Can P1 improve an open cover to an omega-cover in the finite-open game?

Question

Definitions/terms taken from https://www.sciencedirect.com/science/article/abs/pii/S016686411830470X (https://arxiv.org/abs/1806.06001).

In the finite-open game $FO(X)$, during each round $n<\omega$ P1 chooses a finite set $F_n$, then P2 chooses an open set $U_n\supseteq F_n$. P1 wins provided they force P2 to construct an open cover.

If P1 has a winning strategy for this game, can it be improved to force P2 to construct an open $\omega$-cover (each finite set is contained in some single open set from the cover)? I.e. does a winning strategy for P1 in $FO(X)$ imply the existence of a winning strategy for P1 in $\Omega FO(X)$?

In https://www.sciencedirect.com/science/article/abs/pii/S016686411830470X the answer is "yes" for both full-information and predetermined strategies, assuming spaces are $T_{3.5}$ (Thms 15 and 17 items (d) and (e)). Is $T_{3.5}$ (assumed for $C_p$ theory results) necessary though? (We are working to model this result in pi-Base and wish to have minimal assumptions.)

Answer 1

A direct approach doesn't seem immediately clear, but I think the proof in the arXiv version (which contains more detail than the published version, it seems) works in the full-information case without separation axioms. The pre-determined case is shown at Do uncountable spaces admit Markov strategies in Rothberger-style games?. I'll write out some of the details for the full-information case here to the potential benefit of the reader (and perhaps even the writer) as an initial pass to showing that no separation axioms are required (and hopefully to inspire someone else to produce a more direct approach).

First we show that P1 having a winning strategy in $FO(X)$ implies that P1 has a winning strategy in the point-open game $PO(X)$ by adapting Telgarsky's proof in this paper. The basic idea is to unfold P1's strategy in $FO(X)$ to one in $PO(X)$. Despite the fact that Telgarsky states a blanket assumption that all spaces considered in the paper are completely regular, no separation axioms are used in the following proof.

Given $n \in \omega$, suppose we have $\{ F_k : k < n \} \subseteq [X]^{<\omega}$, $\{ M_k : k < n \} \subseteq \omega$, and $\{ \{ U_{k,j} : j \leq M_k \} : k < n \}$ defined. According to P1's strategy in $FO(X)$, there is some $F_n = \{ x_{n,j} : j \leq M_n \} \subseteq X$ where $M_n \in \omega$ that is played in response to $\left\langle \bigcup_{j \leq M_k} U_{k,j} : k < n \right\rangle$. Basically what we do here is let P1 play the enumeration of $F_n$ in $PO(X)$ in response to the appropriately ordered sequence $\langle U_{k,j} : j \leq M_k, k < n \rangle$. So let P1 play $x_{n,0}$ and P2 respond with $U_{n,0} \supseteq x_{n,0}$. Proceed in this way until P1 plays $x_{n,M_n}$ and P2 plays $U_{n,M_n} \supseteq x_{n,M_n}$. Now notice that $F_n \subseteq \bigcup_{j \leq M_n} U_{n,j}$ and so is a valid play in $FO(X)$. This defines a strategy for P1 in $PO(X)$. Note that it is winning for P1 in $PO(X)$ since it was built according to a winning strategy in $FO(X)$.

The duality of $PO(X)$ and the Rothberger game doesn't require separation axioms. The particular component of duality where P1 having a winning strategy in $PO(X)$ implies the existence of a winning strategy for P2 in the Rothberger game is straightforward and clearly requires no separation axioms. In the $n^{\mathrm{th}}$ inning, given an open cover $\mathscr U_n$ of $X$, P2 need only choose $U_n \in \mathscr U_n$ that covers the point $x_n$ played by P1 in $PO(X)$.

For the portion of the proof that P2 having a winning strategy in the Rothberger game implies that P2 has a winning strategy in the $\omega$-Rothberger game $\mathsf G_1(\Omega,\Omega)$, I'll defer to the content of (a) implies (b) in Theorem 15 of the aforementioned paper. This portion relies on two ingredients: Corollary 4.9 of Selective games on binary relations and a bijection $\omega \to \omega^2$. The corollary mentioned apparently requires no separation axioms and states that being strategically Rothberger is preserved by finite products. Then the bijection is used to play the strategy on every finite power of the ground space $X$ to guarantee that an $\omega$-cover is produced in the end.

