The Cantor Set and What Remains After Removal, math that surprises you lesson

Start with the segment from $0$ to $1$ and remove its middle third. Then remove the middle third of each of the two pieces that are left, and keep going forever. Each step keeps only two thirds of the remaining length, so the total length shrinks toward zero. Is anything left at the end? This page builds the Cantor set, shows that it has total length zero and still contains infinitely many points, and uses base three to say exactly which points survive.

Key terms

Cantor set

Definition The middle-thirds Cantor set is built in stages. Stage $0$ is the closed interval $C_0 = [0, 1]$ . To pass from one stage to the next, take each closed interval $[a, b]$ of the current stage, let $d = (b - a)/3$ , and remove its open middle third, the points strictly between $a + d$ and $b - d$ , keeping the two closed outer thirds $[a, a + d]$ and $[b - d, b]$ . Stage $n$ , written $C_n$ , consists of $2^n$ separate closed intervals, each of length $3^{-n}$ , with total length $(2/3)^n$ . The Cantor set $C$ is the set of points that belong to every stage,

C = C_0 \cap C_1 \cap C_2 \cap \cdots.

The total length of the stages tends to $0$ , and $C$ has total length zero, yet $C$ is not empty. It contains $0$ , $1$ and every endpoint of every interval of every stage, because an endpoint is never strictly inside a removed middle third, and it also contains points that are never endpoints, such as $\tfrac14$ . The Cantor set is made of two copies of itself, each scaled by $\tfrac13$ ,

C = \tfrac13 C \cup \left(\tfrac13 C + \tfrac23\right),

where $\tfrac13 C$ is the set of all numbers $x/3$ with $x$ in $C$ , and adding $\tfrac23$ shifts that copy to the right end. It is named after Georg Cantor, who described it in 1883.

In plain words Keep cutting out the middle of every piece. The pieces get so short, so fast, that their total length disappears in the limit. But cutting out an open middle never touches the ends of a piece, and those ends, along with many other points, are never removed.

Example The first two stages are

\begin{aligned} C_1 &= \bigl[0, \tfrac13\bigr] \cup \bigl[\tfrac23, 1\bigr], \\[0.4em] C_2 &= \bigl[0, \tfrac19\bigr] \cup \bigl[\tfrac29, \tfrac13\bigr] \\[0.4em] &\quad \cup \bigl[\tfrac23, \tfrac79\bigr] \cup \bigl[\tfrac89, 1\bigr], \end{aligned}

with total lengths $\tfrac23$ and $\tfrac49$ , and $C_3$ has $8$ intervals of length $\tfrac{1}{27}$ with total length $\tfrac{8}{27}$ . The point $\tfrac14$ lies in the left third $\bigl[0, \tfrac13\bigr]$ . Stretching that third by $3$ to fill $[0, 1]$ sends $\tfrac14$ to $\tfrac34$ , which lies in the right third, and stretching the right third $\bigl[\tfrac23, 1\bigr]$ to fill $[0, 1]$ sends $\tfrac34$ back to $3 \cdot \tfrac34 - 2 = \tfrac14$ . So the pattern left, right, left, right repeats forever, $\tfrac14$ always lands in a kept third, and it is in $C$ . It is never an endpoint, because every endpoint is a fraction whose denominator is a power of $3$ , while $\tfrac14 = m/3^n$ would need $3^n = 4m$ , an odd number equal to an even one.

Open interval

Definition For real numbers $a < b$ , the open interval $(a, b)$ is the set of all $x$ with $a < x < b$ , and the closed interval $[a, b]$ is the set of all $x$ with $a \le x \le b$ . Both have length $b - a$ , but the open interval leaves out its two endpoints and the closed one includes them. Removing an open interval $(c, d)$ from a closed interval $[a, b]$ with $a \le c < d \le b$ leaves the two closed intervals $[a, c]$ and $[d, b]$ , so the points $c$ and $d$ stay. In the Cantor construction every removed middle third is open, so every stage is a union of closed intervals and every endpoint created at any stage stays in the Cantor set. A single point has length zero, so removing or keeping endpoints never changes a length, but it does change which points are left.

In plain words Open means without the ends. Taking out the open middle third of $[0, 1]$ takes out every point strictly between $\tfrac13$ and $\tfrac23$ , so $\tfrac13$ and $\tfrac23$ themselves stay.

