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Let $X$ be a second countable topological space. Then $A \subset X$ is a compact subset of $X$ if and only if every sequence $\{x_{n}\}_{n \in \mathbb{N}} \subset A$ has a convergent sub-sequence with a limit point $x \in A$.

I can prove it when $X$ is a metrizable space, but I'm having trouble when $X$ is only second countable.

I'll appreciate any help. Thanks!

Brian Mac Guire
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    If you try to modify the proof for a metrisable space to work under the assumption of second countability, at which point do you have a problem? – Daniel Fischer Jan 21 '20 at 15:24
  • Hi Daniel. I can prove that if A is compact then every sequence in A has a convergent sub-sequence, but I can't say that it converges to a point in A. A metrizable space is Hausdorff, and so A is also closed, then the limit point of the sub-sequence is in A. Can you help me? Thanks! @DanielFischer – Brian Mac Guire Jan 21 '20 at 15:43
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    I see. So the only problem is that we don't have the assumption that $X$ is Hausdorff? That would guarantee that every limit point of a sequence in $A$ lies in $A$. But that's more than we need, we just need that there's (at least) one in $A$. Trick: You can assume that $A = X$. You can do that because second countability is hereditary, every subspace of a second countable space is also second countable. – Daniel Fischer Jan 21 '20 at 15:50
  • And that's it! yes, that's correct. Thanks so much @DanielFischer – Brian Mac Guire Jan 21 '20 at 16:57

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Suppose $A$ is compact. Suppose $(x_n)$ is a sequence in $A$. We can assume WLOG that $S=\{x_n \mid n \in \Bbb N\}$ is infinite, or we'd have a constant subsequence etc. As $A$ is compact, the infinite set $S$ has a condensation point $p \in A$, i.e. a point such that every neighbourhood of $p$ contains infinitely many points of $S$.

(The proof is simple: if no such point exists we get a quick contradiction with compactness: every point $a \in A$ would then have a neighbourhood $O(a)$ with $O(a) \cap S$ finite and a finite subcover gives a contradiction..)

As this $p$ has a decreasing neighbourhood base (as $X$ is in particular first countable), this allows us to define a subsequence of $x_n$ that converges to $p$ by a straightforward recursion. (Note that we only used first countability here.)

For the reverse: if $A$ obeys the sequential compactness condition, it is easily seen to be countably compact in the sense that every countable cover has a finite subcover (because the covering version is equivalent to the fact that every countably infinite set has a condensation point (or "strong limit point"), see e.g. part of my answer here ) and we can arrange an infinite subset in a bijective sequence so that a limit of a subsequence will be a condensation point of that set. As $X$ is second countable, all subsets of it, so $A$ too, is second countable and hence Lindelöf. And a Lindelöf countably compact subset is compact. QED.

Note that all of this is separation axiom free, BTW.

Henno Brandsma
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