Assume spaces are regular.


A space is [$\sigma$-compact](https://topology.pi-base.org/properties/P000017) if and only if the second player in the [Menger game](https://en.wikipedia.org/wiki/Selection_principle#The_Menger_game) has a winning Markov strategy (relying on only the most recent move of the opponent and the round number) [[source]](https://dml.cz/handle/10338.dmlcz/146790).


The proof there nearly translates directly to this conjecture: a space is [hemicompact](https://topology.pi-base.org/properties/P000111) if and only if the second player in the $K$-Rothberger game has a winning Markov strategy, where the $K$-Rothberger game is played with $k$-covers (each compact set is contained in some open set) and P2 may choose only one set per turn rather than finitely-many.


But there's a gotcha: while in the proof for the Menger game, we had $x\not\in\bigcap\_{\mathcal U}\bigcup\sigma(\mathcal U,n)\Rightarrow x\not\in\bigcup\sigma(\mathcal U,n)$ for some open cover $\mathcal U$, I don't believe it's true that we'd obtain a $k$-cover $\mathcal V$ for which $K\not\subseteq\sigma(\mathcal V,n)$ just because $K\not\subseteq\bigcap\_{\mathcal V}\sigma(\mathcal V,n)$.


So if that cannot be fixed, what counterexample exists?



Does there exist a non-hemicompact regular space for which the 2nd player in the $K$-Rothberger game has a winning Markov strategy?
I believe no such example exists. In fact, finite selections do the trick, so I'll detail the case for both finite and single selections (though it seems the argument for single selections below requires $T\_1$).



> 
> **Claim.** For a regular space $X$, if the second player has a Markov winning strategy in the $k$-Menger game on $X$, then $X$ is hemicompact. (This doesn't require the $T\_1$ assumption.)
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*proof.* According to Theorem 30 of [the arxiv version of this paper](https://arxiv.org/pdf/2102.00296.pdf), the $k$-Menger game on $X$ is equivalent to the Menger game on $\mathbb K(X)$, the space of compact subsets of $X$ with the Vietoris topology. To detail a bit more here, note that Lemma 19 shows that any $k$-cover of $\mathbb K(X)$ can be refined by a $k$-cover on $X$ by sets of the form $[V]$ where $V$ is open in $X$ and $[V]$ is the set of all compact subsets of $V$. In particular, any open cover of $\mathbb K(X)$ can be closed under finite unions to produce a $k$-cover of $\mathbb K(X)$ and then refined by a $k$-cover of $X$ in the indicated way. This will allow us to transfer the Markov winning strategy for the second player in the $k$-Menger game on $X$ to a Markov winning strategy in the Menger game on $\mathbb K(X)$.


According to [Michael's paper on spaces of subsets](https://www.ams.org/journals/tran/1951-071-01/S0002-9947-1951-0042109-4/S0002-9947-1951-0042109-4.pdf), $X$ is regular if and only if $\mathbb K(X)$ is regular (no requirements on $T\_1$ here).


Suppose that the second player has a Markov winning strategy in the $k$-Menger game on a regular space $X$. Then the second player has a Markov winning strategy in the Menger game on $\mathbb K(X)$ in the following way. Given an open cover $\mathscr U$ of $\mathbb K(X)$, we can form the collection of finite unions, which is a $k$-cover. For any selection of finite sets of the form $[V]$ in the specified refinement, we can collect a finite collection of members of $\mathscr U$ whose union covers the union of the finite collection of sets of the form $[V]$. Then, when the second player produces a $k$-cover of $X$, the corresponding finite selections from open covers of $\mathbb K(X)$ form an open cover. So the second player has a winning Markov strategy in the Menger game on $\mathbb K(X)$.


By the equivalence with $\sigma$-compactness mentioned in the question, since $\mathbb K(X)$ is regular, $\mathbb K(X)$ is $\sigma$-compact. By a straightforward argument detailed in Theorem 40 [this paper](https://arxiv.org/pdf/1812.06379.pdf), $X$ is hemicompact. The argument relies on 2.5.2 of [Michael's paper](https://www.ams.org/journals/tran/1951-071-01/S0002-9947-1951-0042109-4/S0002-9947-1951-0042109-4.pdf) which doesn't rely on any separation axiom assumptions. $\square$



> 
> The case for single selections (this portion requires $T\_1$)
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On the other hand, Theorem 39 of [this paper](https://arxiv.org/pdf/1812.06379.pdf) gives a pretty general equivalence with single selections and predetermined strategies for the first player in a point-open-like game. Then using the duality results of [Clontz' Dual Selection Games](https://www.sciencedirect.com/science/article/abs/pii/S0166864120300031) should establish the equivalence with Markov winning strategies for the second player in the $k$-Rothberger game and being hemicompact.


