Baire Category theorem application

Question

(Baire Category Theorem). Let $(X,d)$ be a (non-empty) complete space. Suppose $X=\bigcup_n F_n$ where each $F_n$ is closed. Then there exists $n$ such that $F_n$ has non-empty interior.

Proof Suppose for the sake of a contradiction that for each $n$, $F_n$ has empty interior. In particular, that would imply there exists an element $x_1\in F_1^c$ (otherwise $X$ would be empty). So, we may obtain a closed neighborhood $G_1$ of $x_1$ such that $G_1\cap F_1=\varnothing$, where we can specify that $diamG_1<\frac{1}{2}$ (this exists because a singleton is such an example). Consider $F_1\cup F_2$. Since they have empty interior, we may find $x_2\in X\backslash (F_1\cup F_2)$, with a closed neighborhood $G$ of $x_2$ satisfying $diamG<\frac{1}{2^2}$, such that $G\cap (F_1\cup F_2)=\varnothing$. Set $G_2=G\cap G_1$. This set in particular, is a closed set containing $x_2$......

My question is, why is $G\cap G_1$ a closed neighborhood of $x_2$?

Answer 1

If you already know Baire, suppose that all $F_n$ have empty interior.

Define $D_n = X\setminus F_n$. Then $D_n$ is open (as $F_n$ is closed) and dense (as $\overline{X\setminus F_n}= X\setminus \operatorname{int}(F_n)=X$)

Then Baire's theorem says that $$\bigcap_n D_n = \bigcap (X\setminus F_n) = X\setminus (\bigcup_n F_n) = X\setminus X=\emptyset$$

would be dense in $X$, a contradiction. So the assumption that all $F_n$ have empty interior is incorrect. QED.

Answer 2

You can fix the argument like this: Since $F_1\cup F_2$ is closed, $V=X-(F_1\cup F_2)$ is open, remark that $V$ is dense and open since the interior of $F_1\cap F_2$ is empty. Let $G^0$ be the interior of $G$, $G^0\cap V$ is not empty and open since $G^0$ is open and $V$ is open and dense. Let $x_2\in G^0\cap V$ there exists $B(x_2,r)\subset G^0\cap V,\bar B(x_2,r/2)\subset B(x_2,r)\subset G_0\cap V$. Where $\bar B(x_2,r/2)$ is the closed ball of radius $r/2$. we can take $r<{1\over 2^2}$ and $G_1=\bar B(x_2,r)\subset G\subset V$.