# A proof using Baire Category Theorem

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 $Y$.

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 $B_r'(u)\cap D=\emptyset$.
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 $X=Y$.