Are any open subsets of $\mathbb{R}^N$ $ \sigma$ -compact?

Question

I could find a description which implies that any non-empty open subsets in a euclidean space is $\sigma$-compact in Rudin's "Functional Analysis" (2nd edition) p.33 (in Example 1.44) with no proofs. However, I could not find any proofs or reliable references of this matter. Could you tell me some of these?

Answer 1

Sorry, I could give a proof by myself right after post my question.

Let $N$ be a positive integer, $\mathbb{R}^N$ be the $n$-dimensional Euclidean space and $U$ be any open subsets of $\mathbb{R}^N$. Let a function $d$ be for $x \in \mathbb{R}^N$ and $A \subset \mathbb{R}^N$ $$d(x, A) = \inf_{y \in A}{|x-y|}$$ , where $|x|$ is the usual Euclidean metric. Then, $d(-,A):\mathbb{R}^N \longrightarrow \mathbb{R}$ is continuous for any $A \subset \mathbb{R}^N$. For positive integers $n$, let $K_n$ be $K_n = \{x \in \mathbb{R}^N \mid |x| \le n\} \cap \{x \in \mathbb{R}^N \mid d(x, \mathbb{R}^N \setminus U) \ge 1/n\}$. Then, $K_n$ is compact and $\bigcup_{n=1}^{\infty}{K_n} = U$.

Answer 2

For the point-set topology afficionado, if Example 1.44 of Rudin's FA is the motivation, then the notion of "$\sigma$-compact" being used is more than "$X$ is a countable union of compact subsets", rather it is "$X$ is exhausted by compact subsets", that is $X=\bigcup_{n =1}^\infty K_n$ where each $K_n$ is a compact subset and $K_n \subseteq \text{int}(K_{n+1})$.

If we call the condition that $X$ is a countable union of compact sets "$\omega$-compact" however, then $\sigma$-compact is equivalent to the condition that $X$ is both "$\omega$-compact" and locally compact (i.e., every $x \in X$ has a compact neighbourhood). Indeed if $X$ is exhausted by compact sets it is certainly $\omega$-compact, while if $x\in X$, taking $n$ minimal such that $x \in K_n$, we see that $K_{n+1}$ is a compact neighbourhood of $x$.

If on the other hand $X$ is locally compact and $\omega$-compact, then for each $x\in X$ let $K_x$ be a compact neighbourhood of $x$, so that $x \in V_x= \text{int}(K_x)\subseteq K_x$, and let $X = \bigcup_{n \geq 1} A_n$ where each $C_n$ is compact. Replacing $A_n$ by $\bigcup_{k=1}^n A_k$ we may assume the $A_n$ are nested. Now we claim that for each $n \geq 0$ there is a compact set $K_n$ such that

i) $K_0=\emptyset$ and

ii) for $n \geq 1$, $A_n\cup K_{n-1}\subseteq \text{int}(K_n)$.

Clearly the sequence $(K_n)$ gives an exhaustion of $X$ by compact sets, so it suffices to verify the claim, which follows by induction: $K_0=\emptyset$ and if $n>0$ let $F_{n+1} = A_{n+1} \cup K_n$. Since $F_{n+1}$ is a finite union of compact sets it is compact, and hence as $F_{n+1} \subseteq \bigcup_{x \in F_{n+1}} V_x$, there is a finite subset $I_{n+1}\subseteq F_{n+1}$ such that $F_{n+1}\subseteq \bigcup_{x\in I_{n+1}} V_x$. Thus if $K_{n+1} = \bigcup_{x \in I_{n+1}} K_x$ it follows that $$ A_{n+1}\cup K_n \subseteq \bigcup_{x \in I_{n+1}}\text{int}(K_x) \subseteq \text{int}(\bigcup_{x\in I_{n+1}} K_x) = \text{int}(K_{n+1}) $$ so that $K_{n+1}$ satisfies the required condition.