4
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

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.