Does compact scattered imply finite derivation?

Question

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 $K^{(\alpha)}= \bigcap_{\beta<\alpha}K^{(\beta)}$).

Is the following claim correct?

Claim: If $X$ is compact and scattered then there exists a finite $n \in \mathbb N$ such that, $X^{(n)} = \emptyset$.

Proof: If $X$ is scattered, and $X \neq \emptyset$, then there exists an ordinal $\alpha > 0$ such that, $\xi(X) = \alpha$, (where $\xi(X) = inf \{ \alpha : x^{(\alpha)} = \emptyset \}$ and $X^{(\alpha)}$ is the $\alpha$-derivation of $X$). The set $\{ X \setminus X^{\alpha} : \alpha < \xi(X) \}$ is an open cover $\mathcal W$, of $X$, such that, for each $\alpha < \xi(X)$, there exists $x_\alpha$, such that, $x_{\alpha} \in X^{(\alpha)}$ and for each $\alpha \neq \beta < \xi(X)$, $x_{\alpha} \notin X^{(\beta)}$. $X$ is compact so $\mathcal W$ has a finite subcover $\mathcal W'$. So, the set $\{ \alpha : \alpha < \xi(X) \}$ is finite. This implies that $\xi(X)$ is finite.

Thank you!

What does scattered mean again ? – Rene Schipperus Jun 19 '14 at 13:02 — Rene Schipperus, Jun 19 '14 at 13:02

user642796 · Accepted Answer · 2014-06-22T06:51:49.527

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This is false. Consider the ordinal space $X = \omega_1 + 1 = [0, \omega_1]$. This is a compact scattered space, however it's "Cantr-Bendixon rank" $\xi ( X )$ is infinite; uncountable, even. It's not too hard to show (by induction) that $X^{(\alpha)} \cap \omega_1$ is a club subset of $\omega_1$ for each $\alpha < \omega_1$ (and so $\omega_1 \in X^{(\alpha)}$). It then follows without too much difficulty that $X^{(\omega_1)} = \{ \omega_1 \}$, and so $\xi (X) = \omega_1+1$.

Your proposed "proof" does not work in this instance, since the family $\{ X \setminus X^{(\alpha)} : \alpha < \xi(X) \}$ doesn't even cover $X$ (since $\omega_1 \in X^{(\alpha)}$ for each $\alpha < \omega_1 +1 = \xi (X)$.

As Dave L. Renfro notes in the comments below, you can get away with ordinal spaces much much much shorter than $\omega_1 +1$. Even $\omega^\omega + 1$ will work. This space has Cantor-Bendixon rank $\omega+1$, but is still compact and scattered (all closed ordinal spaces are compact and scattered).

_{Never answer while sick.}

edited Jun 22 '14 at 06:51

answered Jun 19 '14 at 13:29

user642796

52,188

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As far as I can see, $X^{(\omega)}$ is not just ${\omega_1}$. Doesn't it contain, for example, the countable ordinal $\omega^\omega$ (ordinal exponentiation)? I think the derivation process reaches the singleton ${\omega_1}$ only after $\omega_1$ steps. – Andreas Blass Jun 19 '14 at 15:07
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I think you mean ${\omega}^{\omega}$ instead of $\omega_{1}.$ Indeed, as linear orderings $\left(\omega_1 + 1 \right)^{(\beta)} = \omega_1 + 1$ for all $\beta < \omega_{1}.$ Regarding ways to visualize ${\omega}^{\omega},$ see Easy visualizations of small countable ordinals and Intuition for $\omega^\omega$. – Dave L. Renfro Jun 19 '14 at 15:21
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@AndreasBlass: Thanks. I felt there might have been something wrong, but in the midst of a 103 degree fever I simply wasn't thinking straight. – user642796 Jun 22 '14 at 06:55

Does compact scattered imply finite derivation?

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