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:
$X= \bigcup_{n \in \mathbb{N}} \{\overline{B_{n}}: B_{n} \in \mathcal{B}, \overline{B_{n}} \ \textrm{is compact}\}$
Best Answer
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:
Or you could replace that last sentence with something like this:
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.