I am looking for an example of a separable, locally-compact metric space which is not $\sigma$-compact.

At first I thought I could show that if a metric space is separable and locally-compact, then it must be $\sigma$-compact. But I could not show it and I haven't found any theorem that implies that. So I figured there must be an example of a space which is metrizable, separable, locally-compact but not $\sigma$-compact. Clearly such a space cannot be compact, so I am looking for a locally-compact non-compact metric space.

Can someone give me such an example?


Such a space does not exist.

Note that every second-countable locally compact Hausdorff space $X$ is σ-compact.

  • Proof. The family $\mathcal B$ of all open $U \subseteq X$ with compact closure forms a base for $X$ by local compactness. By second-countability some countable subfamily $\mathcal B_0$ of $\mathcal B$ is itself a base for $X$. But then $X = \bigcup \{ \overline U : U \in \mathcal B_0 \}$, a countable union of compact sets, so $X$ is σ-compact.

Now recall that separability is equivalent to second-countability in metric spaces.

  • $\begingroup$ Wait I do have a problem. After you constructed the base from the dense countable subset with compact closure neighborhoods, how do you know it is a base for all the space? The neighborhoods for each point from the dense subset can be arbitrarily small, so how do you even know that the union is all the space? $\endgroup$ – User666x Aug 23 '15 at 9:36
  • $\begingroup$ @AvivEshed I am not using the countable dense set to construct the base, but the knowledge that having a countable dense set means that the metric space is second-countable. It is a fact (though unfortunately not commonly taught) that given any base $\mathcal B$ for a second-countable topological space you can find a countable subfamily $\mathcal B_0$ of $\mathcal B$ which is also a base for $X$. The basic idea for this can be seen in this answer of mine. $\endgroup$ – Meta-мета-μετα-meta-мета-μετα Aug 23 '15 at 11:33
  • $\begingroup$ So for the metric space I will still start with the base consisting of all open sets with compact closure, and then take a countable subfamily of this which is also a base. $\endgroup$ – Meta-мета-μετα-meta-мета-μετα Aug 23 '15 at 11:33
  • $\begingroup$ Awesome! This is mind blowing :) $\endgroup$ – User666x Aug 23 '15 at 12:15

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