# Banach-Mazur game and the Baire property

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?

Thanks!

• 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?
– user762914
May 17, 2020 at 15:53

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.