"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!

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

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].

  • $\begingroup$ One also needs to remark that the closure of a totally bounded set is still totally bounded. $\endgroup$ – Henno Brandsma Jan 18 at 9:22
  • $\begingroup$ @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. $\endgroup$ – Kavi Rama Murthy Jan 18 at 9:28

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