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Let X be a metric space. Show that a subset E of X is compact if and only if every infinite subset of E has a limit point in E. How to prove the converse statement ? i.e. if every infinite subset of E has a limit point in E then E is compact in X. I want to prove from the definition of compact sets.

infintedimensional
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  • What is your definition of compact sets? – uniquesolution Aug 11 '18 at 11:00
  • If every open cover has a finite subcover. – infintedimensional Aug 11 '18 at 11:04
  • See here –  Aug 11 '18 at 11:17
  • Take a sequence in your space. If its range is finite, it has a convergent subsequence. If not, the range is infinite and has a limit point. You can find a subsequence converging to the limit point. Hence, every sequence has a convergent subsequence and the space is compact. –  Aug 11 '18 at 11:40
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    @Math_QED A space can be sequentially compact and not compact. At best, what you have done is to prove that limit point compact implies sequentially compact in a metric space. But even for that you only said "You can find a subsequence ..." without really saying why. –  Aug 11 '18 at 11:48
  • The OP said that $X$ is a metric space (and the question is tagged accordingly), in which the two compactness notions are equivalent. Secondly, I just gave a hint. The details are left as an exercice for the OP. –  Aug 11 '18 at 12:14
  • @Math_QED The question is showing that equivalence, for which there is no hint in your comment. –  Aug 11 '18 at 12:23
  • This question is a duplicate of https://math.stackexchange.com/questions/773484/let-x-be-a-metric-space-in-which-every-infinite-subset-has-a-limit-point-prove?rq=1 –  Aug 11 '18 at 12:36

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First note that by this answer we know that if $X$ limit point compact (i.e. every infinite set has a limit point), then $X$ is Lindelöf (i.e. every open cover has a countable subcover), otherwise $X$ would have a closed and discrete uncountable subspace, contradicting the limit point compactness.

Then as metric spaces are $T_1$, this answer shows that a limit point compact metric $X$ is also countably compact (every countable cover has a finite subcover).

And clearly then $X$ is compact (we reduce an open cover to a countable one via Lindelöfness, and then to a finite one using countable compactness). Also see this answer for an overview of similar ideas.

Henno Brandsma
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