Calculus 2 Honor Project-By Huanmei Zhu; Mentor: Dr. Apfaltrer

The Cantor Set

1. What is the Cantor set?

Cantor set is a set of points that lie on the interval [0,1]. We obtain them by deleting the middle thirds step by step. The Cantor set was introduced by the German mathematician Georg Ferdinand Ludwig Philipp Cantor in 1883, and it was named after him.
GeorgCantor.jpg

2. Construction of the Cantor set.

We construct the Cantor set starting with a line segment [0,1]. Cut it into three identical parts. Delete the middle one. Then, repeat the cutting into thirds and deleting the middle on each of the two remaining segments. Now delete the middle thirds from the remaining four segments. Then, the segments remaining from that, and so forth. A few steps in the construction of the Cantor set are shown in the following picture:

ConstrucTheCantorSetWith0&1.bmp

3. The Cantor set has length 0.

Each of the remaining segments in the process above has a length. For example, the length is 1 on the first line; $\frac{2}{3}$ on the second line; $\frac{4}{9}$ on the next. However, at the end of the process, we will end up with no length. Namely, the Cantor set has length 0. To prove this, we have to go into geometric series, which is a subject of Calculus 3.

A good example of geometric series is:

(1)
\begin{align} 1+\frac{1}{2}\,+\frac{1}{4}\,+\frac{1}{8}\,+\frac{1}{16}\,+... \end{align}

As we let r=$\frac{1}{2}$, we will have

(2)
\begin{equation} 1+r+r^2+r^3+r^4+... \end{equation}

Its sum may be go to $\infty$, or it may converge to an exact value.

In the general geometric series, if r=$\frac{1}{2}$<1, the sum of a geometric series is as follows

(3)
\begin{align} \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\label{geomSum} \end{align}

Below, we use a graphical proof to show this.

GraphQPST.jpg

On the graph above, we can find that the angles RQP and PST are right angles, and the angles PRQ and TPS are equal. So the two triangles PQR and TSP are similar. Using this similarity, we obtain that the ratios ST to PS and PQ to QR are the same, or

(4)
\begin{align} \frac{\overline{ST}}{\overline{PS}}=\frac{\overline{PQ}}{\overline{QR}}, \end{align}

In other words,

(5)
\begin{align} \frac{1+r+r^2+r^3+r^4+r^5+r^6...}{1}=\frac{1}{1-r} \end{align}
(6)
\begin{align} \frac{a}{a}\bullet\frac{1+r+r^2+r^3+r^4+r^5+r^6...}{1}=\frac{1}{1-r} \end{align}
(7)
\begin{align} \frac{a+ar+ar^2+ar^3+ar^4+ar^5+ar^6...}{a}=\frac{1}{1-r} \end{align}

Therefore,

(8)
\begin{align} a+ar+ar^2+ar^3+ar^4+ar^5+ar^6...=\frac{a}{1-r} \end{align}

By the definition of geometric series in Calculus 3, we know that

(9)
\begin{align} \sum_{n=0}^{\infty} ar^n = a+ar+ar^2+...+ar^n+...,\ \ \ \ \ \ \ \ \ \ a\is\not = 0. \end{align}

So,

(10)
\begin{align} \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \end{align}

We use this result to measure the length of the complement of the Cantor set:

(11)
\begin{align} \frac{1}{3}\,+\frac{2}{9}\,+\frac{4}{27}\,+\frac{8}{81}\,+... \end{align}
(12)
\begin{align} =\frac{1}{3}\,(1+\frac{2}{3}\,+\frac{4}{9}\,+\frac{8}{27}\,+...) \end{align}
(13)
\begin{align} =\frac{1}{3}\,[1+\frac{2}{3}\,+(\frac{2}{3})^2\,+(\frac{2}{3})^3\,+...] \end{align}

The series

(14)
\begin{align} 1+\frac{2}{3}\,+(\frac{2}{3})^2\,+(\frac{2}{3})^3\,+... \end{align}

is a geometric series with $a=1$ and $r=\frac{2}{3}$.
Therefore,

(15)
\begin{align} 1+\frac{2}{3}\,+(\frac{2}{3})^2\,+(\frac{2}{3})^3\,+...=\frac{a}{1-r}\,=3. \end{align}

Consequently,

(16)
\begin{align} \frac{1}{3}\,+\frac{2}{9}\,+\frac{4}{27}\,+\frac{8}{81}\,+...=\frac{1}{3}\bullet3=1 \end{align}

The length of the complement of the Cantor set in the interval [0,1] has length 1, and it follows that the Cantor set has length 0.

