How are the Menger and $\Omega$-Menger games related?

Question

Let $G_{fin}(\mathcal A,\mathcal B)$ be a selection game taking place over $\omega$ rounds. In round $n<\omega$, P1 chooses some $A_n\in\mathcal A$, then P2 chooses some finite subcollection $B_n\in[A_n]^{<\aleph_0}$. P2 wins provided $\bigcup\{B_n:n<\omega\}\in\mathcal B$.

Then the Menger game is $G_{fin}(\mathcal O,\mathcal O)$, where $\mathcal O$ is the collection of all open covers for some topological space. And the $\Omega$-Menger game is $G_{fin}(\Omega,\Omega)$, where $\Omega$ is the collection of all open $\omega$-covers for some topological space, where an $\omega$-cover has the property that every finite subset (rather than just singleton) of the space is contained in a single open set from the collection.

Considering the possible full and limited-information strategies for each player, how are these games related? For example, is there a space for which one player has a winning strategy in one game, but the other player has a winning strategy for the other game?

Answer 1

In general, if the second player can win the $\Omega$-Menger game, then they can win the Menger game. You take a given open cover and close it under finite unions. This is how to play the auxiliary game. When the second player makes finite selections, there are corresponding finite selections from the given open covers. If the second player produces an $\omega$-cover, then the corresponding choices form an open cover. This covers any level of strategy definable (where here I also mean P2 being able to beat any strategy type for P1 transfers in the same way; e.g. if P1 doesn't have a winning strategy in the $\Omega$-Menger game, then P1 also doesn't have a winning strategy in the Menger game).

Full-information strategies for P2

As in the paper linked in your hint (Theorem 35), strategically Menger is equivalent to strategically $\Omega$-Menger.

Full-information and pre-determined strategies for P1

According to https://mathoverflow.net/a/443001/57800, it is shown in https://arxiv.org/pdf/1806.10385.pdf that, consistently, a certain class of spaces are Menger if and only if they are Menger in all finite powers. By results of https://www.sciencedirect.com/science/article/pii/S0166864196000752, being Menger in every finite power is equivalent to being $\Omega$-Menger. However, this doesn't seem to answer the question whether being Menger is (consistently) equivalent to being $\Omega$-Menger for all topological spaces. I gather that level of generality is still open. Maybe there is a ZFC example of a Menger space which is not $\Omega$-Menger.

Edit: I forgot to mention that, in the Combinatorics of Open Covers II linked above, a special Luzin set is constructed which is Rothberger (hence Menger) but not $\Omega$-Menger. So it is consistent that the two properties are not equivalent.

The fact that the properties of Menger and $\Omega$-Menger are equivalent to P1 not having a winning strategy in the respective game is well known. I think those results are due to Hurewicz and Scheepers, respectively, but I could be misremembering. At any rate, this collapses the strategic information for P1 (full-information and pre-determined).

Constant strategies for P1

At the level of constant strategies for P1, by techniques from https://www.sciencedirect.com/science/article/abs/pii/S0166864120300031, one can show that P1 not having a constant winning strategy is equivalent to the Lindelof-like principle ${\mathcal A \choose \mathcal B}$; that is, every element of $\mathcal A$ contains a subset which is in $\mathcal B$. The Sorgenfrey line is a well-known example of a Lindelof space which is not an $\epsilon$-space; that is, a space which is Lindelof in all finite powers (a result of https://www.sciencedirect.com/science/article/pii/0166864182900657). These properties of being Lindelof and an $\varepsilon$-space correspond to P1 not having a constant winning strategy in the Menger and $\Omega$-Menger game, respectively.

Answer 2

Elaborating on my hint comment and the proof sketch I made on pi-Base's GitHub here: P2 has a winning Markov strategy (depending on only the most recent move of P2 and the round number) in either game if and only if they have a winning Markov strategy in the other game.

Given a winning Markov strategy $\sigma$ in the $\Omega$-Menger game, and an open cover $\mathcal U$, let $\mathcal U'$ be the closure of $\mathcal U$ under finite unions, noting $\mathcal U'$ must be an $\omega$-cover. Then we want to define a Markov strategy $\tau$ in the Menger game by $\tau'(\mathcal U,n)=\sigma(\mathcal U',n)$. But that's not quite right: $\tau'$ chooses finite subsets of $\mathcal U'$, and we need finite subsets of $\mathcal U$. So let $\tau(\mathcal U,n)$ select finitely-many sets in $\mathcal U$ such that $\bigcup\tau(\mathcal U,n)=\bigcup\tau'(\mathcal U,n)=\bigcup\sigma(\mathcal U',n)$.

Let $\mathcal U_n$ be an open cover for each $n<\omega$, so $\mathcal U_n'$ is an $\omega$-cover for each $n<\omega$. It follows that since $\bigcup\{\sigma(\mathcal U_n',n):n<\omega\}$ is an $\omega$-cover, $\bigcup\{\tau(\mathcal U_n,n):n<\omega\}$ is a cover, showing $\tau$ is a winning Markov strategy for the Menger game.

To move the other direction, let's start with the observation that P2 has a Markov strategy in the Menger game if and only if the space is $\sigma$-relatively-compact, where we adopt third definition from this question: a subset is relatively compact if every cover of the entire space admits a finite subcover for the subset, and $\sigma$-relatively-compact means a countable union of relatively compact subsets. This was established in (Clontz 2017), which mostly riffed on a technique from (Scheepers 1995).

And we'll require this lemma: if $R$ is relatively compact to $X$, then $R^n$ is relatively compact to $X^n$. This follows from another lemma: if $R,S$ are relatively compact to $X,Y$ respectively, then $R\times S$ is relatively compact to $X\times Y$. The standard tube lemma approach works here. Note $\{x\}\times S$ is relatively compact to $\{x\}\times Y$ for each $x\in X$. So given an open cover $\mathcal U$ of $X\times Y$, first choose for each $x\in X$ a finite subcollection $\mathcal F_x$ that covers $\{x\}\times S$. Then for each $\mathcal F_x$, choose an open $V_x\subseteq X$ such that the tube $V_x\times S\subseteq\bigcup \mathcal F_x$. Finally, choose a finite subcollection $\{V_{x_i}:i\leq N\}\subseteq\{V_x:x\in X\}$ covering $R$. Then $\bigcup\{\mathcal F_{x_i}:i\leq N\}$ is a finite subcollection of $\mathcal U$ covering $R\times S$.

So we assume $X$ is $\sigma$-relatively-compact: $X=\bigcup\{R_n:n<\omega\}$ with $R_n$ relatively compact to $X$, where $R_{n+1}\supseteq R_n$. Then we will define a strategy $\sigma(\mathcal U,n)$ for $\omega$-covers $\mathcal U$: let $\mathcal U'=\{U^n:U\in\mathcal U\}$. This is an open cover for $X^n$ since for each tuple we may choose $U\in\mathcal U$ containing the finite set of its elements, so the tuple belongs to $U^n$. Then let $\sigma'(\mathcal U,n)$ be a finite subcollection of $\mathcal U'$ covering $R_n^n$, and then $\sigma(\mathcal U,n)$ can be the corresponding subcollection of $\mathcal U$, which $\omega$-covers $R_n$. It follows that given any sequence of $\omega$-covers $\mathcal U_n$, $\bigcup\{\sigma(\mathcal U_n,n):n<\omega\}$ is an $\omega$-cover, and $\sigma$ is a winning strategy for the $\Omega$-Menger game.