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:
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].
- Each member of Y has non-empty interior.
- Each non-empty open subset of X contains a member of Y.
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
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
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 xn will be excluded by the set Y2n, and so the intersection of all the Yn'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.
A proof can be found in