Cantor set

The Cantor ternary set is a remarkable subset of the real numbers named after German mathematician Georg Cantor who described the set in 1883. It has the same cardinality as $\mathbb{R}$, yet it has zero Lebesgue measure.

The set can be constructed recursively by first considering the unit interval $C_0=[0,1]$. In the next step, we divide the set into three equal parts and discard the open middle set. This leads to the new set $C_1=[0,\frac{1}{3}]\cup [\frac{2}{3},1]$. This procedure is repeated on the two remaining subintervals $[0,\frac{1}{3}]$ and $[\frac{2}{3},1]$, and iteratively on all remaining subsets such that $C_n=\frac{1}{3}{C}_{n-1}\cup (\frac{2}{3}+\frac{1}{3}{C}_{n-1})$. The Cantor set is defined by the limit as $n\to \mathrm{\infty }$, i.e., $\mathcal{C}={\cap }_{n=0}^{\mathrm{\infty }}{C}_n$.

The recursion can be illustrated in python in the following way

  import numpy as np

  def transform_interval(x, scale=1.0, translation=0.0):
      return tuple(map(lambda z: z * scale + translation, x))

  def Cantor(n):
      if n==0: return {(0,1)}
      Cleft = set(map(lambda x: transform_interval(x, scale=1/3), Cantor(n-1)))
      Cright = set(map(lambda x: transform_interval(x, translation=2/3), Cleft))
      return Cleft.union(Cright)

  print(Cantor(1))

{(0.6666666666666666, 1.0), (0.0, 0.3333333333333333)}

The following code displays the first 5 iterations of the algorithm and clearly illustrates the self-similarity property in the sets $C_n$.

  def intervalprint(x, n=240):
     sym = n*[' ']
     val = np.linspace(0, 1, num=n)
     for interval in x:
	idx = np.where((val >= interval[0]) & (val <= interval[1]))[0]
	for i in idx:
           sym[i] = u'\u2500'
     st = ''.join(sym)
     print(st)

  x = list(map(Cantor, [0,1,2,3,4,5]))
  for i in range(0, len(x)): intervalprint(x[i])

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It is clear that $\mathcal{C}$ is bounded (being a subset of the unit interval) and closed since it constructed by removing countably many open intervals from a closed interval. It also has zero measure, meaning that it does not contain any intervals.

To see this, note that in the first step the open middle set $(\frac{1}{3},\frac{2}{3})$ is removed which has measure 1/3. From the remaining two closed sets, $[0,\frac{1}{3}]$ and $[\frac{2}{3},1]$, the open middle sets are removed in the next iteration: $(\frac{1}{3^2},\frac{2}{3^2})$ and $(\frac{2}{3}+\frac{1}{3^2},\frac{2}{3}+\frac{2}{3^2})$, each of measure $\frac{1}{3^2}$. In the $n$th iteration $2^n-1$ open intervals are removed each of length $\frac{1}{3^n}.$ Hence, the Lebesgue measure of the Cantor set is $$m(\mathcal{C})=m([0,1])-\sum _{k=0}^{\mathrm{\infty }}\frac{2^n}{3^{n+1}}=1-\frac{1}{3}\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{2}{3}\right)}^n=1-\frac{1}{3}(1-\frac{2}{3}{)}^{-1}=0.$$

In this sense, we can think of the Cantor set just containing “dust”.

It can also be shown that the Cantor set can be represented as $$C=\{\sum _{i=1}^{\mathrm{\infty }}\frac{x_n}{3^n}\mid x_n\in 0,2\},$$ from which it again follows that the set is uncountable. This can also be seen by considering an arbitrary countable subset $$N=\{n_1,n_2,\dots \}\subseteq \mathcal{C}.$$ In the construction of $\mathcal{C}$, we must have that $n_1$ belongs to one of the two closed intervals in $C_1$. Hence, there is one interval $I_1$ of length $\frac{1}{3}$ such that $n_1\notin I_1$. Similarly, $C_2\cap I_1$ is the union of two new closed intervals of length $\frac{1}{3^2}$, and $n_2$ can at most be in one of these two sets. Let $I_2\subset I_1$ be this interval s.t. $n_2\notin I_2$. Continuing in this fashion we construct a series of intervals $\mathcal{C}\supset I_1\supset I_2\supset \cdots \supset I_k\supset \cdots$, where $I_k$ is one of the closed intervals of $C_n$ of length $\frac{1}{3^n}$ such that $n_k\notin I_k$. The limit $I_{\mathrm{\infty }}={\cap }_{k=1}^{\mathrm{\infty }}{I}_k$ is non-empty (see below), and by construction $N\cap I_{\mathrm{\infty }}=\mathrm{\varnothing }$. Hence given any arbitrary countable subset of $\mathcal{C}$ we can still find elements in $\mathcal{C}$ that does not belong to this subset, and thus there cannot exist a bijection between $\mathbb{N}$ and $\mathcal{C}$ and we conclude that $\mathcal{C}$ is uncountable infinite.

Here we used Cantor’s Intersection Theorem: Let $(X,d)$ be a complete metric space. Let $(I_k{)}_{k\in N}$ be a sequence of decreasing non-empty, closed subsets of $X$ then $\bigcap _{k\in N}{I}_k\ne \mathrm{\varnothing }.$

Proof: For each $k\in N$ choose $x_k\in I_k$. The diameter of $I_k$ is defined as $\mathrm{diam}(I_k)=sup\{𝑑(𝑥,𝑦):𝑥,𝑦\in I_k\}$. We only need to consider the case where $\mathrm{diam}(I_k)\to 0.$ In this case, let $ϵ>0$ then there exists $N>0$ such that $d(x,x^{\prime })<ϵ$ for all $x,x^{\prime }\in I_N$. Let $m,n>N$ then $x_n,x_m\in I_N$ since $I_N\supset I_{N+1}\supset \dots \supset I_m,I_n$, and thus $d(x_n,x_m)<ϵ$. We conclude that $(x_k{)}_{k\in \mathbb{N}}$ is a Cauchy sequence. Since $X$ is complete $x_n\to x$ as $n\to \mathrm{\infty }$ for some $x\in X$. Note that $(x_n{)}_{n=N}^{\mathrm{\infty }}\subset I_N$ and since $I_N$ is closed, the limit point also lies in the set $x\in I_N$. As this holds for any $N$, we conclude that $x\in \bigcap _{k\in N}{I}_k$. This concludes the proof. Note, that in this case the limit point $x$ is in fact the only element. Since if we let $x^{\prime }$ be another element then $x,x^{\prime }\in I_n$ for all $n\in N$ but $d(x,x^{\prime })\le \mathrm{diam}(F_n)\to 0$ as $n\to \mathrm{\infty }$. Hence $d(x,x^{\prime })=0$, and therefore $\bigcap _{k\in N}{I}_k=\{x\}.$