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