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Given a set $P$ of real numbers, its derived set is the set of all accumulation points - $a\in P^\prime$ if every open set containing $a$ also contains an infinite number of points from $P$ (equivalently, at least one point from $P$ different than $a$).

Cantor defined $P^0=P$ and $P^k$ to be the derived set of $P^{k-1}$. Then he had the wonderful idea to continue this construction to $P^\omega = \bigcap_{k} P^k$, and then we have $P^{\omega +1}$ being the derived set of $P^\omega$ and so on.

The question is whether it is interesting to continue into the ordinals in this manner. i.e. is there a set $P$ such that $P^\omega \ne P^{\omega +1}$. Is there an example?

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You can embed any countable ordinal in the rational numbers. I think this can get you interesting results in the derived set of any countable ordinal depth; for instance, a set of order type $\omega^\omega$ should do what you want.

To construct one, start with an $\omega$ between 0 and 1 (take, say, $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots$). Now add an $\omega^2$ between 1 and 2 by squeezing what you did in the first interval between 1 and $1\frac{1}{2}$, then again between $1\frac{1}{2}$ and $1\frac{2}{3}$, etc. Next make an $\omega^3$ in the next interval... you get the point. This gives you a set of order type $\omega^\omega$ but it stretches to infinity - you can squash it all though to reside in a finite interval, and this should get you a new accumulation point at step $\omega+1$.

EDIT: Here's a more explicit version. Let $S_0=\{1\}$ and think of $S_0$ as residing in the half-open interval $(0,1]$. Inductively, suppose you have $S_{n-1}$ constructed in the interval $(n-1,n]$. Construct $S_n$ by linearly mapping the interval $(n-1,n]$ onto the interval $\left(n, n+\frac{1}{2}\right]$ and taking the image of $S_{n-1}$, then again map that interval onto $\left(n+\frac{1}{2}, n+\frac{2}{3}\right]$ and take the image of $S_{n-1}$, and so on $\omega$ times mapping the same set into the intervals $\left(n+\frac{k}{k+1}, n+\frac{k+1}{k+2}\right]$. This gives you $S_n$; now let $S = \cup S_n$ be the union of all those sets.

For example, $S_1 = \{1\frac{1}{2}, 1\frac{2}{3}, 1\frac{3}{4}, \ldots\}$. $S_2$ is made up of infinitely many sequences converging on the points $2\frac{1}{2}, 2\frac{2}{3}$ etc.

You'll observe that $S_0$ has order type $1$, $S_1$ has order type $\omega$, $S_2$ has order type $\omega^2$ and generally $S_n$ has order type $\omega^n$. $S$ therefore has order type $\omega^\omega$.

You'll also observe that $S_0' = \emptyset$ while $S_1' = S_0+1$ (a copy of $S_0$ shifted 1 to the right), $S_2' = S_1+1$ since $S_2$ has accumulation points exactly where $S_1$ has elements (shifted 1 to the right, of course), and so on and so forth. Thus $S^m$ is non-empty for all natural $m$ (in fact, $S_m$ is just $S+m$), however $S^\omega$ is empty.

To remedy that, map the half-line $[0,\infty)$ monotonically onto a finite interval such as $[0,1)$ (e.g. let $t \mapsto \frac{t}{t+1}$) and let $P$ be the image of $S$ under that mapping. $P$ has an accumulation point at $1$, and that accumulation point remains also an accumulation point of $P'$, $P''$ and generally $P^m$ for any natural $m$. Therefore $P^\omega=\{1\}$ while $P^{\omega+1}=\emptyset$ as desired.

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I'm afraid I can't see how it answers my question. Most importantly, what is P? Is this the set you constructed? What are its accumulation points? – Gadi A May 25 '11 at 6:04
Sure, this is P. You can try to figure out the derived sets for yourself but I'll also add more details later. – Alon Amit May 25 '11 at 16:14
Edited. Hope it's clear now. – Alon Amit May 25 '11 at 20:28

For any topological space $X$, the Cantor-Bendixon derivatives of $X$ are defined by transfinite induction:

  1. $X^0 = X$;
  2. $X^{\alpha+1} = (X^{\alpha})'$;
  3. $X^{\gamma} = \cap_{\alpha\lt \gamma}X^{\alpha}$ for limit ordinals $\gamma$.

By simple cardinality arguments, there must exist a least ordinal $\alpha$ such that $X^{\alpha} = X^{\alpha+1}$. This is called the Cantor-Bendixon rank of $X$.

As noted in JDH's answer to this Math Overflow question, every ordinal is the Cantor-Bendixon rank of some topological space.

For $X$ contained in $\mathbb{R}$, the Cantor-Bendixon rank is always countable, but for every countable ordinal $\alpha$, there is a closed subset $X$ of $\mathbb{R}$ with $X^{\alpha}$ a singleton. See for example this write-up by Harold Simmons. So $$\mathrm{sup}\{\text{CB rank}(X)\mid X\subseteq \mathbb{R}\} = \omega_1,$$ where $\omega_1$ is the first uncountable ordinal.

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As to the question of whether this is "interesting", even Cantor's original application on sets of uniqueness for trigonometric series (see Alekos Kechris's wonderful survey here: requires the consideration of closed sets of arbitrarily large countable rank. – Andrés E. Caicedo May 25 '11 at 20:42

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