Conceptual proofs of the Menger and Rothberger games
aa r X i v : . [ m a t h . GN ] A p r CONCEPTUAL PROOFS OF THE MENGER AND ROTHBERGERGAMES
PIOTR SZEWCZAK AND BOAZ TSABAN
Abstract.
We provide conceptual proofs of the two most fundamental theorems concerningtopological games and open covers: Hurewicz’s Theorem concerning the Menger game, andPawlikowski’s Theorem concerning the Rothberger game. Introduction
Topological games form a major tool in the study of topological properties and theirrelations to Ramsey theory, forcing, function spaces, and other related topics. Excellentsurveys [11, 2], and a comprehensive bibliography for further inspection [3], are available.At the heart of the theory of selection principles [13], covering properties are defined by theability to diagonalize, in canonical ways, sequences of open covers. Each of these coveringproperties has an associated two-player game. Often the nonexistence of a winning strategyfor the first player in the associated game is equivalent to the original property, and thisforms a strong tool for establishing results concerning the original property.Allowing few potential exceptions, all proofs of results of this type use, as black boxes, twofundamental theorems. The original proofs of the fundamental theorems are technical anddifficult to digest, and the lack of their understanding has undoubtedly veiled deep insightsthat necessitate variations in the original proofs. We present here new, conceptual proofsof these theorems. These proofs build on the earlier proofs, but are intuitive and can begrasped without the need of technical verifications.2.
The Menger game
A topological space has
Menger’s property S fin (O , O) if, for each sequence of open covers, U , U , . . . , we can select finite sets F ⊆ U , F ⊆ F ,. . . whose union S n F n covers thespace. The symbol O in this notation indicates that we are provided with open covers, andneed to obtain an open cover. These properties were also considered for additional classesof open covers, in the realm of selection principles [13]. Menger’s game G fin (O , O) is a game for two players, Alice and Bob, with an inning pereach natural number n . In each inning, Alice picks an open cover of the ambient space and Bob selects finitely many members from this cover. Bob wins if the sets he selectedthroughout the game cover the space. If this is not the case, Alice wins.If Alice does not have a winning strategy in the game G fin (O , O), then S fin (O , O) holds.The converse implication is a deep theorem of Hurewicz [5, Theorem 10]. This theoremis often used within selection principles, and has applications in diverse contexts, such asD-spaces [1] and additive Ramsey theory [12]. We present here a conceptual proof of thistheorem. The proof uses the same initial simplifications as in Scheepers’s proof of Hurewicz’sTheorem [9, Theorem 13]. We add one further simplification that makes calculations easier,and an appropriate notion that goes through induction, and thus eliminates the necessity totrack the history of the game.
Theorem 1 (Hurewicz) . Let X be a space satisfying S fin (O , O) space. Then Alice does nothave a winning strategy in the game G fin (O , O) .Proof. Fix an arbitrary strategy for Alice in the game G fin (O , O).If there is a play in which Bob covers the space after finitely many steps, then we aredone. Thus, we assume that, in no position, a finite selection suffices, together with theearlier selections, to cover the space. Since the space X satisfies S fin (O , O), it is Lindel¨of.By restricting Bob’s moves to countable subcovers of Alice’s covers, we may assume thatAlice’s covers are countable. Given that, we may assume that Alice’s covers are increasing ,that is, of the form { U , U , . . . } with U ⊆ U ⊆ · · · , and that Bob selects a single set ineach move. Indeed, given a countable cover { U , U , . . . } , we can restrict Bob’s selections tothe form { U , U , . . . , U n } , for n ∈ N . Since Bob’s goal is just to cover the space, we maypretend that Bob is provided covers of the form { U , U ∪ U , U ∪ U ∪ U , . . . } , and if Bob selects an element U ∪· · ·∪ U n , he replies to Alice with the legal move { U , U , . . . ,U n } . If Bob manages to cover the space, this is not due to the unions.Finally, we may assume that for each reply { U , U , . . . } (with U ⊆ U ⊆ · · · ) of Alice’sstrategy to a move U , we have U = U . Indeed, we can transform the given cover into thecover { U, U ∪ U , U ∪ U , . . . } . If Bob chooses U , we provide Alice with the answer U , andif he chooses U ∪ U n , we provide Alice with the answer U n . Since Bob has already chosenthe set U , its addition in the new strategy does not help covering more points.With these simplifications, Alice’s strategy is identified with a tree of open sets, as follows:Alice’s initial move is an open cover { U , U , . . . } . If Bob replies U n , then Alice’s move is { U n, , U n, , . . . } . In general, if Bob replies U σ , for σ ∈ N k , then Alice’s move is an increasingopen cover U σ := { U σ (1) ,...,σ ( k ) , , U σ (1) ,...,σ ( k ) , , . . . } , with U σ = U σ (1) ,...,σ ( k ) , .The proof will reduce to the following concept. HE MENGER AND ROTHBERGER GAMES 3
Definition 2.
A countable cover U of a space X is a tail cover if the set of intersections ofcofinite subsets of U is an open cover of X .Equivalently, a cover { U , U , . . . } is a tail cover if the family n ∞ \ n =1 U n , ∞ \ n =2 U n , . . . o of intersections of cofinal segments of the cover is an open cover. Lemma 3.
