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?


1 Answer 1


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.


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