Given a topological space $\mathcal{X}=(X,\tau)$ and $A\subseteq X$, the Banach-Mazur game on of $A$, $G^{**}(A)$, is the game played as follows:

  • Players $1$ and $2$ alternately play decreasing nonempty open sets $U_0\supseteq V_0\supseteq U_1\supseteq V_1\supseteq ...$.

  • Player $2$ wins iff $\bigcap_{i\in\mathbb{N}} V_i\subseteq A$.

Now a theorem say the following:

Let $\mathcal{X}=(X,\tau)$ be a Polish space. Then:

  • $A$ is comeager iff Player $2$ has a winning strategy in $G^{**}(A)$
  • If $A$ is meager in some non-empty open subset iff Player $1$ has a winning strategy in $G^{**}(A) $

I want to solve the following Kechris' exercise:

Given $X$ a Polish space then $A\subseteq X$ has the Baire property iff for all open $U$ the game $G^{**}(\sim A\cup U )$ is determined (i.e. one of the two players has a winning strategy)

I think that by $\sim A\cup U$ he meant $(X\setminus A)\cup U$, but I'm not sure. I tried to prove this fact but I don't get much further. I would have used the game $G^{**}(\sim(A\Delta U))$ since I want to prove that $A\Delta U$ is meager for some open $U$, but probably the two games (mine and the one given by Kechris) are equivalent for this purpose.

So I think that the way to do this is to show that Player $1$ cannot win every such game, hence there is an $U$ such that Player $2$ wins the game and therefore $A\Delta U$ is meager.

Any help?


  • $\begingroup$ This is a topological version of the game Nim. Aren't you constructing a $G_\delta$ set, so that the limit is comeagre? Or at least, doesn't that connect the sequence of moves to the winning conditions? $\endgroup$
    – user762914
    May 17, 2020 at 15:53

1 Answer 1


Actually, it was your first guess: $$ \mathop{\sim} A\cup U = (\mathop{\sim} A)\cup U = (X\setminus A)\cup U = X \setminus (A\setminus U). $$ The point is that you can always choose a canonical open $U$ such that $U\setminus A$ is meager:

You take for this the selector $$U(A):= \textstyle\bigcup\{ U\text{ open} : U\setminus A\text{ is meager}\}.$$

This choice ensures that Player 1 can't have a winning strategy. Therefore Player 2 has a winning strategy, $A\mathbin{\triangle} U$ is meager and you are done.


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