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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.

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    $\begingroup$ "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) $\endgroup$
    – Cortizol
    Commented Jun 27, 2013 at 21:34

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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.
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