Does there exist a non-hemicompact regular space for which the 2nd player in the $K$-Rothberger game has a winning Markov strategy?

Question

Assume spaces are regular.

A space is $\sigma$-compact if and only if the second player in the Menger game has a winning Markov strategy (relying on only the most recent move of the opponent and the round number) [source].

The proof there nearly translates directly to this conjecture: a space is hemicompact 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?

Answer 1

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.)

proof. According to Theorem 30 of the arxiv version of this paper, 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, $X$ 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, $X$ is hemicompact. The argument relies on 2.5.2 of Michael's paper which doesn't rely on any separation axiom assumptions. $\square$

The case for single selections (this portion requires $T_1$)

On the other hand, Theorem 39 of this paper 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 should establish the equivalence with Markov winning strategies for the second player in the $k$-Rothberger game and being hemicompact.

Answer 2

Writing up a direct proof for Rothberger based upon Caruvana's references.

It's unclear why I didn't think to try it, but it's much easier to think about the K-Rothberger game in terms of its dual - by the techniques of that article, it is dual to the compact-open k-cover game ("KO k-game"), where P1 chooses compact sets, and P2 chooses open neighborhoods of those sets. Then P1 (not P2) wins if P2's choices form a k-cover.

So P2 having a winning Markov strategy in the K-Rothberger game is equivalent to P1 having a winning predetermined strategy in the KO k-game. This is exactly the property where there exists a countable sequence of compact sets $K_n$ such that for any neighborhood assignment $U_n\supseteq K_n$, the assignment forms a k-cover.

So let's assume $T_1$, and we'll show this is equivalent to hemicompactness. Given a witness $K_n$ for hemicompactness, it's immediate that any neighborhood assignment creates a $k$-cover.

So take the countable sequence of compact sets $K_n$ such that for any neighborhood assignment $U_n\supseteq K_n$, the assignment forms a k-cover. We will show that these $K_n$ witness hemicompactness. To see this, suppose not, and let $x_n\in K\setminus K_n$ for some compact set $K$ for all $n<\omega$. Since the space is $T_1$, $U_n=X\setminus\{x_n\}$ is a neighborhood assignment, but $K\not\subseteq U_n$ for all $n<\omega$; thus the neighborhood assignment is not a k-cover, a contradiction.

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And here's a direct proof for Menger that doesn't rely on casting things to the hyperspace of compact subsets (and illustrates how to avoid the issue brought up in my original question).

Hemicompactness directly yields a winning Markov strategy for P2 in the K-Rothberger game, and each such strategy is also a winning Markov strategy for P2 in the K-Menger game.

Let $X$ be regular. We wish to show that a winning Markov strategy $\sigma$ in the K-Menger game implies hemicompactness.

Consider the set $\mathcal R_n$, the collection of all compact sets contained within some element of $\sigma(\mathcal U,n)$ for all k-covers $\mathcal U$. We claim for each compact $K$ that $K\in\mathcal R_n$ for some $n<\omega$. To see this, suppose not. Then for each $n<\omega$ P1 could pick a k-cover $\mathcal U_n$ such that $K$ is not contained within any open set in $\sigma(\mathcal U_n,n)$, defeating $\sigma$, a contradiction.

We claim $\bigcup\mathcal R_n$ is relatively compact. Given an open cover $\mathcal V$ of $X$, let $\mathcal U=\{\bigcup\mathcal F:\mathcal F\in[\mathcal V]^{<\aleph_0}\}$, a k-cover. So denote $\sigma(\mathcal U,n)=\{\bigcup\mathcal F_i:i<N\}$. Consider $\mathcal F=\bigcup\{\mathcal F_i:i<N\}$, a finite subset of $\mathcal V$. Given $x\in\bigcup\mathcal R_n$, pick a compact set $K\in\mathcal R_n$ with $x\in K$. By definition, $K$ is contained in some element $\bigcup\mathcal F_I$ of $\sigma(\mathcal U,n)=\{\bigcup\mathcal F_i:i<N\}$. Thus, $x\in K$ belongs to some element of $\mathcal F_I\subseteq\mathcal F$, showing $\mathcal F$ is a finite subcollection of $\mathcal V$ covering $\bigcup\mathcal R_n$.

Finally, let $K_n=\overline{\bigcup\mathcal R_n}$. By regularity, the closure of this relatively compact set is compact. And for every compact $K$, there exists some $n<\omega$ with $K\in\mathcal R_n$, and therefore $K\subseteq K_n$. Thus $K_n$ witnesses hemicompactness.