I'd like hints only please. I have an analysis book and I could look up the proof myself but I'm trying to prove it myself as an exercise; so giving the full proof would be redundant as well as counterproductive to my own learning.(adsbygoogle = window.adsbygoogle || []).push({});

X is a metric space.

In this other book, K is compact iff every sequence in K has a convergent subsequence. I know that topologically, this is not the standard definiton of compact; some authors would call this sequentially compact.

K is bicompact if every open cover of K admits a finite refinement that also covers K. This is how compactness is usually defined as far as I knew.

Hints on how compact implies bicompact and bicompact implies compact would be very much appreciated.

Right now, I'm working on compact implies bicompact so that is the current priority. I don't need hints on the other direction at this time since I haven't tried it.

A general hint like "prove the contrapositive" or "use contradiction" might be helpful though I've tried the first of the two already. I suppose as I await a reply, I'll try contradiction...

Thanks!

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# Hints only please: compact iff bicompact

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