# Cantor-Bendixson rank of a closed countable subset of the reals, and scattered sets

I am reading the notes in the following link, and I am a bit unclear about the connection between scattered sets, countable sets, and sets for which the Cantor-Bendixson derivative is eventually the empty set.

http://www.cs.man.ac.uk/~hsimmons/DOCUMENTS/PAPERSandNOTES/CB-examples.pdf

On page 3, the author says: A closed set X (of the reals) is scattered if $X^{\alpha}=\emptyset$, where $\alpha$ is the Cantor-Bendixson rank of $X$.

From what I could find, scattered means that any proper subset contains isolated points. However, for a countable subset of the reals, the Cantor-Bendixson-derivatives eventually vanish. As well, if I understand the definition of scattered correctly, a converging sequence together with its limit would be a closed, non-scattered set (the set with only the limit point in it will not have isolated points), but the CB-derivatives still eventually vanish. It seems to me that a closed set of the reals has vanishing CB-derivatives if and only if it is countable.

Am I misreading the definition of scattered?

• Don't forget that what you quoted from page 3 is about closed sets. A countable subset of $\mathbb R$ need not be scattered, as pointed out in Brian M. Scott's answer, but a countable closed set of reals is scattered. Jul 9, 2015 at 17:39
• So closed set of reals are scattered if and only if they are countable, right? Jul 9, 2015 at 17:44
• Yes, because the Cantor-Bendixson analysis can remove only countably many points. So if you start with an uncountable closed set, there will be a perfect (hence uncountable) closed set of points that never get removed. Jul 9, 2015 at 17:48
• @Markus A bit OT, but do you still have the notes you link in the question. The webpage does not load anymore, and I would like to see them. Mar 12, 2020 at 20:34

You’ve misunderstood a couple of things. First, it’s not true that the Cantor-Bendixson derivatives of a countable set of reals necessarily vanish: every C-B derivative of $\Bbb Q$ is $\Bbb Q$, since $\Bbb Q$ has no isolated points to remove.

Secondly, a space is scattered if every subset contains at least one point that is isolated in that subset considered as a space in its own right. A simple sequence with its limit point is scattered: if the limit point is $p$, the point $p$ is isolated in the set $\{p\}$.

It’s true, however, that a closed subset of $\Bbb R$ is scattered (equivalently, has vanishing C-B derivative) if and only if it is countable.

First, no uncountable subset of $\Bbb R$ is scattered. This follows from the fact that $\Bbb R$ is second countable. Let $\mathscr{B}$ be a countable base for the topology, and let $A\subseteq\Bbb R$ be uncountable. Let $$\mathscr{B}_0=\{B\in\mathscr{B}:B\cap A\text{ is countable}\}\;,$$ and let $A_0=A\setminus\bigcup\mathscr{B}_0$. Clearly $\bigcup\mathscr{B}_0$ is countable, so $A_0$ is uncountable. If $x\in A_0$, and $U$ is any open nbhd of $x$, then there is a $B\in\mathscr{B}$ such that $x\in B\subseteq U$. Clearly $B\notin\mathscr{B}_0$, so $B\cap A$ is uncountable, and therefore $B\cap A_0$ is uncountable as well. In particular, $x$ is not isolated in $A_0$. Thus, $A_0$ has no isolated points, and $A$ is not scattered.

Secondly, if $A\subseteq\Bbb R$ is countable and not scattered, then $A$ contains a countable infinite subset $A_0$ with no isolated points. Such a set is order-isomorphic to $\Bbb Q$ and therefore not closed.

• But $\mathbb{Q}$ is not closed. Doesn't the CB-derivative of a closed, countable set always vanish? Jul 9, 2015 at 17:39
• @Markus: Yes, but at that point in your question you didn’t limit yourself to closed sets. Jul 9, 2015 at 17:40
• I see, thank you. So to make sure I understand, vanishing CB-derivative is equivalent with a set being scattered, and there are scattered (and necessarily countable), but not closed subsets of the reals (a converging sequence without the limit, for example). I guess I was confused because the author could as well have used countable instead of scattered, since he was talking about closed sets anyway. Thanks again, this was very helpful! Jul 9, 2015 at 17:59
• @Markus: Yes, vanishing C-B derivative is equivalent to being scattered. Scattered subsets of $\Bbb R$ must be countable, and countable closed subsets must be scattered. You’re welcome! Jul 9, 2015 at 18:01
• @BrianM.Scott: I'm having a bit of trouble seeing the equivalence between scattered and vanishing C-B derivative; I've asked a question here, and would really appreciate your insight. Thanks! Oct 13, 2015 at 16:29

Here's a fun exercise on scattered sets: Show that a subset of the real line is scattered iff it is countable G-delta set (G-delta means countable intersection of open sets).

• I've been trying to solve this for a few days but haven't had any luck, would you mind giving a hint? Brian's answer shows scattered $\implies$ countable, and I know closed $\implies G_\delta$ but can't get much further... Oct 26, 2021 at 11:58