Csdrhrt

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.

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

• The First Uncountable Ordinal

(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 $[0,omega_1]$.)

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.