Equivalence to being a scattered space. Let $X$ be a topological Hausdorff space with Cantor-Bendixson rank $\lambda$. 
A paper I am reading claims that these two are equivalent:


*

*$X$ is scattered.

*$X = \cup_{\alpha < \lambda} I_\alpha$


Where:
$X_0 = X$, $I_\alpha$ is the set of isolated points of $X_\alpha$, $X_{\alpha+1} = X_\alpha - I_\alpha$ and for limit $\alpha$, $X_\alpha = \cap_{\beta < \alpha}X_\beta$. 
How can you prove the equivalence?
 A: You inductive definition of the $X_\alpha$ can be relatively easily shown to be equivalent to setting $X_{\alpha} = X \setminus \bigcup_{\xi < \alpha} I_\xi$ for all $\alpha$.  I will use this instead.

Given any $X$, define construct an $\mathbf{Ord}$-sequence $\langle I_\alpha \rangle_{\alpha \in \mathbf{Ord}}$ by setting $I_\alpha$ to be the set of isolated points of $X \setminus \bigcup_{\xi < \alpha} I_\xi$.  As $\langle \bigcup_{\xi < \alpha} I_\xi \rangle_{\alpha \in \mathbf{Ord}}$ is increasing, there must be a least $\lambda$ such that $\bigcup_{\xi < \alpha} I_\xi = \bigcup_{\xi < \lambda} I_\xi$ for all $\alpha \geq \lambda$, which implies that $I_\lambda = \varnothing$.
If $X$ is scattered, then it must be that $X \setminus \bigcup_{\xi < \lambda} I_\xi = \varnothing$, (since otherwise this set would have isolated points) meaning that $X = \bigcup_{\xi < \lambda} I_\xi$.
If $X = \bigcup_{\alpha < \lambda} I_\alpha$, then given any nonempty $A \subseteq X$ there is a least $\alpha < \lambda$ such that $A \cap I_\alpha \neq \varnothing$, and it follows that every point in $A \cap I_\alpha$ is an isolated point of $A$.
