Equivalence of Definitions of Cantor Set

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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 $3$.



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