Expanding a compact subset of $\mathbb R^n$

Question

Note: A set $X \subset \mathbb{R}^n$ is covering compact if every open cover of $X$ has a finite subcover.

If $U \in \mathbb{R}^n$ is open and $C \subset U$ is covering compact, show that there is a covering compact set $D$ such that $C \subset \text{int}(D)$ and $D \subset U$.

Here's my thinking so far. Let $$ be the set consisting of open balls of radius one around each point $ \in $. Then we have $$ \subset \bigcup_{\beta \in S}{\beta}.$$ Since $⊂$ and $$ is covering compact, there exists a subcover $′⊂$ such that $$⊂ \bigcup_{\beta \in S′}{}.$$ I'm trying to see if this subcover ′ can be turned into a closed, bounded set, but I'm having trouble. Any help?

Answer 1

Hint: This isn't true in general topological spaces so you'll need to use the properties of $\Bbb R_n$. If $C$ is compact, then it also must be closed. Use that fact to put an open ball around every point of $C$ that is bounded away from the complement of $U$. Now use compactness to pick finitely many such balls and take their closure. That's the set you'll want.

I've left a lot of details for you to work out.

Answer 2

For each $p\in C$ take $r_p>0$ such that the open ball $B(p,r_p)$ is a subset of $U.$ Then $\{B(p,r_p/2):p\in C\}$ is an open cover of $C,$ so take a finite $C^*\subset C$ such that $C\subset \cup \{B(p,r_p/2):p\in C^*\}.$ Let $D$ be the closure of $\cup \{B(p,r_p/2):p\in C^*\}.$

Answer 3

$U$ is open so $U'$ is closed. Also, $C\cap U'=\emptyset$ since $C\subset U.$ Since $C$ is compact and the distance between disjoint compact set and closed set in a metric space is positive, $d(C,U')=\epsilon>0.$ Let $$D=\overline{\bigcup_{x\in C} B(x,\epsilon/2)}.$$ Then $D$ is closed, $C\subset D.$ Finally, $D\subset U$ because otherwise $D\cap U'$ would be non-empty and there would be an element $y\in D\cap U'$ so that $d(C,U')\leq d(C,y)\leq\epsilon/2.$

Answer 4

Let $X$ be any regular, locally compact topological space (such as $\mathbb R^n$). Then $X$ has this property: for each compact $C$ and open $U\supseteq C$, there is a compact $D$ with $C\subseteq int(D)\subseteq D\subseteq U$.

To see this, for each $c\in C$ choose a neighborhood $V_c$ of $c$ whose closure is compact (by locally compact) and a subset of $U$ (by regular), that is, where $c\in V_c\subseteq cl(V_c)\subseteq U$.

Then $\{V_c:c\in C\}$ covers $C$, so choose $\{c_n:n<N\}\subseteq C$ with $C\subseteq V=\bigcup\{V_{c_n}:n<N\}$. Then $D=\bigcup\{cl(V_{c_n}):n<N\}\subseteq U$ is compact with $C\subseteq V\subseteq int(D)\subseteq D\subseteq U$.