By Dual Selection Games, $\mathsf G_1(\Omega, \Omega)$ is dual to $\Omega FO(X)$ and this duality doesn't seem to require any separation axioms. So, if P2 has a winning strategy in the $\omega$-Rothberger game, P1 has a winning strategy in $\Omega FO(X)$.

In the final analysis, we see that P1 having a winning strategy in $FO(X)$ does imply a winning strategy in $\Omega FO(X)$ without any separation axioms.

Commentary: To get one example of a more direct approach from $FO(X)$ to $\Omega FO(X)$, I suspect one need only show that the property of P1 having a winning strategy in $FO(X)$ is preserved under finite powers, which should amount to Prop. 4.4 of Selective games on binary relations. Then a bijection $\omega \to \omega^2$ would allow one to run it through all finite powers to guarantee an $\omega$-cover in the end.

Answer 2

A partial answer. Suppose we can prove that that P1's winning strategies for $FO(X)$ are finitely productive: that is, if P1 has a winning [predetermined] strategy for $FO(X)$ and $FO(Y)$, they have a winning [predetermined] strategy for $FO(X\times Y)$. As Caruvana points out, the full-information case seems to be demonstrated by https://doi.org/10.1016/j.topol.2015.05.071 and the predetermined case is a corollary of https://math.stackexchange.com/a/4737306 modulo a brief proof that the product of two topologically countable spaces is topologically countable.

Let $b:\omega^2\to\omega$ be a bijection such that $i<j$ implies $b(m,i)<b(m,j)$. For each $F\in[X^m]^{<\aleph_0}$, let $F'\in[X]^{<\aleph_0}$ be defined by $F'=\bigcup\{\operatorname{range}(t):t\in F\}$, noting $F\subseteq (F')^m$.

First, let $\sigma_m$ be a winning predetermined strategy for $FO(X^m)$ for each $m<\omega$. Define $\sigma(n)=\sigma_{b^{-1}(n)(0)}(b^{-1}(n)(1))'$ for each $n<\omega$; that is, $\sigma(b(m,i))=\sigma_m(i)'$ for each $(m,i)\in\omega^2$.

To see that this is a winning strategy for $\Omega FO(X)$, we should show that P2 must produce an $\omega$-cover. So let $F$ be a finite subset of $X$, and suppose P2 chooses $U_n\supseteq\sigma(n)$ open. Let $\vec x\in X^m$ with $\operatorname{range}(\vec x)=F$. Since $$\sigma_m(i)\subseteq(\sigma_m(i)')^m=\sigma(b(m,i))^m\subseteq U_{b(m,i)}^m$$ and $\sigma_m$ is winning, it defeats the play $U_{b(m,0)}^m,U_{b(m,1)}^m,\dots$ by P2. So there must be $i<\omega$ such that $\vec x\in U_{b(m,i)}^m$. Therefore, $F=\operatorname{range}(\vec x)\subseteq U_{b(m,i)}$.

Finally, let $\sigma_m$ be a winning (full-information) strategy for $FO(X^m)$ for each $m<\omega$. For a sequence $\mathbf{U}=\langle U_0,\dots,U_{n-1}\rangle$ of open subsets of $X$, let $n=b(m,i)$ and $\mathbf{U}'=\langle U_{b(m,0)}^m,\dots,U_{b(m,i-1)}^m\rangle$. Define $\sigma(\mathbf U)=\sigma_m(\mathbf U')'$.

To see that this is a winning strategy for $\Omega FO(X)$, we should show that P2 must produce an $\omega$-cover. So let $F$ be a finite subset of $X$, and suppose P2 chooses $U_n\supseteq\sigma(\langle U_0,\dots,U_{n-1}\rangle)$ open. Let $\vec x\in X^m$ with $\operatorname{range}(\vec x)=F$. Since $$\sigma_m(\mathbf U')\subseteq(\sigma_m(\mathbf U')')^m=\sigma(\mathbf U)^m\subseteq U_{b(m,i)}^m$$ and $\sigma_m$ is winning, it defeats the play $U_{b(m,0)}^m,U_{b(m,1)}^m,\dots$ by P2. So there must be $i<\omega$ such that $\vec x\in U_{b(m,i)}^m$. Therefore, $F=\operatorname{range}(\vec x)\subseteq U_{b(m,i)}$.