0

Let $E \subseteq \mathbb{R}^n$ be an open set, and $K\subseteq E$ compact.

Can I find $K'$ compact such that $K'\subseteq E$ and such that there exists $\epsilon >0$ with the property that $\forall x \in K$, $B_{\epsilon}(x) \subseteq K'$?

My try: $d:=\text{dist}(E^c,K)>0$ because $K$ is compact, $E^c$ is closed and they're disjoint (A and B disjoint, A compact, and B closed implies there is positive distance between both sets).

Therefore I could take $\epsilon:=d/2$ and $K':=\overline{\cup_{x \in K}B_{\epsilon}(x)}$. It is obvious that $K'$ is closed and bounded, however I'm having an hard time proving it is contained in $E$.

Do you perhaps have any hint?

Lilla
  • 2,099

1 Answers1

4

Yes, $r\equiv d(K,\mathbb R^{n} \setminus E) >0$. So we can take $K'=\{x : d (x,K) \leq \frac r 2\}$. $K'$ is contained in $E$, it is compact and we can take $\epsilon =\frac r 2$.

Note that if $x \in K'$then (by compactness of $K$) there exists $y \in K$ such that $d(x,y) \leq \frac r 2$. If $x \notin E$ then $r=d(K, \mathbb R^{n}\setminus E) \leq d(y, x) \leq \frac r 2$ which is a contradiction. This proves that $K'$ is contained in $E$.

Kavi Rama Murthy
  • 311,013