The Cantor Set

By Nick Wan and Elaine Wissuchek

Introduction to the Cantor Set

Cantor set is a special subset of the closed interval [0, 1] invented by a German mathematician Georg Cantor in 1883. In order to construct this set, we need to construct infinitely many subset of $[0, 1]$ inductively and take the intersection of all of them. Specifically:

Let $I_0 := [0, 1]$ . Remove the open middle third $(1/3, 2/3)$ from $I_0$ .

Let $I_1: = [0, 1/3]\cup[2/3, 1]$ , precisely $I_1 = I_0\setminus (1/3, 2/3)$ . Remove middle third open intervals $(1/9, 2/9)$ and $(7/9, 8/9)$ from the respective closed intervals whose union is $I_1$ .

Let $I_2: = [0, 1/9]\cup[2/9, 3/9]\cup[6/9, 7/9]\cup[8/9, 9/9]$ say, $I_2^1\cup I_2^2\cup I_2^3\cup I_2^4$ .

Continue this procedure of removal of open intervals from each closed interval $I^k_j$ where $1\leq k \leq 2^j$ whose union is $I_j$ where $j = 3, 4,..$ and get $I_{(j+1)}$ by taking the union of whatever is left from each $I^k_j$ after the removal of open intervals as before.

Define the Cantor set $C:= \cap_{n=1}^{\infty} I_n$ .

Another way to view the Cantor set is in terms of ternary expansions:

Given $x\in (0,1)$ a real number, there is a sequence of integers $(a_k)^{\infty}_{k=1}, a_k\in \{0,1,2\}$ with $k\in \mathbb{N}$ such that the series $\sum_{k=1}^{\infty} \frac{a_k}{3^k}$ converges to $x$ . In other words, we can write $x$ in a ternary form: $x=0.a_1 a_2 a_3 ...$ (base 3). For example, $\frac{1}{9}=(0.01000...)_3$ .

In the case or our Cantor set, any point $x$ in $C$ has the following ternary expansion:

$x=\sum_{i=1}^{\infty} \frac{d_k}{3^k}$ with all $d_k\in\{0,2\}$ from the set $a_k$ .

For example, the element of the Cantor set $x=\frac{1}{3}$ corresponds to $d_k=(0,2,2,2,2,2,2,...)$ and $x=\frac{2}{3}$ corresponds to $d_k=(2,0,0,0,0,0,0,...)$ .

Uncountability of the Cantor Set by Diagonalization

We can use Cantor’s diagonalization method to show that the Cantor set is not countable.

Let $(v_n)=\{ (d_k)_n : n \in \mathbb{N}\}$ be a sequence so that each $v_n$ represents an element of the Cantor set. Let all $(v_n) \in V$ . Assume that $f:\mathbb{N} \rightarrow V$ is a one-to-one correspondence between the natural numbers and the set $V$ so that the elements of the cantor set are countable.

Consider $(v_n)^* \in V$ where $(v_n)^*=\{ (d_k)_n^* =2$ if $(d_n)_n =0$ and $(d_k)_n^* =0$ if $(d_n)_n =2$ for $(d_k)_n^* \in (v_n)$ and $(d_n)_n \in (v_n)$ . Therefore $(v_n)^*$ will be distinct from any $(v_n) \in V$ in the $k$ th position. Thus for $(v_n)^* \in V$ , we cannot find an $n\in \mathbb{N}$ for which $f(n)=(v_n)^*$ , so $f$ is not onto, a contradiction to the one-to-one correspondence of $f$ . Hence by contradicting the definition of countability, that $\mathbb{N} \sim V$ , $V$ is uncountable. Since all elements of the cantor set are represented in $V$ , the Cantor set is uncountable.

The Cantor Set is Closed

Each $In$ is a union of $2^n$ closed intervals. As a finite union of closed intervals, it is a closed set in $[0, 1]$ . Then $C$ is a closed subset of $[0, 1]$ being the intersection of closed sets $I_n$ for $n = 1, 2..$

The Cantor Set is perfect

A set is considered perfect if the set is closed and all the points of the set are limit points of the set. For each endpoint in the set $C$ there will always exist another point in the set within a deleted neighborhood of some radius $\varepsilon > 0$ on one side of that point since the remaining intervals at each step are being divided into infinitely small subintervals and since the real numbers are infinitely dense. Likewise, for each nonendpoint in the set there will always exist another point in the set within a deleted neighborhood of some radius $\varepsilon > 0$ on both sides of that point. Hence, all deleted neighborhoods of any radius $\varepsilon > 0$ around each point of the set $C$ for which the intersection of that deleted neighborhood and the set are nonempty. Therefore, each point in the set is a limit point of the set, and since the set is closed, the set $C$ is perfect.

Dimension of the Cantor Set

Definition: Dimension

The Topological Definition: A set $S$ is of dimension $k$ when each point $s\in S$ has an $\varepsilon$ neighborhood whose boundaries meet other points in a set of dimension $k-1$ and $k$ is the lease non-negative integer for which this holds.

The empty set has a topological dimension ( $d_T$ ) of -1. Consider a finite set of points $A$ because by the definition of isolated points, they are not limit points, so there exists an $\varepsilon$ neighborhood that does not intersect $A$ . Instead, this $\varepsilon$ neighborhood intersects the empty set, so a finite set of points will have $d_T=0$ .

Notice that we can choose $(\frac{1}{3})^{n+1}<\varepsilon <2(\frac{1}{3})^{n+1}$ for any $n$ so that there exists an $\varepsilon$ neighborhood that does not intersect the Cantor set. Instead, this $\varepsilon$ neighborhood intersects the empty set, so the Cantor set’s topological dimension ( $d_T$ ) is 0.

However, this does not account for the infinite number of points that will still exist within the defined epsilon neighborhood of the point in the cantor set, wherever $\varepsilon \leq (\frac{1}{3})^{n+1}$ . The Topological definition of dimension is further limited because it does not describe the way the cantor set scales well. To fix these problems, we turn to a definition that allows for fractional dimensions.

The Hausdorff-Besicovitch Definition: The exponent $d$ that a scale factor $\frac{1}{k}$ must take so that $k^d=N$ where $N$ is the number of scaled objects needed to create the original object.

When the Cantor set is scaled to $\frac{1}{k}=\frac{1}{3}$ of its original size, it takes $N=2$ of the scaled objects to recreate the original Cantor set. Thus $3^d=2$ . $d=\frac{ln2}{ln3}\approx .631$ The Cantor set’s HB dimension ( $d_{HB}$ ) $\approx .631$ .