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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?

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  • $\begingroup$ 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. $\endgroup$ Jul 9, 2015 at 17:39
  • $\begingroup$ So closed set of reals are scattered if and only if they are countable, right? $\endgroup$
    – Markus
    Jul 9, 2015 at 17:44
  • $\begingroup$ 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. $\endgroup$ Jul 9, 2015 at 17:48
  • $\begingroup$ @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. $\endgroup$
    – Mad Hatter
    Mar 12, 2020 at 20:34

2 Answers 2

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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.

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  • $\begingroup$ But $\mathbb{Q}$ is not closed. Doesn't the CB-derivative of a closed, countable set always vanish? $\endgroup$
    – Markus
    Jul 9, 2015 at 17:39
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    $\begingroup$ @Markus: Yes, but at that point in your question you didn’t limit yourself to closed sets. $\endgroup$ Jul 9, 2015 at 17:40
  • $\begingroup$ 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! $\endgroup$
    – Markus
    Jul 9, 2015 at 17:59
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    $\begingroup$ @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! $\endgroup$ Jul 9, 2015 at 18:01
  • $\begingroup$ @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! $\endgroup$
    – H.Durham
    Oct 13, 2015 at 16:29
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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).

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  • $\begingroup$ 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... $\endgroup$
    – Bcpicao
    Oct 26, 2021 at 11:58

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