# A $\sigma$-compact but not hemicompact space?

All spaces are at least Hausdorff. A topological space $$X$$ is called

• $$\sigma$$-compact if there is a countable sequence $$(K_n)_{n<\omega}$$ of compact subsets of $$X$$ such that $$X=\bigcup_n K_n$$.
• hemicompact if there is a countable sequence $$(K_n)_{n<\omega}$$ of compact subsets of $$X$$ such that for every $$K\subseteq X$$ compact there is $$n\in\omega$$ with $$K\subseteq K_n$$.

In particular a hemicompact space $$X$$ is $$\sigma$$-compact since for every $$x\in X$$ there is $$n$$ with $$\{x\}\subseteq K_n$$, hence $$X=\bigcup_n K_n$$. I'm interested in conditions on $$X$$ that are sufficient to reverse this implication, but I am more interested in an example of a space $$X$$ (with $$X$$ at least Hausdorff, better if completely regular) which is $$\sigma$$-compact but not hemicompact. I have checked the standard sources (Counterexamples in Topology and the pi-base website) but there are no examples of such spaces there, hence my question:

What is an example of an Hausdorff space $$X$$ which is $$\sigma$$-compact but not hemicompact?

A very simple example is the rational numbers. Obviously $$\mathbb{Q}$$ is $$\sigma$$-compact by covering it with singletons. However, I claim it is not hemicompact. Indeed, suppose $$(K_n)$$ is a sequence of compact subsets of $$\mathbb{Q}$$. Each $$K_n$$ does not contain any neighborhood of $$0$$, so we can pick a sequence $$(x_n)$$ where each $$x_n\not\in K_n$$ and $$x_n\to 0$$. Then the set $$\{x_n:n\in\mathbb{N}\}\cup\{0\}$$ is compact but not contained in any $$K_n$$.
More generally, this argument shows that a hemicompact first-countable space must be locally compact (in the weak sense of having a compact neighborhood of every point). So any $$\sigma$$-compact first-countable space that is not locally compact is a counterexample.