More on products of Baire spaces

TL;DR

This paper presents new results on the Baire product problem, showing conditions under which products of certain Baire spaces remain Baire, including almost locally ccc spaces and spaces with specific game-theoretic properties.

Contribution

It introduces novel conditions ensuring that products of Baire spaces are Baire, expanding understanding of Baire space behavior under product operations.

Findings

Product of almost locally ccc Baire spaces is Baire

Product of a Baire space and a $eta$-unfavorable 1st countable space is Baire

New conditions for Baire property preservation in products

Abstract

New results on the Baire product problem are presented. It is shown that an arbitrary product of almost locally ccc Baire spaces is Baire; moreover, the product of a Baire space and a 1st countable space which is $\beta$-unfavorable in the strong Choquet game is Baire.

Equations

$$\mathcal{E}=\mathcal{E}(X)=\mathcal{E}(X,\mathcal{B})=\{(x,U)\in X\times \mathcal{B}:x\in U\}.$$

$$(x_0,V_0),U_0,\dots ,(x_n,V_n),U_n,\dots$$

$$\downarrow {\tau }^n=\{f\in (\tau \setminus \{\varnothing \}{)}^n:f(k+1)\subseteq f(k)\text{whenever}k\le n\},\text{and}$$

$$\downarrow {\tau }^{\omega }=\{f\in (\tau \setminus \{\varnothing \}{)}^{\omega }:f(k+1)\subseteq f(k)\text{whenever}k<\omega \}.$$

$$\mathcal{K}(X)=\{f\in \downarrow {\tau }^{\omega }:\bigcap _{n}f(n)\ne \varnothing \}.$$

$$[f{\upharpoonright }_{n+1}]=\{g\in \mathcal{K}(X):g{\upharpoonright }_{n+1}=f{\upharpoonright }_{n+1}\}.$$

$$[f]=\{g\in \mathcal{K}(X):g{\upharpoonright }_{n+1}=f\}.$$

$${\mathcal{K}}_{\mathcal{B}}^0(X)=\{f\in \mathcal{K}(X)\cap {\mathcal{B}}^{\omega }:(f(n){)}_n\text{is a neighborhood base at each}x\in \bigcap _{n}f(n)\}.$$

$$U_1\times V_{(0,0)}\subseteq O_1\cap [A_0\times V_{(0)}]$$

$$U_1\times V_{(1)}\subseteq O_1\cap [A_0\times V_{\varnothing }\cap W_{y_{\varnothing }}(y_{(0)})];$$

$$U_n={\sigma }_X(A_0,\dots ,A_{n-1}),$$

$$U_n\times V_{t^{-}}\subseteq O_n\cap [A_{n-1}\cap V_t],$$

$$U_n\times V_{t^{+}}\subseteq O_n\cap [A_{n-1}\times V_{s_t}\cap W_{y_{s_t}}(y_{s_t^{⌢}0},\dots ,y_t)].$$

$$H_i\times V_{t_i^{-}}\subseteq O_{n+1}\cap [H_{i-1}\times V_{t_i}],$$

$$H_{i+2^n}\times V_{t_i^{+}}\subseteq O_{n+1}\cap [H_{i+2^n-1}\times V_{s_{t_i}}\cap W_{y_{s_{t_i}}}(y_{s_{t_i}^{⌢}0},\dots ,y_{t_i})].$$

$$U_{n+1}\times V_{t^{-}}\subseteq O_{n+1}\cap [A_n\cap V_t]$$

$$U_{n+1}\times V_{t^{+}}\subseteq O_{n+1}\cap [A_n\times V_{s_t}\cap W_{y_{s_t}}(y_{s_t^{⌢}0},\dots ,y_t)].$$

$$(y_{t^{⌢}k}{)}_k\text{ converges to}{y}_t\text{for each}t\in T.$$

$$k_{B_0}=\min \{k\ge 0:y_{(k)}\in B_0\},z_1=y_{(k_{B_0})},W_1=B_0\cap V_{(k_{B_0})},$$

$$z_n=y_{(k_{B_0},\dots ,k_{B_{n-1}})},\text{and}{W}_n=B_{n-1}\cap V_{(k_{B_0},\dots ,k_{B_{n-1}})}$$

$$k_{B_n}=\min \{k\ge n:y_{t^{⌢}k}\in B_n\},z_{n+1}=y_{t^{⌢}{k}_{B_n}},W_{n+1}=B_n\cap V_{t^{⌢}{k}_{B_n}},$$

$$(z_0,W_0),B_0,(z_1,W_1),\dots ,B_n,(z_{n+1},W_{n+1}),\dots$$

$$\sigma (x,V)=\{\begin{array}{ll}\{x\},& \text{if}x\in D,\\ V,& \text{if}x\in \mathbb{Q}.\end{array}$$

$${\sigma }^{\ast }(\varnothing )=\prod _{i\in I_0}{V}_{0,i}^{\ast }\times \prod _{i\notin I_0}\mathcal{K}(X_i),\text{where}{V}_{0,i}^{\ast }=[(V_{0,i})]$$

$$\sigma (U_0)=\prod _{i\in I_1}{V}_{1,i}\times \prod _{i\notin I_1}{X}_i,$$

$${\sigma }^{\ast }(U_0^{\ast })=\prod _{i\in I_1}{V}_{1,i}^{\ast }\times \prod _{i\notin I_1}\mathcal{K}(X_i),\text{where}$$

