Locally compact Polish space is $\sigma$-compact

compactnessgeneral-topologymetric-spacespolish-spaces

I have recently encountered this result.

  • Let $X$ be $\sigma$-compact, locally compact Hausdorff space and $\mu$ is a Radon measure on $X$. Then the space of continuous functions with compact support is dense in that of $\mu$-integrable functions w.r.t. $\|\cdot\|_{L_1}$. ref

I got that

  • Polish space is not necessarily $\sigma$-compact. ref
  • $\sigma$-compact Polish space is not necessarily locally compact. ref

I would like to ask if the following result holds, i.e.,

Locally compact Polish space is $\sigma$-compact.

Update:

  • As proven in an below answer: locally-compact second-countable space is $\sigma$-compact.

  • From this, every locally-compact second-countable Hausdorff space is a Polish space.

Best Answer

Yes, a locally compact Polish space is $\sigma$-compact.

Since it is Polish, such a space admits a countable base $\mathcal{U} = \{B_i\}_{i=1}^n$ of open sets for the entire space; by local compactness, each element in $\mathcal{U}$ can be taken so that $\overline{B_i}$ is compact. Since $X = \bigcup_{n=1}^\infty \overline{B_i}$, we have exhibited the $\sigma$-compactness of $X$.

Related Question