Suppose, in a Gale-Stewart game, player I and player II choose from $\omega$ in a alternating fashion. If the outcome is in the winning set $W$, then player I wins. Otherwise player II wins. If $W$ is a closed but not open set in $\omega^{\omega}$, is it true that player II must have a winning strategy?
Some thoughts: Firstly, we only have to consider the case when $W$ is not countable, and we know $|W| = \mathfrak{c}$. The problem is reduced to how to characterize closed but not open sets with the cardinality equals the continuum. I don't know how to do this. I can only come up with some examples, say $A^{\omega}$, provided $A \subset \omega$ and $|A| \geq 2$, set of all permutations, and set of all injections from $\omega$ to $\omega$. It seems to me it holds in these examples. Is it true in general?
Consider the set $W$ of those sequences $x$ in $\omega^\omega$ in which $x_{2n}=0$ for all $n$; the even-numbered components $x_{2n}$ are those chosen by player I. So Player I has a trivial winning strategy: Just play 0 at every move. But $W$ is closed and not open.