A Baire space is a topological space with the following property: for each countable collection of open dense sets ${\displaystyle \{U_{n}\}_{n=1}^{\infty }}$, their intersection ${\displaystyle \textstyle \bigcap _{n=1}^{\infty }U_{n}}$ is dense.

The Baire Category Theorem states: Every complete metric space is a Baire space.

How to prove that:

Let $(X,d)$ be a complete metric space and $\{F_n\}_{n\in \mathbb{N}}$ a collection of closed subsets of $X$. With the help of Baire Category Theorem prove that if $Y\subset X$ is a closed subset and $X=\bigcup_{n=1}^\infty F_n$, then $U=\bigcup_{n=1}^\infty\operatorname{int}_Y (Y\cap F_n)$ is dense in $Y$.


2 Answers 2


Since $X$ is complete and $Y$ is a closed subset of $X$, $Y$ is complete. And since, for each $n\in\mathbb N$, $F_n$ is a closed subset of $X$, $Y\cap F_n$ is a closed subset of $Y$. Finally,$$X=\bigcup_{n\in\mathbb N}F_n\implies Y=\bigcup_{n\in\mathbb N}Y\cap F_n.$$So, this expresses $Y$ as a countable union of closed subsets of $Y$. Let $D$ be the closure, in $Y$, of $\bigcup_{n\in\mathbb N}\operatorname{int}_Y(Y\cap F_n)$. I will prove that $D=Y$. Suppose otherwise. Let $y\in Y\setminus D$. Since $Y\setminus D$ is an open set, there is some $r>0$ such that the closed ball $B_r'(y)$ doesn't intersect $D$. Since $B_r'(y)$ is a closed subset of $X$, it is a complete metric space. Since we also have$$B_r'(y)=\bigcup_{n\in\mathbb N}B_r'(y)\cap F_n$$and each $B_r'(y)\cap F_n$ is a closed subset of $B_r'(y)$, it follows from the Baire category theorem that, for some $n\in\mathbb N$, $\operatorname{int}_Y\bigl(B_r'(y)\cap F_n\bigr)\neq\emptyset$. Take $z\in\operatorname{int}_Y\bigl(B_r'(y)\cap F_n\bigr)$. Then$$z\in\bigcup_{n\in\mathbb N}\operatorname{int}_Y\bigl(B_r'(y)\cap F_n\bigr)\subset\bigcup_{n\in\mathbb N}\operatorname{int}_Y\bigl(Y\cap F_n\bigr).$$But then $z\in D$, which cannot be, because $z\in B_r'(y)$ and $B_r'(u)\cap D=\emptyset$.

  • $\begingroup$ Which version of the definition of the Baire Space do you use on the prove? $\endgroup$
    – Emo
    Nov 26, 2017 at 16:07
  • $\begingroup$ @Emin The one that you mentioned in the statement of the problem. $\endgroup$ Nov 26, 2017 at 16:09
  • $\begingroup$ I am not noticing a countably collection of open dense sets on your proof. $\endgroup$
    – Emo
    Nov 26, 2017 at 16:14
  • $\begingroup$ @Emin That's because I used the fact that the assertion “the intersection of every countable family of open dense sets is dense” is equivalent to the assertion “the union of every countable family of closed subsets with empty interior has empty interior”. $\endgroup$ Nov 26, 2017 at 16:22

Hint: First, show that $U$ is nonempty. Then, given any nonempty open $V\subseteq X$, apply the same reasoning to $X=Y=V$ to conclude that $U\cap V$ is nonempty. Then apply the conclusion to $X=Y$.


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