We now prove the result in (16)
geometrically:

(17)
\begin{align} \frac{1}{3}\,+\frac{2}{9}\,+\frac{4}{27}\,+\frac{8}{81}\,+... \end{align}
(18)
\begin{align} =\sum_{n=0}^{\infty}\frac{2^n}{3^{n+1}} \end{align}
(19)
\begin{align} =\frac{1}{3}\sum_{n=0}^{\infty}\frac{2^n}{3^n} \end{align}
(20)
\begin{align} =\frac{1}{3}\sum_{n=0}^{\infty}(\frac{2}{3})^n \end{align}

because,

(21)
\begin{align} S_n=\sum_{k=0}^{n} ar^k, \ \ \ \ \ \ \ \ \ 0<|r|<1 \end{align}
(22)
\begin{equation} =ar^0+ar^1+ar^2+...+ar^{n-1}+ar^n \end{equation}
(23)
\begin{align} rS_n=r\sum_{k=0}^{n}ar^k=ar^1+ar^2+ar^3+...+ar^n+ar^{n+1} \end{align}

Therefore,

(24)
\begin{equation} S_{n}-rS_{n}=ar^0-ar^{n+1}=a(1-r^{n+1}) \end{equation}
(25)
\begin{equation} (1-r)S_{n}=a(1-r^{n+1}) \end{equation}
(26)
\begin{align} S_{n}=a\bullet\frac{1-r^{n+1}}{1-r} \end{align}

When $n\to \infty$,

(27)
\begin{align} S_{n}=a\bullet\frac{1-0}{1-r}=\frac{a}{1-r} \end{align}

Hence, the measure of the complement of the Cantor set

(28)
\begin{align} \frac{1}{3}\,+\frac{2}{9}\,+\frac{4}{27}\,+\frac{8}{81}\,+... \end{align}
(29)
\begin{align} =\frac{1}{3}\sum_{n=0}^{\infty}(\frac{2}{3})^n \end{align}
(30)
\begin{align} =\frac{1}{3}\bullet \frac{1}{1-\frac{2}{3}} \end{align}
(31)
\begin{align} =\frac{\frac{1}{3}}{\frac{1}{3}} \end{align}
(32)
\begin{equation} =1 \end{equation}

We proved that the Cantor set has length 0 in two different ways, and it seems that we are left with nothing, but are we? The endpoints of the intervals remain. These are 0, 1, 1/3, 2/3, 1/9, 2/9 7/9, 8/9, … . Are there any additional points left? We answer this question in the next section.

4. How many points are left (in the Cantor set) after we delete the middle thirds?

After deleting the middle thirds, an uncountable set remains. To prove this, we have to show that the cardinalitiy of the interval [0,1] and the Cantor set are actually equal.

What is the cardinality? Roughly speaking, the cardinality is the number of elements of a set.

Finding the cardinality of a finite set is easy, we just need to count the number of elements. But what about the cardinality of an infinite set? For example, $\mathbb{N}=${1, 2, 3, 4, 5, …}, or $\mathbb{Z}=${ …, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, …}, or even worse, the real number interval $[0,1]$.

Cantor and other mathematicians extend the notion of cardinality to include sets of infinite size by using bijective functions. We will briefly go over the definition of a function, and define the meaning of a bijective function, in order to find the cardinality of our set of interest, the Cantor set.

1) What is a function?

A function is a rule that assigns to every value x in a set called the domain, only one element in another set called the codomain.

An injective function

An injective function is a function f that associates to distinct arguments (elements of the domain), distinct values in the codomain. It is also called a one-to-one function. An example of an injective function is shown below.

injective.png

A surjective function

A surjective function is a function f, such that for every y in the codomain, there is at least one x in the domain with $f(x)=y$. It is also called an onto function. The picture shown on the next is an example of a surjective function.

surjective.png

A bijective function

A bijective function is a function f with the property that for every y in the codomain, there is exactly one x in the domain such as f(x)=y. It is both one-to-one and onto. The following picture is an example of a bijective function.

bijective.png

If there is a function g from a set A to another set B that is bijective (both one-to-one and onto) we say the two sets have the same cardinality. For example, the set $A={a, b, c}$ and the set $B={1,2,3}$ have the same cardinality. Indeed, the function g, which maps a to 1, b to 2 and c to 3 is bijective. Therefore the set A has the same cardinality as B, namely cardinality 3.

This definition of cardinality holds both for finite and for infinite sets, even though it is not as intuitive for infinite sets, as we show below.

2) Which set is bigger, {0, 1, 2, 3, 4, 5, 6, 7, 8, …}, or { …, -4, -3, -2, -1, 0, 1, 2, 3, 4, …}

Intuitively, it seems Z={ …, -4, -3, -2, -1, 0, 1, 2, 3, 4, …} is twice as big as N={0, 1, 2, 3, 4, 5, 6, 7, 8, …}, because it contains both the positive and the negative integers and 0. In fact, we can find a bijective function between both sets.

With this knowledge, we may draw the relationship between set N and set Z as follows:

N&Z.jpg

We find that there is a function

(33)
\begin{align} g(n)=\frac{n}{2} \end{align}

that is bijective if n is even; and there is another function

(34)
\begin{align} g(n)=\frac{n+1}{-2} \end{align}

that is bijective too if n is odd. So, the given sets N and Z have the same cardinality, namely, a countably infinite cardinality (because we find a bijection between the set - Z in this case - and the natural numbers N).

Cantor shows that there are sets (like the real interval [0,1]) that have an infinite number of points, but for which one cannot find a bijection to the natural integers. In fact, sets, like this one, have just too many numbers; therefore, we cannot find an "enumeration" for them. We say that such sets have uncountably infinite cardinalities.

