Equivalence of Definitions of Cantor Set
Theorem
The following definitions of the concept of the Cantor Set $\CC$ are equivalent::
$(1)$: As a Limit of Intersections
Define, for $n \in \N$, subsequently:
- $\map k n := \dfrac {3^n - 1} 2$
- $\ds A_n := \bigcup_{i \mathop = 1}^{\map k n} \openint {\frac {2 i - 1} {3^n} } {\frac {2 i} {3^n} }$
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:
- $\CC_n := \closedint 0 1 \setminus A_n$
The Cantor set $\CC$ is defined as:
- $\ds \CC = \bigcap_{n \mathop = 1}^\infty \CC_n$
$(2)$: From Ternary Representation
Consider the closed interval $\closedint 0 1 \subset \R$.
The Cantor set $\CC$ 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 $\CC$ 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 $n \in \N$:
- $C_n := \ds \bigcup S_n$
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 $\CC$ is defined as its limit, that is:
- $\ds \CC := \bigcap_{n \mathop \in \N} C_n$
Proof
Let $\CC_n$ be defined as in $(1)$.
Let $x \in \closedint 0 1$.
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 Sequence:
- $\ds 1 = \sum_{n \mathop = 0}^\infty \frac 2 3 \paren {\frac 1 3}^n$
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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Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $1$