$$V_{1,i}^{\ast }=\{\begin{array}{ll}[(U_{0,i}(0),\dots ,U_{0,i}(m_{0,i}),V_{1,i})],& \text{if}i\in J_0,\\ [(V_{1,i})],& \text{if}i\in I_1\setminus J_0.\end{array}$$

$$U_k^{\ast }=\prod _{i\in J_k}{U}_{k,i}^{\ast }\times \prod _{i\notin J_k}\mathcal{K}(X_i)$$

$${\sigma }^{\ast }(U_0^{\ast },\dots ,U_k^{\ast })=\prod _{i\in I_{k+1}}{V}_{k+1,i}^{\ast }\times \prod _{i\notin I_{k+1}}\mathcal{K}(X_i)$$

$$V_{k+1,i}^{\ast }=\{\begin{array}{ll}[(U_{k,i}(0),\dots ,U_{k,i}(m_{k,i}),V_{k+1,i})],& \text{if}i\in J_k,\\ [(V_{k+1,i})],& \text{if}i\in I_{k+1}\setminus J_k\end{array}$$

$$\prod _{i\in I_{k+1}}{V}_{k+1,i}\times \prod _{i\notin I_{k+1}}{X}_i=\sigma \left(U_0,\dots U_k\right).$$

$${\sigma }^{\ast }(\varnothing ),U_0^{\ast },\dots ,U_n^{\ast },{\sigma }^{\ast }(U_0^{\ast },\dots ,U_n^{\ast }),\dots$$

$$U_{0,i}^{\ast }=\{\begin{array}{ll}[(V_{0,i}(0),\dots ,V_{0,i}(m_{0,i}),U_{0,i})],& \text{for all}i\in I_0,\\ [(U_{0,i})],& \text{for all}i\in J_0\setminus I_0,\end{array}$$

Full text

1. introduction

A topological space is a Baire space provided countable intersections of dense open subsets are dense [13]. If the product $X\times Y$ is Baire, then $X,Y$ must be Baire; however, the converse is not true in general. Indeed, Oxtoby [20] constructed, under CH, a Baire space with a non-Baire square, and various absolute examples followed (see [4], [8], [22], [23]). As a result, there has been a considerable effort to find various completeness properties for the coordinate spaces to get Baireness of the product (cf. [17], [10], [20], [13], [1], [29], [8], [22], [9], [31], [3], [19], [18]). There have been two successful approaches in solving the product problem: given Baire spaces $X,Y$, either one adds some condition to $Y$ (such as 2nd countability [20], the uK-U property [9], having a countable-in-itself $\pi$-base [31]), or strengthens completeness of $Y$ (to Čech-completeness, (strong) $\alpha$-favorability [1],[29], or more recently, to hereditary Baireness [3],[19],[18]). It is the purpose of this paper to generalize these product theorems, as well as, show how a new fairly weak completeness property of $\beta$*-unfavorability in the strong Choquet game * [24],[6],[27] can be added to the list of spaces giving a Baire product.

Since Baire spaces can be characterized via the Banach-Mazur game, it is not surprising that topological games have been applied to attack the Baire product problem. Our results continue in this line of research (precise definitions will be given in the next section); in the games two players take countably many turns in choosing objects from a topological space $X$: in the strong Choquet game [2, 14] player $\beta$ starts, and always chooses an open set $V$ and a point $x\in V$, then player $\alpha$ responds by choosing an open set $U$ such that $x\in U\subseteq V$, next $\beta$ chooses an open set $V^{\prime}$ and a point $x^{\prime}$ with $x^{\prime}\in V^{\prime}\subseteq U$, etc. Player $\alpha$ wins if the intersection of the chosen open sets is nonempty, otherwise, $\beta$ wins.

The strong Choquet game provides a useful unifying platform for studying completeness-type properties, as the following two celebrated theorems demonstrate in a metrizable space $X$:

• •

$Ch(X)$ is $\alpha$-favorable if and only if $X$ is completely metrizable [2],

• •

$Ch(X)$ is $\beta$-unfavorable if and only if $X$ is hereditarily Baire (i.e. the nonempty closed subspaces of $X$ are Baire) [6, 27, 24].

The Banach-Mazur game $BM(X)$ [13] (also called the Choquet game [14]) is played as $Ch(X)$, except that both $\beta,\alpha$ choose open sets only. In a topological space $X$, $BM(X)$ is $\beta$-unfavorable iff $X$ is a Baire space [21, 16, 26]; consequently, if $BM(X)$ is $\alpha$-favorable, then $X$ is a Baire space.

To put our results in perspective, recall that $X\times Y$ is a Baire space if $X$ is a Baire topological space and

• •

either $Y$ is a topological space such that $BM(Y)$ is $\alpha$-favorable (in particular, if $Ch(Y)$ is $\alpha$-favorable) [29],

• •

or $Y$ is a hereditarily Baire space which is metrizable [19], or more generally, 1st countable $T_{3}$ space [18].

Since there are spaces which are $\alpha$-favorable in the strong Choquet game but are not hereditarily Baire (the Michael line [7]), as well as metric hereditarily Baire spaces, which are not $\alpha$-favorable in the Banach-Mazur game (a Bernstein set [7]), being $\beta$-unfavorable in the strong Choquet game is distinct from both hereditary Baireness as well as being $\alpha$-favorable in the strong Choquet game, thus, it is natural to ask the status of this property in the Baire product problem. Our main result in Section 3 (Theorem 3.2) implies the following:

Theorem 1.1**.**

Let $X$ be a Baire space, $Y$ be a 1st countable topological space such that $Ch(Y)$ is $\beta$-unfavorable. Then $X\times Y$ is a Baire space.

