Is closure of a compact set itself?

Question

I came across this question:

If a point $a$ and a compact set $B$ have $d(a,B)=\inf\{d(a,b):d\in B\}=0$, does it mean $a$ is in the closure of $B$? If it is, can we say since $B$ is compact, its closure is itself so $a$ is also contained in $B$? Any help is appreciated.

Of course.
Since the set is compact then it is closed,and its closure is equal to the set. — Marios Gretsas, Sep 29 '19 at 00:01

score 1 · Accepted Answer · answered Sep 29 '19 at 00:01

1

In a metric space(more generally in a Hausdorff topological space) any compact set is closed. Hence a compact set is its own closure.

answered Sep 29 '19 at 00:01

Kavi Rama Murthy

311,013

Is closure of a compact set itself?

1 Answers1