Is it true that $A$ is scattered?

Question

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

Define for each ordinal $\alpha$, $A_{\alpha} = X \setminus X^{(\alpha)}$.

Let $A = \bigcup A_{\alpha}$. Is it true that $A$ is scattered?

Thank you!

Answer 1

Yes, almost by definition. But let's introduce some additional notation. For each $\alpha$ let $I_\alpha$ be the set of isolated points of $X^{(\alpha)}$ (so then $X^{(\alpha)} = I_\alpha \sqcup X^{(\alpha+1)}$ (disjoint union), and $A_\alpha = \bigcup_{\xi < \alpha} I_\xi$). It follows that $A = \bigcup_\alpha I_\alpha$, and for each $\alpha$ each point $x \in I_\alpha$ has an open neighbourhood $U$ in $X$ such that $U \cap X^{(\alpha)} = \{ x \}$, which in particular means that

1. $U \cap I_\alpha = \{ x \}$; and
2. $U \subseteq \bigcup_{\xi \leq \alpha} I_\xi\;(= A_{\alpha+1})$.

So given any nonempty $B \subseteq A$, that the least $\alpha$ such that $B \cap I_\alpha \neq \varnothing$, and then find an appropriate open neighbourhood $U$ of some $x \in B \cap I_{\alpha}$. It will follow that $U \cap B$ is a singleton.