4
$\begingroup$

I was curious about the relations between second-countable, separable, Lindelöf and $\sigma$-compact topologies in the context of metric spaces.

I am aware of the following implications in general topological spaces:

  • second-countable $\Rightarrow$ separable $\not \Rightarrow$ Lindelöf, $\;$ [thanks bof]
  • $\sigma$-compact $\Rightarrow$ Lindelöf
  • second-countable + locally compact $\Rightarrow$ $\sigma$-compact

as well as the reversed implications in the case of metric spaces:

  • Lindelöf $\Leftrightarrow$ separable $\Leftrightarrow$ second-countable

Since all the proofs I've seen so far require the LC condition I assume it is not true in general that second-countable topological spaces are $\sigma$-compact (although seeing an actual counterexample would be nice).
So what about metrizable topological spaces?


Ideas so far:
If we can proof that every subset of a $\sigma$-compact space is again $\sigma$-compact, then this would follow from the fact, that every separable metric space is homeomorphic to a subset of the Hilbert cube (which is compact). $\;$[debunked by bof]

$\endgroup$
  • 2
    $\begingroup$ The set of irrational numbers is a separable metric space which is not $\sigma$-compact. (Every $\sigma$-compact subset of the irrational numbers is meager.) Hilbert space is another example. $\endgroup$ – bof Nov 12 '18 at 11:10
  • 2
    $\begingroup$ By the way, separable does not imply Lindelöf in general topological spaces. For example, the Sorgenfrey plane is separable but not Lindelöf. $\endgroup$ – bof Nov 12 '18 at 11:12
  • $\begingroup$ Every closed subset of a $\sigma$-compact space is again $\sigma$-compact. Not all subspace, as witnessed by the rational and the irrationals assubspaces of the reals. $\endgroup$ – Henno Brandsma Nov 12 '18 at 11:55
  • $\begingroup$ @bof I also wanted to disprove local compactness. $\endgroup$ – Henno Brandsma Nov 12 '18 at 12:25
4
$\begingroup$

A counterexample is the "Baire space" $\mathcal{N} = \mathbb{N}^{\mathbb{N}}$. This is one of the main examples of a Polish space: a separable, completely metrizable space.

One fact about this space is that all compact subsets have empty interior, that is, all compact subsets are nowhere dense. By the Baire Category Theorem, $\mathcal{N}$ is not the countable union of nowhere dense subsets, and so together with the above fact it cannot be $\sigma$-compact.

See also this question and its answer for more details:

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.