Banach–Mazur game

Let ${\displaystyle Y}$ be a non-empty topological space, ${\displaystyle X}$ a fixed subset of ${\displaystyle Y}$ and ${\displaystyle {\mathcal {W}}}$ a family of subsets of ${\displaystyle Y}$ that have the following properties:

• Each member of ${\displaystyle {\mathcal {W}}}$ has non-empty interior.
• Each non-empty open subset of ${\displaystyle Y}$ contains a member of ${\displaystyle {\mathcal {W}}}$.

Players, ${\displaystyle P_{1}}$ and ${\displaystyle P_{2}}$ alternately choose elements from ${\displaystyle {\mathcal {W}}}$ to form a sequence ${\displaystyle W_{0}\supseteq W_{1}\supseteq \cdots .}$

${\displaystyle P_{1}}$ wins if and only if

${\displaystyle X\cap \left(\bigcap _{n<\omega }W_{n}\right)\neq \emptyset .}$

Otherwise, ${\displaystyle P_{2}}$ wins. This is called a general Banach–Mazur game and denoted by ${\displaystyle MB(X,Y,{\mathcal {W}}).}$

• ${\displaystyle P_{2}}$ 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

${\displaystyle \bigcap _{n<\omega }W_{n}\neq \emptyset .}$

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 ${\displaystyle X\cap (\cap _{n<\omega }J_{n})\neq \emptyset }$ and ${\displaystyle P_{2}}$ wins if and only if ${\displaystyle X\cap (\cap _{n<\omega }J_{n})=\emptyset }$. This game is denoted by ${\displaystyle MB(X,J).}$

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.

• Choquet game

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

• "Banach–Mazur game", Encyclopedia of Mathematics, EMS Press, 2001 [1994]