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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 $(\mathbb{R}, \tau^{*})$.

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 $\langle\mathcal U_0,U_0,\mathcal U_1,U_1,\dots,\mathcal U_n,U_n,\dots\rangle$, the winner is Player II if $X\subseteq\bigcup_{n\in\omega}U_n$, and Player I otherwise.

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  • $\begingroup$ 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. $\endgroup$ – user 987 Apr 24 at 3:58
  • $\begingroup$ For a general overview of some topological games, you could see the following article arxiv.org/pdf/1306.5463.pdf $\endgroup$ – user 987 Apr 24 at 4:11
  • $\begingroup$ $(\mathbb{R},\tau^{*})$ is compact, because if you consider $\{U_{\alpha}\}_{\alpha \in\Lambda }$ an open cover of $\mathbb{R}$, then $\mathbb{R}\setminus U_{\alpha}$ is compact for every $\alpha\in\Lambda$, note that each $U_{\alpha}$ is open in the usual topology. Then for some $\alpha_{0}\in\Lambda$ we can cover by a finite number $\mathbb{R}\setminus U_{\alpha_{0}}$ $\endgroup$ – user 987 Apr 24 at 4:22
  • $\begingroup$ You are right, only that the property of being Lindelof helps to consider only open covers that are countable of the space. $\endgroup$ – user 987 Apr 24 at 4:27
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Here is a winning strategy for Player I. (So Player II of course has no winning strategy.)

In inning $n\in\omega$, Player I chooses $$\mathcal U_n=\{(a,b)\cup(-\infty,-10)\cup(10,\infty):a,b\in\mathbb R,\ 0\lt b-a\lt2^{-n}\}.$$

Clearly $\mathcal U_n\subseteq\tau^*$ and $\mathcal U_n$ covers $\mathbb R$.

For each $n\in\omega$, Player II chooses a set $U_n\in\mathcal U_n$.

Let $\lambda$ denote Lebesgue measure. Then $\lambda(U_n\cap[0,3])\lt2^{-n}$, whence $$\lambda\left(\bigcup_{n\in\omega}\left(U_n\cap[0,3]\right)\right)\lt\sum_{n=0}^\infty2^{-n}=2,$$ whence $[0,3]\not\subseteq\bigcup_{n\in\omega}U_n$.

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