limit points of a countable set

Question

Let $X$ be a complete space and $A\subset X$ be a closed countable set. Let $A^{'}$ denote the set of all limit points of $A.$ Prove that there exists an integer $n$ such that $A^{n}=\emptyset?$ where $A^{1}=A^{'}$ and $A^{n+1}=(A^{n})^{'}.$ Using Cantor's intersection theorem, I showed that if in addition $A^{'}$ is bounded, then $A^{n}=\emptyset$ for some integer $n.$ But I failed to prove the result without this condition.

Answer 1

The process you are considering here is the Cantor-Bendixson-derivative. You claim that every closed countable subset of a complete space has finite Cantor-Bendixson-rank, but that is not true. A countable Polish space can have any countable ordinal as its CB-rank.

Once you know how to built an example for all $\beta < \alpha$, you can built an example for $\alpha$ by arranging the examples as converging to a fresh point.