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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Let us take any set $E$ in a metric space. Is it possible to find a $E'$ and $E''$ such that $E$ and $E'$ are non empty but $E''$ is empty.

If so, Let us now say $E^n$ is $n$th derived set of $E$. Is a generalization possible when we can find a set $E$ such that $E^n$ is nonempty but $E^{n+1}$ is empty?

P.S. By arbitrary I mean it is at least possible to find one such metric space and one such Set $E$, where the first statement might be satisfied.

share|cite|improve this question
What do you call "unit ball in the plane"? Is it $\,S^1:=\{(x,y)\in\Bbb R^2\;;\;x^2+y^2=1\}\,$? Because if it is then $\,E=E'=E''\,$, if I'm not wrong...Wait, the derived set is not the boundary but the set of all limit points (the same as cluster or accumulation points, in this case) of $\,E\,$ I wrong? – DonAntonio Jan 9 '13 at 13:20
(Assuming that by $E^\prime$ you are denoting the Cantor-Bendixson derivative of $E$.) In an arbitrary metric space?! Please give an example of a subset of $E$ of $\mathbb{N}$ (which is metric under the discrete metric) with $E^\prime \neq \emptyset$. – inactive... for now Jan 9 '13 at 13:22
@Freddy: this was not an example for derivatives, only for boundaries (written 2 times above already) – Ilya Jan 9 '13 at 13:38
I see @Freddy , yet Ilya already explained that he thought $\,E'\,$ is the boundary of $\,E\,$ , not what we thought, I guess. – DonAntonio Jan 9 '13 at 13:38
@Freddy , then I think we should first focus on $\,N=1\,$ and if we find an example there then we could try to generalize to higher dimensions. If for the apparently easier one we can't find an example I don't think there's point to look higher... – DonAntonio Jan 9 '13 at 13:41
up vote 4 down vote accepted

In case the metric space has at least one cluster point, you can take as $E$ any sequence that clusters around that point and doesn't cluster anywhere else. The derived set of the sequence will be just the cluster point and obviously the derived set of the one-element set is empty. E.g., take $(1/n)_{n \geq 1} \subset \mathbb R$.

You can iterate this process to higher dimension at least in $\mathbb R^n$. For example, take $E = (1/n, 1/m)_{n,m \geq 1} \subset \mathbb R^2$. Then you have $E' = (1/n, 0)_{n \geq 1} \cup (0, 1/m)_{m \geq 1}$ and $E'' = {(0,0)}$. And so on.

share|cite|improve this answer
Thank You. It took a while to understand that generalization was for derivative rather than merely for dimension. – 007resu Jan 9 '13 at 14:05
Similar sets in the case of the metric space ${\mathbb R}$ (indeed, the following sets have "the same" linear orderings as those Marek gave) can be obtained by looking at the zeros of $\sin \left( 1/x \right),$ the zeros of $\sin \left( \frac{1}{\sin \left( 1/x \right)} \right),$ etc. More formally, let $f_{1}(x) = \sin \left( \frac{1}{x} \right)$ and, for each integer $n \geq 1,$ let $f_{n+1}(x) = \left( \frac{1}{f_{n}(x)} \right),$ and consider the sets of zeros of these functions. – Dave L. Renfro Jan 9 '13 at 15:20

$\newcommand{\Cantor}{\mathcal{C}}$A very nice metric space to accomplish this is $\Cantor = \{0,1 \}^{\mathbb{N}}$ consisting of all infinite binary sequences. (I'll take $\mathbb{N} = \{ 0 , 1 , 2 , \ldots \}$ in the sequel.) A metric for this space is given as follows: for $\mathbf{x} = ( x_n )_{n \in \mathbb{N}} , \mathbf{y} = ( y_n )_{n \in \mathbb{N}} \in \mathcal{N}$ we define $$d ( \mathbf{x} , \mathbf{y} ) = \begin{cases} 0, &\text{if }\mathbf{x} = \mathbf{y} \\ 2^{-(n+1)}, &\text{if }\mathbf{x} \neq \mathbf{y}, n = \min \{ n \in \mathbb{N} : x_n \neq y_n \}. \end{cases}$$

A basis for this space is obtained by taking all sets of the form $$[s] = \{ \mathbf{x} \in \Cantor : s \sqsubset \mathbf{x} \}$$where $s$ is a finite binary sequence (and $s \sqsubset \mathbf{x}$ means that $s$ is an initial segment of $\mathbf{x}$).

(This space is homeomorphic to the usual Cantor ternary set, and is, unsurprisingly, called the Cantor space.)

A purely formal construction of the required sets is as follows: For each $n$ let $$E_n = \{ \mathbf{x} \in \Cantor : \mathbf{x} \text{ has at most }n \text{ }1\text{s} \}$$ These sets are easily seen to be closed. (If $\mathbf{y} \notin E_n$, then $\mathbf{y}$ has at least $n+1$ $1$s, so let $\ell$ be the coordinate of the $(n+1)$st $1$, and let $s$ be the initial segment of $\mathbf{y}$ up to (and including) the $\ell$th coordinate. Then every $\mathbf{z} \in [s]$ contains at least $n+1$ $1$s, and is not in $E_n$.)

We claim that $( E_{n+1} )^\prime = E_n$. Note that if $\mathbf{x} \in E_{n+1} \setminus E_n$, then $\mathbf{x}$ has exactly $n+1$ $1$s in it. Let $\ell$ denote the coordinate of the $(n+1)$st $1$ in $\mathbf{x}$, and let $s$ denote the initial segment of $\mathbf{x}$ up to, and including, the $\ell$th coordinate. Then $E_{n+1} \cap [s] = \{ \mathbf{x} \}$, and so $\mathbf{x}$ is an isolated point of $E_{n+1}$. This implies that $(E_{n+1})^\prime \subseteq E_n$. Conversely, if $\mathbf{x} \in E_n \subseteq E_{n+1}$ then given any initial segment $s$ of $x$ we can find a coordinate $\ell > \mathrm{length} (s)$ such that $x_\ell = 0$. Define $\mathbf{y}$ so that $$y_n = \begin{cases} x_n, &\text{if }n \neq \ell \\ 1, &\text{if }n = \ell. \end{cases}$$ Then $\mathbf{y} \in E_{n+1} \cap [s]$ and is distinct from $\mathbf{x}$. Therefore $\mathbf{x} \in E_{n+1}^\prime$.

(One can go quite a bit further and construct, for each $\alpha < \omega_1$ a closed $E \subseteq \Cantor$ such that $E^{(\alpha)} \neq \emptyset$ but $E^{(\alpha+1)} = \emptyset$.)

The nice thing about $\Cantor$ is that if $X$ is any nonempty perfect Polish space (i.e. a separable complete metric without isolated points) includes a closed homeomorphic copy of $\Cantor$, and so every nonempty perfect Polish space has closed subsets of the kind desired.

share|cite|improve this answer

Your Answer


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.