The principle $S_1(\mathcal O,\mathcal O)$ versus the game $G_1(\mathcal O,\mathcal O)$

Question

Given a topological space $X$, Let $\mathcal O$, denote the set of all open covers of $X$.

We say that a space $X$ satisfies $S_1(\mathcal O,\mathcal O)$, if for every sequence of open covers $\{ \mathcal U_n : n \in \mathbb N \}$, there exists a sequence of open sets $\{ U_n : U_n \in \mathcal U_n \text{ for each } n \in \mathbb N \}$, which is an open cover of $X$.

Let $G_1(\mathcal O,\mathcal O)$ be the game with the following rules:

1. There are two players: ONE and TWO.

2. At each odd step $2n-1$, where $n \in \mathbb N$, ONE picks an open cover $\mathcal U_{2n-1}$ of $X$.

3. At each even step $2n$, where $n \in \mathbb N$, TWO responds by picking an open set $U_{2n} \in \mathcal U_{2n-1}$

4. If the sequence $\{ U_{2n} : n \in \mathbb N \}$ is an open cover of $X$ then TWO wins.

5. If the sequence $\{ U_{2n} : n \in \mathbb N \}$ is not an open cover of $X$ then ONE wins.

6. A strategy for ONE, is a function $F$, that, given any initial finite sequence of moves, $\{ \mathcal U_1, U_2, \mathcal U_3, U_4,...,U_{2n-2} \}$ of the game, produces the next move $\mathcal U_{2n-1}$.

7. A winning strategy for ONE, is a strategy $F$, such that $\bigcup_{n=1}^{\infty} U_{2n}$ is never a cover of $X$.

8. A strategy for TWO, is a function $F$, that, given any initial finite sequence of moves, $\{ \mathcal U_1, U_2, \mathcal U_3, U_4,...,\mathcal U_{2n-1} \}$ of the game, produces the next move $U_{2n}$

9. A winning strategy for TWO, is a strategy $F$, such that $\bigcup_{n=1}^{\infty} U_{2n}$ is allways a cover of $X$.

We say that a topological space $X$ satisfies $G_1(\mathcal O,\mathcal O)$ if TWO has a winning strategy for $G_1(\mathcal O,\mathcal O)$ in $X$.

I am trying to prove the claim below and have mannaged to prove only one direction. Any help?

This claim is some lighter version of theorem 3 in this article.

Claim: $X$ satisfies $S_1(\mathcal O,\mathcal O)$ if and only if TWO has a winning strategy for $G_1(\mathcal O,\mathcal O)$ in $X$.

Proof: First direction: Suppose that TWO has a winning strategy for $G_1(\mathcal O,\mathcal O)$ in $X$. Then, Given a sequence $\mathcal U_{2n-1} = \{ \mathcal U_1, \mathcal U_3, \mathcal U_5, ... \}$ of open covers of $X$, Start with $\{ \mathcal U_1 \}$, and let $U_2 = F(U_1)$, where $F$ is the winning strategy for TWO. Then, take $U_4 = F(\mathcal U_1,U_2,\mathcal U_3)$, at the $n$'th step, $U_{2n} = F(\mathcal U_1,U_2, \mathcal U_3,U_4,...,\mathcal U_{2n-1})$. It is clear that $\bigcup U_{2n}$ is an open cover of $X$ since $F$ is a winning strategy. Second direction: Suppose that $X$ satisfies $S_1(\mathcal O,\mathcal O)$. Then, for every sequence of open covers $\mathcal U_n$ of $X$, there is a sequence $\{ U_n : U_n \in \mathcal U_n, n \in \mathbb N \}$, such that $\bigcup U_n$ is an open cover of $X$. We have to show that TWO has a winning strategy in the game $G_1(\mathcal O,\mathcal O)$. So, how can I build this strategy?

Thank you!

Answer 1

First, let's note that the selection principle $S_1(\mathcal A,\mathcal B)$ is equivalent to ONE lacking a winning "predetermined" strategy in the selection game $G_1(\mathcal A,\mathcal B)$ (which uses the round number but ignores the moves of TWO). https://arxiv.org/abs/1809.10783

Your claim that $S_1(\mathcal O,\mathcal O)$ is equivalent to TWO having a winning strategy in $G_1(\mathcal O,\mathcal O)$ is false: a Lusin set is an example of a space where neither player has a winning strategy for $G_1(\mathcal O,\mathcal O)$. Therefore $S_1(\mathcal O,\mathcal O)$ holds while TWO lacks a winning strategy for $G_1(\mathcal O,\mathcal O)$. http://www.ams.org/journals/proc/1995-123-11/S0002-9939-1995-1273523-1/home.html