TITLE: Derived sets and ordinals
QUESTION [7 upvotes]: Given a set $P$ of real numbers, its derived set is the set of all accumulation points - $a\in P^\prime$ if every open set containing $a$ also contains an infinite number of points from $P$ (equivalently, at least one point from $P$ different than $a$).
Cantor defined $P^0=P$ and $P^k$ to be the derived set of $P^{k-1}$. Then he had the wonderful idea to continue this construction to $P^\omega = \bigcap_{k} P^k$, and then we have $P^{\omega +1}$ being the derived set of $P^\omega$ and so on.
The question is whether it is interesting to continue into the ordinals in this manner. i.e. is there a set $P$ such that $P^\omega \ne P^{\omega +1}$. Is there an example?

REPLY [6 votes]: For any topological space $X$, the Cantor-Bendixon derivatives of $X$ are defined by transfinite induction:

$X^0 = X$;
$X^{\alpha+1} = (X^{\alpha})'$;
$X^{\gamma} = \cap_{\alpha\lt \gamma}X^{\alpha}$ for limit ordinals $\gamma$.

By simple cardinality arguments, there must exist a least ordinal $\alpha$ such that $X^{\alpha} = X^{\alpha+1}$. This is called the Cantor-Bendixon rank of $X$.
As noted in JDH's answer to this Math Overflow question, every ordinal is the Cantor-Bendixon rank of some topological space. 
For $X$ contained in $\mathbb{R}$, the Cantor-Bendixon rank is always countable, but for every countable ordinal $\alpha$, there is a closed subset $X$ of $\mathbb{R}$ with $X^{\alpha}$ a singleton. See for example this write-up by Harold Simmons. So 
$$\mathrm{sup}\{\text{CB rank}(X)\mid X\subseteq \mathbb{R}\} = \omega_1,$$
where $\omega_1$ is the first uncountable ordinal.