Prove that the union of sequentially compact spaces is sequentially compact

Asked Oct 03 '22 at 14:09

Active Oct 03 '22 at 14:09

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I'm teaching myself metric spaces, and was wondering how to do this.

I have tried starting with two sequentially compact spaces (A and B) and then proving the general finite case using induction. Since we've defined sequentially compact as a metric space where any sequence has a convergent subsequence, I've tried to split into cases and consider when happens at the tail of the sequence, but am not having much success.

asked Oct 03 '22 at 14:09

yw_2003

1

First, look for a subsequence than lies entirely in one of the two spaces. – David Mitra Oct 03 '22 at 14:13
And remember that a subsequence of a subsequence is also a subsequence of the original sequence. – Oct 03 '22 at 14:15
Are you referring to a finite union? – רוי רז Oct 03 '22 at 14:15
@רוירז Yes, a finite union – yw_2003 Oct 03 '22 at 14:31
@DavidMitra Thanks! What would my cases be then? – yw_2003 Oct 03 '22 at 14:34

Prove that the union of sequentially compact spaces is sequentially compact

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