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



Rothberger game and Meager in itself sets
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:


<https://math.stackexchange.com/questions/3200075/compact-complement-topology-and-rothberger-game/3200150#3200150>


