Odd Corollary in Baby Rudin Chapter 2, Question 28 [duplicate]

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 CardonaMar 12 '15 at 18:51

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.
• 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. – Tom Nov 19 '12 at 13:33
• 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. – Gautam Shenoy Nov 19 '12 at 13:36