# Cantor Set

## Definition of Cantor Set

For each $P_i$ is non empty compact set and $P_0 \supset P_1 \supset P_2 \cdots$. Define $P$ as follows: \begin{align} P := \cap_{n=0}^\infty P_n \end{align} By the theorem 3.2.7, $P$ is nonempty compact set. We say that $P$ is the Cantor Set.

## Properties of Cantor Set

(1) $P \neq \emptyset$ and $P$ is compact

(2)$P$ contains all the end points of $J_{n,k} \text{ for all } n=0,1,2,\ldots, k=1,2,\ldots, 2^n$.

(3) Every point of $P$ is a limit point of $P$.

<proof> Let $p \in P$ and let $\epsilon >0$ be given. Choose $m \in \mathbb{N}$ such that $\frac{1}{3^n} < \epsilon$. Sine $p \in P_m$, $p \in J_{m,k}$ for some $k$. $J_{m,k}$ can be written as follows: $J_{m,k} := [\frac{x_k}{3^m}, \frac{x_k +1}{3^m}]$ for some $0 < x_k <3^m$. So, we have $J_{m,k} \subset N_\epsilon (p)$ since the length of the intervals $J_{m,k} = \frac{1}{3^m} < \epsilon$. It implies that $P \cap N^{\prime} (p)$ has at least one of the end points of $J_{m,k}.$

$\therefore p$ is a limit point of $P$. $\tag*{\square}$ (4) The sum of the lengths of the intervals removed is 1. In other words the Cantor set $P$ is measure zero set with respect to Lebesque measure.

<proof> \begin{align} \begin{split} \frac{1}{3} + 2\times \frac{1}{3^2} + 2^2 \times \frac{1}{3^3} + \cdots &= \sum_{n=1}^\infty \frac{2^{n-1}}{3^n} \\ &= \frac{1}{3} \sum_{n=1}^\infty \frac{2^{n-1}}{3^n}\\ &=\frac{1}{3} \cdot \frac{1}{1-\frac{2}{3}} \\ &=1 \end{split} \end{align} $\tag*{\square}$

(5) $P$ contains no intervals <proof> Let be an interval $[a,b]$ for some $a,b \in \mathbb{R}$. Choose $n \in \mathbb{N}$ such that $\frac{1}{3^n} < b-a$. Then any $[a,b] \not\subset J_{n,k}$. Since $P$ is countable intersection of $P_i$ and each $P_i = \cup_{i=1}^{2^i} J_{i,k}$, $[a,b] \not\subset P.$ $\tag*{\square}$

(6) Fore each $x \in [0,1], x = 0.n_1n_2n_3\cdots$ is the ternary expansion of $x$. $x \in P \iff n_k \in \{0, 2\}$ <proof> $\Rightarrow$ If $x=0$, it is trivial case. Suppose $x\neq 0$.

By Remark 1.6.3, $x$ has an infinite expansion of $x$. As described in Definition 1.6.1 we choose the largest integers among 0,1,2 such that $\frac{n_1}{3} < x$. Since $\frac{1}{3} \lneq x \lneq \frac{2}{3}$, $n_1 \neq 1$. Since $x\neq0$, $n_1= 1$. Now suppose $n_k \neq 1$ Then choose the largest integer $n_{k+1}$ among 0, 1,2 such that $\frac{n_1}{3} + \cdots + \frac{n_k}{3^{n_k}} + \frac{n_{k+1}}{3} < x.$ Since $x \in P \subset P_{k+1}, x \in J_{k+1,m}$ for some $m$. Since we cut out the first middle third of $J_{k,i} \text{ for } i=1, \ldots, 2^k$ to construct $J_{k+1,j}$ for $j=1, \ldots, 2^{k+1}$, $n_{k+1} \neq 1$.

$\therefore n_k \neq 0$ for all $k=1,2,\ldots$

$\Leftarrow$ Let $x =0.n_1n_2\ldots$ where for all $n_i$ is 0 or 1. If $x$ has a finite ternary expansion, we can rewrite it with infinite ternary expansion by Remark 1.6.3.

