Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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