Difference between revisions of "Cantor set"

Latest revision as of 15:32, 18 June 2020

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: $\newcommand{\cantor}[1]{#1\phantom{#1}#1}\cantor{\cantor{\cantor{\cantor{.}}}}$ $$\newcommand{\cantor}[1]{#1\phantom{#1}#1}\cantor{\cantor{\cantor{\cantor{.}}}}$$

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-

This article is a stub. Help us out by expanding it.

@@ Line 21: / Line 21: @@
 Another equivalent representation for <math>\mathcal{C}</math> is: Start with the interval <math>[0,1]</math>, then scale it by <math>\frac{1}{3}</math>. Then join it with a copy shifted by <math>\frac{2}{3}</math>, and repeat ''ad infinitum''.
-Using this representation, <math>\mathcal{C}</math> can be rendered in [[LaTeX]]: <cmath>\newcommand{\cantor}{#1\phantom{#1}#1}\cantor{\cantor{\cantor{\cantor{.}}}}</cmath>
+Using this representation, <math>\mathcal{C}</math> can be rendered in [[LaTeX]]: <cmath>\newcommand{\cantor}[1]{#1\phantom{#1}#1}\cantor{\cantor{\cantor{\cantor{.}}}}</cmath>
+<code>
+$$\newcommand{\cantor}[1]{#1\phantom{#1}#1}\cantor{\cantor{\cantor{\cantor{.}}}}$$
+</code>
+A distorted version of <math>\mathcal{C}</math> can be found by repeatedly applying the function <math>f(x)=a(x-\frac{1}{2})^2+1-\frac{a}{4},a>4</math>, and keeping the values of x for which the values always remain bounded. This constructs <math>\mathcal{C}</math> by repeatedly removing the middle. This works since for <math>x\notin [0,1]</math> the values will always diverge, and the values of <math>x</math> for which <math>f(x)\in [0,1]</math> is the union of intervals <math>\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]</math>, which are disjoint when <math>a>4</math>. -- EVIN-
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Difference between revisions of "Cantor set"

Latest revision as of 15:32, 18 June 2020