The proof works for finite products, but it does not naturally extend to infinite products, so we will separately consider the infinite product case in Section 4, using the idea of a Krom space ([15],[11]), to obtain:

Theorem 1.2**.**

Let $I$ be an index set, and $X_{i}$ be an almost locally ccc Baire space (defined in Section 3) for each $i\in I$ . Then $\prod_{i}X_{i}$ is a Baire space.

2. Preliminaries

Unless otherwise noted, all spaces are topological spaces. As usual, $\omega$ denotes the non-negative integers, and $n\geq 1$ will be considered as sets of predecessors $n=\{0,\dots,n-1\}$. Let $\mathcal{B}$ be a base for a topological space $X$, and denote

[TABLE]

In the strong Choquet game $Ch(X)$ players $\beta$ and $\alpha$ alternate in choosing $(x_{n},V_{n})\in\mathcal{E}$ and $U_{n}\in\mathcal{B}$, respectively, with $\beta$ choosing first, so that for each $n<\omega$, $x_{n}\in U_{n}\subseteq V_{n}$, and $V_{n+1}\subseteq U_{n}$. The play

[TABLE]

is won by $\alpha$, if $\bigcap_{n}U_{n}(=\bigcap_{n}V_{n})\neq\emptyset$; otherwise, $\beta$ wins.

A strategy in $Ch(X)$ for $\alpha$ (resp. $\beta$) is a function $\sigma:\mathcal{E}^{<\omega}\to\mathcal{B}$ (resp. $\sigma:\mathcal{B}^{<\omega}\to\mathcal{E}$) such that $x_{n}\in\sigma((x_{0},V_{0}),\dots,(x_{n},V_{n}))\subseteq V_{n}$ for all $((x_{0},V_{0}),\dots,(x_{n},V_{n}))\in\mathcal{E}^{<\omega}$ (resp. $\sigma(\emptyset)=(x_{0},V_{0})$ and $V_{n}\subseteq U_{n-1}$, where $\sigma(U_{0},\dots,U_{n-1})=(x_{n},V_{n})$, for all $(U_{0},\dots,U_{n-1})\in\mathcal{B}^{n}$, $n\geq 1$). A strategy $\sigma$ for $\alpha$ (resp. $\beta$) is a winning strategy, if $\alpha$ (resp. $\beta$) wins every play of $Ch(X)$ compatible with $\sigma$, i.e. such that $\sigma((x_{0},V_{0}),\dots,(x_{n},V_{n}))=U_{n}$ for all $n<\omega$ (resp. $\sigma(\emptyset)=(x_{0},V_{0})$ and $\sigma(U_{0},\dots,U_{n-1})=(x_{n},V_{n})$ for all $n\geq 1$). We will say that $Ch(X)$* is $\alpha$-, $\beta$-favorable*, respectively, provided $\alpha$, resp. $\beta$, has a winning strategy in $Ch(X)$.

The Banach-Mazur game $BM(X)$ [13] is played similarly to $Ch(X)$, the only difference is that both $\beta,\alpha$ choose open sets from a fixed $\pi$-base of $X$. Winning strategies, $\alpha$-, and $\beta$-favorability of $BM(X)$ can be defined analogously to $Ch(X)$.

In the Gruenhage game $G(X)$ [12] given a point $x\in X$, at the $n$-th round Player I picks an open neighborhood $U_{n}$ of $x$, and Player II chooses $x_{n}\in U_{n}$. Player I wins if the sequence $(x_{n})_{n}$ converges to $x$, otherwise, Player II wins; $x$ is a $W$-point [25], provided Player I has a winning strategy $W_{x}:X^{<\omega}\to\{\text{open neighborhoods of}\ x\}$ in $G(X)$ at $x$.

Given a topological space $(X,\tau)$, consider the ultrametric space $\tau^{\omega}$, where $\tau$ has the discrete topology. For every $n<\omega$ denote

[TABLE]

The Krom space [15, 11] of $X$ is defined as

[TABLE]

Note that a base of neighborhoods at $f\in\mathcal{K}(X)$ is $\{[f\restriction_{n+1}]:n<\omega\}$, where

[TABLE]

Put differently, a base for $\mathcal{K}(X)$ is the family of all sets $[f]$, $f\in\bigcup_{n}\downarrow\tau^{n}$ where, if $n<\omega$ and $f\in\downarrow\tau^{n}$, then

[TABLE]

Given a base $\mathcal{B}$ for $X$, we will also consider the following subspace of $\mathcal{K}(X)$:

[TABLE]

Krom’s Theorem [15, Theorem 3] states that for topological spaces $X,Y$, $X\times Y$ is a Baire space iff $X\times\mathcal{K}(Y)$ is a Baire space iff $\mathcal{K}(X)\times\mathcal{K}(Y)$ is a Baire space.

Given a set $S$, $s\in S$, $n<\omega$, and $f\in S^{n}$, the notation $f^{\frown}s$ stands for the $g\in S^{n+1}$ for which $g\restriction_{n}=f\restriction_{n}$, and $g(n)=s$.