Let n be a natural number. Define V n := S σ ∈ N n U σ . Then the family V n is a tailcover of X .Proof. The proof is by induction on n .The open cover V = U () is increasing, and thus the set of cofinite intersections is again V , an open cover of X .Let n be a natural number. For brevity, enumerate V n = { V , V , . . . } , and V n +1 = ∞ [ k =1 { V k , V k , . . . } , where V k = V k ⊆ V k ⊆ · · · . We assume, inductively, that the family V n is a tail cover of X . Let V be a cofinite subsetof V n +1 . For each natural number k , let m k be the minimal natural number with V km k ∈ V .Then T ( V ∩ { V k , V k , . . . } ) = V km k for all k , and m k = 1 for all but finitely many naturalnumbers k . Let I := { k : m k = 1 } , a cofinite subset of N . We have \ V = \ k ∈ N ( V ∩ { V k , V k , . . . } ) = \ k ∈ N V km k = \ k ∈ I V k ∩ \ k ∈ N \ I V km k , Since V n is a tail cover, the set T k ∈ I V k is open. The remaining part is a finite intersectionof open sets. Thus, the set T V is open.Let x ∈ X . Since V n is a tail cover, the set I := { k : x ∈ V k } is cofinite. For k ∈ N \ I ,let m k be the minimal natural number with x ∈ V km k . Then x ∈ T k ∈ I V k ∩ T k ∈ N \ I V km k and,as we saw, the latter set is an intersection of a cofinite subset of the family V n +1 . (cid:3) For each n , let V ′ n be the set of intersections of cofinite subsets of V n . Applying theproperty S fin (O , O) to the sequence V ′ , V ′ , . . . , Bob obtains cofinite sets W n ⊆ V n such that X = S n T W n . In the n -th inning, Alice provides Bob with a cover that is an infinitesubset of the family V n . Since the family W n is cofinite in V n , Bob can choose an element V n ∈ V n ∩ W n . Then X = S n V n , and Bob wins.This completes the proof of Hurewicz’s Theorem. (cid:3) P. SZEWCZAK AND B. TSABAN
To treat Rothberger’s game, we need a result slightly stronger than Hurewicz’s. Theoriginal proof of the following result, due to Pawlikowski [7, Lemma 1], is much more technicalthan the combination of the present proofs of Theorem 1 and Corollary 4.We recall that Menger’s property S fin (O , O) is preserved by countable unions: Given acountable union of Menger spaces, and a sequence of open covers, we can split the sequenceof covers into infinitely many disjoint subsequences, and use each subsequence to cover oneof the given Menger spaces.
Corollary 4 (Pawlikowski) . Let X be a space satisfying S fin (O , O) . For each strategy forAlice in the game G fin (O , O) , there is a play according to this strategy, ( U , F , U , F , . . . ) , such that for each point x ∈ X we have x ∈ S F n for infinitely many n .Proof. We apply a reduction of Scheepers, originally used to prove the analogous theoremfor the game considered in Section 3 [10, Theorem 3].The product space X × N , a countable union of Menger spaces, satisfies S fin (O , O). Wedefine a strategy for Alice in the game G fin (O , O), played on the space X × N . Let U beAlice’s first move in the original game. Then, in the new game, her first move is˜ U := { U × { n } : U ∈ U , n ∈ N } . If Bob selects a finite set ˜
F ⊆ ˜ U , we take the set F := n U ∈ U : there is n with U × { n } ∈ ˜ F o as a move in the original game. Then Alice replies with a cover V , and we continue in thesame manner. By Hurewicz’s Theorem, there is a play( ˜ U , ˜ F , ˜ U , ˜ F , . . . )in the new game, with S n ˜ F n a cover of X × N . Consider the corresponding play in theoriginal strategy, ( U , F , U , F , . . . ) . Let x ∈ X . There is a natural number n with ( x, ∈ S ˜ F n . Then x ∈ S F n . The set F := ( k ∈ N : there is U with U × { k } ∈ n [ i =1 ˜ F i ) is finite. Let m be a natural number greater than all elements of the set F . There is anatural number n with ( x, m ) ∈ S ˜ F n . Then x ∈ S F n , and n < n . Continuing in asimilar manner, we see that x ∈ S F n for infinitely many n . (cid:3) HE MENGER AND ROTHBERGER GAMES 5
Remark . A cover of a space is large if each point is covered by infinitely many membersof the cover. Let Λ be the family of all large covers of an ambient space X . We have S fin (O , O) = S fin (Λ , Λ) ([9, Corollary 5], [6, Theorem 1.2]). With some initial simplificationsof the considered strategies, Corollary 4 implies that a topological space X satisfies S fin (Λ , Λ)if and only if Alice does not have a winning strategy in the corresponding game G fin (Λ , Λ).In fact, these results are essentially identical.3.
