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.

(Rudin's Principles of Mathematical Analysis, 3rd ed., Chapter 2, Problem 28)

Now I've managed to hit upon a proof using the preceding problem (which can be generalised) as follows:

Let $X$ be a separable metric space, $E \subset X$, $E$ is uncountable, and let $P$ be the set of all condensation points of $E$. Prove that $P$ is perfect and that at most countably many points of $E$ are not in $P$. In other words, show that $P^c \cap E$ is at most countable.

Now what bothers me is what Rudin states as the corollary to the result he asks us to prove in Problem 28. Here's what he states:

Every countable closed set in $\mathbb{R}^k$ has isolated points.

How to prove this from the original result?

Can we also state the following?

Every countable closed set in a separable metric space has isolated points.


marked as duplicate by dustin, John Gowers, Robert Cardona, Did real-analysis Mar 14 '15 at 19:15

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    $\begingroup$ What about the rationals? Always ask yourself about the rationals. $\endgroup$ – Carl Mummert Nov 10 '14 at 2:49
  • $\begingroup$ @Carl, I'm not able to see your point. So could you please elaborate? $\endgroup$ – Saaqib Mahmood Nov 11 '14 at 14:17
  • $\begingroup$ @SaaqibMahmuud the rationals are a dense separable set wich is not closed so it doesnt have any isolated points. $\endgroup$ – Aram Nov 11 '14 at 14:31
  • $\begingroup$ @Saaqib Mahmuud: If we consider just the rationals as their own separable metric space, they are a countable closed set without isolated points. $\endgroup$ – Carl Mummert Nov 11 '14 at 14:40
  • $\begingroup$ @Aram: see above, please $\endgroup$ – Carl Mummert Nov 11 '14 at 14:41

Every countable closed set in $\mathbb{R}^k$ has isolated points.

HINT: Negate it (that is there is no isolated point) then surround two different points by a closed ball then pick $2$ different points inside each ball surround again the $4$ by a smaller closed ball (that is none of the points is inside any of the others balls), and repeat, then apply the Nested Intersection Theorem (Remember a closed subset of a complete space is complete) to any sequence of nested balls and deduce you have a bijection of your limit points with $\{0,1\}^{\aleph} \sim\aleph^{\aleph}$


Suppose your set doesnt have any isolated points, then it is a perfect set of $\mathbb{R}^k$ but this would mean that is not countable, contradiction.

  • $\begingroup$ I'm afraid I didn't get your hint. Moreover, I would like to deduce this fact as a corollary to the main result whose proof is required in Prob. 28, Chap. 2 in Rudin's text. $\endgroup$ – Saaqib Mahmood Nov 11 '14 at 14:16
  • $\begingroup$ @SaaqibMahmuud I'll edit, but you should try to do it, not trying to understand it, since it was a hint for you to expand upon. $\endgroup$ – Aram Nov 11 '14 at 14:33
  • $\begingroup$ @SaaqibMahmuud I edited it. $\endgroup$ – Aram Nov 14 '14 at 1:22

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