3. Finite Baire products

We will say that a space is almost locally ccc, provided every open set contains an open ccc subspace i.e., if the space has a $\pi$-base of open ccc subspaces. This property is strictly weaker than being almost locally uK-U [31] (see [9, Examples 1,2]), as well as having a countable-in-itself $\pi$-base [31] (i.e., having a $\pi$-base each member of which contains countably many members of said $\pi$-base – this property is also termed having a locally countable pseudo-base in [20]), which are known to produce Baire products (see [20, Theorem 2], [9, Property 1] [31, Proposition 4]). Since these properties all coincide in Baire metric spaces (see [31, Proposition 3]), a simple observation about Krom spaces immediately yields a generalization of these Baire product theorems:

Theorem 3.1**.**

Let $X,Y$ be a Baire spaces, and $Y$ be almost locally ccc. Then $X\times Y$ is a Baire space.

Proof.

First note that $\mathcal{K}(Y)$ has a countable-in-itself $\pi$-base: indeed, let $f\in\downarrow\tau^{n}$ for some $n<\omega$, choose $U\subset f(n)$ which is ccc, and define $f_{0}=f^{\frown}U$. Consider a pairwise disjoint open partition $\{[g]:g\in J\}$ of $[f_{0}]$, where $J\subset\bigcup_{n<\omega}\downarrow\tau^{n}$. For each $g\in J$ let $n_{g}<\omega$ be such that $g\in\downarrow\tau^{n_{g}}$; then $\{g(n_{g}):g\in J\}$ is a pairwise disjoint open partition of $U$, which must be countable, and so is $\{[g]:g\in J\}$; thus, $\mathcal{K}(Y)$ is an almost locally ccc metric space, and so it has a countable-in-itself $\pi$-base.

It follows from Krom’s theorem that $\mathcal{K}(Y)$ is a Baire space, moreover, by [20, Theorem 2], $X\times\mathcal{K}(Y)$ is a Baire space, which it turn implies $X\times Y$ is a Baire space by Krom’s theorem. ∎

An approach involving the strong Choquet game yields a different kind of generalization of Baire product theorems (cf. [1], [29],[19],[18],[3]):

Theorem 3.2**.**

Let $X$ be a Baire space and $Y$ have a dense set of $W$-points and $Ch(Y)$ be $\beta$-unfavorable. Then $X\times Y$ is a Baire space.

Proof.

Denote by $\tau_{X},\tau_{Y}$ the nonempty open subsets of $X,Y$, respectively. Let $\{O_{n}:n<\omega\}$ be a decreasing sequence of dense open subsets of $X\times Y$, and pick $U\in\tau_{X},V\in\tau_{Y}$. If $y\in Y$ is a $W$-point, denote by $W_{y}$ a winning strategy for the open-set picker in the Gruenhage game at $y$.

Define a tree $T\subset\omega^{<\omega}$ as follows: let $T_{0}=\{\emptyset\}$ be the root of the tree, and $T_{1}=\{(0)\}$ its first level; further, given level $T_{n}$ for some $n\geq 1$, and $t=(t_{0},\dots,t_{k})\in T_{n}$, define the immediate successors of $t$ as $t^{-}=(t_{0},\dots,t_{k},0)$ and $t^{+}=(t_{0},\dots,t_{k}+1)$, and put $T_{n+1}=\{t^{-},t^{+}:t\in T_{n}\}$. For example, if $t=(0)\in T_{1}$, then $t^{-}=(0,0),\ t^{+}=(1)$; further, if $t=(0,0)\in T_{2}$, then $t^{-}=(0,0,0),\ t^{+}=(0,1)$, and if $t=(1)\in T_{2}$, then $t^{-}=(1,0),\ t^{+}=(2)$, etc. It follows that each $t\in T_{n}$ for $n\geq 1$ has a ”source“ $s_{t}\in T_{k}$ for some $0\leq k<n$ such that $t=s_{t}^{\frown}(n-k-1)$; in other words, $s_{t}$ is the node where the last minus-branching occurs before $t$; so, $s_{(0)}=\emptyset,s_{(0,0)}=(0),s_{(1)}=\emptyset,s_{(0,0,0)}=(0,0),s_{(0,1)}=(0),s_{(1,0)}=(1),s_{(2)}=\emptyset$, etc.

We will define a strategy $\sigma_{X}$ for $\beta$ in $BM(X)$: first, pick a $W$-point $y_{\emptyset}\in V$ and denote $V_{\emptyset}=V$. Then choose $U_{0}\in\tau_{X}$, $V_{(0)}\in\tau_{Y}$ so that $U_{0}\times V_{(0)}\subseteq O_{0}\cap U\times V$, pick a $W$-point $y_{(0)}\in V_{(0)}$ and put $\sigma_{X}(\emptyset)=U_{0}$. Given $A_{0}\in\tau_{X},A_{0}\subseteq U_{0}$, find $U_{1}\in\tau_{X}$ and $V_{t}\in\tau_{Y}$ for each $t\in T_{2}$ so that

[TABLE]

moreover, pick a $W$-point $y_{t}\in V_{t}$ for each $t\in T_{2}$, and put $\sigma_{X}(A_{0})=U_{1}$.

