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 ?
Best Answer
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.