# Does locally compact separable Hausdorff imply $\sigma$-compact?

Is it true that every locally compact separable Hausdorff space is $\sigma$-compact?

-
If I remember correctly this is true for locally compact groups - I believe it is in Serge Lang "Real analysis" in the sections on the Haar measure, or in Loomis book "Abstract Harmonic Analysis". –  AD. Jan 4 '12 at 16:21

The following is counterexample 65 in Counterexamples in Topology, the rational sequence topology. Let $X=\mathbb{R}$ and for each $r\in\mathbb{R}$, pick a sequence $(q_n^r)$ of rationals converging to $r$. Let the basis of the topology be all sets of the form $\{q_n^r,q_{n+1}^r,\ldots\}\cup\{r\}$ for some $r$ and $n$ and all sets of the form $\{q\}$ with $q$ rational. The space is separable, since every basic open set contains a rational number. The space is Hausdorff, since sequences converging to different points are eventually in disjoint neighborhoods. The space is locally compact, since $\{q_1^r,q_2^r,\ldots\}\cup\{r\}$ is a compact neighborhood of $r$. But a compact set can contain only finitely many irrational numbers, and since the set of irrational numbers is uncountable, the space is not $\sigma$-compact.
This is of course a Mrówka $\Psi$-space: the sets $$\Big\{\{q_n^r:r\in\mathbb{Z}^+\}:r\in\mathbb{R}\setminus\mathbb{Q}\Big\}$$ are an almost disjoint family in $\mathbb{Q}$. (It’s one of several slick ways of constructing such a family of cardinality $2^\omega$.) –  Brian M. Scott Jan 5 '12 at 0:53
Note that Jones' lemma implies that this space is not normal. @BrianM.Scott for a proper $\Psi$, shouldn't we have m.a.d. family; I'm not sure the rational sequence topology is homeomorphic to a Mrówka space based on a m.a.d. family (the maximality gives pseudocompactness). Do you know? –  Henno Brandsma Jan 5 '12 at 18:53
@Henno: I’ve seen the term used both ways; I tend to give it the more general sense. Here $\mathbb{Z}$ is a closed discrete set of isolated points, which you won’t have in a $\Psi$ in the narrow sense. –  Brian M. Scott Jan 6 '12 at 0:36
It’s not true in general. Let $\mathscr{A}\,$ be an uncountable family of pairwise almost disjoint infinite subsets of $\omega$ (i.e., distinct members of $\mathscr{A}\,$ have finite intersections). Let $X=\omega\cup\mathscr{A}\,$. Points of $\omega$ are isolated, and for each $A\in\mathscr{A}\,$, $$\Big\{\{A\}\cup (A\setminus F):F\subseteq \omega\text{ is finite}\Big\}$$ is a local base at $A$. ($X$ is an example of a Mrówka $\Psi$-space.) It’s easy to see that $X$ is Tikhonov, zero-dimensional, separable, and locally compact, but $$\Big\{\{A\}\cup\omega:A\in\mathscr{A}\,\Big\}$$ is an open cover of $X$ with no countable subcover, so $X$ is not Lindelöf. However, a locally compact space is Lindelöf iff it’s $\sigma$-compact, so $X$ cannot be $\sigma$-compact. (Of course it’s easy enough to see directly that $X$ is not $\sigma$-compact: if $X=\bigcup_n Y_n$, some $Y_n\cap\mathscr{A}\,$ must be infinite, and then $Y_n$ cannot be compact.)