# Cantor set

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The Cantor set $\mathcal{C}$ is a subset of the real numbers that exhibits a number of interesting and counter-intuitive properties. It is among the simplest examples of a fractal. Topologically, it is a closed set, and also a perfect set. Despite containing an uncountable number of elements, it has Lebesgue measure equal to $0$.

The Cantor set can be described recursively as follows: begin with the closed interval $[0,1]$, and then remove the open middle third segment $(1/3,2/3)$, dividing the interval into two intervals of length $\frac{1}{3}$. Then remove the middle third of the two remaining segments, and remove the middle third of the four remaining segments, and so on ad infinitum. $[asy] int max = 7; real thick = 0.025; void cantor(int n, real y){ if(n == 0) fill((0,y+thick)--(0,y-thick)--(1,y-thick)--(1,y+thick)--cycle,linewidth(3)); if(n != 0) { cantor(n-1,y); for(int i = 0; i <= 3^(n-1); ++i) fill( ( (1.0+3*i)/(3^n) ,y+0.1)--( (1.0+3*i)/(3^n) ,y-0.1)--( (2.0+3*i)/(3^n) ,y-0.1)--( (2.0+3*i)/(3^n) ,y+0.1)--cycle,white); } } for(int i = 0; i < max; ++i) cantor(i,-0.2*i); [/asy]$

Equivalently, we may define $\mathcal{C}$ to be the set of real numbers between $0$ and $1$ with a base three expansion that contains only the digits $0$ and $2$ (including repeating decimals).

Another equivalent representation for $\mathcal{C}$ is: Start with the interval $[0,1]$, then scale it by $\frac{1}{3}$. Then join it with a copy shifted by $\frac{2}{3}$, and repeat ad infinitum.

Using this representation, $\mathcal{C}$ can be rendered in LaTeX: $$\newcommand{\cantor}{#1\phantom{#1}#1}\cantor{\cantor{\cantor{\cantor{.}}}}$$  $$\newcommand{\cantor}{#1\phantom{#1}#1}\cantor{\cantor{\cantor{\cantor{.}}}}$$ 

A distorted version of $\mathcal{C}$ can be found by repeatedly applying the function $f(x)=a(x-\frac{1}{2})^2+1-\frac{a}{4},a>4$, and keeping the values of x for which the values always remain bounded. This constructs $\mathcal{C}$ by repeatedly removing the middle. This works since for $x\notin [0,1]$ the values will always diverge, and the values of $x$ for which $f(x)\in [0,1]$ is the union of intervals $\left[0,\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{1}{a}}\right]\cup \left[\frac{1}{2}+\sqrt{\frac{1}{4}-\frac{1}{a}},1\right]$, which are disjoint when $a>4$. -- EVIN-