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Let $X$ be a metric space. If every infinite subset of $X$ has at least one accumulation point, then $X$ is compact.

Does anyone know how to prove that?

Nate Eldredge
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Darkmaster
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  • If every infinite subset has an accumulation point, then every sequence has a converging subsequence and thus the space is sequentially compact. You can then use Lebesgue lemma or something like that. If you google 'sequentially compact implies compact', you can find detailed proofs. – n7kvz Jun 08 '19 at 21:26
  • @skullph Is it possible to demostrate that proposition using open covers? – Darkmaster Jun 08 '19 at 21:33
  • You can first show that a sequentially compact metric space is totally bounded, i.e. for r>0 given, you can cover your space by a finite number of balls with radius r. Now you can use this fact to construct your finite subcover of a given open cover. https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Compact has multiple proofs – n7kvz Jun 08 '19 at 21:47

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$X$ is totally bounded. That follows from the argument in this question and also, $X$ is complete, because a Cauchy sequence (its set of values, rather) will have an accumulation point, hence a convergent subsequence and so it is itself convergent to that point. And a totally bounded complete metric space is compact.

Henno Brandsma
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Take a sequence $(a_n)_{n\in\Bbb N}$ in $X$. Then $A:=\{a_n~\vert~n\in\Bbb N\}$ is an infinite subspace of $X$ so it must have an accumulation point $x\in X$. For $n\geq 1$, $A$ must intersect $B(x,1/n)-\{x\}$. Take $\varphi(n)\in\Bbb N$ such that $$a_{\varphi(n)}\in B(x,1/n)-\{x\}$$ and do this by induction to make sure that $\varphi$ is strictly increasing. Then $(a_{\varphi(n)})_{n\in\Bbb N}$ is a subsequence of $(a_n)_{n\in\Bbb N}$ converging to some limit $x$, so $X$ is compact.

Adam Chalumeau
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  • Is it possible to demostrate that proposition using open covers? – Darkmaster Jun 08 '19 at 21:29
  • @Darkmaster probably but you have to use that $X$ is a metric space somewhere, you can't only use abstract covers . Sequential characterization is better suited in this case. – Adam Chalumeau Jun 08 '19 at 21:36