Prove a closed set is a countable union of compact sets, in $\mathbb{R}^n$.

Question

I'm studying Sard's theorem and I want to know why is true that, in $\mathbb{R}^n$, every closed subset can be expressed as a countable union of compact sets. Thank you, :)

If you denote $X$ your set, and $X_i={ x \in X : |x|<i }$, then $X_i$ are compact (closed and bounded) and $$X= \bigcup_i X_i$$ — Crostul, May 03 '18 at 17:46

score 10 · Accepted Answer · answered May 03 '18 at 17:55

10

Consider $\bar{B}(0,n)$, the closed ball with center $0$ and radius $N$. If $C$ is a closed set, consider $$ C_n=C\cap\bar{B}(0,n) $$ Then $C_n$ is a closed subset of $\bar{B}(0,n)$, which is compact, hence $C_n$ is compact. Obviously $$ C=\bigcup_{\substack{n\in\mathbb{N}\\n>0}}C_n $$

answered May 03 '18 at 17:55

egreg

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Thanks a lot! – Alexandre Muñoz May 03 '18 at 19:38

Prove a closed set is a countable union of compact sets, in $\mathbb{R}^n$.

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