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Let $(M,d)$ be a metric space. Show that $M$ is compact if and only if all infinite open cover of $M$ have a proper subcover.

We remember: If $\mathcal{U}$ is a cover of $M$, then a proper subcover is a proper subset $\widehat{\mathcal{U}}$ of $\mathcal{U}$ such that it is cover of $M$.

Remark: The diection $(\Rightarrow)$ is trivial. The problem is the direction $(\Leftarrow)$, I think hypotheses are missing, but I have not been able to find a counterexample or prove it.

This exercise I found in a book that for me is not very reliable since I have found other exercises with errors.

William Elliot
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Diego Fonseca
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In addition to the answer provided in this post, if one can take the true statement that every metric space that is limit point compact (in other words, every infinite subset of $M$ admits a limit point in $M$) is compact as given, then one can proceed as follow:

Let $E$ be an infinite subset of $M$, suppose no point $x$ of $M$ is a limit point of $E$ then $E$ is closed and every point $x$ of $E$ has a neighborhood $V_x$ that contains no other point of $E$. $\mathcal{U} = \{V_x:x\in E\} \cup \{E^c\}$ is then an infinite open cover of $M$ and the only possible proper subcover of $\mathcal{U}$ is $\{V_x:x\in E\}$, which has no proper subcover.

Project Book
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  • Actually, $\mathcal{U}$ does not have proper subcovers. – Diego Fonseca Oct 21 '17 at 03:24
  • @DiegoFonseca I can redefine one of the $V_x$ to contain $E^c$. – Project Book Oct 21 '17 at 03:32
  • This is probably the intended idea, to use the limit point compactness equivalence for metric spaces. The linked (in the comments) idea by Brian is also nice: take any open cover, it has a point finite refinement (and as all point finite covers have an irreducible subcover which is finite) we can find a finite subcover of the original cover. This only uses that metric spaces are metacompact. – Henno Brandsma Oct 21 '17 at 12:13