A proof can be found here, but it seems that it uses AC. I would like to know if there is a proof for this fact without AC.

I came up with this question after seeing a proof of sequential compactness $\implies$ Compactness for metrizable topological spaces without using generalized Lebesgue's Number Lemma for sequentially compact metric spaces. Both proofs can be found here.


1 Answer 1


The result cannot be proved without the axiom of choice. For instance, let $X$ be an infinite set which contains no countably infinite subset (the existence of such a set is consistent with ZF), and give $X$ the discrete metric. Then $X$ is sequentially compact since there are no sequences in $X$ which take infinitely many different values so every sequence has a constant subsequence. But $X$ is not Lindelöf, since the open cover of $X$ by singletons has no countable subcover.

  • $\begingroup$ May I know how do we construct such X? Thanks. $\endgroup$ Mar 28, 2019 at 4:57
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    $\begingroup$ That's a very long story if you're not familiar with forcing. See math.stackexchange.com/a/1396679/86856 for a sketch of the construction; you can find the details, for instance, in Example 15.52 of Jech's Set Theory (3rd edition). $\endgroup$ Mar 28, 2019 at 5:18
  • $\begingroup$ That is indeed a good answer, but just curious, does the statement "sequential compactness implies compactness for metric spaces" rely on the axiom of countable choice? Is there a proof without it? $\endgroup$ Mar 28, 2019 at 5:27
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    $\begingroup$ @WilliamSun: Since compact is stronger than Lindelöf, if you need choice to prove sequentially compact metric spaces are Lindelöf you also need it to prove they are compact. $\endgroup$ Mar 28, 2019 at 5:29

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