5
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ "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
    Jun 27 '13 at 21:34
7
$\begingroup$

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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.