈ d∈D Vd , there exist d and k such that x ∈ Vd and d ∈ Dk . Note that if a ∈ D ∩ Vd , then a ∈ Dj for some j ≤ k. Therefore, Vd ∩ D is finite. Since we are assuming that X is T1 , this means that D has no accumulation points. 2 So now we are going to implement the same idea, but for Menger spaces: Proposition 6.4 (Aurichi [4]). Every Menger space is D. Proof. Given an o.n.a. Vx : x ∈ X for X, we will define a strategy for Alice in Gfin (O, O). By Theorem 2.32, since the space is Menger, such a strategy cannot be a winning one and we can apply the same reasoning as we did in the σ-compact case. In other words, we are going to define a strategy that is good for what we want, but cannot be good for winning the game, since this is impossible by hypothesis. In the 0-th inning, Alice will play {Vx : x ∈ X}. Then Bob will choose a finite D0 ⊆ X — meaning that  he chooses {Vx : x ∈ D0 }. Let us call U0 = x∈D0 Vx . In the next inning, Alice plays {U0 ∪Vx : x ∈ X\U0 }. Note that this is a cover of X. Then, we can think that Bob picks D1 ⊆ X\U0 where the actual move in  the game is {U0 ∪ Vx : x ∈ D1 }. Then we define U1 = x∈D1 Vx and continue like this. Since this strategy cannot be a winning strategy, there is a way for Bob to play that gives us the same situation (i)–(iii) as in   the previous proof. Hence D = k∈ω Dk is closed discrete and X = x∈D Vx . 2

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Acknowledgements We would like to thank Piotr Szewczak, Boaz Tsaban and Lyubomyr Zdomskyy, the organizers of the conference Frontiers of Selection Principles, for inviting us to teach the minicourse on which these notes were based. We are also grateful to the referees for their helpful suggestions and remarks on a previous version of these notes. References [1] A.V. Arkhangel’ski˘ı, Frequency spectrum of a topological space and the classification of spaces, Dokl. Akad. Nauk SSSR 206 (1972) 265–268, English translation in Sov. Math. Dokl. 13 (1972) 1185–1189. [2] A.V. Arhangel’skii, R.Z. Buzyakova, Addition theorems and D-spaces, Comment. Math. Univ. Carol. 43 (4) (2002) 653–663. [3] L.F. Aurichi, Examples from trees, related to discrete subsets, pseudo-radiality and ω-boundedness, Topol. Appl. 156 (2009) 775–782. [4] L.F. Aurichi, D-spaces, topological games, and selection principles, Topol. Proc. 36 (2010) 107–122. [5] L.F. Aurichi, A. Bella, R.R. 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