For any countable ordinal $\alpha$, there is some closed set of reals whose Cantor-Bendixson rank is $\alpha$

Question

I want to prove that for any countable ordinal $\alpha$, there is some closed set $C\subset \mathbb R$ such that the Cantor-Bendixson rank of $C$ is $\alpha$.

I have not been able to create a successful construction and am out of ideas right now, but I still believe this should be true. Could anybody give a proof (or counterexample)?

Does it work?: Recursively take $C_1 = {0}\cup {1/n : n<\omega}$ and take $C_{\alpha+1}$ as replacing all isolated points of $C_\alpha$ to a sufficiently small copy of $C_1$. For limit $\delta$, take $C_\delta$ the union of all $C_\alpha$ for $\alpha<\delta$. — Hanul Jeon, Feb 02 '18 at 09:01
@HanulJeon I think that may work. But I don't see how to work out the details. Could you expand please? — CuriousKid7, Feb 02 '18 at 09:12
I realize that my construction does not guarantee $C_\omega$ is a closed set. — Hanul Jeon, Feb 02 '18 at 09:21
Related: Embedding ordinals in $\mathbb{Q}$ AND Countable ordinals are embeddable in the rationals $\Bbb Q$ -- proofs and their use of AC AND A canonical injection $f:\alpha\to[0,1]$ for an arbitrary countable ordinal $\alpha$. — Dave L. Renfro, Feb 02 '18 at 12:00

Henno Brandsma · Accepted Answer · 2018-02-03T07:29:32.863

For a countable ordinal $\alpha$, consider the (compact) space $X_\alpha = \omega^\alpha+1$ (ordinal exponentiation of course, so this is a countable ordinal) in the order topology. It's classical that the Cantor Bendixson rank of $X_\alpha$ is exactly $\alpha$. You could probably show it by transfinite induction. This paper has more details.

And every countable ordinal in the order topology is metrisable (second countable and normal) and embeds into $\mathbb{Q}$ (which contains an order isomorphic copy of any countable ordered space) and thus into the the reals.

This question and its answers give more info again.

The rank is $\alpha+1$, not $\alpha$. – Junyan Xu Apr 13 '18 at 01:43 — Junyan Xu, Apr 13 '18 at 01:43

For any countable ordinal $\alpha$, there is some closed set of reals whose Cantor-Bendixson rank is $\alpha$

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