"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!
On
Case 1 Assume that $A$ is a closed set : If $x_\alpha$ is a sequence in $A$, then for $1$ there is a compact set $A_1$ s.t. $d(x_\alpha,y_\alpha )< 1 $ for some $y_\alpha \in A_1$. Since $y_\alpha\rightarrow y\in A_1$, then $|y-x_\alpha| \leq |y-y_\alpha | + |y_\alpha - x_\alpha | \leq 2 $ for infinitely many $\alpha$.
Here we choose any $z_1=x_\alpha$ and repeatedly we have $z_k$ i.e. $|z_i-z_j|\leq \frac{4}{N} $ for all $i,\ j\geq N$. Hence $z_k$ is a Cauchy sequence so that $z_k\rightarrow z\in A$. Hence $A$ is compact.
Case 2 Assume that $A$ is not closed : For $\varepsilon >0$ and any $\overline{x}$ in the closure $\overline{A}$ there is $ x\in A$ s.t. $d( x,\overline{x})< \varepsilon$. Hence $$ d( \overline{x},x_\varepsilon )\leq d(\overline{x},x) + d(x,x_\varepsilon )\leq 2\varepsilon $$ for some $x_\varepsilon\in A_\varepsilon$. Hence we complete the proof.
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].