# Equivalence of Definitions of Cantor Set

## Contents

## Theorem

The following definitions of the concept of **the Cantor Set $\mathcal C$** are equivalent::

### $1$: As a Limit of Intersections

Define, for $n \in \N$, subsequently:

- $\map k n := \dfrac {3^n - 1} 2$

- $\displaystyle A_n := \bigcup_{i \mathop = 1}^{\map k n} \openint {\frac {2 i - 1} {3^n} } {\frac {2 i} {3^n} }$

Since $3^n$ is always odd, $\map k n$ is always an integer, and hence the union will always be perfectly defined.

Consider the closed interval $\closedint 0 1 \subset \R$.

Define:

- $\mathcal C_n := \closedint 0 1 \setminus A_n$

The **Cantor set** $\mathcal C$ is defined as:

- $\displaystyle \mathcal C = \bigcap_{n \mathop = 1}^\infty \mathcal C_n$

### $2$: From Ternary Representation

Consider the closed interval $\closedint 0 1 \subset \R$.

The **Cantor set** $\mathcal C$ consists of all the points in $\closedint 0 1$ which can be expressed in base $3$ without using the digit $1$.

From Representation of Ternary Expansions, if any number has two different ternary representations, for example:

- $\dfrac 1 3 = 0.10000 \ldots = 0.02222$

then at most one of these can be written without any $1$'s in it.

Therefore this representation of points of $\mathcal C$ is unique.

### $3$: As a Limit of a Decreasing Sequence

Let $\map {I_c} \R$ denote the set of all closed real intervals.

Define the mapping $t_1: \map {I_c} \R \to \map {I_c} \R$ by:

- $\map {t_1} {\closedint a b} := \closedint a {\dfrac 1 3 \paren {a + b} }$

and similarly $t_3: \map {I_c} \R \to \map {I_c} \R$ by:

- $\map {t_3} {\closedint a b} := \closedint {\dfrac 2 3 \paren {a + b} } b$

Note in particular how:

- $\map {t_1} {\closedint a b} \subseteq \closedint a b$
- $\map {t_3} {\closedint a b} \subseteq \closedint a b$

Subsequently, define inductively:

- $S_0 := \set {\closedint 0 1}$
- $S_{n + 1} := \map {t_1} {C_n} \cup \map {t_3} {C_n}$

and put, for all $n \in \N$:

- $C_n := \displaystyle \bigcup S_n$

Note that $C_{n + 1} \subseteq C_n$ for all $n \in \N$, so that this forms a decreasing sequence of sets.

Then the **Cantor set** $\mathcal C$ is defined as its limit, that is:

- $\mathcal C := \displaystyle \bigcap_{n \mathop \in \N} C_n$

## Proof

Let $\mathcal C_n$ be defined as in $(1)$.

Let $x \in \left[{0 \,.\,.\, 1}\right]$.

We need to show that:

- $x$ can be written in base $3$ without using the digit $1$ if and only if:
- $\forall n \in \Z, n \ge 1: x \in C_n$

First we note that from Sum of Infinite Geometric Progression:

- $\displaystyle 1 = \sum_{n \mathop = 0}^\infty \frac 2 3 \left({\frac 1 3}\right)^n$

that is:

- $1 = 0.2222 \ldots_3$

Thus any real number which, expressed in base $3$, ends in $\ldots 10000 \ldots$ can be expressed as one ending in $\ldots 02222 \ldots$ by dividing the above by an appropriate power of $3$.

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{II}: \ 29: \ 1$