# Compact complement topology and Rothberger game.

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.

• 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. Apr 24, 2019 at 3:58
• For a general overview of some topological games, you could see the following article arxiv.org/pdf/1306.5463.pdf Apr 24, 2019 at 4:11
• $(\mathbb{R},\tau^{*})$ 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 $\mathbb{R}\setminus U_{\alpha_{0}}$ Apr 24, 2019 at 4:22
• You are right, only that the property of being Lindelof helps to consider only open covers that are countable of the space. Apr 24, 2019 at 4:27

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 $$[0,3]\not\subseteq\bigcup_{n\in\omega}U_n$$.