# Cantor-Bendixson theorem proof

I am looking for a proof of Cantor-Bendixson theorem involving transfinite numbers (I am interested only in the case of real line).

I fact, I have already seen one but I have a trouble in understanding it. It was in the book "Baire - theory of discontinued functions". Baire firstly defines transfinite numbers in usual way ($\omega , \omega^\omega$ and so on) and defines $\Omega = \operatorname{lim}[\omega, \omega^\omega, \omega^{\omega^\omega}, \dots]$. After that he uses notion of derived set. He defines $P^\Omega$ as the intersection of all $P^{\omega^{\dots}}$.

When he proves that $P^\Omega$ is perfect he uses the fact : if interval $[C;D]$ doesn't has any points of $P^\Omega$ inside (thought it may have them as endpoints) there is a transfinite number $\alpha < \Omega$ such that $P^\alpha$ doesn't has any points inside $[C;D]$. And the proof relies on the fact that for every sequence of numbers $\alpha _{i} < \Omega$ there exists the transfinite number $\beta$ such that it is bigger than any $\alpha$ and less than $\Omega$. Which is not true, since those numbers might be $\omega, \omega^\omega, \omega^{\omega^\omega}, \dots$ .

Maybe there is some way to overcome those difficulties? (after that the proof s quite simple, so it's enough to only resolve this problem for a complete proof)

Thank you very much!

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As Asaf mentions in his answer, the upper bound of the ordinals you appear to be using is not $\omega_1$ (the least uncountable ordinal), but actually $\varepsilon_0$ (epsilon-naught), which is itself countable. One can construct examples of closed subsets of $\mathbb{R}$ whose Cantor-Bendixson rank is greater than $\varepsilon_0$, meaning that this exact approach (as determined by your definition of $\Omega$) cannot work (although the idea behind the proof certainly does work). –  Arthur Fischer Jan 5 '13 at 14:42

I am very doubtful that the proof of the Cantor Bendixson Theorem using Cantor Bendixson derivatives need needs to use the fact that every subsequence of $\{\omega, \omega^\omega, ...\}$ has a larger element in the sequence, which is not true as you pointed out.

A usual proof of the Cantor Bendixson Theorem using derivative goes as follows: For any set $X \subset \mathbb{R}$, let $X'$ be the set of limit points of $X$. Define by transfinite recursion on the the ordinals,

$X_0 = X$

$X_{\alpha + 1} = X_\alpha'$

$X_\lambda = \bigcap_{\gamma < \lambda} X_\gamma$ when $\lambda$ is a limit ordinals.

The claim is that there exists a countable ordinals $\beta$ (i.e. $\beta < \omega_1$) such that $X_\beta = X_{\beta + 1}$.

To prove this, fix $(U_n)_{n \in \mathbb{N}}$ to be a countable basis for $\mathbb{R}$. For any closed set $F$, define $N(F) = \{n \in \omega : U_n \cap F \neq \emptyset\}$. Since $F$ is closed, $\mathbb{R} - F = \bigcup_{n \notin N(F)} U_n$. Thus for any two closed sets $F \neq E$, $N(F) \neq N(E)$. Moreover, if $F \subseteq E$, then $N(F) \subseteq N(E)$. Hence it has been shown that if $F \subsetneq E$, then $N(F) \subsetneq N(E)$.

Now applying this to the sequence of closed sets, $(X_\alpha)$. For any $\gamma < \alpha$ such that $X_\alpha \subsetneq X_\gamma$, one has $N(X_\alpha) \subsetneq N(X_\gamma)$. Since $N(X_\alpha)$ are subsets of $\mathbb{N}$ and $\mathbb{N}$ is countable, you can not have a uncountable $\beta$ such that for all $\eta < \xi < \beta$, $N(X_\xi) \subsetneq N(X_\eta)$.

Hence it has been shown that there exists a countable $\beta$ such that $X_\beta = X_{\beta + 1}$. It is clear that $X_\beta$ has no isolated points. Thus $X_\beta$ is your perfect kernel.

There is a easier proof of Cantor Bendixson that does not use ordinals. Note that a condensation point is a point $x$ such that every neighborhood of $X$ is uncountable. Let $Z$ denote the set of condensation points of $X$. The claim is that $X = Z \cup C$ where $C$ is countable and $Z$ is perfect. Let $C = X - Z$. Let $U_n$ be a countable basis for $X$ (i.e. take a countable basis for $\mathbb{R}$ and intersect it with your closed set $X$). By definition of $Z$, $C$ is the union of all $U_n$ such that $U_n$ is countable. A countable union of countable set is countable, hence, $C$ is countable. $Z$ is perfect because for any $x \in X$, take any open set $x \in V$. By definition of $x \in Z$, $V$ contains uncountable many points. $C = X - Z$ is countable, so $V$ actually contains uncountable many points of $Z$. Thus $X = Z \cup C$, where $Z$ is perfect and $C$ is countable.

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I should remark, and it is implicit in my answer, that $\Omega$ was often used to denote $\omega_1$, the set of all countable ordinals. Since the OP did not quote the exact words from the book, I tend to believe that he did not realize that $\Omega$ was the set of countable ordinals, rather than $\varepsilon_0$. –  Asaf Karagila Jan 5 '13 at 14:01

But $\Omega$ is used to denote $\omega_1$, not $\varepsilon_0$. It is the first uncountable ordinal.

It follows that every sequence is bounded by some $\beta$, and the proof follows as wanted.

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