Assume that for some $n\geq 1$, and given $A_{0},\dots,A_{n-1}\in\tau_{X}$, we have constructed $U_{n}\in\tau_{X}$ along with $(y_{t},V_{t})\in\mathcal{E}(Y)$ for each $t\in T_{n+1}$ so that each $y_{t}$ is a $W$-point,

[TABLE]

and for each $t\in T_{n}$

[TABLE]

Let $A_{n}\in\tau_{X},A_{n}\subseteq U_{n}$ be given, and denote $T_{n+1}=\{t_{1},\dots,t_{2^{n}}\}$. Using density of $O_{n+1}$ repeatedly, we can define a decreasing sequence $\{H_{i}\in\tau_{X}:i\leq 2^{n+1}\}$, where $H_{0}=A_{n}$, as well as $V_{t_{i}^{-}},V_{t_{i}^{+}}\in\tau_{Y}$ so that for all $1\leq i\leq 2^{n}$

[TABLE]

Then for $U_{n+1}=H_{2^{n+1}}$ and each $t\in T_{n+1}$ we have

[TABLE]

Pick a $W$-point $y_{t}\in V_{t}$ for each $t\in T_{n+2}$, and define $\sigma_{X}(A_{0},\dots,A_{n})=U_{n+1}$, which concludes the definition of $\sigma_{X}$. Notice that, by (2),

[TABLE]

Since $X$ is a Baire space, there is play $U_{0},A_{0},\dots,U_{n},A_{n},\dots$ of $BM(X)$ compatible with $\sigma_{X}$ that $\beta$ looses, i.e. there is some $x\in\bigcap_{n}U_{n}$.

We will define a strategy $\sigma_{Y}$ for $\beta$ in $Ch(Y)$. First, put $\sigma_{Y}(\emptyset)=(z_{0},W_{0})$, where $z_{0}=y_{\emptyset}$, and $W_{0}=V_{\emptyset}$. Let $B_{0}\in\tau_{Y}$ with $z_{0}\in B_{0}\subseteq W_{0}$ be given. Using (3), we can define

[TABLE]

and put $\sigma_{Y}(B_{0})=(z_{1},W_{1})$. Assume that $\sigma_{Y}(B_{0},\dots,B_{n-1})=(z_{n},W_{n})\in\mathcal{E}(Y)$ have been defined for some $n\geq 1$ and $B_{0},\dots,B_{n-1}\in\tau_{Y}$ so that

[TABLE]

for appropriate $k_{B_{i}}\geq i$ for each $i<n$. Let $B_{n}\in\tau_{Y}$ be such that $z_{n}\in B_{n}\subseteq W_{n}$, then for $t=(k_{B_{0}},\dots,k_{B_{n-1}})$, $(y_{t^{\frown}k})_{k}$ converges to $z_{n}=y_{t}$ by (3), so we can define

[TABLE]

and put $\sigma_{Y}(B_{0},\dots,B_{n})=(z_{n+1},W_{n+1})$.

Since $\sigma_{Y}$ cannot be a winning strategy for $\beta$ in $Ch(Y)$, there is a play

[TABLE]

of $Ch(Y)$ compatible with $\sigma_{Y}$ that $\beta$ looses. Then there is some $y\in\bigcap_{n}W_{n}\subseteq\bigcap_{n}V_{(k_{B_{0}},\dots,k_{B_{n}})}$, so (1) and (2) imply that $(x,y)\in U\times V\cap\bigcap_{n}O_{n}$, and we are done. ∎

The proof of Theorem 1.1 immediately follows, which in turn implies the following (recall, that a space is $R_{0}$ [5], when every open subset contains the closure of each of its points):

Corollary 3.3**.**

Let $X$ be a Baire space, and $Y$ a 1st countable hereditarily Baire $R_{0}$-space. Then $X\times Y$ is a Baire space.

Proof.

It suffices to note that a 1st countable hereditarily Baire $R_{0}$-space $Y$ is $\beta$-unfavorable in $Ch(Y)$ by [32, Corollary 3.8.]; thus, Theorem 1.1 applies. ∎

Remark 3.4*.*

Some of the results in [18] are similar in flavor to the above results, in particular, [18, Theorem 4.4] states, that if $X$ is a Baire space, and $Y$ is a $T_{3}$-space possessing a rich family $\mathcal{F}$ of Baire spaces (i.e. $\mathcal{F}$ consists of nonempty separable closed Baire subspaces of $X$ such that if $Y\subset X$ is separable, then $Y\subseteq F$ for some $F\in\mathcal{F}$, moreover, $\overline{\bigcup_{n<\omega}F_{n}}\in\mathcal{F}$ whenever $\{F_{n}:n<\omega\}\subset\mathcal{F}$), then $X\times Y$ is a Baire space. The next example shows that our results are different (although overlapping), since spaces that are $\beta$-unfavorable in the strong Choquet game are not directly connected to spaces having rich Baire families. Indeed, there exists a $T_{1}$-space $X$ with no rich Baire family which is $\alpha$-favorable in $Ch(X)$: to see this, let $\mathbb{Q}$ be the rationals and $D$ an uncountable set. Define $X=\mathbb{Q}\cup D$, let elements of $D$ be isolated, and a neighborhood base at $q\in\mathbb{Q}$ be of the form $I\cup D\setminus C$, where $I\subseteq\mathbb{Q}$ is a Euclidean open neighborhood of $q$, and $C\subset D$ is countable. Then

$\bullet$ $X$ is strongly $\alpha$-favorable: define a tactic $\sigma$ for $\alpha$ in $Ch(X)$ as follows

[TABLE]

Each play of $Ch(X)$ compatible with $\sigma$ contains an element of $D$ in the intersection, so $\sigma$ is a winning tactic for $\alpha$.

