Why does "Totally bounded" need "Complete" in order to imply "Compact"?
Shouldn't the definition of totally bounded imply the existence of a convergent subsequence of every sequence?
[Math] Why does Totally bounded need Complete in order to imply Compact
general-topologymetric-spaces
Best Answer
No, total boundedness of $\langle X,d\rangle$ implies that every sequence in $X$ has a Cauchy subsequence. Completeness of $\langle X,d\rangle$ implies that every Cauchy sequence in $X$ actually has a limit point in $X$ and therefore converges. The two together therefore imply that every sequence in $X$ has a convergent subsequence, i.e., that $X$ is sequentially compact. Finally, there is a theorem that a metric space is sequentially compact if and only if it’s compact, so total boundedness plus completeness imply compactness.