A topological space is countably compact if every countable open cover has a finite sub cover. Prove that a metric space is countably compact if and only if every infinite sequence in $X$ has a convergent subsequence.
[Math] Prove that a metric space is countably compact if and only if every infinite sequence in $X$ has a convergent subsequence.
compactnessgeneral-topologymetric-spaces
Best Answer
HINT: Suppose first that $X$ is not countably compact. Then there is a countable open cover $\mathscr{U}=\{U_n:n\in\Bbb N\}$ of $X$ that has no finite subcover. For $n\in\Bbb N$ let $V_n=\bigcup_{k\le n}U_k$. Note that $V_n\subseteq V_{n+1}$ for each $n\in\Bbb N$.
Now suppose that $\langle x_n:n\in\Bbb N\rangle$ is a sequence in $X$ with no convergent subsequence, and let $D=\{x_n:n\in\Bbb N\}$.