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If $X$ is a complete metric space such that $X =\bigcup\limits_{n {\in \mathbb{N}}}F_n$, where each $F_n$ is closed, prove that
$\bigcup\limits_{ n {\in \mathbb{N}}}$$F_n^o$ is dense in $X$. ($F_n^o$ denotes the interior of $F_n$.)

I've tried a few ways to prove it, and I'm pretty sure that it uses the Baire Category Theorem somehow, but I'm not sure.

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Well, suppose not. Then there exists an open set $G \subset X$ such that $G \cap \bigcup_{n \in \mathbb{N}}F_n^o = \emptyset$. Hence there exists a subset $E \subset G$ such that $\bar{E} \subset G$ so that $\bar{E} \cap \bigcup_{n \in \mathbb{N}}F_n^o = \emptyset$, where $\bar{E}$ denotes the closure of $E$.

Now observe the following set-theoretic property: $(\bar{A} \cap B)^o \subset \bar{A} \cap B^o$. To see this, let $x \in (\bar{A} \cap B)^o.$ Then there exists $\varepsilon \gt 0$ such that $B_\varepsilon(x) \subset \bar{A} \cap B$. Thus $x$ is an interior point of $B$ and a point of $\bar{A}$. That is, $x \in \bar{A} \cap B^o$.

Going back to the problem at hand, we have that for each $n \in \mathbb{N}$, $\bar{E} \cap F_n^o = \emptyset$. But then each $(\bar{E} \cap F_n)^o = \emptyset$ (since we just showed $(\bar{E} \cap F_n)^o \subset \bar{E} \cap F_n^o$). Thus for each $n$, $\bar{E} \cap F_n$ is nowhere dense (a set is nowhere dense iff its closure has empty interior).

Now, we can write $\bar{E} = \bar{E} \cap X = \bar{E} \cap \bigcup_{n \in \mathbb{N}}F_n = \bigcup_{n \in \mathbb{N}} \bar{E} \cap F_n$. But we just demonstrated that $\bar{E} \cap F_n$ is nowhere dense for each $n \in \mathbb{N}$, so therefore $\bar{E}$ is of the first category. But $\bar{E}$ is a closed subset of a complete metric space and is thus complete, and therefore is of the second category. Hence we have a contradiction, and so our assumption that $\bigcup_{n \in \mathbb{N}}F_n^o$ wasn't dense in $X$ was wrong.

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    $\begingroup$ You should state that G and E are non-empty....+1 $\endgroup$ Dec 22, 2016 at 3:42
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    $\begingroup$ You can shorten this by eliminating $E,$ using the fact that the open set $G$ is a completely metrizable subspace of $X.$ But then someone would demand that you prove it...... Good A. $\endgroup$ Jan 5, 2017 at 17:06

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