# Covering compact sets

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 $$D \subset U$$.

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

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

• What have you tried? Also, I'm not sure the phrase "covering compact" is standard though I think I know what it means. It might be helpful for you to define it. – Robert Shore Feb 21 at 5:08
• I don't know whether that statement is true in general topological spaces, but it looks like you're working in $\Bbb R^n$. Is that correct? Rather than putting your attempts in comments, it's better form to put them directly into your question. That way people scanning the question list can see that you've made an effort before making the decision whether to click through. – Robert Shore Feb 21 at 5:16
• Yes, i'm working in $\mathbb{R}^n$. – Mingle McPringle Feb 21 at 5:20
• Please edit your question to be self-contained by adding that fact and by adding your attempts to solve the proble, as set forth in your comments. – Robert Shore Feb 21 at 5:23
• okay, i moved it – Mingle McPringle Feb 21 at 5:27

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.
• Do you mean put an open ball around every point of $C$ that is bounded away from the complement of $C$? – Mingle McPringle Feb 21 at 5:33
• No, that won't be possible for points on the boundary of $C$. – Robert Shore Feb 21 at 7:48
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 $$\cup \{B(p,r_p/2):p\in C^*\}.$$