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Given a Tychonoff space $X$, let $\Omega$ be the set of all $\omega$-coverings of $X$, where by $\omega$-cover we mean a collection $\mathcal{U}$ of open sets of $X$ such that for any $F\in[X]^{<\omega}$ there exists $U\in\mathcal{U}$ such that $F\subset U$.

Now, let G$_1(\Omega,\Omega)$ be the game between Player I and Player II defined as follows: for every inning $n<\omega$, the Player I chooses an $\omega$-cover $\mathcal{U}_n$, and then Player II picks $U_n\in\mathcal{U}_n$; Player II wins iff $\{U_n:n\in\omega\}\in\Omega$.

Clearly, if the Player II has a winning strategy in the game $\mathrm{G}_1(\Omega,\Omega)$, then the Player I does not have a winning strategy in the same game.

I would like to know any references with examples of spaces where the game $\mathrm{G}_1(\Omega,\Omega)$ is undetermined, i.e., such that both players do not have winning strategies in the game $\mathrm{G}_1(\Omega,\Omega)$.

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  • $\begingroup$ Is there an easy example of an uncountable space in which Player I does not have a winning strategy? Player II wins in any countable space by simply diagonalizing against $[X]^{<\omega}$ $\endgroup$ – Carl Mummert Apr 28 '17 at 0:17
  • $\begingroup$ @CarlMummert None that I am aware of. Here, the authors shows that under CH there exists a Lusin set satisfying $\mathrm{S}_1(\Omega,\Omega)$, which is equivalent to Player I does not having a winning strategy. $\endgroup$ – Renan Maneli Mezabarba Apr 28 '17 at 0:48
  • $\begingroup$ Thanks, that is helpful. I notice that they specifically require their covers to be countable, and require $\omega$-covers not to contain $X$, neither of which is required here. I don't know yet if that makes a difference. $\endgroup$ – Carl Mummert Apr 28 '17 at 0:53
  • $\begingroup$ Well, there's no loss of generality in supposing that any $\omega$-cover has a countable $\omega$-subcover (if not, then the Player I get a winning strategy by simply playing always the same $\omega$-cover). Same thing with the assumption about $X$: it is not smart to Player I to play $\omega$-covers containing $X$, so we way exclude this in the definition without loss of generality. $\endgroup$ – Renan Maneli Mezabarba Apr 28 '17 at 0:58
  • $\begingroup$ I see what you mean about $X$, but is it true in general that every $\omega$-cover has a countable $\omega$-subcover? Or, for which spaces is this the case? Otherwise, as you say, Player I would have a winning strategy, but that's what I am hoping to see. $\endgroup$ – Carl Mummert Apr 28 '17 at 1:01

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