Please see my Edited version at the end of the post.



My definition of Cantor Set is just like that of wikipedia.

That is, $C=[0,1]\setminus \bigcup_{i=1}^{\infty} \bigcup_{k=0}^{3^{i-1}-1} (\frac{3k+1}{3^i}, \frac{3k+2}{3^i})$.

With this definition, i have shown that $C$ is compact, perfect, equipotent with $2^{\aleph_0}$ and contains no openset. (i.e. Basic properties of Cantor set)

I preferred this definition to another since this definition is simple and strictly written in first-order logic.

Let $C_n = [0,1]\setminus \bigcup_{i=1}^n \bigcup_{k=0}^{3^{i-1}-1} (\frac{3k+1}{3^i},\frac{3k+2}{3^i})$.

Then $\bigcap_{n\in \mathbb{N}} C_n = C$.

Here, how do i prove that $C_n$ is a disjoint union of $2^n$ intervals, each of length $3^{-n}$?

(To make it clear, intervals here refer to closed connected sets)

=========================== EDIT:

This is not actually i meant, but this is exactly the same as what i wanted to prove anyway..

Let $A_0=B_0=[0,1]$. Define $\{A_n\}$ recursively such as; $A_{n+1}=\frac{A_n}{3} \cup (\frac{2}{3} + \frac{A_n}{3})$.

Now, define $B_n=[0,1]\setminus \bigcup_{i=1}^n \bigcup_{k=0}^{3^{i-1}-1} (\frac{3k+1}{3^i},\frac{3k+2}{3^i})$, $\forall n\in \mathbb{Z}^+$.

How do i prove $A_n=B_n$?

  • $\begingroup$ I feel very uncomfortable to define Cantor Set as "Removing the middle part constantly", since it doesn't seem to be written in first-order logic (even though it is).. Please, if 'removing the middle part' can be precisely written in mathematical language, let me know.. $\endgroup$
    – Katlus
    Nov 29, 2012 at 15:38
  • $\begingroup$ The Cantor set is not a disjoint countable union of intervals, but its complement is. Because the Cantor set is an intersection of closed sets its complement is open. And every open set in $\mathbb R$ is a disjoint union of open intervals. $\endgroup$
    – kahen
    Nov 29, 2012 at 15:42
  • $\begingroup$ @kahen I'm sorry that the title is wrong but i didn't mean Cantor Set itself but $C_n$ at the end of my post. I don't know how to call this $C_n$ $\endgroup$
    – Katlus
    Nov 29, 2012 at 15:45
  • $\begingroup$ Well, you could use recursion if you can show that $C_n$ is made out of two small $C_{n-1}$s. $\endgroup$ Nov 29, 2012 at 15:53
  • $\begingroup$ @Karolis That is exactly what i tried, but i found it really hard to prove.. Please help $\endgroup$
    – Katlus
    Nov 29, 2012 at 15:55

2 Answers 2


I don't have enough reputation to post a comment, so i'll write it here.

There's a little typo in the definition of the Cantor-set (first formula), the interval should be $\left(\frac{3k+1}{3^i},\frac{3k+2}{3^i}\right)$

I think you can prove it by induction using the recursive formula of the wikipedia article : $C_n=\frac{C_{n-1}}{3}\cup\left(\frac{2}{3}+\frac{C_{n-1}}{3}\right)$ I hope this is helpful.

EDIT : I was typing when other comments came up. So here are some details you need.

By induction, assume that $C_{n-1}$ is the disjoint union of $2^{n-1}$ intervals in $[0,1]$ each of them with lenght $\frac{1}{3^{n-1}}$.

On one hand, $A_n:=\frac{C_{n-1}}{3}$ is a subset of $[0,\frac{1}{3}]$ and is again a disjoint union of $2^{n-1}$ intervals but of lenght $\frac{1}{3}\cdot\frac{1}{3^{n-1}}=\frac{1}{3^{n}}$

On the other hand, $B_n:=\frac{2}{3}+\frac{C_{n-1}}{3}$ is a subset of $[\frac{2}{3},1]$ and is as well a disjoint union of $2^{n-1}$ intervals of lenght $\frac{1}{3^{n}}$.

Conclusion : $C_n$ is the union of $A_n$ and $B_n$ which is disjoint and hence composed of $2\cdot2^{n-1}=2^n$ intervals of the desired lenght.

  • $\begingroup$ @Thaibaut My main problem is i cannot prove the above recursive formula. (That is, $C_{n+1}=\frac{C_n}{3} \cup (\frac{2}{3} + \frac{C_n}{3})$). And i think that definition of $C_n$ in my post and that in wikipedia are different. I believe wikipedia defined $C_n$ as 'removing the middle' argument. What i want to show is that $C_n$ in wikipedia and $C_n$ in my post are actually equal. $\endgroup$
    – Katlus
    Nov 29, 2012 at 16:23
  • $\begingroup$ See my edited post. $\endgroup$
    – Katlus
    Nov 29, 2012 at 16:41
  • $\begingroup$ Sorry about the missunderstanding then. I'll think about it a little and reply later. $\endgroup$ Nov 29, 2012 at 17:10
  • 1
    $\begingroup$ ajur.uni.edu/v5n2/Soltanifar%20pp%209-12.pdf $\endgroup$
    – Katlus
    Nov 29, 2012 at 20:25

In $\mathbb{R}^2$, for any two Cantor sets $A$ and $B$ we can find a homeomorphism $h \colon \mathbb{R}^2 \to \mathbb{R}^2$ which carries $h(A)$ onto $B$. This essentially says they're equivalently embedded. I'm not sure this directly answers your question but in $\mathbb{R}^2$, talk of `the' Cantor set isn't necessarily correct, however upto homeomorphism of the ambient space we do only have one equivalence class of Cantor set.


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