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

Question

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.

I'm not sure how to do that, please help me. Thank you so much.

hint: Every metric space is a perfectly normal space. – M.Sina Mar 01 '13 at 03:08 — M.Sina, Mar 01 '13 at 03:08

score 15 · Accepted Answer · answered Mar 01 '13 at 03:08

15

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 $x\in F$.

answered Mar 01 '13 at 03:08

David Moews

16,651

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 01 '13 at 03:54
1

Yes, that's right. – David Moews Mar 01 '13 at 03:57

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

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