# $X$ is locally compact, Hausdorff, and has a countable family of compact subsets covering $X$, then $X$ is second-countable

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.

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.

(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 $$[0,\omega_1]$$.)

• So my conclusion is false? – IntegrateThis Nov 15 '18 at 5:32
• @IntegrateThis Correct: it is not true that every locally compact, σ-compact Hausdorff space is second-countable. – stochastic randomness Nov 15 '18 at 5:55
• +1............ $[0,\omega_1]$ is not even first-countable. – DanielWainfleet Nov 15 '18 at 7:52
• Why dooes the base have to include singletons and not larger open sets that could encompass them – IntegrateThis Nov 18 '18 at 3:51
• @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.) – stochastic randomness Nov 18 '18 at 6:08

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.

• 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. – bof Nov 15 '18 at 6:04