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Suppose $X$ is locally compact, Hausdorff, and has a countable family of compact subsets whose union is all of $X$. I want to show that $X$ is second-countable. Suppose the family of countable compact subsets are labelled $C_i$, $i \in \Bbb N$, Then there are a couple of ideas that come to mind but I am having trouble putting it all together.

1) Taking the complements of the $C_i$ (problem: what if an element is in all of them, also the complements might not satisfy the requirements of a basis)

2) Using local compactness to say that every point has a compact set containing it with an open neighborhood (also containing it), but how will this relate to the $C_i$ ?

Any hints appreciated.

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It is not even true that compact Hausdorff spaces are second-countable. (Of course, a compact Hausdorff space is locally compact and can be covered by a countable family of compact subsets.)

An example is the ordinal space $[ 0 , \omega_1 ]$ (where $\omega_1$ denotes the least uncountable ordinal).

  • As a linearly-ordered space, $[0 , \omega_1 ]$ is Hausdorff.
  • Since $[0 , \omega_1]$ has a greatest element, it is compact (this is essentially because there are no infinite strictly decreasing sequences of ordinals).
  • It is not second-countable because any base must include the singletons $\{ \alpha \}$ where $\alpha < \omega_1$ is a successor ordinal, and there are uncountably many of these.

More information about this space can be found on the following post on Dan Ma's Topology Blog:

(If you want a non-compact example of a locally compact σ-compact Hausdorff space that is not second-countable you can just take a disjoint union of countably-infinite copies of $[0,\omega_1]$.)

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  • $\begingroup$ So my conclusion is false? $\endgroup$ – IntegrateThis Nov 15 '18 at 5:32
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    $\begingroup$ @IntegrateThis Correct: it is not true that every locally compact, σ-compact Hausdorff space is second-countable. $\endgroup$ – stochastic randomness Nov 15 '18 at 5:55
  • $\begingroup$ +1............ $[0,\omega_1]$ is not even first-countable. $\endgroup$ – DanielWainfleet Nov 15 '18 at 7:52
  • $\begingroup$ Why dooes the base have to include singletons and not larger open sets that could encompass them $\endgroup$ – IntegrateThis Nov 18 '18 at 3:51
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    $\begingroup$ @IntegrateThis Note that given a $\alpha < \omega_1$ the open interval $( \alpha , \alpha+2)$ is the singleton $\{ \alpha+1 \}$, so these singletons are open. If $\mathcal{B}$ is a base, for every such $\alpha+1$ there must be some $U \in \mathcal{B}$ with $\alpha+1 \in U \subseteq \{ \alpha+1 \}$, meaning $U = \{ \alpha + 1 \}$. (More generally, call an open subset $U$ of a topological space $X$ minimally open if there is no open set $V$ satisfying $\emptyset \subsetneq V \subsetneq U$. Then a base for a topological space must include all of the minimally open subsets.) $\endgroup$ – stochastic randomness Nov 18 '18 at 6:08
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I think the following qualifies as a counterexample . . .

Let $k$ be a cardinal number with $k > c$, where $c=|\mathbb{R}|$.

Let $I=[0,1]$ with the usual topology, and let $X$ be the product of $k$ copies of $I$.

Then

  • $X$ is compact and Hausdorff.$\\[4pt]$
  • $X$ is locally compact.$\\[4pt]$
  • $X$ is covered by itself, so $X$ qualifies as being covered by a countable union of compact sets.$\\[4pt]$
  • The cardinality of the topology on $X$ is greater than $c$ (since $k > c$).$\\[4pt]$

But $X$ is not second countable.

Suppose instead that $X$ had a countable base, $B$ say.

Then since every open subset of $X$ is the union of some subset of $B$, it would follow that the cardinality of the topology on $X$ is at most the cardinality of the power set of $B$, hence at most $c$, contradiction.

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    $\begingroup$ Right. And $k\gt\mathfrak c$ is more than you need: if $k$ is an uncountable cardinal, then $I^k$ is not even first countable. If $k\gt\mathfrak c$ then $I^k$ is not separable. $\endgroup$ – bof Nov 15 '18 at 6:04

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