Note: This question is being posted primarily as a reference to include for https://topology.pi-base.org/ following the guidelines for references as spelled out at https://github.com/pi-base/data/blob/master/CONTRIBUTING.md.
A Fortissimo space is defined as follows, as in https://en.wikipedia.org/wiki/Counterexamples_in_Topology: let $Y$ be an uncountable set and suppose $\infty \not\in Y$. Then let $X = Y \cup \{ \infty \}$ have the topology $\{ U \subseteq X : U \subseteq Y \vee (\infty \in U \wedge X \setminus U \text{ is countable} \}.$ When the uncountable set $Y$ is the continuum, this is https://topology.pi-base.org/spaces/S000022.
A space is Rothberger https://en.wikipedia.org/wiki/Rothberger_space if it satisfies the selection principle $\mathsf S_1(\mathcal O, \mathcal O)$: for every sequence $\langle \mathscr U_n : n \in \omega \rangle$ of open covers of $X$, there is a selection $U_n \in \mathscr U_n$ for each $n \in \omega$ so that $\{ U_n : n \in \omega \}$ is a cover of $X$.
It is easy to see that any Fortissimo space is Rothberger since you take the initial selection to be any neighborhood of $\infty$. Then you run through an enumeration of the countable complement of that neighborhood.
There are games naturally corresponding to selection principles as discussed in the Wikipedia article on Rothberger spaces linked above. Indeed, the procedure described above is a winning strategy for the second player in the Rothberger game.
One can also define stronger strategies like Markov strategies for the second player which only depend on the current play by the first player, and no other previous information.
One can also investigate other cover types like $\omega$-cover and $k$-covers. $\omega$-covers (resp. $k$-covers) of a space $X$ are open covers so that every finite (resp. compact) subset of $X$ is contained in a member of the cover and $X$ itself is not a member of the cover. Since $X$ is anticompact (https://topology.pi-base.org/properties/P000136), the $\omega$-covers and $k$-covers coincide.
Can the discussion above be improved to Markov strategies and/or include $\omega$-covers?
Best Answer
The Fortissimo space $X$ has the property that the second player has a winning strategy in the game $\mathsf G_1(\Omega, \Omega)$, but no winning Markov strategy. The game $\mathsf G_1(\Omega, \Omega)$ is similar to the Rothberger game, except that the first player is to play $\omega$-covers and the second player is aiming to make an $\omega$-cover with their selections.
According to https://doi.org/10.1016/j.topol.2019.07.008, for Tychonoff spaces ($X$ is Tychonoff, as asserted in Counterexamples in Topology), having a Markov strategy in the Rothberger game is equivalent to being countable. Since $X$ is uncountable, there is no Markov strategy for the second player.
For $n \in \omega$, suppose we have $A_j = \{ x_{j,\ell} : \ell \in \omega \} \subseteq Y$ for each $j < n$ defined. Consider $F_n = \{ x_{j,\ell} : j < n ,\ell \leq n \} \cup \{ \infty \}$, a finite subset of $X$. Given an $\omega$-cover $\mathscr U_n$ of $X$, let $U_n \in \mathscr U_n$ be so that $F_n \subseteq U_n$. Since $\infty \in U_n$, we can let $A_n = \{ x_{n,\ell} : \ell \in \omega\} = X \setminus U_n$. This defines the strategy for the second player. (Notice that the assumption that each $A_j$ was previously defined for $j < n$ is what keeps this a full-information strategy.)
To see that this strategy is a winning strategy, consider $\{ U_n : n \in \omega \}$ built according to the strategy with the auxiliary sets $F_n$ and $A_n$. Let $F \subseteq X$ be finite. Since $F$ is finite, there is some $k \in \omega$ so that $F \cap \bigcup_{n\in\omega} A_n \subseteq F_k$. To see that $F\subseteq U_k$, note that any $x \in F \cap \bigcup_{n\in\omega} A_n$ has the property that $x \in F_k \subseteq U_k$; on the other hand, $x \in F \setminus \bigcup_{n \in \omega} A_n \subseteq \bigcap_{n\in\omega} U_n \subseteq U_k$.