Does $\xi(X) = \alpha + 1$ imply that $X^{(\alpha)}$ is a discrete space?

Let $K$ be a (Hausdorff) scattered topological space and for each ordinal $\alpha$ denote by $K^{(\alpha)}$ the $\alpha$th derivative of $K$ by the Cantor-Bendixson derivation (i.e., define transfinitely: $K^{(0)} = K$; for each ordinal $\alpha$, let $K^{(\alpha+1)}$ be the set of nonisolated points in $K^{(\alpha)}$ when $K^{(\alpha)}$ is equipped with the subspace topology); for $\alpha$ a limit ordinal, set $K^{(\alpha)}= \bigcap_{\beta<\alpha}K^{(\beta)}$).

Also define: $\xi(X) = \inf\{ \alpha : X^{(\alpha)} = \emptyset \}$ if some $\alpha$ exists such that $X^{(\alpha)} = \emptyset$, and $\xi(X) = \infty$ otherwise.

My question is: Suppose that $\xi(X) = \alpha + 1$. Does this imply that $X^{(\alpha)}$ is a discrete space?

Thank you!

Yes, essentially by definition. If $X^{(\alpha)}$ is not discrete, then some point of $X^{(\alpha)}$ must have the property that each of its neighbourhoods contains another element of $X^{(\alpha)}$ (or, more succinctly, is not an isolated point of the set). Then it must be that $X^{(\alpha+1)}$ is nonempty, contradicting the definition of $\xi(X)$.