Someone know examples of topological spaces of first category and in which Player II has a winning strategy in the Rothberger game?

Remember that:

The Rothberger game on a topological space $X$ is played according to the following rules:

In each inning $n\in\omega$, Player I chooses an open cover $\mathcal{U_n}$ of $X$, and then Player II picks an open set $U_{n}\in\mathcal{U}_{n}$. At the end of the play $\langle \mathcal{U}_{0}, U_{0}, \mathcal{U}_{1}, U_{1}, ..., \mathcal{U}_{n}, U_{n}, .... \rangle $. The winner is Player II if $X\subseteq\bigcup_{n\in\omega}U_n$, and Player I otherwise.



Again, it seems the right order topology on $\Bbb R$ works. Meagerness is verified here. The basis elements of this topology are $B_y=\{x\in\Bbb R\mid x>y\}$. At the first stage, choose an open set $U_1$ containing $-1$. It must be either an open ray $B_y$ such that $y<-1$, or $U_1=\Bbb R$. In the latter case, we've won. At the $n$-th stage, choose $U_n$ containing $-n$, the same reasoning goes through. It's easy to see $\Bbb R= \bigcup_{n\in\omega}\{x\in \Bbb R\mid x>-n\}\subset\bigcup_{n\in \omega}U_n$.

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  • $\begingroup$ @user987 I will check it out when I'm available. Or you can ask a new question to seek other's answers. $\endgroup$ – YuiTo Cheng Apr 24 '19 at 3:40
  • $\begingroup$ @user987 Actually, $\Bbb Q$ is meager and countable. And every countable space is Rothberger. (if you don't care countability) $\endgroup$ – YuiTo Cheng Apr 24 '19 at 3:50
  • $\begingroup$ thanks YuiTo Cheng, of course but I was looking for uncountable topological spaces. At first I thought that if the space is meager in itself and Player II has winning strategy in the Rothberger game, then the space had to be countable. $\endgroup$ – user 987 Apr 24 '19 at 3:56

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