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.

In fact it's trivial, just consider the sub-level sets {d_F<1/n} of the distance function from F — Pietro Majer, Jan 30 '25 at 17:54

David Moews · Accepted Answer · 2024-10-26T18:15:39.703

18

If $\displaystyle{x\in\bigcap_{n=1}^{\infty}B_n}$, 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 $x\in F$.

edited Oct 26 '24 at 18:15

answered Mar 01 '13 at 03:08

David Moews

17,341

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 $G_\delta$

1 Answers1

Linked