$\bullet$ $X$ has no rich Baire family: indeed, for every separable $S\supseteq\mathbb{Q}$ we have $S=\mathbb{Q}\cup C$ for some countable $C\subset D$. It follows that if $I\subseteq\mathbb{Q}$ is a Euclidean open neighborhood of some $q\in\mathbb{Q}$, then it is also an open set in $S$ (since $I=S\cap(I\cup D\setminus C)$), and of the 1st category in $S$, thus, $S$ is not a Baire space.

Remark 3.5*.*

It is known that hereditary Baireness is not a stand-alone topological property that gives a Baire product since, under (CH), there is a hereditarily Baire space with a non-Baire square [28]; however, it is an open question weather $X\times Y$ is Baire if $X$ is Baire and $Ch(Y)$ is $\beta$-unfavorable.

4. Infinite Baire products

The following is the arbitrary product version of Krom’s Theorem:

Theorem 4.1**.**

Let $I$ be an index set. Then $\prod_{i\in I}X_{i}$ is a Baire space if and only if $\prod_{i\in I}\mathcal{K}(X_{i})$ is a Baire space.

Proof.

Denote $X=\prod_{i\in I}X_{i}$ and $X^{*}=\prod_{i\in I}\mathcal{K}(X_{i})$.

$\bullet$ Assume that $\beta$ has a winning strategy $\sigma$ in $BM(X)$, and define a strategy $\sigma^{*}$ for $\beta$ in $BM(X^{*})$ as follows: if $\sigma(\emptyset)=\prod_{i\in I_{0}}V_{0,i}\times\prod_{i\notin I_{0}}X_{i}$ for some finite $I_{0}\subset I$ and $V_{0,i}\in\mathcal{B}_{i}$, define

[TABLE]

If $U_{0}^{*}\subseteq\sigma^{*}(\emptyset)$ is $\alpha$’s response in $BM(X^{*})$, then $U_{0}^{*}=\prod_{i\in J_{0}}U_{0,i}^{*}\times\prod_{i\notin J_{0}}\mathcal{K}(X_{i})$ for some finite $J_{0}\supset I_{0}$, and for all $i\in J_{0}$, $U_{0,i}^{*}=[(U_{0,i}(0),\dots,U_{0,i}(m_{0,i}))]$ for some decreasing $X_{i}$-open $U_{0,i}(0),\dots,U_{0,i}(m_{0,i})$ and $m_{0,i}\geq 0$, where $U_{0,i}(0)=V_{0,i}$ for all $i\in I_{0}$. Denote $U_{0}=\prod_{i\in J_{0}}U_{0,i}(m_{0,i})\times\prod_{i\notin J_{0}}X_{i}$ and let

[TABLE]

where $I_{1}\supseteq J_{0}$ is finite, $V_{1,i}\in\mathcal{B}_{i}$ for each $i\in I_{1}$ and $V_{1,i}\subseteq U_{0,i}(m_{0,i})$ whenever $i\in J_{0}$. Define

[TABLE]

Proceeding inductively, we can define $\sigma^{*}$ so that whenever $k<\omega$, and

[TABLE]

is given for some finite $J_{k}$, and for all $i\in J_{k}$, $U_{k,i}^{*}=[(U_{k,i}(0),\dots,U_{k,i}(m_{k,i}))]$ for decreasing $X_{i}$-open $U_{k,i}(0),\dots,U_{k,i}(m_{k,i})$ and $m_{k,i}\geq 0$, then

[TABLE]

have been chosen, where $I_{k+1}\supseteq J_{k}$ is finite, and

[TABLE]

is such that

[TABLE]

where $U_{j}=\prod_{i\in J_{j}}U_{j,i}(m_{j,i})\times\prod_{i\notin J_{j}}X_{i}$ for all $j\leq k$. We will show that $\sigma^{*}$ is a winning strategy for $\beta$ in $MB(X^{*})$: indeed, take a play of $MB(X^{*})$

[TABLE]

compatible with $\sigma^{*}$, and assume there is some $f\in\bigcap_{n}\sigma^{*}(U_{0}^{*},\dots,U_{n}^{*})$. Then for each $i\in I$, $f(i)\in\mathcal{K}(X_{i})$ so we can pick some $x_{i}\in\bigcap_{n}f(i)(n)$. Moreover, if $i\in I_{k}$ for a given $k<\omega$, then $x_{i}\in\bigcap_{n\geq k}V_{n,i}$, so $(x_{i})_{i\in I}\in\prod_{i\in I_{k}}V_{k,i}\times\prod_{i\notin I_{k+1}}X_{i}$, thus, $(x_{i})_{i\in I}\in\bigcap_{n}\sigma(U_{0},\dots U_{k})$ which is impossible, since $\sigma$ is a winning strategy for $\beta$ in $BM(X)$.