Example Removing $\bigl(\tfrac13, \tfrac23\bigr)$ from $[0, 1]$ leaves $\bigl[0, \tfrac13\bigr] \cup \bigl[\tfrac23, 1\bigr]$ . The point $\tfrac12$ is strictly between $\tfrac13$ and $\tfrac23$ , so it is removed at the first step. The point $\tfrac13$ is never removed, since at every later step it is the right endpoint of the interval that contains it, first $\bigl[\tfrac29, \tfrac13\bigr]$ , then $\bigl[\tfrac{8}{27}, \tfrac13\bigr]$ , and so on, and the removed middle of an interval never contains its endpoints.

Measure zero

Definition A set $E$ of real numbers has measure zero, or length zero, if for every positive number $\varepsilon$ there is a finite or endless list of intervals that covers $E$ , meaning every point of $E$ lies in at least one of them, and whose lengths add up to less than $\varepsilon$ . Each stage $C_n$ covers the Cantor set with $2^n$ intervals of total length $(2/3)^n$ , and because $\tfrac23 < 1$ this total drops below any given $\varepsilon$ once $n$ is large enough. So the Cantor set has measure zero. Measure zero does not mean empty or finite. A single point, every finite set and every set whose elements can be arranged in one endless list, such as the rational numbers, have measure zero, and the Cantor set has measure zero although its points cannot be arranged in such a list at all.

In plain words A set has measure zero when it can be hidden inside intervals whose total length is as small as you like. Points have no length, so even infinitely many of them can take up no length at all.

Example To cover the Cantor set with intervals of total length less than $0.01$ , stage $12$ works. Its $4096$ intervals each have length $\tfrac{1}{531441}$ , and their total is

\left(\tfrac23\right)^{12} = \frac{4096}{531441} \approx 0.0077.

Stage $11$ is not enough, since its total is $\tfrac{2048}{177147} \approx 0.0116$ . The removed intervals, $1$ of length $\tfrac13$ , then $2$ of length $\tfrac19$ , then $4$ of length $\tfrac{1}{27}$ and so on, have total length

\begin{aligned} &\tfrac13 + \tfrac29 + \tfrac{4}{27} + \cdots \\[0.4em] &\quad = \frac{1/3}{1 - 2/3} = 1, \end{aligned}

the whole length of $[0, 1]$ , and still they do not remove every point. Measure zero depends on removing enough. If step $k$ removed from each of the $2^{k-1}$ intervals only an open middle piece of length $1/4^k$ , the removed lengths would add up to $\tfrac12$ , and the points that remain, a set often called a fat Cantor set, would have total length $\tfrac12$ .

Ternary expansion

Definition A ternary expansion, or base-three expansion, of a number $x$ in $[0, 1]$ is an endless string of digits $(0.a_1 a_2 a_3 \ldots)_3$ , each digit $a_k$ equal to $0$ , $1$ or $2$ , such that

\begin{aligned} x &= \sum_{k=1}^{\infty} \frac{a_k}{3^k} \\[0.4em] &= \frac{a_1}{3} + \frac{a_2}{9} + \frac{a_3}{27} + \cdots. \end{aligned}

Every $x$ in $[0, 1]$ has at least one ternary expansion. A number has exactly two when it is a fraction $m/3^n$ strictly between $0$ and $1$ , one ending in endless $0$ s and one ending in endless $2$ s, as in $\tfrac13 = (0.1000\ldots)_3 = (0.0222\ldots)_3$ , and every other number has exactly one. The first digit tells which third of $[0, 1]$ contains $x$ . An expansion beginning with $0$ puts $x$ in $\bigl[0, \tfrac13\bigr]$ , one beginning with $2$ puts it in $\bigl[\tfrac23, 1\bigr]$ , and $x$ lies in the removed open middle third exactly when every expansion of $x$ begins with $1$ . Each later digit does the same inside the third chosen before. So $x$ belongs to the Cantor set exactly when it has at least one ternary expansion that uses only the digits $0$ and $2$ . Such an expansion is unique, so the points of the Cantor set match the endless strings of $0$ s and $2$ s one to one, and the reasoning of Cantor’s diagonal argument, applied to these strings, shows that they cannot all be listed.

