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

Question

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?

Answer 1

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.

Answer 2

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).