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 '20 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.