The Rothberger game
The definitions of
Rothberger’s property S (O , O) and the corresponding game G (O , O)are similar to those of S fin (O , O) and G fin (O , O), respectively, but here we select one elementfrom each cover. Here too, if Alice does not have a winning strategy then the space satisfies S (O , O). The converse implication was established by Pawlikowski [7, p. 279], improvingconsiderably over partial results of Galvin [4, Corollary 4] and Rec law [8, Corollary 2].We provide a conceptual proof of Pawlikowski’s Theorem. We first isolate an argument inPawlikowski’s proof, that does not involve games.
Lemma 6 (Pawlikowski) . Let X be a space satisfying S (O , O) . Let F , F , . . . be nonemptyfinite families of open sets such that, for each point x ∈ X , we have x ∈ S F n for infinitelymany n . Then there are elements U ∈ F , U ∈ F , . . . such that the family { U , U , . . . } covers the space X .Proof. For each natural number n , let U n be the family of all intersections of n open setstaken from distinct members of the sequence F , F , . . . . Then U n is an open cover of X .By the property S (O , O), there are elements V ∈ U , V ∈ U , . . . that cover the space X .Extend the set V to an element of some family F n . We can extend V to an element of some other family F n , and so on. We obtain a selection of at most element from each family F n ,that covers X . We can extend our selection to have an element from each family F n . (cid:3) For a natural number k and families of sets U , . . . , U k , let U ∧ · · · ∧ U k := { U ∩ · · · ∩ U k : U ∈ U , . . . , U k ∈ U k } . Theorem 7 (Pawlikowski) . Let X be a space satisfying S (O , O) . Then Alice does not havea winning strategy in the game S (O , O) .Proof. Fix an arbitrary strategy for Alice in the Rothberger game G (O , O). Since S (O , O)spaces are Lindel¨of, we may assume that each cover in the strategy is countable. Let N < ∞ be the set of finite sequences of natural numbers. We index the open covers in the strategyas U σ = { U σ, , U σ, , . . . } , P. SZEWCZAK AND B. TSABAN for σ ∈ N < ∞ , so that U = { U , U , . . . } is Alice’s first move, and for each finite sequence k , . . . , k n of natural numbers, U k ,...,k n is Alice’s reply to the position( U , U k , U k , U k ,k , U k ,k , . . . , U k ,...,k n ) . For finite sequences τ, σ ∈ N n , we write τ ≤ σ if τ ( i ) ≤ σ ( i ) for all i = 1 , . . . , n . Wedefine a strategy for Alice in the Menger game S fin (O , O). Alice’s first move is U , her firstmove in the original strategy. Assume that Bob selects a finite subset F of U . Let m bethe minimal natural number with F ⊆ { U , . . . , U m } . Then, in the Menger game, Alice’sresponse is the joint refinement U ∧ · · · ∧ U m . Assume that Bob chooses a finite subset F of this refinement. Let m be the minimal natural number such that F refines all sets { U i, , . . . , U i,m } , for i = 1 , . . . , m . Then Alice’s reply is the joint refinement V τ ≤ ( m ,m ) U τ .In general, Alice provides a cover of the form V τ ≤ σ U τ , for σ ∈ N < ∞ , Bob selects a finitefamily refining all families { U τ, , . . . , U τ,m } for τ ≤ σ , whit the minimal natural number m ,and Alice replies V τ ≤ ( σ,m ) U τ .By Pawlikowski’s Theorem (Corollary 4), there is a play( U , F , ^ k ≤ m U k , F , ^ ( k ,k ) ≤ ( m ,m ) U k ,k , . . . ) , according to the new strategy, such that every point of the space is covered infinitely oftenin the sequence S F , S F , . . . . By Lemma 6, we can pick one element from each set F n and cover the space. There is k ≤ m such that the first picked element is a subset of U k .There is k ≤ m such that the second picked element is a subset of U k ,k , and so on. Thenthe play ( U , U k , U k , U k ,k , . . . )is in accordance with Alice’s strategy in the Rothberger game, and is won by Bob. (cid:3) Acknowledgments.
We thank Micha l Machura and Jialiang He for helpful discussions thathelped us understand Pawlikowski’s proof better. The first named author thanks the secondfor his hospitality during 2016, where the present proof of Hurewicz’s Theorem was obtained.The second named author thanks the first for his hospitality beyond the conference
Frontiersof Selection Principles (Warsaw, 2017), where the main breakthroughs leading to the presentproof of Pawlikowski’s Theorem were made. A preliminary version of this paper was quotedin a survey of Aurichi and Dias [2], and this led to a “flood” of requests by colleagues tosee these notes, and this urged us to complete this paper. We thank our colleagues for theirenthusiasm.
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Selection Principle , en.wikipedia.org/wiki/Selection_principle Piotr Szewczak, Institute of Mathematics, Faculty of Mathematics and Natural ScienceCollege of Sciences, Cardinal Stefan Wyszy´nski University in Warsaw, Warsaw, Poland,and Department of Mathematics, Bar-Ilan University, Ramat Gan, Israel
E-mail address : [email protected] URL : http://piotrszewczak.pl Boaz Tsaban, Department of Mathematics, Bar-Ilan University, Ramat Gan, Israel
E-mail address : [email protected] URL ::