Closed set as a countable intersection of open sets Let's take a metric space. Then any closed set can be written as a countable intersection of open sets.
How can I prove that?
 A: Hint: Let $A\subseteq X$ be closed. For $\epsilon > 0$ let 
$$ U_\epsilon(A) = \{x \in X \mid \def\dist{\mathop{\rm dist}}\dist(x,A) < \epsilon \}$$
where $\dist(x, A) := \inf_{y \in A} d(x,y)$. What can you say about the sets $U_\epsilon(A)$, what is $\bigcap_{\epsilon > 0} U_\epsilon(A)$?
A: Let $A\subseteq X$ be closed. For all $n\in \mathbb N$ define $$U_n=\bigcup _{a\in A} B(a,\frac{1}{n}).$$
$U_n$ is open as a union of open balls. We prove that $A=\bigcap _{n\in \mathbb N} U_n$. 
Clearly  $A \subseteq \bigcap _{n\in \mathbb N} U_n$. 
To prove $A \supseteq \bigcap _{n\in \mathbb N} U_n$ we take $x\notin A$ and show that $x\notin \bigcap U_n$.
Since $A$ is closed, $A^C$ is open, therefore $\exists n \in \mathbb N$ such that $B(x,\frac{1}{n}) \cap A=\emptyset$. That is, for all $a\in A$: $a\notin B(x,\frac{1}{n})$; and thus for all $a\in A$: $x\notin B(a,\frac{1}{n}) \Longrightarrow x\notin \bigcup_{a\in A} B(a,\frac{1}{n}) \Longrightarrow x\notin U_n \Longrightarrow  x\notin \bigcap U_n$. 
A: Let $D(A,\varepsilon)=\left\{ y\in M | d(A,y)<\varepsilon \right\}$, where $d(A,y)=\inf \left\{ d(z,y)|z\in A  \right\}$. This set is open .Define a sequence of $\varepsilon$ as $\varepsilon_{n}=\frac{1}{n}$. Claim $A=\cap_{n=1}^{\infty}D(A,\varepsilon_{n})$. Proof: If $x\in A$, then $d(A,x)=0$ and so $x \in D(A,\varepsilon_{n})$ for all $n\in \mathbb{N}$ $\Rightarrow$ $A\subset \cap_{n=1}^{\infty}D(A,\varepsilon_{n})$. Conversely if $x\in \cap_{n=1}^{\infty}D(A,\varepsilon_{n})$ then $x\in D(A,\varepsilon_{n})$ for all $n\in \mathbb{N}$ $\Rightarrow$ $\forall  \varepsilon>0$ $D(x,\varepsilon)\cap A\setminus \left\{ x \right\}\neq \emptyset$ and so $x$ is an accumulation point of $A$. $A$ is, however, closed so contains all its accumulation points, so $x\in A$ $\Rightarrow$ $\cap_{n=1}^{\infty}D(A,\varepsilon_{n})\subset A$.
A: About the openness of $G_n=\{x\in M|d(x,F)<\frac 1n\}$, where $n\in \Bbb N, (M,d)$ our metric space and $F\subset M$ a closed set; let $y\in G_n$ and $ε:=\frac 1n -d(y,F)>0$. If $z\in B^d_ε(y)$, we have $d(z,F)\leq d(z,x)$, for all $x\in G_n$; so $d(z,F)\leq d(z,x)\leq d(z,y)+d(y,x)\implies d(z,F)\leq d(z,y)+d(y,F)<ε+d(y,F)=\frac 1n$, thus $z\in G_n$, which implies that $B^d_ε(y)\subset G_n$, so $G_n$ is open.
