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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?

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.

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 $\textsf{BM}(X)$.

Thanks

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?

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 $X$.

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 $\textsf{BM}(X)$.

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.

Thanks

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 $\textsf{BM}(X)$.

Thanks