The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds

For $S\subseteq \mathbb{R}$ write

$\frac{1}{3}S:=\{x\cdot 3^{-1}:x\in S\},\frac{2}{3}+\frac{1}{3}S:=\{2\cdot 3^{-1}+x\cdot 3^{-1}:x\in S\},$

and let $F:\mathcal{P}(\mathbb{R})\to \mathcal{P}(\mathbb{R})$ be

$F(S):=\frac{1}{3}S\cup (\frac{2}{3}+\frac{1}{3}S)\mathrm{.}$

By the recursion theorem (The recursion theorem), applied to the set $\mathcal{P}(\mathbb{R})$, the starting element $[0,1]$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) and the function $F$, there is a unique family $(C_n{)}_{n\in \mathbb{N}}$ of subsets of $\mathbb{R}$ with

$C_0=[0,1],C_{n+1}=F(C_n)=\frac{1}{3}{C}_n\cup (\frac{2}{3}+\frac{1}{3}{C}_n)(n\in \mathbb{N})\mathrm{.}$

The Cantor middle-thirds set is

$C:={\bigcap }_{n\in \mathbb{N}}{C}_n\mathrm{.}$

The first step really is the removal of the open middle third. Directly from the clauses,

$C_1=\frac{1}{3}[0,1]\cup (\frac{2}{3}+\frac{1}{3}[0,1])=[0,\frac{1}{3}]\cup [\frac{2}{3},1]=[0,1]\setminus (\frac{1}{3},\frac{2}{3}),$

the middle equality because $x\mapsto x\cdot 3^{-1}$ is an order isomorphism of $\mathbb{R}$ onto itself with inverse $x\mapsto 3x$ (Ordered field, Sign rules for products and monotonicity of multiplication), and the last because $0\le x\le 1$ splits, by totality of the order, into $x\le \frac{1}{3}$, $\frac{1}{3}<x<\frac{2}{3}$ and $x\ge \frac{2}{3}$. The recursion then performs the same operation inside each of the two scaled copies, which is what "removing the open middle thirds" names.

Every $C_n$ lies in $[0,1]$, by induction on $n$ (The principle of mathematical induction): $C_0=[0,1]$; and if $C_n\subseteq [0,1]$ then $\frac{1}{3}{C}_n\subseteq [0,\frac{1}{3}]$ and $\frac{2}{3}+\frac{1}{3}{C}_n\subseteq [\frac{2}{3},1]$, so $C_{n+1}\subseteq [0,1]$ (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication). The same computation shows that the two halves of $C_{n+1}$ are disjoint, the first lying in $[0,\frac{1}{3}]$ and the second in $[\frac{2}{3},1]$, and $\frac{1}{3}<\frac{2}{3}$ (The multiplicative identity is positive).

The family is nested, $C_{n+1}\subseteq C_n$ for every $n$, again by induction. For $n=0$ this is $C_1=[0,\frac{1}{3}]\cup [\frac{2}{3},1]\subseteq [0,1]$. And $F$ is monotone, in the sense that $S\subseteq T$ implies $F(S)\subseteq F(T)$, directly from the displayed description of $F$; so $C_{n+1}\subseteq C_n$ gives $C_{n+2}=F(C_{n+1})\subseteq F(C_n)=C_{n+1}$. Consequently $C={\bigcap }_n{C}_n\subseteq C_m$ for every $m$, and ${\bigcap }_n{C}_{n+1}={\bigcap }_n{C}_n=C$.

Powers. Here $3^{-n}$ means $(3^{-1}{)}^n$, the integer power of Integer powers $a^m$, so that $3^0=1$, $3^{-(n+1)}\cdot 3=3^{-n}$ and $3^{-n}>0$ for every $n$ (Laws of integer exponents, Complete ordered field (least-upper-bound property)).

• Why the self-similar recursion rather than a description by digits. The clause $C_{n+1}=F(C_n)$ is a single application of The recursion theorem to one explicitly given function on $\mathcal{P}(\mathbb{R})$, so nothing is selected at any stage and no listing of the $2^n$ intervals making up $C_n$ has to be constructed. Every structural property below is then proved by induction on $n$ through $F$. The description by ternary digits is a theorem about $C$, not its definition, and it is The Cantor set is exactly the set of ${\sum }_{k\ge 1}{a}_k{3}^{-k}$ with every $a_k\in \{0,2\}$, and this gives a bijection with $\{0,1{\}}^{\mathbb{N}}$.

• $C$ is not empty. $0\in C_n$ for every $n$, by induction: $0\in [0,1]$, and $0\in C_n$ gives $0=0\cdot 3^{-1}\in \frac{1}{3}{C}_n\subseteq C_{n+1}$. Likewise $1\in C$, since $1\in C_n$ gives $1=\frac{2}{3}+1\cdot \frac{1}{3}\in C_{n+1}$. So $C$ contains at least the two endpoints; that it is in fact uncountable is The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points.

• The construction with a different proportion. Replacing "middle third" by an interval of length $4^{-n}$ removed at stage $n$ produces a set that is closed, has empty interior and is not of measure zero (The Smith-Volterra-Cantor set: the same construction removing, at stage $n\ge 1$, an open middle interval of length $4^{-n}$ from each of the $2^{n-1}$ remaining intervals). So none of the qualitative properties of $C$ proved on this page is a consequence of its being nowhere dense, and the two constructions are kept apart deliberately.