How to show that complete and pseudocompact implies sequentially compact (in a metric space)?

Asked Oct 12 '23 at 03:09

Active Oct 12 '23 at 03:09

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I have proved that pseudocompact implies completeness in a metric space, which is, as I understand it, a step to proving pseudocompactness implies compactness. How do I show complete and pseudocompact implies (sequentially) compact?

asked Oct 12 '23 at 03:09

jzintegral97

1

Prove pseudocompact implies total boundedness. See also A metric space is compact iff it is pseudocompact. – user10354138 Oct 12 '23 at 07:30

How to show that complete and pseudocompact implies sequentially compact (in a metric space)?

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