FIP nonempty intersection implies sequentially compact

Asked Aug 05 '16 at 23:38

Active Sep 25 '18 at 10:17

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How do you prove that if closed sets have the FIP in a metric space $K$ then their intersection is nonempty, then the set is also sequentially compact?

I've seen proofs that FIP nonempty intersection is equivalent to compactness and proofs that compactness is equivalent to sequential compactness, but how do you prove it directly?

The converse is easy, but I don't see how to prove this direction..

edited Aug 06 '16 at 00:56

asked Aug 05 '16 at 23:38

Valerie

"FIP nonempty intersection" is indeed equivalent to compactness, but compactness neither implies nor is implied by sequential compactness. You could maybe find counterexamples to both implications here on MSE. – Stefan Hamcke Aug 06 '16 at 00:16
See here for counterexamples. – Stefan Hamcke Aug 06 '16 at 00:19
@StefanHamcke Supposed to be a metric space – Valerie Aug 06 '16 at 00:48
@StefanHamcke is the answer below correct? it doesn't seem to be – Valerie Aug 06 '16 at 00:53
Okay. Then it is better to mention in your question that you are dealing with metric spaces, so people don't get confused. – Stefan Hamcke Aug 06 '16 at 00:55

FIP nonempty intersection implies sequentially compact

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