Difference between revisions of "Cantor set"
(4 intermediate revisions by the same user not shown) | |||
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''. | 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{.}}}</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- | ||
{{stub}} | {{stub}} |
Latest revision as of 16:32, 18 June 2020
The Cantor set 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
.
The Cantor set can be described recursively as follows: begin with the closed interval , and then remove the open middle third segment
, dividing the interval into two intervals of length
. 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]](http://latex.artofproblemsolving.com/8/e/7/8e7092d9d51c41b46f2c5eb4f1c7114e576d8d25.png)
Equivalently, we may define to be the set of real numbers between
and
with a base three expansion that contains only the digits
and
(including repeating decimals).
Another equivalent representation for is: Start with the interval
, then scale it by
. Then join it with a copy shifted by
, and repeat ad infinitum.
Using this representation, can be rendered in LaTeX:
$$\newcommand{\cantor}[1]{#1\phantom{#1}#1}\cantor{\cantor{\cantor{\cantor{.}}}}$$
A distorted version of can be found by repeatedly applying the function
, and keeping the values of x for which the values always remain bounded. This constructs
by repeatedly removing the middle. This works since for
the values will always diverge, and the values of
for which
is the union of intervals
, which are disjoint when
. -- EVIN-
This article is a stub. Help us out by expanding it.