I've finished all the questions in Chapter 2 of Principles of Mathematical Analysis by Walter Rudin (self study), but I have a question about Q.28, which reads:

Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable. (Corollary: Every countable closed set in $\mathbb{R}^k$ has isolated points.)

It's easy to answer the question, given what's proved in Q.27, but the corollary is a bit weird. It looks to me as though it is just an immediate consequence of the fact that non-empty perfect sets in $\mathbb{R}^k$ are uncountable, which is proved in the main text. I don't see what it has to do with what is proved in this question.

Given how meticulous the book is, I suspect the apparent non-sequiteur means I'm missing something.


marked as duplicate by dustin, user147263, Krish, John Gowers, Robert Cardona Mar 12 '15 at 18:51

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An isolated point by def is a point that belongs to the set that is not a limit point. Suppose the closed countable set had no isolated points, then EVERY point of it is a limit point. Thus the set is perfect. But a non-empty perfect set in $\mathbb{R}^n$ is uncountable. Contradiction.

The separable space here is $\mathbb{R}^n$. The corollary is definitely sequitur.

  • $\begingroup$ But that doesn't use the main body of the question, to which it is supposed to be a corollary: "every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable". $\endgroup$ – Tom Nov 19 '12 at 13:00
  • $\begingroup$ To be clear, I'm not saying I have difficulty seeing that the corollary is true - it's obviously true. I have difficulty seeing in what way it is considered to be a "corollary" of Q.28. $\endgroup$ – Tom Nov 19 '12 at 13:04
  • $\begingroup$ It does. The theorem says ANY closed set can be decomposed as such. But for a "closed" countable set, the perfect part is forced empty due to cardinality reasons. This means that a closed countable set is NOT perfect (nor does it have a subset that is nonempty perfect) and so it has isolated pts. $\endgroup$ – Gautam Shenoy Nov 19 '12 at 13:18
  • $\begingroup$ There's nothing in the result proved in Q.28 to say the countable part can't itself be perfect. An example is $\mathbb{Q}$, considered as a metric space in its own right. You can decomposing the whole of $\mathbb{Q}$ (closed) into the empty set and $\mathbb{Q}$ itself, which is both countable and perfect. To prove the corollary you need the additional fact that perfect subsets of $\mathbb{R}$ are uncountable. However, that is all you need (it's trivial to prove the corollary from it alone) and I still can't see how the decomposition in Q.28 contributes in any way. $\endgroup$ – Tom Nov 19 '12 at 13:33
  • $\begingroup$ My analysis was for $R^k$, not all separable metric spaces. In $R^k$, countable sets are not perfect. But I get what you are saying. The fact that perfect sets are uncountable was indeed enough to arrive at the conclusion and the theorem could have been bypassed. $\endgroup$ – Gautam Shenoy Nov 19 '12 at 13:36

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