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Prove that a metrizable space is countably compact iff it is compact.

($\Rightarrow$)

I let $\{O_i\}$ be a countable open cover for $(X,T)$ with a finite subcover. Let $\{U_i\}$ be an uncountable open cover (because otherwise it's countable and then trivial). From here I'm stuck. I figure that I can add an uncountable number of additional open balls to the countable set to make in an uncountable set or select a countable subset of the uncountable set, but I'm not sure how I'd do this. To prove this without sequential compactness is what I'm looking for.

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  • $\begingroup$ Possible duplicate: math.stackexchange.com/questions/431146/… $\endgroup$ – Maik Pickl Jun 6 '16 at 12:57
  • $\begingroup$ I'm trying to prove this without knowing anything about sequential compactness. $\endgroup$ – Oliver G Jun 6 '16 at 13:21
  • $\begingroup$ But in metric space sequential compactness and compactness are equivalent. Or do you mean that you have not learned about sequential compactness yet? $\endgroup$ – Maik Pickl Jun 6 '16 at 13:38
  • $\begingroup$ Yes, I want to try to prove this given that I have not learned about sequential compactness yet. Sequential compactness is the next type of compactness I'll be learning about in the book I'm reading, but I want to prove this without skipping ahead. $\endgroup$ – Oliver G Jun 6 '16 at 13:44
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I don’t know a direct argument along the lines that you’ve tried to start. I would prove it in steps, as follows.

  • Show that if $X$ is countably compact, then $X$ does not have an infinite, closed, discrete subset.
  • Show that for each $n\in\Bbb Z^+$ there is a finite $F_n\subseteq X$ such that $\left\{B\left(x,\frac1n\right):x\in F_n\right\}$ covers $X$.
  • Show that $\left\{B\left(x,\frac1n\right):n\in\Bbb Z^+\text{ and }x\in F_n\right\}$ is a countable open base for $X$.
  • Prove that in fact every open cover of $X$ has a countable subcover (i.e., $X$ is Lindelöf).
  • Conclude that $X$ is compact.

I’ll leave it at that for now in case you’d like to try to fill in the details.

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    $\begingroup$ I don’t suppose that the downvoter would care actually to be helpful and explain what problem he or she perceives here? $\endgroup$ – Brian M. Scott Jun 6 '16 at 17:25

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