I am looking for a direct proof (not by contradiction) that a compact metric space is sequentially compact, ie constructing a converging subsequence from any sequence.
Thanks
2026-04-03 07:34:01.1775201641
A direct proof that a compact metric space is sequentially compact
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For each $n$ cover the space $X$ by balls of radius $1/n$. Choose finite subcovers for each $n$. Then there is at least one set in each cover with infinitely many sequence elements. Use this to create a subsequence which is Cauchy.