# Prove that a metric space is countably compact if and only if every infinite sequence in $X$ has a convergent subsequence.

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.

• "every infinite sequence in X has a convergent subsequence." If some space have this property we say that space is sequentially compact space (for information only) Jun 27 '13 at 21:34

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 $n\in\Bbb N$.
• Show that $\{V_n:n\in\Bbb N\}$ is a countable open cover of $X$ with no finite subcover.
• Let $M=\{n\in\Bbb N:V_n\subsetneqq V_{n+1}\}$. Show that $M$ is infinite.
• For each $n\in M$ let $x_n\in V_{n+1}\setminus V_n$. Show that $\langle x_n:n\in M\rangle$ has no convergent subsequence.
Now suppose that $\langle x_n:n\in\Bbb N\rangle$ is a sequence in $X$ with no convergent subsequence, and let $D=\{x_n:n\in\Bbb N\}$.
• Show that each $x\in X$ has an open neighborhood whose intersection with $D$ is finite. Conclude that $D$ is a closed subset of $X$.
• Show that for each $x\in D$ there is an $\epsilon_x>0$ such that $B(x,\epsilon_x)\cap D=\{x\}$, i.e., that $D$ is a discrete set.
• Show that $\{X\setminus D\}\cup\{B(x,\epsilon_x):x\in D\}$ is a countable open cover of $X$ with no finite subcover.