Alternative definition of the Cantor Set

Question

The Cantor Set $C$ can be defined as taking the interval $C_0=[0,1]$

then $C_1=[0, \frac{1}{3}] \cup [\frac{2}{3},1]$

then $C_2=[0, \frac{1}{9}] \cup [\frac{2}{9},\frac{1}{3}] \cup [\frac{4}{9},\frac{5}{9}] \cup [\frac{2}{3},\frac{7}{9}] \cup [\frac{8}{9},1]$

and so on... Basically this definition involves splitting up the interval and then taking every other segment then $C$ is the intersection of all these so it still creates the same Cantor set in the end.

My question is what is the formula for $C_n$?

See:https://math.stackexchange.com/questions/2461773/what-is-the-cantor-set-how-do-i-write-it-mathematically — 1ENİGMA1, Dec 01 '17 at 12:36
The formula given in your linked post wouldn't give $C_2$ in my post — dahaka5, Dec 01 '17 at 12:41
I know it is not the usual way that the Cantor set is constructed, but surely after you do this and then take the intersection of all them you will get the same Cantor set no? — dahaka5, Dec 01 '17 at 12:50
The intersection of $C_1$ and $C_2$ gives the intervals of the second step of the „usual“ method, so yes. — T'x, Dec 01 '17 at 12:54

score 2 · Accepted Answer · answered Dec 01 '17 at 12:42

2

$C_n$ is the union over $[k/3^n,(k+1)/3^n]$ where $k = 0,2,...$ up to the largest even number less than $3^n$.

answered Dec 01 '17 at 12:42

T'x

578

Your formula is incorrect with $n=2$. – Akira Jan 21 '19 at 07:10
what do you think is incorrect? – T'x Jan 21 '19 at 07:17
For $n=2$, according to your formula: $C_2= \bigcup_{k=0}^{\lfloor 3^2/2\rfloor}\left[\frac{2k}{3^2},\frac{2k+1}{3^2}\right]= \bigcup_{k=0}^{4}\left[\frac{2k}{9},\frac{2k+1}{9}\right]=[0,1/9] \cup [2/9,3/9] \cup [4/9,5/9] \cup [6/9,7/9]$ $\cup [8/9,1]$. We know that $[4/9,5/9]$ is not an interval of $C_2$. – Akira Jan 21 '19 at 07:24
be careful! to get the cantor set you still need to take the intersections of all $C_n$, which is also stated in the question. – T'x Jan 21 '19 at 07:27
But you said "$C_n$ is the union over..." – Akira Jan 21 '19 at 07:28
yes, of course, since after that the intersection gives $C$. – T'x Jan 21 '19 at 07:29
It will be much more clearer if you edit your answer to reflect that. – Akira Jan 21 '19 at 07:30
I don't see any problem, since the original question asks for the definition of $C_n$ and not the cantor set $C$. – T'x Jan 21 '19 at 08:24

CiaPan · Answer 2 · 2017-12-01T13:12:29.797

2

It's $$\bigcap_{n=0}^\infty\left(\bigcup_{i=0}^{\lfloor 3^n/2\rfloor}\left[\frac{2i}{3^n},\frac{2i+1}{3^n}\right]\right)$$

edited Dec 01 '17 at 13:12

answered Dec 01 '17 at 13:04

CiaPan

13,049

Hm, I think you are missing some intervals? Whats with $[8/9,1]$ or $[26/27,1]$? – T'x Dec 01 '17 at 13:11
@T'x You're right. Sometimes I write faster than I think. And sometimes I submit faster than I read what I wrote. – CiaPan Dec 01 '17 at 13:14

Stéphane Jaouen · Answer 3 · 2024-03-28T18:29:38.463

"The formula" is the one we want to give. To understand the idea of the following construction, I refer to my answer here.

Let $$C_0=D_0:=[0,1]$$ $$C_1=D_1=[0,\frac13]\cup[\frac23,1]$$

The idea I'm going to try to formalize is that if we enlarge three times from step to step, give or take, the non-integer part of $x$ must belong to $C_1$.

Let then$$D_2:=\{x\in [0,1]|9x-\left\lfloor 9x\right\rfloor\in C_1\}$$ $$C_2:=C_1\cap D_2$$ and, by induction, $$\forall n\ge3, D_n:=\{x\in [0,1]|3^nx-\left\lfloor 3^nx\right\rfloor\in C_1\}$$ $$\boxed{C_n:=C_{n-1}\cap D_n}$$

Alternative definition of the Cantor Set

3 Answers3

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