0
$\begingroup$

In the question asked here, it is claimed in a comment and answer of Brian M. Scott that a subset of $\mathbb{R}$ is scattered if and only if it has vanishing Cantor-Bendixson derivative.

I thought that this should be true in general for a separable metric space, but I could only prove it for closed sets. Indeed I found a scattered subset of $\mathbb{R}^2$ which has uncountable closure and hence nonvanishing derivative.

Thus my question is: is there something special about $\mathbb{R}$, or have I misunderstood a definition?

Briefly, the example is the subset of $[0,1]^2$ given by the union of subsets of horizontal lines of height $1/n$, with each containing $n$ equidistributed points.

$\endgroup$
5
  • $\begingroup$ I'm inclined to say that you've misunderstood a definition, as what Brian said should be true of all topological spaces under appropriate definitions. But as you haven't indicated what definitions you are using (or the example you came up with) it's difficult to say for certain. $\endgroup$
    – user642796
    Oct 13, 2015 at 14:52
  • $\begingroup$ What's your example? Have you double-checked the definition of "scattered"? $\endgroup$ Oct 13, 2015 at 14:57
  • $\begingroup$ Well, my definitions are as in the question linked. In particular I take the definition of the C-B as given in the linked article on that question, which I now notice requires the set to be closed. So I think my confusion stems from the fact that a subspace may not contain all of its limits points, and so when taking limit points it needs to be clarified whether to include the ones not in the set but present in the ambient space. I'll add the example now. $\endgroup$
    – H.Durham
    Oct 13, 2015 at 15:00
  • $\begingroup$ Regarding your example, what are the isolated points (as per the definition of "scattered") for the subset $[0,1] \times \{0\}?$ $\endgroup$ Oct 13, 2015 at 15:08
  • $\begingroup$ That set is not a subset, since all the points must have positive height. Perhaps the example is not clear? $\endgroup$
    – H.Durham
    Oct 13, 2015 at 16:19

1 Answer 1

3
$\begingroup$

You’ve misunderstood the definition of Cantor-Bendixson rank. The Cantor-Bendixson rank of a set is calculated on the basis of the points in that set; if it happens to be a subset of some ambient space, treat it as a subspace, not as a subset. The Cantor-Bendixson derivative of your set is therefore empty: it consists entirely of isolated points. It has no limit points in the set itself.

$\endgroup$
7
  • $\begingroup$ There seems to be a difference in the definition of the derived set: whether or not it includes limit points outside the set. $\endgroup$
    – Bcpicao
    Oct 20, 2021 at 17:52
  • $\begingroup$ @Bcpicao: One considers only points in the set. If you look at the Wikipedia definition of the C-B rank, you’ll see that it defines it only for spaces, not for subsets of spaces, so the derivative is being taken only in the set in question treated as a space in its own right. $\endgroup$ Oct 20, 2021 at 21:11
  • $\begingroup$ Thanks for the prompt reply. From the examples on the "Derived set" wikipedia page, one clearly does not treat the set as subspaces, but uses the topology of the parent set (having in general no partial order between $A$ and $A'$, whereas for you $A'\subset A$). Why would one restrict the definition of the C-B derivative to using the subspace topology? (I can ask this as a separate question, if you find this interesting enough) $\endgroup$
    – Bcpicao
    Oct 21, 2021 at 11:24
  • $\begingroup$ Just realised that in most articles that define it "the broad way", they just take a closed set and from that point on both definitions coincide. $\endgroup$
    – Bcpicao
    Oct 21, 2021 at 11:46
  • $\begingroup$ @Bcpicao: Those are simply example of taking the derived set in general. Note, however, that the section of the article on the Cantor-Bendixson rank defines only the Cantor-Bendixson rank of a space, not of an arbitrary subset of a space. We can of course still talk about the C-B rank of a subset of a space, but we want it to be an inherent property of the subset, one that does not depend on how that subset is embedded in the ambient space, so we simply look at the subset as a space in its own right and calculate the C-B rank of that space. $\endgroup$ Oct 21, 2021 at 21:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.