Banach–Mazur game

Properties

• ${\displaystyle P_2}$ has a winning strategy if and only if ${\displaystyle X}$ is of the first category in ${\displaystyle Y}$ (a set is of the first category or meagre if it is the countable union of nowhere-dense sets).
• If ${\displaystyle Y}$ is a complete metric space, ${\displaystyle P_1}$ has a winning strategy if and only if ${\displaystyle X}$ is comeager in some non-empty open subset of ${\displaystyle Y.}$
• If ${\displaystyle X}$ has the Baire property in ${\displaystyle Y}$, then ${\displaystyle MB(X,Y,\mathcal{W})}$ is determined.
• The siftable and strongly-siftable spaces introduced by Choquet can be defined in terms of stationary strategies in suitable modifications of the game. Let ${\displaystyle BM(X)}$ denote a modification of ${\displaystyle MB(X,Y,\mathcal{W})}$ where ${\displaystyle X=Y,\mathcal{W}}$ is the family of all non-empty open sets in ${\displaystyle X}$ and ${\displaystyle P_2}$ wins a play ${\displaystyle (W_0,W_1,\cdots )}$ if and only if $${\displaystyle \bigcap _{n<\omega }{W}_n\ne \mathrm{\varnothing }.}$$ Then ${\displaystyle X}$ is siftable if and only if ${\displaystyle P_2}$ has a stationary winning strategy in ${\displaystyle BM(X).}$

• A Markov winning strategy for ${\displaystyle P_2}$ in ${\displaystyle BM(X)}$ can be reduced to a stationary winning strategy. Furthermore, if ${\displaystyle P_2}$ has a winning strategy in ${\displaystyle BM(X)}$, then ${\displaystyle P_2}$ has a winning strategy depending only on two preceding moves. It is still an unsettled question whether a winning strategy for ${\displaystyle P_2}$ can be reduced to a winning strategy that depends only on the last two moves of ${\displaystyle P_1}$.
• ${\displaystyle X}$ is called weakly ${\displaystyle \alpha }$-favorable if ${\displaystyle P_2}$ has a winning strategy in ${\displaystyle BM(X)}$. Then, ${\displaystyle X}$ is a Baire space if and only if ${\displaystyle P_1}$ has no winning strategy in ${\displaystyle BM(X)}$. It follows that each weakly ${\displaystyle \alpha }$-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 arises when ${\displaystyle Y=J=[0,1]}$ and ${\displaystyle \mathcal{W}}$ consist of all closed intervals in the unit interval. Then ${\displaystyle P_1}$ wins if and only if ${\displaystyle X\cap ({\cap }_{n<\omega }{J}_n)\ne \mathrm{\varnothing }}$ and ${\displaystyle P_2}$ wins if and only if ${\displaystyle X\cap ({\cap }_{n<\omega }{J}_n)=\mathrm{\varnothing }}$. This game is denoted by ${\displaystyle MB(X,J).}$

A simple proof: winning strategies

It is natural to ask for what sets ${\displaystyle X}$ does ${\displaystyle P_2}$ have a winning strategy in ${\displaystyle MB(X,Y,\mathcal{W})}$. Clearly, if ${\displaystyle X}$ is empty, ${\displaystyle P_2}$ has a winning strategy, therefore the question can be informally rephrased as how "small" (respectively, "big") does ${\displaystyle X}$ (respectively, the complement of ${\displaystyle X}$ in ${\displaystyle Y}$) have to be to ensure that ${\displaystyle P_2}$ has a winning strategy. The following result gives a flavor of how the proofs used to derive the properties in the previous section work:

Proposition. ${\displaystyle P_2}$ has a winning strategy in ${\displaystyle MB(X,Y,\mathcal{W})}$ if ${\displaystyle X}$ is countable, ${\displaystyle Y}$ is T₁, and ${\displaystyle Y}$ has no isolated points.

Proof. Index the elements of X as a sequence: ${\displaystyle x_1,x_2,\cdots .}$ Suppose ${\displaystyle P_1}$ has chosen ${\displaystyle W_1,}$ if ${\displaystyle U_1}$ is the non-empty interior of ${\displaystyle W_1}$ then ${\displaystyle U_1\setminus \{x_1\}}$ is a non-empty open set in ${\displaystyle Y,}$ so ${\displaystyle P_2}$ can choose ${\displaystyle \mathcal{W}\ni W_2\subset U_1\setminus \{x_1\}.}$ Then ${\displaystyle P_1}$ chooses ${\displaystyle W_3\subset W_2}$ and, in a similar fashion, ${\displaystyle P_2}$ can choose ${\displaystyle W_4\subset W_3}$ that excludes ${\displaystyle x_2}$. Continuing in this way, each point ${\displaystyle x_n}$ will be excluded by the set ${\displaystyle W_{2n},}$ so that the intersection of all ${\displaystyle W_n}$ will not intersect ${\displaystyle X}$.

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

A stronger result relates ${\displaystyle X}$ to first-order sets.

Proposition. ${\displaystyle P_2}$ has a winning strategy in ${\displaystyle MB(X,Y,\mathcal{W})}$ if and only if ${\displaystyle X}$ is meagre.

This does not imply that ${\displaystyle P_1}$ has a winning strategy if ${\displaystyle X}$ is not meagre. In fact, if ${\displaystyle Y}$ is a complete metric space, then ${\displaystyle P_1}$ has a winning strategy if and only if there is some ${\displaystyle W_i\in \mathcal{W}}$ such that ${\displaystyle X\cap W_i}$ is a comeagre subset of ${\displaystyle W_i.}$ It may be the case that neither player has a winning strategy: let ${\displaystyle Y}$ be the unit interval and ${\displaystyle \mathcal{W}}$ be the family of closed intervals in the unit interval. 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 the unit interval for which the Banach–Mazur game is not determined.