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In Rudin's Principles of Mathematical Analysis, he gives the following proof:

2.37 Theorem: If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in K.

Proof: If no point of $K$ were a limit point of $E$, then each $q\in K$ would have a neighborhood $V_q$ which contains at most one point of $E$. It is clear that no finite subcollection of ${V_q}$ can cover $E$; and the same is true of $K$, since $E\subset K$. This contradicts the compactness of $K$.

The way I understand this proof is as follows: If each $V_q$ is open, then $\cup \{ V_q \}$ would be an open cover for $K$ (and $E$ as well, since $E \subset K$). Since $E$ is an infinite set of points, it follows that no finite subcollection of ${V_q}$ can cover $E$ (since there is at most one point in each ${V_q}$, and a finite collection of points cannot cover an infinite collection of points).

So it seems to me that this whole proof hinges on the fact that each set $V_q$ is an open set? Earlier in Rudin we prove that the union of open sets is open. So to get an open cover for $E$ we would need to take the union of open sets. Why are these single-point sets considered open? I thought single-point sets were closed:

  1. Is a single point in euclidean space open, closed, neither or both?
  2. One Point Set Closed

Or is there something I'm missing here?

laichzeit0
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    The $V_q$ are not singletons, $V_q\cap E$ are – Alessandro Codenotti Nov 11 '20 at 13:50
  • Can't believe I didn't realize that. You are right. Sometimes I wish Rudin had just a little bit more "meat" to it, or some pictures. I feel like I'm deciphering Hieroglyphics, alone, in the dark, with a box of matches. – laichzeit0 Nov 11 '20 at 13:54
  • To answer the title, single-point sets are not open, since their complement is not closed. But you cannot say, “a single point set is not open because it is closed.”. There are subsets of metric spaces that are both open and closed. It just happens so that the only subsets of $\mathbb{R}$ with the usual metric that are both open and closed are $\emptyset$ and $ \mathbb{R}$ – Adam Rubinson Nov 11 '20 at 14:22
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    Take your time with Rudin, it's generally worth the effort. –  Nov 11 '20 at 14:24
  • I agree with @JustinYoung that you should take your time, and I'd add that you could devote some of that time to drawing pictures to help you visualize the text. – Andreas Blass Nov 11 '20 at 16:22

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