In plain words A ternary expansion is a set of directions for finding a point. Each digit says which third to go into, $0$ for the left, $1$ for the middle and $2$ for the right. The middle thirds are the parts that get removed, so the points that survive are exactly the ones that can be reached without ever going into a middle third.

Example The expansion $(0.020202\ldots)_3$ says left, right, left, right and so on. Its value is a geometric series with first term $\tfrac29$ and ratio $\tfrac19$ ,

\begin{aligned} &\frac{2}{9} + \frac{2}{81} + \frac{2}{729} + \cdots \\[0.4em] &\quad = \frac{2/9}{1 - 1/9} = \frac14, \end{aligned}

so $\tfrac14$ is in the Cantor set, as found above. In the same way $\tfrac34 = (0.2020\ldots)_3$ is in it. The number $\tfrac12 = (0.111\ldots)_3$ has only this one expansion, so it is not. The endpoint $\tfrac13$ shows why the rule says at least one expansion. Its expansion $(0.1000\ldots)_3$ contains a $1$ , but its other expansion $(0.0222\ldots)_3$ , whose value is $\tfrac{2/9}{1 - 1/3} = \tfrac13$ , uses only $0$ and $2$ , so $\tfrac13$ is in the Cantor set, as the construction says.

Check yourself

How many intervals make up stage $5$ , how long is each one, and how much total length has been removed by then?

Show the answer

Each step doubles the number of intervals and divides their length by $3$ , so stage $5$ has $2^5 = 32$ intervals, each of length $3^{-5} = \tfrac{1}{243}$ . Their total length is $\tfrac{32}{243} \approx 0.132$ , so the removed length is

1 - \frac{32}{243} = \frac{211}{243} \approx 0.868.

Adding the removed pieces step by step gives the same value, since $\tfrac13 + \tfrac29 + \tfrac{4}{27} + \tfrac{8}{81} + \tfrac{16}{243} = \tfrac{211}{243}$ .

The numbers $\tfrac79$ and $\tfrac49$ have the ternary expansions $(0.21000\ldots)_3$ and $(0.11000\ldots)_3$ , which both contain the digit $1$ . Decide whether each number belongs to the Cantor set.

Show the answer

The digit $1$ in these expansions alone decides nothing, because membership asks whether some expansion avoids the digit $1$ . Each of these numbers is a fraction with denominator $9 = 3^2$ , so each has a second expansion ending in endless $2$ s.

For $\tfrac79$ the second expansion is $(0.20222\ldots)_3$ , whose value is $\tfrac23 + \tfrac{2/27}{1 - 1/3} = \tfrac23 + \tfrac19 = \tfrac79$ . It uses only $0$ and $2$ , so $\tfrac79$ is in the Cantor set. It is the right endpoint of the interval $\bigl[\tfrac23, \tfrac79\bigr]$ of stage $2$ .

For $\tfrac49$ the second expansion is $(0.10222\ldots)_3$ , which still begins with $1$ . Both expansions begin with $1$ , so $\tfrac49$ lies in the open middle third $\bigl(\tfrac13, \tfrac23\bigr)$ and is removed at the first step.

Change the rule so that the closed middle third of each piece is removed, endpoints included. Which of the points $0$ , $\tfrac13$ and $\tfrac14$ survive, and what happens to the total length?

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The point $\tfrac13$ is an endpoint of the first closed middle third $\bigl[\tfrac13, \tfrac23\bigr]$ , so it is removed at the first step, and what is left is $\bigl[0, \tfrac13\bigr) \cup \bigl(\tfrac23, 1\bigr]$ , which is no longer a union of closed intervals.

A closed middle third is the open middle third together with its two endpoints, and these endpoints are fractions strictly between $0$ and $1$ whose denominators are powers of $3$ . The point $0$ is never strictly inside an open middle third, as in the original construction, and it is not strictly between $0$ and $1$ , so it survives. The point $\tfrac14$ is never inside an open middle third either, because it is in the Cantor set, and it is not a fraction with a power of $3$ as denominator, so it survives as well.

The total length after $n$ steps is still $(2/3)^n$ , because the extra points removed have no length, and the set that survives still has measure zero. The change removes the endpoints of all the removed middle thirds and keeps every other point of the Cantor set.

The Cantor Set and What Remains After Removal

Key terms

Cantor set

Open interval

Measure zero

Ternary expansion

Check yourself

How this lesson connects

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