Cantor Set

TheCantor setissetof points lying on a line segment. It is created by taking some interval, for instance $[0,1],$and removing the middle third $\left(\frac{1}{3},\frac{2}{3}\right)$, then removing the middle third of each of the two remaining sections $\left(\frac{1}{9},\frac{2}{9}\right)$and $\left(\frac{7}{9},\frac{8}{9}\right)$, then removing the middle third of the remaining four sections, and so on ad infinitum.

It is a closed set consisting entirely of boundary points, and is an important counterexample in set theory andgeneral topology. When learning aboutcardinality, one is first shown subintervals of thereal numbers, $\mathbb{R}$, as examples of uncountably infinite sets. This might suggest that any uncountably infinite subset of $\mathbb{R}$must contain an interval; such an assertion isfalse. A counterexample to this claim is the Cantor set $\mathcal{C} \subset [0,1]$, which is uncountable despite not containing any intervals. In addition, Cantor sets are uncountable, may have 0 or positiveLebesgue measures, and are nowhere dense. Cantor sets are the only disconnected, perfect, compact metric space up to ahomeomorphism.

The Cantor set is constructed by removing increasingly small subintervals from $[0,1]$. In the first step, remove $\left(\frac13, \frac23\right)$from $[0,1]$. In the second step, remove $左(\ frac19 \ frac29 \ \) \杯\离开(\ frac79,\frac89\right)$from what remains after the first step. In general, on the $n^\text{th}$step, remove $\left(\frac{1}{3^n}, \frac{2}{3^n}\right) \cup \left(\frac{4}{3^n}, \frac{5}{3^n}\right) \cup \cdots \cup \left(\frac{3^n - 2}{3^n}, \frac{3^n - 1}{3^n}\right)$from what remains after the $(n-1)^\text{th}$step. After all $\mathbb{N}$steps have been taken, what remains is the Cantor set $\mathcal{C}$.

This construction can be formalized as follows. Let $\mathcal{C}_0 = [0,1]$. Then, after the first step, what remains is $\mathcal{C}_1 = \left[0,\frac13\right] \cup \left[\frac23,1\right]$. After the second step, what remains is $\mathcal{C}_2 = \left[0,\frac{1}{9}\right] \cup \left[\frac{2}{9}, \frac{1}{3}\right] \cup \left[\frac{2}{3},\frac{7}{9}\right] \cup \left[\frac{8}{9}, 1\right].$Continuing in this manner, one obtains an infinite collection of sets $\{\mathcal{C}_i\}_{i=0}^{\infty}$such that $\mathcal{C}_i \subset \mathcal{C}_{i-1}$for all $i\ge 1$. Then, one defines the Cantor set to be $\mathcal{C} = \bigcap_{i=0}^{\infty} \mathcal{C}_i.$Intuitively, this makes sense; since $\bigcap_{i=0}^{n} \mathcal{C}_i = \mathcal{C}_n$for all $n\ge 1$, taking an infinite intersection would provide the "limiting set" in this situation.

After seven iterations of the Cantor set's construction. Going from top to bottom, the image depicts the sets $\mathcal{C}_0$through $\mathcal{C}_6$.^[1]

There is an alternate characterization that will be useful to prove some properties of the Cantor set:

$\mathcal{C}$consists precisely of the real numbers in $[0,1]$whosebase-3expansions only contain the digits 0 and 2.

Base-3 expansions, also called ternary expansions, represent decimal numbers on using the digits $0,1,2$. For instance, the number 3 in decimal is 10 in base-3. Fractions in base-3 are represented in terms of negative powers of 3, where a fraction $N$in decimal can be represented as $N = c_1b^{-1}+c_2b^{-2}+c_3b^{-3}+\cdots+c_n{b^{-n}}+\cdots,$where $0≤ c_i <3$, and the fraction is written in base-3 as $N = 0.c_1c_2c_3...$.
For instance, $\frac14 = 0.25 = 0\times 3^{-1} + 2\times 3^{-2} + 0\times 3^{-3} + 2\times 3^{-4}+\cdots = 0.\overline{02}_3$.

Similarly, decimal fractions can be converted by multiplying by $3$, taking the result mod 3, then multiplying the remainder by $3$, and then taking mod 3 of that... etc. until the remainder is 0: $\begin{aligned} 0.25 \times 3 = \mathbf{0} + 0.75,\ \ c_1 &= 0\\ 0.75 \times 3 = \mathbf{2} + 0.25,\ \ c_2 &= 2\\ 0.25 \times 3 = \mathbf{0}+ 0.75,\ \ c_3 &= 0\\ 0.75 \times 3 = \mathbf{2} + 0.25,\ \ c_4 &= 2\\ &\vdots\\ &\Rightarrow 0.0202..._3. \end{aligned}$

