I'm trying to solve an exercise that says
Show that a locally compact space is $\sigma$-compact if and only if is separable.
Here locally compact means that also is Hausdorff. I had shown that separability imply $\sigma$-compactness but I'm stuck in the other direction.
Assuming that $X$ is $\sigma$-compact it seems enough to show that a compact Hausdorff space is separable. However I don't have a clue about how to do it.
My first thought was try to show that a compact Hausdorff space is first countable, what would imply that it is second countable and from here the proof is almost done. However it seems that my assumption is not true, so I'm again in the starting point.
Some hint will be appreciated, thank you.
EDIT: it seems that the exercise is wrong. Searching in the web I found a "sketch" for a proof that a compact Hausdorff space is not separable:
Another natural example: take more than |R| copies of the unit interval and take their product. This is compact Hausdorff (Tychonov theorem) but not separable (proof not too hard, but omitted).
Hope this helped,
My knowledge of topology is little and the exercise appear in a book of analysis (this is a part of the exercise 18 on page 57 of Analysis III of Amann and Escher.)
My hope is that @HennoBrandsma (an user of this web) appear and clarify the question :)