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Let $(X,d)$ be compact metric space and $C$ be the set of condensation points of $X$. Following the notations here, is the equality $\bigcap\limits_{n \geq 1} X^{(n)}=C$ true ?

I succeeded in showing the inclusion $C \subset \bigcap\limits_{n \geq 1} X^{(n)}$ but the other inclusion seems to be more difficult.

Have you an idea?

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What is $\,X^{(n)}\,$ , again? –  DonAntonio Aug 7 '12 at 13:42
    
The notations are given in the second link I gave: $X'$ is the derived set and for all $n \geq 2$, $X^{(n)}=(X^{n-1})'$. –  Seirios Aug 7 '12 at 13:52

2 Answers 2

up vote 4 down vote accepted

Consider the ordinal $X=\omega^\omega+1$ with order topology. It is a successor ordinal, so it is compact. It is also countable, so it is metrizable, and has no condensation points (it can be seen either using abstract fact that a second-countable compact Hausdorff space is Polish, or the fact that any countable ordinal embeds into rationals as a totally ordered set).

On the other hand, notice that $X'$ is the set of limit ordinals within $X$. Similarly, $X''$ is the set of limits of limit ordinals, etc., so for any $n$, we have $\omega^n\in X^{(n)}$, and for $X^{(\omega)}=\bigcap_n X^{(n)}$ we have $\omega^\omega\in X^{(\omega)}$, so $X^{(\omega)}\neq C=\emptyset$.

On the other hand, by Cantor-Bendixson theorem, we can see that for any compact metrizable space (indeed, any Polish space), there exists a countable ordinal $\alpha$ such that $X^{(\alpha)}=X^{(\infty)}=P=C$, which is $\omega+1$ in the above case (the minimal such $\alpha$ is called the Cantor-Bendixson rank, and is of interest in e.g. model theory, where it is one of the interpretations of the Morley rank).

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I agree with your counter-example and that $P=C \subset X^{(\beta)}$ for any countable ordinal $\beta$, but how do you show the existence of such a convenient $\alpha$? –  Seirios Aug 7 '12 at 16:42
    
@Seirios: until it stabilizes, the sequence $X^{(\alpha)}$ is strictly decreasing, so it drops at least one point at every step, so if it did not stabilize for uncountably many steps, the perfect kernel would have uncountable complement, which is a contradiction with the Cantor-Bendixson theorem. –  tomasz Aug 7 '12 at 17:43
    
I understood, thank you. –  Seirios Aug 7 '12 at 20:14

In fact, the inclusion can be deduced from the property: For all non empty perfect closed subset $P \subset X$, there exist two non empty perfect closed and disjoint subsets $P_1,P_2 \subset P$. Thus, one may construct an injection from $\{0,1\}^{\mathbb{N}}$ into any neighborhood of an element of $\bigcap\limits_{n \geq 1 } X^{(n)}$.

EDIT: As tomasz said, the construction works only if $X$ has finitely many isolated points.

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Not really. You are assuming that the intersection is perfect, which need not be the case. What you say is a proof that a perfect set is of cardinality $\mathfrak c$, which is true, but you can't use it until you give a (positive) answer to your question. I don't believe that it is true, now that I reflected on my mistake. ;) I'm trying to construct a tangible counterexample now. –  tomasz Aug 7 '12 at 14:56

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