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Let $S$ be a compact subset of $(\mathbb{R}^n, d_2)$ where $d_2$ is the Euclidean metric. For all $r > 0$, we write $S_r = \bigcup\limits_{s \in S} \overline{B}_r(s)$.

I want to show that $S_r$ is a closed set but I am a little stuck. I know that a finite union of closed sets is closed, but in this case is it necessarily true that $S$ is finite?

zerxee
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  • Can you obtain a finite subset $I$ of $S$ such that $\bigcup_{s\in I}\overline{B}_r(s)$ is equal to the union taken over the whole $S$? – Javi Feb 24 '19 at 21:38
  • To answer your last question, no, an arbitrary compact subset need not be finite (for example a closed ball of positive radius has infinitely many points). – Javi Feb 24 '19 at 21:40
  • I don't know if you've learned about the facts that I used in my post or not. If there's anything you don't understand please feel free to ask. – BigbearZzz Feb 24 '19 at 22:19

1 Answers1

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First, I'll show that if $S$ is compact then $S_r$ is actually equal to the set $$ F = \{ x\in \Bbb R^n : \text{dist}(x,S) \le r\}, $$ where $\text{dist}(x,S):= \inf_{y\in S} d(x,y)$.

It is clear that $S_r \subset F$. As for the converse, we consider an arbitrary $x\in F$. Let $\text{dist}(x,S)=d \le r$, we can find a sequence $y_n\in S$ such that $$ \lim_{n\to\infty} d(x,y_n) = d. $$ By compactness, we can find a subsequence that converges to a point $y\in S$. This point satisfies $d(x,y)=d$, hence $x\in \overline B_r(y)\subset S_r$. This shows that $F\subset S_r$ and we conclude that $S_r = F$.

It is well-known that the function $\text{dist}(\cdot,S):\Bbb R^n\to[0,\infty)$ is a continuous function. Since $F$ is the inverse image of $[0,r]$, which is a closed set, under the mapping $\text{dist}(\cdot,S)^{-1}$, $F$ is a closed set.

BigbearZzz
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