If $A\subset \mathbb R^n$ compact and $U$ open such that $A\subset U$, why is there an open $B$ such that $A\subset B\subset \bar B\subset U$ where $\bar B$ compact ?

I know that since $\mathbb R^n$ is normal, and $A$ is copact (in particular closed), then there is an open $B\supset A$ such that $$A\subset B\subset \bar B\subset U,$$

but why $\bar B$ can be compact ? I was thinking that since $A$ is compact, it's totally bounded, i.e. for all $\varepsilon>0$ there are $x_1,...,x_n$ such that $$A\subset \bigcup_{i=1}^n B(x_i,\varepsilon)=:F.$$ Then $$A\subset (F\cap U)\subset (\bar F\cap U)\subset U,$$ so ok $F\cap U$ is open. There is no reason for $\bar F\cap U$ to be compact except if $\bar F\subset U$ since $\bar F$ compact. I could also consider $\overline{F\cap U}$ but there is no reason here to have $\overline{F\cap U}\subset U$. So how can I manage to find a $\bar B$ compact ?

  • $\begingroup$ In your title it is not mentioned that $\overline B$ is demanded to be compact. That is inconsistent with your question in the body. $\endgroup$ – drhab Aug 25 '18 at 10:52
  • $\begingroup$ @drhab: simply because I can't put more that 150 characters in the title. $\endgroup$ – user386627 Aug 25 '18 at 10:55

Since $A$ is compact it is bounded so $A\subseteq B(0,r)$ for some $r>0$.

Then $A\subseteq U\cap B(0,r)$ where $U\cap B(0,r)$ is an open set so by normality of $\mathbb R^n$ an open set $B$ exists with $A\subseteq B\subseteq\overline B\subseteq U\cap B(0,r)\subseteq U$.

Then $\overline B$ is closed and bounded, hence is compact.

  • $\begingroup$ great answer :) (at the end it wasn't that complicated). Thank you very much. $\endgroup$ – user386627 Aug 25 '18 at 10:57
  • $\begingroup$ You are very welcome. $\endgroup$ – drhab Aug 25 '18 at 11:02

Since $A$ is compact, there exists $\delta>0$ such that $d(A,\partial U)\geq \delta$. Then the closed balls $\Big\{ B(a,\frac{\delta}{2}) \Big\}_{a\in A}$ cover $A$, and their closure is each contained in $U$. By compactness there, exist a finite subcover $\Big\{ B(a_j,\frac{\delta}{2}) \Big\}_{j=1}^m$ and define $B=\underset{j=1}{\overset{m}{\cup}}B(a_j,\frac{\delta}{2})$. Then:

$A\subseteq B\subset\overline{B}\subseteq \underset{j=1}{\overset{m}{\cup}} \overline{B(a_j,\frac{\delta}{2}) } \subset U $.

Then $\overline{B}$ is a closed bounded subset in $\mathbb{R}^n$ and therefore compact.

  • $\begingroup$ good answer too. But Why such a $\delta$ exist ? Is it the $\delta$ lebesgue number ? $\endgroup$ – user386627 Aug 25 '18 at 10:58
  • $\begingroup$ Because the distance between a compact and closed set is positive, which I call $\delta$. $\endgroup$ – Keen-ameteur Aug 25 '18 at 11:19

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