If $S$ is a set and $S'$ denotes its Cantor-Bendixson derivative; then why isn't it just the case that $S''=S'$? In general, we have
$$S' \supset S'' \supset S^{'''}\supset \cdots$$
but my (apparently wrong) intuition is that the set of accumulation points for a set of accumulation points is the same set; much in the same way that the closure of the closure of a set is the closure of the original set.
Are there necessary and sufficient conditions such that $S'=S''$?
Can you please provide a simple counter example $S$ such that $S'\neq S''$?
thank you
For a very simple example, let $S=\{0\}\cup\{1/n:n\in\mathbb{Z}_+\}$. Then $S'=\{0\}$, and $S''=\emptyset$. The key point is that $S'$ is obtained from $S$ not just by adding limit points but also by removing isolated points. Removing those isolated points may cause points that were limit points to no longer be limit points.
There isn't really any interesting characterization of when $S'=S''$ that comes to mind. It's true when it's true: when every point of $S'$ is a limit point of $S'$ (in other words, when $S'$ has no isolated points).