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Difference between revisions of "Cantor set"
 
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The '''Cantor set''' <math>\mathcal{C}</math> is a [[subset]] of the [[real number]]s that exhibits a number of interesting and counter-intuitive properties. It is among the simplest examples of a [[fractal]].
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The '''Cantor set''' <math>\mathcal{C}</math> is a [[subset]] of the [[real number]]s that exhibits a number of interesting and counter-intuitive properties. It is among the simplest examples of a [[fractal]]. [[Topology|Topologically]], it is a [[closed set]], and also a [[perfect set]]. Despite containing an [[uncountable]] number of elements, it has [[Lebesgue measure]] equal to <math>0</math>.  
  
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The Cantor set can be described [[recursion|recursively]] as follows: begin with the set [0,1] and then remove the ([[open interval | open]]) middle third, dividing the [[interval]] into two intervals of length <math>\frac{1}{3}</math>.  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''.
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The Cantor set can be described [[recursion|recursively]] as follows: begin with the [[closed interval]] <math>[0,1]</math>, and then remove the [[open interval | open]] middle third segment <math>(1/3,2/3)</math>, dividing the [[interval]] into two intervals of length <math>\frac{1}{3}</math>.  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''.
  
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Equivalently, we may define <math>\mathcal{C}</math> to be the set of real numbers between <math>0</math> and <math>1</math> with a [[base number | base]] three expansion that contains only the digits <math>0</math> and <math>1</math>.
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<center><asy>
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int max = 7; real thick = 0.025;
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void cantor(int n, real y){
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if(n == 0) fill((0,y+thick)--(0,y-thick)--(1,y-thick)--(1,y+thick)--cycle,linewidth(3));
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if(n != 0) {
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  cantor(n-1,y);
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  for(int i = 0; i <= 3^(n-1); ++i)
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  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);
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}
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}
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for(int i = 0; i < max; ++i)
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cantor(i,-0.2*i);
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</asy></center>
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Equivalently, we may define <math>\mathcal{C}</math> to be the set of real numbers between <math>0</math> and <math>1</math> with a [[base number | base]] three expansion that contains only the digits <math>0</math> and <math>2</math> (including [[0.999...|repeating decimals]]).
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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''.
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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>
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<code>
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$$\newcommand{\cantor}[1]{#1\phantom{#1}#1}\cantor{\cantor{\cantor{\cantor{.}}}}$$
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</code>
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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-
  
 
{{stub}}
 
{{stub}}

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-

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