# A Characterization of Baire Spaces

We present a useful characterization of Baire spaces. A Baire space is a topological space $X$ in which the conclusion of Baire category theorem holds, that is, for each countable family $\left\{U_1,U_2,U_3,\cdots \right\}$ of open and dense subsets of $X$, the intersection $\bigcap \limits_{n=1}^\infty U_n$ is dense in $X$. This definition is equivalent to the statement that every non-empty open subset of $X$ is of second category in $X$. An elementary discussion of Baire spaces is found in this blog post. Baire spaces can also be characterized in terms of the Banach-Mazur game (see Theorem 1 in this post). We add one more characterization in terms of point-finite open cover and locally finite open family (from [2] and [3]). We prove the following theorem.

A collection $\mathcal{S}$ of subsets of a space $X$ is said to be point-finite if every point in the space $X$ belongs to at most finitely many members of $\mathcal{S}$. A collection $\mathcal{S}$ of subsets of $X$ is said to be locally finite at the point $x \in X$ if there is an open set $V \subset X$ such that $x \in V$ and $V$ meets at most finitely many members of $\mathcal{S}$. The collection $\mathcal{S}$ is said to be locally finite in the space $X$ if it is locally finite at every $x \in X$. For any terms and concepts not explicitly defined here, refer to [1] (Engelking) or [4]) (Willard).

Theorem
Let $X$ be a space. The following conditions are equivalent.

1. $X$ is a Baire space.
2. For any point-finite open cover $\mathcal{U}$ of $X$, the set $D=\left\{x \in X: \mathcal{U} \text{ is locally finite at } x \right\}$ is a dense set in $X$.
3. For any countable point-finite open cover $\mathcal{U}$ of $X$, the set $D=\left\{x \in X: \mathcal{U} \text{ is locally finite at } x \right\}$ is a dense set in $X$.

Proof
$1 \Rightarrow 2$
Let $\mathcal{U}$ be a point-finite open cover of $X$. Let $O$ be a non-empty open subset of $X$. We wish to show that $O \cap D \ne \varnothing$ where $D$ is the set defined in condition 2. For each $n$, define

$\displaystyle . \ \ \ \ \ F_n=\left\{x \in O: x \text{ belongs to exactly n members of } \mathcal{U} \right\}$.

Note that $O=\bigcup \limits_{n=1}^\infty F_n$. Since $X$ is a Baire space, $O$ must be of second category in $X$. None of the sets $F_n$ can be a nowhere dense set. Thus for some $n$, $F_n$ has non-empty interior. Choose some non-empty open set $W$ such that $W \subset F_n$.

Pick $y \in W$. Since $y \in F_n$, let $U_1,U_2,\cdots,U_n$ be the $n$ members of $\mathcal{U}$ that contain $y$. Let $V=W \cap \bigcap \limits_{j=1}^n U_n$. Note that $V \subset W \subset F_n \subset O$. Observe that $V$ is a non-empty open set that meets exactly $n$ members of $\mathcal{U}$. Therefore $\mathcal{U}$ is locally finite at points of $V$, leading to the conclusion that $V \subset D$ and $O \cap D \ne \varnothing$.

The direction $2 \Rightarrow 3$ is immediate.

$3 \Rightarrow 1$
Suppose condition 3 holds. We claim that $X$ is a Baire space. Suppose not. Let $U$ be a non-empty open subset of $X$ such that $U=\bigcup \limits_{n=1}^\infty K_n$ where each $K_n$ is nowhere dense in $X$. Let $\mathcal{U}$ be defined as the following:

$\displaystyle . \ \ \ \ \ \mathcal{U}=\left\{X \right\} \cup \left\{U_n: n=1,2,3,\cdots\right\}$,

where $U_n=U - (\overline{K_1} \cup \cdots \cup \overline{K_n})$. Clearly, $\mathcal{U}$ is a point-finite open cover of $X$. By condition 3, $D$ is dense in $X$ ($D$ is defined in condition 3). In particular, $U \cap D \ne \varnothing$. Choose $y \in U \cap D$. Since $\mathcal{U}$ is locally finite at $y$, we can choose some open set $V \subset U$ such that $y \in V$ and such that $V$ meets only finitely many $U_j$, say only up to $U_1,\cdots, U_m$ (so $V \cap U_j = \varnothing$ for all $j > m$).

On the other hand, all sets $K_j$ are nowhere dense. So we can choose some open set $V_0 \subset V$ such that $V_0$ misses the nowhere dense set $\overline{K_1} \cup \cdots \cup \overline{K_m} \cup \overline{K_{m+1}}$. In particular, this means that $V_0 \cap U_{m+1} \ne \varnothing$, contradicting that $V \cap U_j = \varnothing$ for all $j > m$. So $X$ must be a Baire space if condition 3 holds. $\blacksquare$

Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Fletcher, P., Lindgren, W. F., A note on spaces of second category, Arch, Math., 24, 186-187, 1973.
3. McCoy, R. A., A Baire space extension, Proc. Amer. Math. Soc., 33, 199-202, 1972.
4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.