Condensation points and derived sets 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?
 A: 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).
A: 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.
