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]

  • 3
    $\begingroup$ 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, 2018 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, 2018 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. $\endgroup$ Nov 12, 2018 at 11:55
  • $\begingroup$ @bof I also wanted to disprove local compactness. $\endgroup$ Nov 12, 2018 at 12:25
  • $\begingroup$ Any infinite-dimensional separable Banach space is a counterexample. This follows from the fact that the unit ball of a Banach space $E$ is compact iff $E$ is finite dimensional, so a compact subset of an infinite dimensional Banach space has empty interior. Then we just apply the Baire category theorem. $\endgroup$ Nov 11, 2019 at 11:02

1 Answer 1


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:


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