13
$\begingroup$

I want to show that every Hausdorff, locally compact and second countable is $\sigma$-compact.

I'm having trouble writing this rigorously. Can we proceed as follows:

Let $X$ be a locally compact Hausdorff space. Let $\mathcal{B}=\{B_{n}: n \in \mathbb{N}\}$ be a countable basis, then for each $x \in X$ find $U \subseteq X$ open such that $x \in U$ and $\overline{U}$ is compact. Then choose $B_{n} \in \mathcal{B}$ such that $x \in B_{n} \subseteq U$. Since $\overline{U}$ is compact then $\overline{B_{n}}$ is compact. So it seems that $X$ is the countable union of the closure of $B_{n}'s$. Can we simply write this as:

$X= \bigcup_{n \in \mathbb{N}} \{\overline{B_{n}}: B_{n} \in \mathcal{B}, \overline{B_{n}} \ \textrm{is compact}\}$

$\endgroup$
3
  • 5
    $\begingroup$ Using this work you can probably prove a standard lemma: in a second countable space, every cover has a countable subcover. $\endgroup$ Aug 14, 2011 at 5:30
  • 4
    $\begingroup$ every open cover has a countable subcover $\endgroup$
    – GEdgar
    Aug 14, 2011 at 13:03
  • $\begingroup$ the last line for the index it might not be the whole of $N$ but rather some subset of $N$. $\endgroup$
    – Uncool
    Mar 20, 2019 at 13:19

1 Answer 1

20
$\begingroup$

The argument is basically just fine, but the notation in that last line is a bit off: when you write $\bigcup_{n\in\mathbb{N}}$, you’re saying that you want to take the union of some sets indexed by natural numbers. What follows $\bigcup_{n\in\mathbb{N}}$ should be a typical one of these sets with index $n$. If you knew that every member of $\mathcal{B}$ was compact, for instance, you could write $X = \bigcup_{n\in\mathbb{N}}\operatorname{cl}B_n$. What you want here is simply $X = \bigcup \{\operatorname{cl}B:B \in \mathcal{B}\text{ and}\operatorname{cl}B\text{ is compact}\}$.

If fact, you never needed to index $\mathcal{B}$ in the first place. It’s perfectly fine to say this:

  • Let $\mathcal{B}$ be a countable base. Since $X$ is locally compact, for each $x\in X$ there is an open nbhd $U_x$ of $x$ with compact closure, and since $\mathcal{B}$ is a base for $X$, there is some $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq U_x$. Clearly each $B_x$ has compact closure, and $X = \bigcup\limits_{x\in X}\operatorname{cl}B_x$. There are only countably many distinct members of $\mathcal{B}$, so there are only countably many different sets $B_x$, and we have therefore expressed $X$ as the union of countably many compact subsets, as desired.

Or you could replace that last sentence with something like this:

  • $\{B_x:x\in X\} \subseteq \{B \in \mathcal{B}:\operatorname{cl}B\text{ is compact}\} \subseteq \mathcal{B}$, so $\{B_x:x\in X\}$ is countable, and $X = \bigcup\limits_{x\in X}\operatorname{cl}B_x$ therefore expresses $X$ as a countable union of compact subsets.

There are many other perfectly good ways to do it; I’m just trying to give you some idea of what’s possible, since you’re working on the proof-writing as much as you are on the mathematics itself.

$\endgroup$
4
  • $\begingroup$ thank you again! I believe you meant "Since $X$ is locally compact" instead of $X$ is $\sigma$-compact in the first line of your paragraph. $\endgroup$
    – user10
    Aug 14, 2011 at 6:54
  • $\begingroup$ @user10: I did indeed; thanks. I’ve fixed it. $\endgroup$ Aug 14, 2011 at 7:09
  • $\begingroup$ If I understand this correctly, the fact that it is a Hausdorff space here is superfluous. $\endgroup$
    – Dusan
    Aug 14, 2011 at 11:05
  • 6
    $\begingroup$ @Dusan: Requiring $X$ to be Hausdorff avoids having to worry about which definition of local compactness is being used; this Wikipedia article has some details. $\endgroup$ Aug 15, 2011 at 16:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .