What is the Cantor Set? How do I write it mathematically?

Question

I would like some clarification about the Cantor Set:

• What are the elements in the Cantor Set?
• How do I write the Cantor Set in mathematical terms (i.e in a summation)? I have seen online a formula but I do not understand how they got it so would be grateful if you could explain why it is this formula too.

EDIT: The formula I have seen online is: $$\sum_{i=0}^\infty (a(i)/3^i) $$ where a(i) is either 0 or 2.

Answer 1

The Cantor Set $\mathcal C$ can be defined by the following recursive sequence of sets:

$$\mathcal C_0 = [0, 1]$$ $$\mathcal C_{n+1}= \frac{\mathcal C_n}3\cup\left(\frac23+\frac{C_n}3\right)$$ $$\mathcal C := \bigcap^\infty_{n=0}\mathcal C_n$$

This definition should reflect the intuitive notion of what the Cantor Set is and how it can be constructed.

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To explain this further, I will explain the math in laymen's terms.

Start with the closed unit interval, $[0, 1]$.

The next iteration is the previous one divided by three and the previous one divided by three with its origin shifted to be at $\frac23$. This yields $[0, \frac13]\cup[\frac23, 1]$, then $[0, \frac19]\cup[\frac29, \frac13]\cup[\frac23, \frac79]\cup[\frac89, 1]$, etc.

The set itself is the set of points that are present in every iteration of this process.

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In case you were wondering how the definition you gave ties into mine (which I find much more intuitive), Here is an explanation:

Observe that as $\mathcal C_n$ is iterated, the edges of each interval ($0, \frac13, \frac23, \frac19, $ etc) remain in the set.

All of these numbers can be expressed by the sum $$\sum^\infty_{n=1}\frac{a_n}{3^n}$$ where the sequence $a$ is all $0$s and $2$s.

For example, $\frac13$ is the sum when $a_n = \langle 0, 2, 2, 2, 2, \cdots\rangle$, or $\frac29+\frac2{27}+\frac2{81}+\cdots$

I'll leave it to you to find the reason for this connection, but a hint is that it has to do with the $\frac23$ being added.

Answer 2

The elements of the Cantor set are real numbers between $0,1$. The set is formed by iteratively and infinitely cutting all intervals in thirds and removing the middle third.

1. $E_0 = [0,1]$
2. $E_1 = [0, \frac 13] \cup [\frac 23, 1]$
3. $E_2 = [0, \frac 19]\cup [\frac 29, \frac 13]\cup[\frac 23,\frac 79]\cup [\frac 89, 1]$
4. .....

The Cantor set is what remains when you do this an "infinite number of times". If I can abuse notation $CantorSet = E_{\infty}$ or $\lim_{n\rightarrow \infty} E_n$ except... well, that's an abuse of notation. The proper description would be $\cap_{i=0}^{\infty} E_i$ (but that can be intuitively confusing).

Thus all intervals of the form $(\frac {3k+1}{3^m}, \frac{3k+2}{3^m})$ are removed (and no intervals are left.)

S0 the set is $[0,1] \setminus \{(\frac {3k+1}{3^m}, \frac{3k+2}{3^m})|k,m\in \mathbb Z^+\}$

Or ... $\{\sum_{i=0}^\infty (a(i)/3^i)|\{a_i\} \in \{0,2\}^{\infty}\}$ (where $\{0,2\}^{\infty}$ is the set of all infinite sequences containing $0$ or $2$).

If we use decimal base three notation it would be all numbers between $0$ and $1$ that have only $0$ and $2$ in their base three decimal expansion. (But as and infinite number of $2$ is the same as a terminating $1$ this can be restated as all real number between $0$ and $1$ who either have no decimal $1$ or only have a single occurrence of $1$ in its terminating position.)

I think (someone will correct me if I am mistaken) the Cantor-ish set of all the real numbers with only $0$ and $5$ in their base $10$ decimal expansion is topologically equivalent to the Cantor set. This is the set you get by

1. $G_1 = [0,1]$
2. $G_2 = [0,.1]\cup [.5,1]$
3. $G_3 = [0,0.01]\cup [.05,.1]\cup[.5,.51]\cup [.55, 1]$
4. ....

Is the same process, just with a nice base 10 result.

Answer 3

Let us define the Cantor representation of a real number to be the ternary expansion of that number where we don't allow a suffix of $1000\cdots$ or $12222\cdots$. So we wouldn't write $1$ as $1.0000\cdots$, we would write it as $0.2222\cdots$. Likewise, $2/3$ is $0.2000\cdots$ not $0.1222\cdots$.

The Cantor set consists of all real numbers in $[0,1]$ whose Cantor representation does not contain any $1$s. We can build the Cantor set by defining the sets $C_n$ of all real numbers in $[0,1]$ whose Cantor representation does not contain a $1$ in the first $n$ digits after the decimal point. The intersection $\bigcap C_n$ will be the Cantor set.

The set $C_0$ is the entire interval, $[0,1]$.

The set $C_1$ consists of all the numbers

• between $0 = 0.0000\cdots$ and $1/3 = 0.0222\cdots$ and
• between $2/3 = 0.2000\cdots$ and $1 = 0.2222\cdots$.

Thus $C_1 = [0,\frac13] \cup [\frac23, 1]$.

The set $C_2$ consists of the numbers

• between $0 = 0.0000\cdots$ and $1/9 = 0.00222\cdots$ and
• between $2/9 = 0.02000\cdots$ and $1/3 = 0.02222\cdots$ and
• between $2/3 = 0.20000\cdots$ and $7/9 = 0.20222\cdots$ and
• between $8/9 = 0.22000\cdots$ and $1 = 0.22222\cdots$.

So $C_2 = [0,\frac19] \cup [\frac29, \frac13] \cup [\frac23, \frac79] \cup [\frac89, 1]$.

In general, $C_n$ is obtained from $C_{n-1}$ by removing the middle third of each interval in $C_{n - 1}$.

Answer 4

That formula isn't all that helpful. The most useful property is $\mathcal C= \frac{\mathcal C}3\cup\left(\frac23+\frac{\mathcal C}3\right)$, i.e. the Cantor set is the union of two subsets obtained by shrinking/shifting of the Cantor set. That's called self-similarity.