# Show the closure of a subset of a complete metric space is compact.

"Let $$A$$ be a subset of a complete metric space. Assume that for all $$ε > 0$$, there exists a compact set $$A_ε$$ so that $$∀ x ∈ A, d(x, A_ε)<ε$$. Show that $$A$$'s closure is compact."

I am trying to prove it with "Cauchy sequences in complete metric space are convergent" and "sequential compact set is compact". But I do not know how to prove all sequence in A(or A's closure?) has a cauchy subsequence.

Any help or advice would be greatly appreciated. Thank you in advance!

• I would suggest taking the intersection of all sets $A_\varepsilon$. At first glance, this looks like the closure of $A$. – Lubin Jan 18 at 6:07
• @Lubin why would any finite subfamily of these $A_\varepsilon$ intersect? – Henno Brandsma Jan 18 at 9:20
• @HennoBrandsma, it seems, after a second glance, that my suggestion is worthless. Thanks for encouraging me to think. – Lubin Jan 18 at 14:51

## 1 Answer

I think the best proof is to use total boundedness. Let $$\epsilon >0$$. Then $$A_{\epsilon /2}$$ is compact, hence totally bounded. You can cover it by a finite number of open balls of radius $$\epsilon /2$$. Call these balls $$B(x_i,\epsilon /2), 1\leq i \leq n$$. Then verify that $$A$$ is covered by the balls $$B(x_i,\epsilon ), 1\leq i \leq n$$. It follows that $$A$$ is totally bounded. Since $$X$$ is complete it follows that the closure of $$A$$ is compact. [ My comment below may be useful here].

• One also needs to remark that the closure of a totally bounded set is still totally bounded. – Henno Brandsma Jan 18 at 9:22
• @HennoBrandsma I am using a theorem which says that the closure of a subset $A$ of a complete metric space is compact iff $A$ is totally bounded. Yes, the proof of this uses the fact that closure of totally bounded set is totally bounded. – Kavi Rama Murthy Jan 18 at 9:28