# A closed set in a metric space is $G_\delta$

How do I prove that a closed set $F$ in the metric $(X,d)$ is $G_\delta$.

Let $n\in \mathbb{N}$.

I consider $B_n={F}=\bigcup_ {x\in F} B(x,{1\over n})$, which is a collection of an open ball. Then I guess what I have to show next is the intersection of all these open balls is the closed set F.

If $x\in \bigcap_n B_n$, then for each $n$, by the definition of $B_n$, there is some $x_n\in F$ with $d(x_n,x)<1/n$. Then $(x_n)$ converges to $x$, so as $F$ is closed $x\in F$.
• so this shows that $\bigcap {B_n} \subset F$ right? Is $F\subset \bigcap {B_n}$ a fact since $F \subset B_n$ for each n? – Akaichan Mar 1 '13 at 3:54