# Banach-Mazur game and infinite products

Studying the article "Games that involve set theory or topology" of Marion Scheepers, I found the following result

Theorem 46 Let $$\{(X_{i}, \tau_{i}) : i\in I \}$$ be a family of topological spaces. If for every countable subset $$J$$ of $$I$$, Player I does not have a winning strategy in the game $$\textsf{BM} (\prod_{j\in J}X_{j}, \square_{j\in J} \tau_{j} ),$$ then Player I does not have a winning strategy in $$\textsf{BM} (\prod_{i\in I}X_{i}, \prod_{i\in I} \tau_{i} ).$$

Where $$\textsf{BM}(X)$$ denotes the Banach-Mazur game played on a topological space $$X$$.

So far I had no success in proving the theorem, the author only states the result without mentioning any reference. I would like to know if you could give me some idea about it.

Remember that The Banach-Mazur game on $$X$$, $$\textsf{BM}(X)$$, is played as follows: Players I and II play an inning per positive integer. In the $$n$$-th inning Player I chooses a nonempty open set $$A_n$$; Player II responds with a nonempty open set $$B_n \subseteq A_n$$. Player I must also obey the rule that for each $$n$$, $$A_{n+1} \subseteq B_n$$. A play $$A_1, B_1, \cdots A_n, B_n$$ is won by Player II if $$\bigcap_{n\in\omega}T_n \not=\emptyset$$; otherwise, Player I wins.

One of the most important results in this regard is the following

Theorem A nonempty topological space $$X$$ is a Baire space if and only if Player I has no winning strategy in the Banach-Mazur game $$\textsf{BM}(X)$$.

Some remarks

As Taras Banakh noted, there is a theorem for Tychonoff powers,

Theorem Let $$\kappa\geq \omega$$. If $$X^{\omega}$$ is Baire, then $$X^{\kappa}$$ is Baire, where the powers are considered in the Tychonoff product.

When I studied the result, which appears in Kunen and Fleissner's article, only the characterization using games from Baire's spaces is used for the proof.

In fact, if $$\kappa>\omega$$, they show that, if $$X^{\kappa}$$ is not Baire then $$X^{\omega}$$ is not Baire, for this, let $$\sigma$$ be a winning strategy for Player I in $$\textsf{BM}(X^{\kappa})$$. We're going to build a winning strategy $$\tilde{\sigma}$$ for Player I in $$\textsf{BM}(X^{\omega})$$.

Citing what the article by Fleissner and Kunen mentions: The ideia is this: Player I relabels the index set as he goes along so that he is in effect playing according to $$\sigma$$ in $$X^{\kappa}$$

Now, if we assume Scheepers' theorem we have the following

Corollary Let $$X$$ be a topological space, if for all cardinal $$\kappa$$, the box power $$X^{\kappa}$$ is a Baire space then the Tychonoff power $$X^{\omega}$$ is a Baire space.

I think this can help solve the next question

Can one prove in ZFC that if a box product of a collection of Baire spaces is Baire, then its Tychonoff product is Baire?

About products of Baire spaces and spaces with countable cellularity, in Baire spaces - R. C. Haworth, R. A. McCoy, I studied the following result.

Theorem Let $$\{X_{\alpha} : \alpha\in A \}$$ be a family of Baire spaces such that the product of any countable subcollection is a Baire space and such that $$\prod_{\alpha\in A} X_{\alpha}$$ has the countable chain condition. Then $$\prod_{\alpha\in A} X_{\alpha}$$ is a Baire space.

• The theorem is a well-known game characterization of Baire spaces due to Oxtoby. It can be found in the book of Kechris "Classical Descriptive Set Theory". – Taras Banakh Sep 22 '19 at 13:50
• Thank you but the result that I still could not prove is Theorem 46. It is also known that the reciprocal is not valid because in the article Barely Baire Spaces of W. Fleissner and K. Kunen there is a space $X$ such that the Tychonoff power $X^{\omega_{1}}$ is Baire but the box power $X^{\omega_{1}}$ is not Baire. – Gabriel Medina Sep 23 '19 at 1:02
• Well on page 210 of Scheepers' article, he mentions the relationship between the box topology, Tychonoff product and the Banach-Mazur game and finally comments on Theorem 46. – Gabriel Medina Sep 23 '19 at 19:27

## 1 Answer

As Taras Banakh noted, there is a theorem for Tychonoff powers,

Theorem Let $$\kappa\geq \omega$$. If $$X^{\omega}$$ is Baire, then $$X^{\kappa}$$ is Baire, where the powers are considered in the Tychonoff product.

Now, if we assume Scheepers' theorem we have the following

Corollary Let $$X$$ be a topological space, if for all cardinal $$\kappa$$, the box power $$X^{\kappa}$$ is a Baire space then the Tychonoff power $$X^{\omega}$$ is a Baire space.

I think this can help solve the next question

Can one prove in ZFC that if a box product of a collection of Baire spaces is Baire, then its Tychonoff product is Baire?

• The implication ($X^\omega$ is Baire $\Rightarrow$ $X^\kappa$ is Baire) holds only for spaces with countable cellularity. – Taras Banakh Sep 28 '19 at 16:22
• About products of Baire spaces and spaces with countable cellularity, in Baire spaces - R. C. Haworth, R. A. McCoy, I studied the following result. Theorem Let $\{X_{\alpha} : \alpha\in A \}$ be a family of Baire spaces such that the product of any countable subcollection is a Baire space and such that $\prod_{\alpha\in A} X_{\alpha}$ has the countable chain condition. Then $\prod_{\alpha\in A} X_{\alpha}$ is a Baire space. – Gabriel Medina Sep 28 '19 at 18:08