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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.

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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.

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    "In $\mathbb{R}^k$, countable sets are not perfect." - yes, that's why I had to use $\mathbb{Q}$ as a metric space in its own right to make the point. I was just trying to show that the decomposition in Q.28 was extraneous and that what really matters is the uncountability of perfect sets in $\mathbb{R}^k$.2012-11-19