Does a second-countable, locally compact, Hausdorff space admit a countable basis of pre-compact open sets? Let $X$ be a such space. I know that for a Hausdorff space, being locally compact is equivalent to having a basis of pre-compact open sets. But how do I prove that $X$ can be covered by countably many such sets, by using second-countability of X? Thank you.
 A: If we have a countable base $\mathcal{B}$ for $X$ then any other base $\mathcal{C}$ of $X$ has a countable subfamily that is also a base. See here e.g.
We can apply this to the base of pre-compact sets that exists in a locally compact Hausdorff space.
Finally note that a base is in particular a cover of $X$ (for every $x$ there is at least one base element containing it, as $X$ itself is open and so a union of base sets).
Final remark: the closures of the cover elements of course still cover $X$ and show that $X$ is then $\sigma$-compact.
A: Every locally compact Hausdorff space is (completely) regular. Every second-countable regular Hausdorff space is metrisable, and hence separable. Any locally compact metrisable space admits a compatible metric, each of whose (proper) open balls is precompact.
Thus if $X$ is locally compact, Hausdorff and second-countable, then it is Polish. It admits a compatible, but possibly noncomplete, metric $d$ for which there is a countable subset $\{x_n\}_{n\in\mathbb{N}}$ such that $\{B_{\frac{1}{m}}(x_n)\mid m,n\in\mathbb{N}\}$ is a countable base of precompact sets for $X$.
