iterated limit sets of a countable subset of real numbers Let $A\subset \mathbb{R}$ be a closed subset, and $A'$ be the sets of limit points. We know that if $A$ is a countable set, $A'$ is a proper subset of $A$. Is it possible to find a subset( closed and countable) so that $A^{(n)}$ is non-empty for any $n\in \mathbb N$? We denote by $A^{(n)}=(A^{(n-1)})'$.
 A: Yes, just take a copy of the ordinal $\omega^\omega$ in the reals. This has Cantor Bendixson rank exactly $\omega$. 
One way to see this is first to understand how to make a closed set last for exactly $n$ steps. A convergent sequence lasts one step; a convergent sequence of convergent sequences lasts two steps, and so on. At each step, take a convergent sequence of sets of the previous type. Put them together by considering a copy of the $n^{th}$ set in the interval $[n,n+1]$. In this way, applying $n$ derivatives the desired set does this separately in each interval, and the $n^{th}$ interval last for $n$ steps before disappearing. So the whole set will last for $\omega$ steps. If you want a compact example, you can squeeze the example into an open interval and add a limit point (which will add one final isolated point, disappearing in one additional step).
If you think about it, the ordinal $\omega^\omega$ is just like this, since $\omega^\omega$ is the limit of the ordinals $\omega^n$, which lasts for $n$ iterations of the derivative. 
Every countable ordinal has an order isomorphism into the rationals, by Cantor's universality theorem for the countable endless dense linear order. By taking the closure of such an order, you get countable closed sets in the reals of any desired Cantor Bendixson rank.
It may be interesting to note that this kind of of problem was the origin of Cantor's discovery of the ordinals, since before he knew the ordinals, he was playing with sets like this, and they lead naturally to the concept of ordinal.
