The general Banach-Mazur game is defined as follows:

DefinitionWe have a topological spaceXand a familyYof subsets ofXsatisfying the following properties:

A typical example is to let

- Each member of
Yhas non-empty interior.- Each non-empty open subset of
Xcontains a member ofY.

Xequal the unit interval [0, 1] and letYconsist of all closed intervals [a,b] contained in [0, 1].The game proceeds as follows. Let

Cbe any subset ofX. There are two players, Player I and Player II. Player I begins by choosing a memberY_{1}ofY. Player II responds by choosing a memberY_{2}ofYsuch thatY_{2}is contained inY_{1}. Then Player I responds by choosingY_{3}inYwithY_{3}contained inY_{2}, 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 T_{1}, is this necessary?, does anyone have proof not assuming this?] Let the elements of the complement of *C* be *x*_{1}, *x*_{2}, *x*_{3},... Suppose *Y*_{1} has been chosen by Player I. Let *U*_{1} be the (non-empty) interior of *Y*_{1}. Then *U*_{1} - {*x*_{1}} is a non-empty open set in *X*, so Player II can choose a member *Y*_{2} of *Y* contained in *U*_{1} - {*x*_{1}}. After Player I chooses a subset *Y*_{3} of *Y*_{2}, Player II can choose a member *Y*_{4} of *Y* contained in *Y*_{3} that excludes the point *x*_{2} 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,

LetXbe a topological space, letYbe a family of subsets ofXsatisfying the two properties above, and letCbe any subset ofX. Then Player II has a winning strategy for the Banach-Mazur game if and only if the complement ofCinXis meagre.

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