Prove that if a metric space is sequentially compact, it is also compact.

Question

anonymous · Answer

Definitions:
A metric space is sequentially compact iff every sequence has a convergent subsequence.
A metric space is compact iff every open cover has a finite subcover.

anonymous · Answer

The other direction (if a metric space is compact then it is sequentially compact) was proved by using the theorem that every infinite subset of a compact set has a limit point.

anonymous · Answer

Might as well make a note that I am looking for a nudge in the right direction moreso than a complete answer.

anonymous · Answer

See http://www.econ.brown.edu/fac/Mark_Dean/Maths_RA5_10.pdf

anonymous · Answer

Wasn't really looking for a full proof of it, I can always find a full proof on ProofWiki. Trying to prove this mostly by myself, if possible.

anonymous · Answer

Thanks, though.

anonymous · Answer

The proof is not a one line proof anyway.

anonymous · Answer

Can you please help me in a bit @eliassaab and @nbouscal???? http://openstudy.com/study#/updates/4fc5ef77e4b022f1e12d65a1 thank you!!

anonymous · Answer

The nice proof is the one using the Lebesgue covering Lemma can be found on http://www.math.toronto.edu/~mat1300/oldnotes/metricspaces.pdf