In base-3, some points have more than one notation:
$\frac13 \times 3 = \mathbf{1} + 0,\ c_1 = 1$. So $\frac13 = 0.1_3.$But we also have the following: $\开始{对齐}\ frac13 \ * 3 = 0.3333……\ *3 = \mathbf{0} + 0.9999...,\ c_1 &= 0\\ 0.9999... \times 3 = \mathbf{2} + 0.9999...,\ c_2 &= 2\\ 0.9999... \times 3 = \mathbf{2} + 0.9999...,\ c_3 &= 2\\ &\vdots \end{aligned}$So, $\frac13 = 0.\overline{3} = 0.1_3 = 0.0\overline{2}_3.$This sometimes means there is ambiguity in $\mathcal{C}$, for if $\mathcal{C}$expansions only contain the digits 0 and 2, then $\mathcal{C}$contains $0.0\overline{2}_3$but not $0.1_3$even though they are the same. In this case, it's said to contain $0.0\overline{2}_3 = \frac13,$which implies it contains $0.1_3$because of the multiple notations.

Suppose $x\in [0,1]$contains only the digits 0 and 2 in its base-3 expansion. Let $x_n$be the truncation of $x$at $n$places after the decimal point. For example, if $x = 0.020202\cdots_{3}$, then $x_1 = 0$, $x_2 = 0.02_{3}$, $x_4 = 0.0202_{3}$, etc. Certainly, the sequence $x_n$converges to $x$as $n\to\infty$. In particular, for every $n\ge 1$, we have $x_n \le x \le x_n + \frac{1}{3^n}.$Note that the numbers in $[0,1]$

• whose base-3 expansions go on for exactly $n$digits after the decimal point and
• which use only the digits 0 and 2

are precisely theleftendpoints of the intervals whose union is $\mathcal{C}_n$. Thus, the interval $\left[x_n, x_n + \frac1{3^n}\right]$is contained in $\mathcal{C}_n$. It follows that $x\in \mathcal{C}_n$for all $n\ge 0$, and hence $x\in \mathcal{C}$. $_\square$

Conversely, suppose $x\in \mathcal{C}$. Then $x\in \mathcal{C}_n$for all $n\ge 0$. Note that the numbers in $\mathcal{C}_n$are precisely those whose $n^\text{th}$truncation (i.e., the number obtained by taking only the first $n$digits after the decimal point) uses only 0 and 2 as digits in base 3. It follows that every truncation of $x$uses only 0 and 2 as digits. This implies $x$uses only 0 and 2 in its base-3 expansion. $_\square$

From this theorem, the proofs of the following two properties follow:

$\mathcal{C}$does not contain any subintervals of $[0,1]$.

Let $[a,b] \subset [0,1]$be an arbitrary interval. Write $a = 0.a_1 a_2 a_3 \cdots_{3}$and $b = 0. b_1 b_2 b_3 \cdots_{3}.$If some $a_i$equals 1, then by the previous theorem, we know $a\not \in \mathcal{C}$. Similarly, if some $b_i$equals 1, we know $b\not\in \mathcal{C}$. Otherwise, suppose $k$is the smallest index such that $a_k \neq b_k$. Our casework forces $a_k = 0$and $b_k = 2$, so $0.a_1 a_2 a_3 \cdots a_{k-1} 1_{3} \in [a,b]$, and it follows that $[a,b] \not\subset \mathcal{C}$. $_\square$

$\mathcal{C}$is uncountable.

Define a function $f: \mathcal{C} \to [0,1]$as follows. If $x = 0. x_1 x_2 x_3 \cdots_{3}$is an element in $\mathcal{C}$, set $f(x) = 0 . (x_1 / 2) (x_2 / 2) (x_3 / 2) \cdots _{2}.$In other words, take the ternary expansion of $x$, replace every digit $2$with $1$, and consider the result as a binary number.

This function issurjective, since any element $y = 0.y_1 y_2 y_3 \cdots_{2}$of $[0,1]$has $f\big(0. (2y_1) (2y_2) (2y_3) \cdots_{3}\big) = y$.

Now, suppose $\mathcal{C}$is countable. Write $\mathcal{C} = \{c_1, c_2, c_3, \cdots\}$. For each $y\in [0,1]$, let $g(y)$equal the index $j$of an element $c_j$with $f(c_j) = y$(remark: in defining this function $g$, we are implicitly using theaxiom of choice). Then, we may construct a sequence $\{d_i\}_{i\in \mathbb{N}}$such that $d_{g(y)} = y$for all $y \in [0,1]$. This constitutes a bijection between $[0,1]$and $\mathbb{N}$, so it follows that $[0,1]$is countable. Contradiction!

Hence, we conclude $\mathcal{C}$is uncountable. $_\square$

1. 127 "rect", .Cantor ternary set, in seven iterations. Retrieved September 5th, 2016, fromhttps://commons.wikimedia.org/wiki/File:Cantor_set_in_seven_iterations.svg