By "$E$ is countably compact", I mean that every countable open cover of $E$ has a finite subcover. By "$E$ is compact", I mean that every open cover of $E$ has a finite subcover. Let $M$ be a metric space. Prove that if a subset $E$ of $M$ is countably compact, then it is compact.

I am looking for a proof that just uses the machinery of metric space topology, and doesn't appeal to general topological spaces, if this is possible.


This is tied in to a standard set of compactness results for metric spaces. Here the usual sequence of equivalent properties is as follows:

  • Countable compactness
  • Limit point compactness (every infinite set has a limit point)
  • Sequential compactness (every sequence has a convergent subsequence)
  • Compactness

These are all equivalent for metric spaces. I'm afraid I don't know off the top of my head a more direct proof than proving the chain of implications one at a time. I checked in Munkres, which proves that the latter three are equivalent (which should be in any standard text), and the only thing remaining to show is the implication (countable compactness) $\implies$ (limit point compactness), which uses the following modified argument.

You must prove that every set $A$ without limit points is finite. We can assume that $A$ is countable without loss of generality (If every countably infinite set has a limit point, then all infinite sets have limit points). Now for each element $a\in A$, there is a neighborhood $U_a$ of $a$ containing no other points of $A$ (since $a$ is not a limit point of $A$). Since $A$ contains all of its limit points tautologically, $A$ must be closed, which in turn implies that $A$ is countably compact. Therefore the cover $\{U_a\}$ has a finite subcover, implying that $A$ is finite.

  • $\begingroup$ It’s no more direct, but one can bypass sequential compactness: limit point compact implies totally bounded, which implies separable, which implies second countable, which implies Lindelöf. $\endgroup$ – Brian M. Scott Apr 4 '15 at 14:50
  • $\begingroup$ Your proof seems good, except it doesn't make sense to show that $A$ is closed, therefore countably compact, when countable compactness is already assumed. However, I was looking for some intuition as to why countable compactness implies compactness, and this intuition is kind of obscured through the chain of implications: Countable Compactness $\Rightarrow$ Limit Point Compactness $\Rightarrow$ Sequential Compactness $\Rightarrow$ Compactness. Is there any way of doing this more directly, or at least explaining an intuition for it? $\endgroup$ – Joshua Meyers Apr 4 '15 at 16:58
  • $\begingroup$ We need to show that our set $A \subset E$ is countably compact assuming that $E$ is countably compact. As to the intuition, I would also be interested in seeing something less formal. $\endgroup$ – Rolf Hoyer Apr 4 '15 at 17:53
  • $\begingroup$ It is not always true that $A$ contains all of it's limit points. Consider a sequence in $E=[0,1]$. Then $A$, the sequence itself, is infinite but doesn't have a limit point. $\endgroup$ – abcdef Apr 4 '15 at 22:01
  • $\begingroup$ In the argument we were assuming that $A$ was a set without any limit points. $\endgroup$ – Rolf Hoyer Apr 4 '15 at 22:10

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