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I'm looking for a set $A \subset \mathbb{R}$ such that $\bigcap^\infty_{n=0} A^{(n)} $ is a perfect set (i.e $X'=X$) but $\forall n \in \mathbb{N}$ the set $A^{(n)}$ isn't perfect (where $X^{(n)}=(X^{(n-1)})'$ and $X^{(0)}=X$).

nowhere dense
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Yes. For every countable ordinal we can find a set of real numbers whose Cantor-Bendixson rank is that ordinal.

To see this, first note that every countable ordinal embeds (order-wise, and thus topologically wise) into the rational numbers, and so into the real numbers. So it suffices to find an ordinal whose rank is $\omega$.

What does that mean? It means that it takes $\omega$ steps to remove all the points (since countable ordinals can't have a non-empty kernel). If $\omega$ has rank $1$ and $\omega^2$ has rank $2$, and so on, then $\omega^\omega$ has rank $\omega$ (note: this is ordinal exponentiation!).

If you want something with a non-empty kernel, pick an embedding of $\omega^\omega$ into $[0,1]$, say $S$ and take $S\cup[2,3]$ to be your set.

Asaf Karagila
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  • Hi Asaf, question: isn't the Cantor-Bendixson rank of $\omega$ equal to 2? You can embed $\omega$in$\mathbb{Q}$ as the sequence ${1/n} \cup {0}$. Its first derived set is the singleton ${0}$. Its second derived set is the empty set. And that suggests that the CB-rank should be 2, not 1. (Finite sets have CB-rank 1.) I can see your argument showing how every successor ordinal arising as a CB-rank. // This is also the same comment as https://math.stackexchange.com/questions/2632501/for-any-countable-ordinal-alpha-there-is-some-closed-set-of-reals-whose-cant#comment5644129_2633068 – Willie Wong May 05 '20 at 16:31
  • No, that's $\omega+1$, since it contains the limit. $\omega$ is a sequence without its limit. Also, I was meaning to write you an email with all kind of news. The most important of which, I guess, is that I finally got to try Whistle Pig, it was great. – Asaf Karagila May 05 '20 at 16:32
  • Oh, are you embedding $\omega$ as ${1, 2, 3, \ldots}$? This way it will have CB rank 1. But what about on a finite interval? You talked about embedding into $[0,1]$, but you cannot embed $\omega$ into $\mathbb{Q}\cap [0,1]$ without having a limit point. Are you sure you can embed $\omega^\omega$ into $[0,1]$ such that its CB rank is $\omega$ and not $\omega + 1$? – Willie Wong May 05 '20 at 16:54
  • (I'm obsessing over this because of https://mathoverflow.net/questions/359158/coming-up-with-a-rigorous-definition-for-a-riemann-like-sum-which-is-easier-to-c ) – Willie Wong May 05 '20 at 16:57
  • Any embedding of $\omega$ is not going to have the limit itself. The embedding is not closed. The closure would be an embedding of $\omega+1$, assuming that the limit exist. Also, of course I can arrange for the limit to not exist... map $n$ to the $n$th decimal expansion of your favourite irrational in the interval. :-) – Asaf Karagila May 05 '20 at 17:11
  • I guess the issue is that I want to abuse the notion of CB-Rank a bit. I don't just want to talk about the image of $\omega$ as a topological subspace. I want to think about it as a subset. (In the linked question in my comment the ambient interval is important.) In any case, I found a pdf where Andres claims that the CB rank of a (closed) subset of [0,1] cannot be a limit ordinal. Maybe I'll just tweet him. – Willie Wong May 05 '20 at 18:41
  • That is true, a closed subset. But if you want to talk about an arbitrary countable subset, as an image of $\omega$, then it can be any set of rationals, sure. For a closed subset you will always have a last step, which is the "last step of limits". But if you talk about an arbitrary countable set, in what sense you differ between ${1/n\mid n>1}$ from ${0}\cup{1/n\mid n>1}$? – Asaf Karagila May 05 '20 at 18:48
  • I am not sure we are agreeing or disagreeing. My point is simply that as a topological space with the obvious induced topology, ${1/n}$ has CB-rank 1. Considered as a subset of of $[0,1]$, however, ${1/n}$ and ${0} \cup {1/n}$ both have CB-rank 2. (The CB-derivative of non-closed set does not need to be contained in the set.) – Willie Wong May 05 '20 at 18:58
  • Ohh, you're thinking about them as subsets, not as spaces. Okaaaay. That was unclear to me. :-) – Asaf Karagila May 05 '20 at 18:59
  • Yes, that's how I interpreted the original question here too, actually. Hence my confusion. – Willie Wong May 05 '20 at 19:00
  • I am very new in ordinal numbers, and I have two questions. Is Cantor-Bendixson rank of a countable subset of real numbers countable? And given any countable ordinal, can we find a countable subset of the real whose Cantor-Bendixson rank is the given ordinal? – Random Feb 28 '23 at 10:05
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    @Random: The first one is obvious, since we are removing points so at most after countably many steps we had to remove all of them (or find a set without isolated points). The second one is not so obvious, but still yes. First you need to show that you can always find a countable ordinal with a given rank, and then show that any countable ordinal embeds into the real numbers. – Asaf Karagila Feb 28 '23 at 10:35