Banach–Mazur game

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In general topology, set theory and game theory, a BanachMazur game is a topological game played by two players, trying to pin down elements in a set (space).  The concept of a Banach–Mazur game is closely related to the concept of Baire spaces.  This game was the first infinite positional game of perfect information to be studied. It was introduced by Mazur as problem 43 in the Scottish book, and Mazur's questions about it were answered by Banach.

Definition and properties[edit]

In what follows we will make use of the formalism defined in Topological game.  A general Banach–Mazur game is defined as follows: we have a topological space Y, a fixed subset X \subset Y, and a family W of subsets of Y that satisfy the following properties.

  • Each member of W has non-empty interior.
  • Each non-empty open subset of Y contains a member of W.

We will call this game MB(X,Y,W).  Two players, P_1 and P_2, choose alternatively elements W_0, W_1, \cdots of W such that W_0 \supset W_1 \supset \cdots.  The player P_1 wins if and only if X \cap (\cap_{n<\omega} W_n) \neq \emptyset.

The following properties hold.

  • P_2 \uparrow MB(X,Y,W) if and only if X is of the first category in Y (a set is of the first category or meagre if it is the countable union of nowhere-dense sets).
  • Assuming that Y is a complete metric space, P_1 \uparrow MS(X,Y,W) if and only if X is comeager in some nonempty open subset of Y.
  • If X has the Baire property in Y, then MB(X,Y,W) is determined.
  • Any winning strategy of P_2 can be reduced to a stationary winning strategy.
  • The siftable and strongly-siftable spaces introduced by Choquet can be defined in terms of stationary strategies in suitable modifications of the game.  Let BM(X) denote a modification of MB(X,Y,W) where X=Y, W is the family of all nonempty open sets in X, and P_2 wins a play (W_0, W_1, \cdots) if and only if \cap_{n<\omega} W_n \neq \emptyset.  Then X is siftable if and only if P_2 has a stationary winning strategy in BM(X).
  • A Markov winning strategy for P_2 in BM(X) can be reduced to a stationary winning strategy.  Furthermore, if P_2 has a winning strategy in BM(X), then she has a winning strategy depending only on two preceding moves.  It is still an unsettled question whether a winning strategy for P_2 can be reduced to a winning strategy that depends only on the last two moves of P_1.
  • X is called weakly \alpha-favorable if P_2 has a winning strategy in BM(X).  Then, X is a Baire space if and only if P_1 has no winning strategy in BM(X).  It follows that each weakly \alpha-favorable space is a Baire space.

Many other modifications and specializations of the basic game have been proposed: for a thorough account of these, refer to [1987].  The most common special case, called MB(X,J), consists in letting Y = J, i.e. the unit interval [0,1], and in letting W consist of all closed intervals [a,b] contained in [0,1].  The players choose alternatively subintervals J_0, J_1, \cdots of J such that J_0 \supset J_1 \supset \cdots, and P_1 wins if and only if X \cap (\cap_{n<\omega} J_n) \neq \emptysetP_2 wins if and only if X\cap (\cap_{n<\omega} J_n) = \emptyset.

A simple proof: winning strategies[edit]

It is natural to ask for what sets X does P_2 have a winning strategy.  Clearly, if X is empty, P_2 has a winning strategy, therefore the question can be informally rephrased as how "small" (respectively, "big") does X (respectively, the complement of X in Y) have to be to ensure that P_2 has a winning strategy.  To give a flavor of how the proofs used to derive the properties in the previous section work, let us show the following fact.

Fact: P_2 has a winning strategy if X is countable, Y is T1, and Y has no isolated points.

Proof: Let the elements of X be x_1, x_2, \cdots.  Suppose that W_1 has been chosen by P_1, and let U_1 be the (non-empty) interior of W_1.  Then U_1 \setminus \{x_1\} is a non-empty open set in Y, so P_2 can choose a member W_2 of W contained in this set.  Then P_1 chooses a subset W_3 of W_2 and, in a similar fashion, P_2 can choose a member W_4 \subset W_3 that excludes x_2.  Continuing in this way, each point x_n will be excluded by the set W_{2n}, so that the intersection of all the W_n will have empty intersection with XQ.E.D

The assumptions on Y are key to the proof: for instance, if Y=\{a,b,c\} is equipped with the discrete topology and W consists of all non-empty subsets of Y, then P_2 has no winning strategy if X=\{a\} (as a matter of fact, her opponent has a winning strategy).  Similar effects happen if Y is equipped with indiscrete topology and W=\{Y\}.

A stronger result relates X to first-order sets.

Fact: Let Y be a topological space, let W be a family of subsets of Y satisfying the two properties above, and let X be any subset of YP_2 has a winning strategy if and only if X is meagre.

This does not imply that P_1 has a winning strategy if X is not meagre.  In fact, P_1 has a winning strategy if and only if there is some W_i \in W such that X \cap W_i is a comeagre subset of W_i.  It may be the case that neither player has a winning strategy: when Y is [0,1] and W consists of the closed intervals [a,b], the game is determined if the target set has the property of Baire, i.e. if it differs from an open set by a meagre set (but the converse is not true).  Assuming the axiom of choice, there are subsets of [0,1] for which the Banach–Mazur game is not determined.

References[edit]

[1957] Oxtoby, J.C. The Banach–Mazur game and Banach category theorem, Contribution to the Theory of Games, Volume III, Annals of Mathematical Studies 39 (1957), Princeton, 159–163

[1987] Telgársky, R. J. Topological Games: On the 50th Anniversary of the Banach–Mazur Game, Rocky Mountain J. Math. 17 (1987), pp. 227–276.[1] (3.19 MB)

[2003] Julian P. Revalski The Banach-Mazur game: History and recent developments, Seminar notes, Pointe-a-Pitre, Guadeloupe, France, 2003-2004 [2]