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I'm studying Sard's theorem and I want to know why is true that every open set can be expressed as a countable union of compact sets. Thank you!

José Carlos Santos
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Alexandre Muñoz
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Let $\Omega$ be an open set and $\partial \Omega$ denote its boundary. Set $K_{n} = \{x \in \Omega: \text{dist}(x, \partial \Omega) \geq \frac{1}{n}\} \cap \overline{B(0,n)}$, where dist is the distance of the point $x$ from the boundary. Each $K_{n}$ is closed and bounded and therefore compact and clearly $\cup_{n = 1}^{\infty} K_{n} = \Omega$.

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Let $A$ be an open subce of $\mathbb{R}^n$. For each $m\in\mathbb N$, let$$K_m=\left\{x\in A\,\middle|\,d\left(x,A^\complement\right)\geqslant\frac1m\wedge\|x\|\leqslant m\right\}.$$Then each $K_m$ is compact and$$A=\bigcup_{m=1}^\infty K_m.$$

José Carlos Santos
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  • Thanks!!!!!!!:) – Alexandre Muñoz May 04 '18 at 13:22
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    This might be obvious, but could you explain a little bit why $K_m$ is closed? In particular, is there a way to see it without using the limit of a sequence in a closed set stays in the closed set? – Mathematics_Beginner Jan 04 '22 at 22:04
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    @Mathematics_Beginner $f:= d( \cdot, A^c)$ is a continuous function. $K_m$ equals the inverse image $f^{-1}([\frac{1}{m} , \infty))$ intersection with the closed ball ${ || x|| \le m }$. Because $[\frac{1}{m} , \infty)$ is closed, and $f$ continuous, then $f^{-1}([\frac{1}{m} , \infty))$ is closed. Finally $K_m$, being the intersection of two closed sets, namely again $f^{-1}([\frac{1}{m} , \infty))$ and ${ || x|| \le m }$ is itself closed. So probably the most difficult part to show is that $f : x \mapsto d(x, A^c)$ is a continuous function, so of course I didn't prove that part... Sorry – hasManyStupidQuestions Feb 11 '22 at 18:13
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    The fact that $A$ is open means that $A^c$ is closed, which is important for showing that $f$ is continuous, cf. these two related questions: https://math.stackexchange.com/questions/83505/distance-to-a-closed-set and https://math.stackexchange.com/questions/944659/distance-to-a-closed-set-is-continuous/944934 Cf. e.g. this question for why the closed ball ${ || x || \le m }$ really is closed: https://math.stackexchange.com/questions/661759/a-closed-ball-in-a-metric-space-is-a-closed-set – hasManyStupidQuestions Feb 11 '22 at 18:15
  • This result still remains valid if we replace $K_m$ for $(K_m)^\circ$, where $A^\circ$ represents the interior of $A$? – Rodolfo Ferreira de Oliveira Jan 30 '24 at 17:59
  • @RodolfoFerreiradeOliveira Yes, it does. – José Carlos Santos Jan 30 '24 at 18:05
  • So, every open set can be write by a countable union of a open sets $A_m$ with Lipschitz boundary and $A_m \subset A_{m+1}$ for every $m$, right? – Rodolfo Ferreira de Oliveira Jan 30 '24 at 18:10
  • @RodolfoFerreiradeOliveira I don't know the meaning of “Lipschitz boundary”. – José Carlos Santos Jan 30 '24 at 18:12
  • a domain which its boundary is a graph of a lipschitz map . For instance, in construction above, the boundary of every $K_m$ is the curve ${x \in A: d(x,A^c) = \dfrac{1}{m} \ \mbox{e} \ |x|=m}$. Since the distance function is lipschitz, the boundary of $K_m$ is Lipschitz – Rodolfo Ferreira de Oliveira Jan 30 '24 at 18:15
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@harmonicuser has a nice explicit construction for $\mathbb{R}^n$. But as you use the tag general topology:

In fact the statement holds in all hereditarily Lindelöf locally compact Hausdorff spaces $X$ (of which the Euclidean spaces are a prominent example): if $O\subseteq X$ is open, by local compactness plus Hausdorffness we find for each $x \in O$ an open subset $U_x$ with $\overline{U_x}$ compact and $\overline{U_x} \subseteq O$. As $O$ is Lindelöf we reduce the open cover $\{U_x: x \in O\}$of $O$ to a countable subcover $\{U_x: x \in N\}$ of $O$ (where $N \subseteq O$ is countable). Then $O = \cup_{x \in N} \overline{U_x}$ shows that $O$ is $\sigma$-compact.

If $X$ is a space then every $O$ being $\sigma$-compact implies that $X$ is hereditarily Lindelöf (so that condition is necessary) but not necessarly that $X$ is locally compact (as witnessed by $\mathbb{Q}$).

Henno Brandsma
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