$\bullet$ Assume that $\beta$ has a winning strategy $\sigma^{*}$ in $BM(X^{*})$, and define a strategy $\sigma$ for $\beta$ in $BM(X)$ as follows: if $\sigma^{*}(\emptyset)=\prod_{i\in I_{0}}V_{0,i}^{*}\times\prod_{i\notin I_{0}}\mathcal{K}(X_{i})$, where for all $i\in I_{0}$, $V_{0,i}^{*}=[(V_{0,i}(0),\dots,V_{0,i}(m_{0,i}))]$, define $\sigma(\emptyset)=\prod_{i\in I_{0}}V_{0,i}(m_{0,i})\times\prod_{i\notin I_{0}}X_{i}$. Let $U_{0}=\prod_{i\in J_{0}}U_{0,i}\times\prod_{i\notin J_{0}}X_{i}$ be $\alpha$’s response in $BM(X)$. Then $J_{0}\supseteq I_{0}$ is finite and $U_{0,i}\subseteq V_{0,i}(m_{0,i})$ for all $i\in I_{0}$. Define

[TABLE]

and let

[TABLE]

where $V_{1,i}^{*}=[(V_{1,i}(0),\dots,V_{1,i}(m_{1,i}))]$ whenever $i\in I_{1}$. Define

[TABLE]

Proceeding inductively, assume that whenever $k\geq 1$, and $j<k$, then $\sigma(U_{0},\dots,U_{j})=\prod_{i\in I_{j+1}}V_{j,i}(m_{j,i})\times\prod_{i\notin I_{j+1}}X_{i}$ is defined, and let $U_{k}=\prod_{i\in J_{k}}U_{k,i}\times\prod_{i\notin J_{k}}X_{i}$ be $\alpha$’s next step in $BM(X)$. Then $J_{k}\supseteq I_{k}$ is finite and $U_{k,i}\subseteq V_{k,i}(m_{k,i})$ for all $i\in I_{k}$. Define

[TABLE]

and let

[TABLE]

where $V_{k+1,i}^{*}=[(V_{k+1,i}(0),\dots,V_{k+1,i}(m_{k+1,i}))]$ whenever $i\in I_{k+1}$. Define

[TABLE]

We will show that $\sigma$ is a winning strategy for $\beta$ in $BM(X)$: take a play

[TABLE]

compatible with $\sigma$, and assume there is some $(x_{i})_{i\in I}\in\bigcap_{n}\sigma(U_{0},\dots,U_{n})$. For all $k<\omega$ and $i\in I_{k}$, define a decreasing sequence of $X_{i}$-open sets $f(i)$ so that $f(i)\restriction_{n}=(V_{n,i}(0),\dots,V_{n,i}(m_{n,i}))$ for all $n<\omega$, moreover, if $i\in I\setminus\bigcup_{k}I_{k}$, put $f(i)=(X_{i})_{n<\omega}$. Then for each $i\in I$, $x_{i}\in\bigcap_{n}f(i)(n)$, so $f(i)\in\mathcal{K}(X_{i})$, thus, $f\in X^{*}$. Moreover, $f\in\bigcap_{n}\sigma^{*}(U_{0}^{*},\dots,U_{n}^{*})$, which is impossible, since the play

[TABLE]

is compatible with $\sigma^{*}$. ∎

As a consequence, we have

Proof of Theorem 1.2.

Since $\mathcal{K}(X_{i})$ is a Baire space with a countable-in-itself $\pi$-base (see the proof of Theorem 3.1), then $\prod_{i\in I}\mathcal{K}(X_{i})$ is a Baire space by [31, Theorem 5], and so is $\prod_{i}X_{i}$ by our Theorem 4.1. ∎

Recall that $X$ has a base of countable order (BCO) $\mathcal{B}$ [30], provided each strictly decreasing sequence of members of $\mathcal{B}$ containing some $x\in X$ forms a base of neighborhoods at $x$.

Theorem 4.2**.**

Let $I$ be an index set, and for each $i\in I$, $X_{i}$ be an $R_{0}$ hereditarily Baire space with a BCO . Then $\prod_{i}X_{i}$ is a Baire space.

Proof.

For each $i\in I$, choose a BCO $\mathcal{B}_{i}$ for $X_{i}$ and prove that $\mathcal{K}^{0}_{\mathcal{B}_{i}}(X_{i})$* is a dense hereditarily Baire subspace of $\mathcal{K}(X_{i})$*: as for density, take a decreasing $h\in\mathcal{B}_{i}^{k}$, $k<\omega$, choose $g\in\mathcal{K}^{0}_{\mathcal{B}_{i}}(X_{i})$ with $g(0)\subset h(k)$, and define

[TABLE]

Then $f\in[h]\cap\mathcal{K}_{\mathcal{B}_{i}}^{0}(X_{i})$, so $\mathcal{K}_{\mathcal{B}_{i}}^{0}(X_{i})$ is dense in $\mathcal{K}(X_{i})$,

