# Rothberger game and Meager in itself sets

On $$(\mathbb{R}, \tau)$$ the euclidean space of real numbers, we define a new topology by letting $$\tau^{*}=\{X\subseteq \mathbb{R}: X=\emptyset \hspace{0.1cm}\mbox{or}\hspace{0.1cm}\mathbb{R}\setminus X\hspace{0.1cm}\mbox{is}\hspace{0.1cm} \mbox{compact} \hspace{0.1cm} \mbox{in}\hspace{0.1cm} (\mathbb{R}, \tau) \}$$, it is known that $$(\mathbb{R}, \tau^{*})$$ is Lindelöf, meager in itself, my question is if Player II has a winning strategy in the Rothberger game played in $$(\mathbb{R}, \tau^{*})$$.

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.

Thank you

• Crosspost from MSE Apr 24, 2019 at 6:38
• Crossposted too soon. This question has already been answered on math.SE. Player I has a winning strategy, not Player II.
– bof
Apr 24, 2019 at 10:05
• Since the question has been crossposted from math.SE, should I crosspost the easy answer from math.SE too?
– bof
Apr 24, 2019 at 10:07
• @bof I think it is more consistent with the trends/customs to close this question and avoid unnecessary duplication of content Apr 24, 2019 at 14:27
• I'm voting to close this question for the reason given in my previous comment Apr 24, 2019 at 14:27

The game is a win for Player I, not Player II. A winning strategy for Player I is to choose, in inning $$n$$, the open cover $$\mathcal U_n=\{(a,b)\cup(-\infty,-10)\cup(10,\infty):a,b\in\mathbb R,\ 0\lt b-a\lt2^{-n}\}.$$ If more details are wanted, see the answer I posted to the same question at math.SE: