Editing Cantor set (section)

==Construction and formula of the ternary set==
The Cantor ternary set <math>\mathcal{C}</math> is created by iteratively deleting the [[open interval|''open'']] middle third from a set of line segments. One starts by deleting the open middle third <math display="inline">\left(\frac{1}{3}, \frac{2}{3}\right)</math> from the [[interval (mathematics)|interval]] <math>\textstyle\left[0, 1\right]</math>, leaving two line segments: <math display="inline">\left[0, \frac{1}{3}\right]\cup\left[\frac{2}{3}, 1\right]</math>.  Next, the open middle third of each of these remaining segments is deleted, leaving four line segments: <math display="inline">\left[0, \frac{1}{9}\right]\cup\left[\frac{2}{9}, \frac{1}{3}\right]\cup\left[\frac{2}{3}, \frac{7}{9}\right]\cup\left[\frac{8}{9}, 1\right]</math>.
The Cantor ternary set contains all points in the interval <math>[0,1]</math> that are not deleted at any step in this [[ad infinitum|infinite process]]. The same construction can be described recursively by setting
: <math>C_0 := [0,1]</math>
and
: <math>C_n := \frac{C_{n-1}} 3 \cup \left(\frac 2 {3} +\frac{C_{n-1}} 3 \right) = \frac13 \bigl(C_{n-1} \cup \left(2 + C_{n-1} \right)\bigr)</math>
for <math>n \ge 1</math>, so that
: <math>\mathcal{C} :=</math> [[Set-theoretic limit#Monotone sequences|<math>{\color{Blue}\lim_{n\to\infty}C_n}</math>]] <math>= \bigcap_{n=0}^\infty C_n = \bigcap_{n=m}^\infty C_n </math> &thinsp; for any &thinsp; <math>m \ge 0</math>.

The first six steps of this process are illustrated below.

[[Image:Cantor set in seven iterations.svg|729px|class=skin-invert|
Cantor ternary set, in seven iterations]]

Using the idea of self-similar transformations, <math>T_L(x)=x/3,</math> <math>T_R(x)=(2+x)/3</math> and <math>C_n =T_L(C_{n-1})\cup T_R(C_{n-1}),</math> the explicit closed formulas for the Cantor set are<ref>{{cite journal | first=Mohsen | last=Soltanifar | title=A Different Description of A Family of Middle-a Cantor Sets | journal=American Journal of Undergraduate Research | volume=5 | issue=2 | pages=9–12 | date=2006 | doi=10.33697/ajur.2006.014| doi-access=free }}</ref>
: <math>\mathcal{C}=[0,1] \,\setminus\, \bigcup_{n=0}^\infty \bigcup_{k=0}^{3^n-1} \left(\frac{3k+1}{3^{n+1}},\frac{3k+2}{3^{n+1}} \right)\!,</math>
where every middle third is removed as the open interval <math display="inline">\left(\frac{3k+1}{3^{n+1}},\frac{3k+2}{3^{n+1}}\right)</math> from the [[closed interval]] <math display="inline">\left[\frac{3k+0}{3^{n+1}},\frac{3k+3}{3^{n+1}}\right] = \left[\frac{k+0}{3^n},\frac{k+1}{3^n}\right]</math> surrounding it, or
: <math>\mathcal{C}=\bigcap_{n=1}^\infty \bigcup_{k=0}^{3^{n-1}-1} \left( \left[\frac{3k+0}{3^n},\frac{3k+1}{3^n}\right] \cup \left[\frac{3k+2}{3^n},\frac{3k+3}{3^n}\right] \right)\!,</math>
where the middle third <math display="inline">\left(\frac{3k+1}{3^n},\frac{3k+2}{3^n}\right) </math> of the foregoing closed interval <math display="inline">\left[\frac{k+0}{3^{n-1}},\frac{k+1}{3^{n-1}}\right] = \left[\frac{3k+0}{3^n},\frac{3k+3}{3^n}\right]</math> is removed by intersecting with <math display="inline">\left[\frac{3k+0}{3^n},\frac{3k+1}{3^n}\right] \cup \left[\frac{3k+2}{3^n},\frac{3k+3}{3^n}\right]\!.</math>

This process of removing middle thirds is a simple example of a [[finite subdivision rule]].  The complement of the Cantor ternary set is an example of a [[fractal string]].

[[File:Cantor set binary tree.svg|400px]]

In arithmetical terms, the Cantor set consists of all [[real number]]s of the [[unit interval]] <math>[0,1]</math> that do not require the digit 1 in order to be expressed as a [[Ternary numeral system|ternary]] (base 3) fraction. As the above diagram illustrates, each point in the Cantor set is uniquely located by a path through an infinitely deep [[binary tree]], where the path turns left or right at each level according to which side of a deleted segment the point lies on. Representing each left turn with 0 and each right turn with 2 yields the ternary fraction for a point. ''Requiring'' the digit 1 is critical: <math display="inline">\frac{1}{3}</math>, which is included in the Cantor set, can be written as <math display="inline">0.1</math>, but also as <math display="inline">0.0\bar{2}</math>, which contains no 1 digits and corresponds to an initial left turn followed by infinitely many right turns in the binary tree.

=== Mandelbrot's construction by "curdling" ===
In ''[[The Fractal Geometry of Nature]]'', mathematician [[Benoit Mandelbrot]] provides a whimsical thought experiment to assist non-mathematical readers in imagining the construction of <math>\mathcal{C}</math>. His narrative begins with imagining a bar, perhaps of lightweight metal, in which the bar's matter "curdles" by iteratively shifting towards its extremities. As the bar's segments become smaller, they become thin, dense slugs that eventually grow too small and faint to see.<blockquote>CURDLING: The construction of the Cantor bar results from the process I call curdling. It begins with a round bar. It is best to think of it as having a very low density. Then matter "curdles" out of this bar's middle third into the end thirds, so that the positions of the latter remain unchanged. Next matter curdles out of the middle third of each end third into its end thirds, and so on ad infinitum until one is left with an infinitely large number of infinitely thin slugs of infinitely high density. These slugs are spaced along the line in the very specific fashion induced by the generating process. In this illustration, curdling (which eventually requires hammering!) stops when both the printer's press and our eye cease to follow; the last line is indistinguishable from the last but one: each of its ultimate parts is seen as a gray slug rather than two parallel black slugs.<ref name=":3" /></blockquote>