$\therefore x \in P.$ $\tag*{\square}$

(7) $P$ is uncountable <proof> Construct a function $f: P \rightarrow [0,1]$. We want to show that the $f$ is surjective. If so, the cardinality of $P$ is greater than or equal to $[0,1]$.

Since every $x \in P$, $x=0.n_1 n_2 \cdots$ where $n_k \neq 1$ with ternary expansion. Then the function $f$ is defined by replacing all the occurrences of 2 with 1.

For every $a \in [0,1]$, represent it with binary in binary form. Then there is $b \in P$ such that $f(b) = a$ because replacing all the occurrences of 1 with 2 of $a$ is ternary expression of $x$ for some $x \in P$.

$\therefore$ the cardinality of $P$ is greater than or equal to $[0,1]$.

Moreover, we already know that $P \subset [0,1]$.

$\therefore$ the cardinality of the two sets are equal, i.e. $P$ is uncountable. $\tag*{\square}$

## Definition 1.6.1

Let $n_1 \in \{0,1,2\}$ be the largest integer such that $\frac{n_1}{3} <x,$ where $0<x\leq1$. Having chosen $n_1, \ldots, n_k$ let $n_{k+1} \in \{0,1,2\}$ be the largest integer such that $\frac{n_1}{3} + \frac{n_2}{3^2} + \cdots + \frac{n_k}{3^k} + \frac{n_{k+1}}{3} < x.$ The expression $.n_1n_2n_3\cdots$ is called the ternary expansion of $x$.

## Theorem 1.6.3

Let $x \in \mathbb{R}$ with $0<x\leq1$, and with $\{n_k\}$ as defined in Definition 1.6.1, let \begin{align} E = \{ \frac{n_1}{3} + \cdots + \frac{n_k}{3^k}: k =1,2,\ldots \}. \end{align} Then $\sup E =x$.

<proof> Since $E \neq \emptyset$ and is bounded above by $x$, there exists $\sup E$ by the least upper bound property. Let $\alpha = \sup E$. Since $x$ is an upper bound, $\alpha \leq x$.

Suppose $\alpha \lneq x$. Let $k$ be the smallest positive integer such that

\begin{align} \frac{1}{3^k} < x-\alpha \quad \text{or} \quad \alpha + \frac{1}{3^k} < x \end{align} But \begin{align} \frac{n_1}{3} + \cdots + \frac{n_k}{3} + \frac{1}{3^k} = \frac{n_1}{3} + \cdots + \frac{n_k + 1}{3} <x. \end{align} If $n_k=0$ or 1, this contradicts the choice of $n_k$. If $n_k=2$, then we have \begin{align} \frac{n_1}{3} + \cdots + \frac{n_{k-1}}{3^{k-1}} < x. \end{align}

If any $n_j, 1\leq j\leq k-1$ is 0 or 1, we have a contradiction to the choice of $n_j$. If all the $n_j=2$, then we obtain \begin{align} \frac{2}{3} + \frac{2}{3^2} + \cdots + \frac{2}{3^k} &= \frac{2}{3}\cdot \frac{1}{1-\frac{1}{3}} \\ &<x \leq 1 \end{align} which is also a contradiction. $\tag*{\square}$

## Remark 1.6.4

For a given $0<x \leq1$, its ternary expansion $.n_1n_2n_3\cdots$ is not unique. The finite expansion can be expressed with infinite expansion. If x has a finite expansion, \begin{align} x = \frac{n_1}{3} + \cdots + \frac{n_k}{3^k}, \quad n_k \in \{1,2\} \end{align} Alternatively, it has infinite expansion as follows:

\begin{align} x = \frac{n_1}{3} + \cdots + \frac{0}{3^k} + \sum_{m=k+1}^\infty \frac{2}{3^m} \quad \text{ when } n_k = 1, \end{align} or

\begin{align} x = \frac{n_1}{3} + \cdots + \frac{1}{3^k} + \sum_{m=k+1}^\infty \frac{2}{3^m} \quad \text{ when } n_k = 2. \end{align}

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