3) So then, what is the cardinality of the Cantor set?

Using the same procedure as above, we show that there is a function f from the Cantor set C to the interval [0,1] that is bijective, in order to prove that their cardinalities are actually equal (namely, uncountably infinite cardinality).

The numbers, such as $0$, $\frac{1}{9}$, $\frac{2}{9}$, $\frac{1}{3}$, $\frac{2}{3}$, $\frac{7}{9}$, $\frac{8}{9}$, $\frac{1}{4}$, $\frac{3}{10}$, and $1$, are in the Cantor set that are not the endpoints of deleted intervals.
ConstrucTheCantorSetWithNumbers.JPG

4) Decimal Expansion

The decimal expansion of $\frac{1}{4}$ is 0.25. It can be shown as:

onefourthbase10.bmp
(35)
\begin{align} 0.25=2\bullet\frac{1}{10}+5\bullet(\frac{1}{10})^2+... \end{align}
(36)
\begin{align} =2\times10^{-1}+5\times10^{-2}+... \end{align}

Other examples are $\frac{3}{10}= (0.3)_{10}$, $\frac{1}{3}= (0.33333...)_{10}$. They are represented in base 10.

5) Binary expansion

Similarly, every number in [0,1] can be written in binary expansion as well (namely, with base 2).

(37)
\begin{align} \frac{1}{4}= (0.01000...)_{2} \end{align}

or

(38)
\begin{equation} (0.001111111...)_{2}, \end{equation}
onefourthbase2.bmp

where the first and the second expansions coincide because of the properties of geometric series that we mentioned above.
Namely,

(39)
\begin{equation} (0.001111111...)_{2} \end{equation}
(40)
\begin{align} = \frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+... \end{align}
(41)
\begin{align} = \frac{1}{2^3}\sum_{n=0}^\infty\frac{1}{2^n} \end{align}
(42)
\begin{align} = \frac{1}{2^3}\bullet\frac{1}{1-\frac{1}{2}} \end{align}
(43)
\begin{align} = \frac{1}{2^3}\bullet2 \end{align}
(44)
\begin{align} = \frac{1}{2^2} \end{align}
(45)
\begin{align} = \frac{1}{4} \end{align}
(46)
\begin{equation} = (0.01000...)_{2} \end{equation}
(47)
\begin{align} 0= (0.00000...)_{2}, 1= (0.11111...)_{2}, \frac{1}{3}= (0.01010101...)_{2} \end{align}

As another example, let's express in binary expansion

(48)
\begin{align} \frac{21}{32}= \frac{16}{32}+ \frac{4}{32}+\frac{1}{32} \end{align}
(49)
\begin{align} = 1\times 2^{-1} +0\times 2^{-2}+1\times 2^{-3}+0\times2^{-4}+1\times2^{-5} \end{align}
(50)
\begin{equation} = (0.10101...)_{2} \end{equation}

6) Tertiary expansion

Also, every number in [0,1] can be written in base 3.

(51)
\begin{align} \frac{1}{4}=(0.02020202...)_{3} \end{align}
(52)
\begin{align} \frac{2}{3}=(0.122...)_{3}=(0.20...)_{3} \end{align}
(53)
\begin{equation} 0=(0.000...)_{3}, 1=(0.222...)_{3} \end{equation}
(54)
\begin{align} \frac{1}{3}=(0.10...)_{3}=(0.022...)_{3} \end{align}
onefourthbase3.bmp

In section 2, we obtained the Cantor set by deleting the middle thirds recursively. The numbers we deleted are therefore numbers in [0,1] whose tertiary expansion have ones (the number 1) at some point in the expansion base 3. In other words, they would be in a middle third at that point. Therefore, the numbers left in the Cantor set have only zeroes and two's in their tertiary expansion.

For this reason, we know that numbers such as $(0.022...)_{3}, (0.200...)_{3}, (0.222...)_{3}$ are in the Cantor set, while $(0.1200...)_{3}$ is not. Therefore we can describe a bijection between the Cantor set and the interval [0,1] by the figure below:

CantorSettoFrom0to1

According to the graph above, we find that there is a function

(55)
\begin{align} f(c)=\frac{c}{2}, \end{align}

where c is an element of the Cantor set, and the function $f(c)$ is bijective function between C and [0,1].

This bijection can be represented in the following graph:
graphofcantorset&[0,1].gif

Conclusion

The Cantor set has the same cardinality as the closed interval [0,1]. Namely, the Cantor set is an uncountably infinite set that has as many points as the interval [0,1], the interval we used to construct the Cantor set. Yet, the Cantor set has length 0!

Bibliography

  1. Cantor set, available at http://en.wikipedia.org/wiki/Cantor_set
  2. Ensley and Crawley, Discrete Mathematics, Hoboken, NJ, Wiley, p. 249-300.
  3. Georg Cantor, available at http://www.bigpedia.com/encyclopedia/Georg_Cantor
  4. Larson, Hostetler and Edwards, Calculus, Boston, MA., Houghton Mifflin, 1998, p. 558-641.
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