Banach-Mazur game

In mathematics, in particular in general topology and set theory, a Banach-Mazur game is a game played between two players, trying to pin down elements in a set (space). The concept of a Banach-Mazur game is closely related to the technical concept of Baire spaces.

The general Banach-Mazur game is defined as follows:

Definition   We have a topological space X and a family Y of subsets of X satisfying the following properties:

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

A typical example is to let X equal the unit interval [0, 1] and let Y consist of all closed intervals [a, b] contained in [0, 1]. The game proceeds as follows. Let C be any subset of X. There are two players, Player I and Player II. Player I begins by choosing a member Y₁ of Y. Player II responds by choosing a member Y₂ of Y such that Y₂ is contained in Y₁. Then Player I responds by choosing Y₃ in Y with Y₃ contained in Y₂, and so on. In this way, we obtain a decreasing sequence of sets

<math>\cdots \subseteq Y_3 \subseteq Y_2 \subseteq Y_1 \subseteq X<math>

where Player I has chosen the sets with odd index and Player II has chosen the sets with even index. Player II wins the game if

<math>\bigcap_{n=1}^{\infty} Y_n \subseteq C.<math>

In other words, Player II wins if, after all the (infinitely many) choices are made, the set that remains lies entirely in C.

The question is then, "For what sets C does Player II have a winning strategy?" Clearly, if C = X, Player II has a winning strategy. So the question can be informally given as, "How 'big' does the set C have to be to insure that Player II has a winning strategy?", or equivalently, "How 'small' does the complement of C in X have to be in order to insure that Player II has a winning strategy?"

Let us show that Player II has a winning strategy if the complement of C in X is countable. [I am also assuming that X is T₁, is this necessary?, does anyone have proof not assuming this?] Let the elements of the complement of C be x₁, x₂, x₃,... Suppose Y₁ has been chosen by Player I. Let U₁ be the (non-empty) interior of Y₁. Then U₁ – {x₁} is a non-empty open set in X, so Player II can choose a member Y₂ of Y contained in U₁ – {x₁}. After Player I chooses a subset Y₃ of Y₂, Player II can choose a member Y₄ of Y contained in Y₃ that excludes the point x₂ in a similar fashion. Continuing in this way, each point x_n will be excluded by the set Y_2n, and so the intersection of all the Y_n's will be contained in C.

A set is "small" in a set-theoretic sense if it is countable. The corresponding notion for topology is the concept of a set being of the first category or meagre. A set is meagre if it is the countable union of nowhere dense sets. It turns out that this is just how small the complement of C has to be in order for Player II to have a winning strategy. In other words,

Let X be a topological space, let Y be a family of subsets of X satisfying the two properties above, and let C be any subset of X. Then Player II has a winning strategy for the Banach-Mazur game if and only if the complement of C in X is meagre.

Reference

A proof can be found in

• 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