To show that $\mathcal{K}^{0}_{\mathcal{B}_{i}}(X_{i})$ is a hereditarily Baire space, we will use that, by [32, Corollary 3.9], in spaces with a BCO, hereditary Baireness is equivalent to $\beta$-unfavorability in the strong Choquet game: indeed, assume that $\sigma_{i}^{*}$ is a winning strategy for $\beta$ in $Ch(\mathcal{K}_{\mathcal{B}_{i}}^{0}(X))$, and define a strategy $\sigma_{i}$ for $\beta$ in $Ch(X_{i})$ as follows: if $\sigma_{i}^{*}(\emptyset)=(f_{0},V_{0}^{*})$ for some $f_{0}\in\mathcal{K}_{\mathcal{B}_{i}}^{0}(X)$ and $V_{0}^{*}=[f_{0}\restriction_{m_{0}}]\cap\mathcal{K}_{\mathcal{B}_{i}}^{0}(X)$, where $m_{0}\geq 1$, then pick $x_{0}\in\bigcap_{n}f_{0}(n)$, choose $V_{0}\in\mathcal{B}_{i}$ so that $x_{0}\in V_{0}\subsetneq f_{0}(m_{0}-1)$, if $f_{0}(m_{0}-1)$ is not a singleton, and $V_{0}=f_{0}(m_{0}-1)$, if $f_{0}(m_{0}-1)$ is a singleton, and define $\sigma_{i}(\emptyset)=(x_{0},V_{0})$. If $x_{0}\in U_{0}\subseteq V_{0}$ for some $U_{0}\in\mathcal{B}_{i}$, let $n_{0}\geq m_{0}$ be such that $f_{0}(n_{0})\subseteq U_{0}$, and consider $\sigma_{i}^{*}([f_{0}\restriction_{n_{0}+1}]\cap\mathcal{K}_{\mathcal{B}_{i}}^{0}(X))=(f_{1},V_{1}^{*})$, where $f_{1}\in\mathcal{K}_{\mathcal{B}_{i}}^{0}(X)$ and $V_{1}^{*}=[f_{1}\restriction_{m_{1}}]\cap\mathcal{K}_{\mathcal{B}_{i}}^{0}(X)$ for some $m_{1}\geq n_{0}+1$. Pick $x_{1}\in\bigcap_{n}f_{1}(n)$, choose $V_{1}\in\mathcal{B}_{i}$ so that $x_{1}\in V_{1}\subsetneq f_{1}(m_{1}-1)$, if$f_{1}(m_{1}-1)$ is not a singleton, and $V_{1}=f_{1}(m_{1}-1)$, if $f_{1}(m_{1}-1)$ is a singleton, and define $\sigma_{i}(U_{0})=(x_{1},V_{1})$. Proceeding inductively, assume that for a given $k\geq 1$ and all $j<k$, $\sigma_{i}(U_{0},\dots,U_{j})=(x_{j+1},V_{j+1})$ has been defined, along with $V_{j+1}^{*}=[f_{j+1}\restriction_{m_{j+1}}]\cap\mathcal{K}_{\mathcal{B}_{i}}^{0}(X)$ and $m_{j+1}>n_{j}\geq m_{j}$ so that $f_{j}(n_{j})\subseteq U_{j}$ and $\sigma_{i}^{*}([f_{0}\restriction_{n_{0}+1}]\cap\mathcal{K}_{\mathcal{B}_{i}}^{0}(X),\dots,[f_{j}\restriction_{n_{j}+1}]\cap\mathcal{K}_{\mathcal{B}_{i}}^{0}(X))=(f_{j+1},V_{j+1}^{*})$, where $V_{j}$ is either a singleton or a proper subset of $f_{j}(m_{j}-1)$. Let $U_{k}\in\mathcal{B}_{i}$ be such that $x_{k}\in U_{k}\subseteq V_{k}$, and find $n_{k}\geq m_{k}$ such that $f_{k}(n_{k})\subseteq U_{k}$ Consider

[TABLE]

where $f_{k+1}\in\mathcal{K}_{\mathcal{B}_{i}}^{0}(X)$ and $V_{k+1}^{*}=[f_{k+1}\restriction_{m_{k+1}}]\cap\mathcal{K}_{\mathcal{B}_{i}}^{0}(X)$ for some $m_{k+1}\geq n_{k}+1$. Pick $x_{k+1}\in\bigcap_{n}f_{k+1}(n)$, choose $V_{k+1}\in\mathcal{B}_{i}$ so that $x_{k+1}\in V_{k+1}\subsetneq f_{k+1}(m_{k+1}-1)$, if $f_{k+1}(m_{k+1}-1)$ is not a singleton, and $V_{k+1}=f_{k+1}(m_{k+1}-1)$, if $f_{k+1}(m_{k+1}-1)$ is a singleton, and put $\sigma_{i}(U_{0},\dots,U_{k})=(x_{k+1},V_{k+1})$. We will show that $\sigma_{i}$ is a winning strategy for $\beta$ in $Ch(X_{i})$: indeed, let

[TABLE]

be a play of $Ch(X_{i})$ compatible with $\sigma_{i}$, and assume $\bigcap_{k}V_{k}\neq\emptyset$. Define $f\in\downarrow\mathcal{B}_{i}^{\omega}$ as follows: for all $k<\omega$ and $m_{k-1}\leq p<m_{k}$ put $f(p)=f_{k}(p)$ (for completeness, let $m_{-1}=0$). Then $\bigcap_{p}f(p)=\bigcap_{k}f(m_{k}-1)=\bigcap_{k}f_{k}(m_{k}-1)\supseteq\bigcap_{k}V_{k}$, so $f\in\mathcal{K}_{\mathcal{B}_{i}}^{0}(X)$, since by the construction of $\sigma_{i}$, $\{V_{n}:n<\omega\}$ is either a strictly decreasing sequence of elements of $\mathcal{B}_{i}$, or a singleton. Moreover, $f\restriction_{m_{k}}=f_{k}\restriction_{m_{k}}$, thus, $f\in\bigcap_{k}V_{k}^{*}$, which is impossible, since

[TABLE]

is a play of $Ch(\mathcal{K}_{\mathcal{B}_{i}}^{0}(X))$ compatible with $\sigma_{i}^{*}$.

It now follows from [3, Theorem 1.1] that $\prod_{i\in I}\mathcal{K}^{0}_{\mathcal{B}_{i}}(X_{i})$ is a Baire space, which is also dense in $\prod_{i\in I}\mathcal{K}(X_{i})$, which it turn implies that $\prod_{i\in I}X_{i}$ is a Baire space by Theorem 4.1. ∎