Is there a way to connect filter convergence to the condition for player 1/2 to win a Banach-Mazur game in an if (and only if) fashion? Thanks! Details below...
A Banach-Mazur game is defined as follows. We operate in a topological space $Y$ and then consider a fixed subset $X \subset Y$ and a family $\mathcal{W}$ of subsets such that each member has non-empty interior and each non-empty subset of $Y$ contains a member of $\mathcal{W}$.
Say two players alternate and choose elements from $\mathcal{W}$ such that, $$ W_0 \supseteq W_1 \supseteq \cdots $$ We say that the second player wins iff, $$ X \cap \left( \bigcap_{n < \omega} W_n \right) = \emptyset $$ otherwise, the first player wins.
It seems we could also work in an alternative paradigm. Say $Y$ has no isolated points, $X$ is countable, and every ultrafilter converges to at least one point then the player two wins. However, I don't know how to proceed further or even if what I'm trying to do makes sense.