Consider the following theorem known as Baire's Category Theorem (BCT).
Theorem.[BCT] A non-empty complete metric space $X$ is not a countable union of nowhere dense sets.
I am interested on how to prove the following using BCT.
The intersection of every countable collection of dense open subsets of a complete metric space is dense.
My draft: Let $(A_k)$ be a sequence of dense open sets in $X$. Then for each $k$, we get
$$
A_k \mbox{ is dense }\Leftrightarrow \overline{A_k}=X \Leftrightarrow \mbox{int }A_{k}^c=\varnothing.
$$
Since $A_k$ is also open, we get
$$
\mbox{int }\overline{A_{k}^c}=\mbox{int }A_{k}^c=\varnothing
$$
and so $A_{k}^c$ is nowhere dense. We now consider the set $$\bigcup_{k=1}^{\infty}A_{k}^c.\qquad \qquad \qquad(*)$$
I am convinced that upon taking the complement of (*), the resulting set must be dense. But I don't know how to verify that step. Any tips?
Best Answer
Let me give you some extra tools to use, here. Some details are left to you to verify.
Definitions: A subset $A$ of a topological space $X$ is said to be nowhere-dense (in $X$) if for every non-empty open subset $U$ of $X$ there is a non-empty open subset $V$ of $X$ such that $V\subseteq U$ and $V\cap A=\emptyset$. (You should be able to prove that this is equivalent to your definition of nowhere-dense.) A topological space $X$ is said to be a Baire space if every countable union of nowhere-dense subsets of $X$ has empty interior.
Lemma 1: $X$ is a Baire space if and only if every countable intersection of open dense subsets of $X$ is dense in $X$.
Proof: Suppose $X$ is a Baire space and $\{G_n\}_{n=1}^\infty$ is a countable collection of open dense sets, so each $X\setminus G_n$ is nowhere dense. (Why?) Thus, for any non-empty open set $U$, we have that $$U\nsubseteq\bigcup_{n=1}^\infty (X\setminus G_n)=X\setminus\left(\bigcap_{n=1}^\infty G_n\right),$$ so there must be some point of $\bigcap_{n=1}^\infty G_n$ in $U$. Thus, $\bigcap_{n=1}^\infty G_n$ is dense.
Suppose countable unions of open dense sets are dense, and let $A_n$ be nowhere dense for each $n$, so each $X\setminus\overline{A_n}$ is open and dense. (Why?) Take any non-empty open $U$, so that $U$ intersects $$\bigcap_{n=1}^\infty\left(X\setminus\overline{A_n}\right)=X\setminus\left(\bigcup_{n=1}^\infty\overline{A_n}\right)\subseteq X\setminus\left(\bigcup_{n=1}^\infty A_n\right),$$ so cannot be a subset of $\bigcup_{n=1}^\infty A_n$. Thus, $X$ is a Baire space. $\Box$
Lemma 2: $X$ is a Baire space if and only if every countable union of closed nowhere-dense subsets of $X$ has empty interior. (I leave the proof to you. Lemma 1 and DeMorgan's Laws should help.)
Lemma 3: If $X$ is a non-empty complete metric space with metric $d:X\times X\to\Bbb R$, and $F$ is a non-empty closed subset of $X$, then $F$ is a complete metric space with metric $d_F$ defined by $d_F(x,y)=d(x,y)$ for all $x,y\in F$. (I leave the proof of this to you.)
Proposition: Assuming BCT holds, every non-empty complete metric space is a Baire space.
Proof: Suppose by way of contradiction that $X$ is a non-empty complete metric space, but isn't a Baire space. Then by Lemma 2, there is some countable collection $\{A_n\}_{n=1}^\infty$ of closed nowhere-dense subsets of $X$ whose union doesn't have empty interior--meaning there is some non-empty open $U$ such that $$U\subseteq\bigcup_{n=1}^\infty A_n.$$ Now, in particular, we can take a non-empty open set $V$ such that $\overline V\subseteq U.$ (Why?) Then $$\overline V\subseteq\bigcup_{n=1}^\infty A_n,$$ so $$\overline V=\overline V\cap\left(\bigcup_{n=1}^\infty A_n\right)=\bigcup_{n=1}^\infty(\overline V\cap A_n).\tag{#}$$ But $\overline V$ is a non-empty complete metric space by Lemma 3, so by BCT, $\overline V$ is not a countable union of closed nowhere-dense subsets of $\overline V$. But each $\overline V\cap A_n$ is closed and nowhere-dense in $\overline V$ (why?), so $(\#)$ gives us the desired contradiction. $\Box$