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Let $M \subset \mathbb{R}^n$ be a compact subset of $\mathbb{R}^n$ and $r>0$. Prove that:

$\bigcup_{p \in M} \overline{B}(p,r)$

Is compact.

I know that this problem has an answer here: union of closed balls centered around points of a compact set

But I'm looking for an alternative proof.

My attempt:

$\bigcup_{p \in M} B(p, \epsilon)$ is an open covering of $M$ and since $M$ is compact, there exist $p_1,...p_n$ such that $M \subseteq \bigcup_{i=1}^n B(p_i, r)$. I was trying to show that $\bigcup_{p \in M} B(p, \epsilon) = \subseteq \bigcup_{i=1}^n B(p_i, r)$ so that $\bigcup_{p \in M} \overline{B}(p, \epsilon) = \bigcup_{i=1}^n \overline{B}(p_i, r)$ and since $\bigcup_{i=1}^n \overline{B}(p_i, r)$ is the finite union of closed and bounded sets It is closed and bounded and therefore compact. But I'm having some trouble trying to prove this. Is there another way to solve the problem?

MC2
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  • Are you just trying to use a less-metric argument, or do you want specifically to find a finite cover? – Keen-ameteur Nov 29 '23 at 12:57
  • I'm just looking for an alternative argument to the linked solution (not neccesariliy with a less metric argument or finite covers) – MC2 Nov 29 '23 at 14:16
  • Well, since bounded closed sets in $\mathbb{R}^n$ are compact, you can use that your set is essentially $M+\overline{B}(0,r)$. See for example in https://math.stackexchange.com/questions/515496/sum-of-closed-and-compact-set-in-a-tvs/1407904#1407904 – Keen-ameteur Nov 29 '23 at 14:19

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