# Fractals: Cantor Sets

I just had my math teacher explain to me how you can have fractional dimensions less than one. He showed me an example of a "Cantor" set (I'm not exactly sure what one is, which is why I'm asking!) So, it looked like the Seirpenski triangle, where it was a line from 0 to 1, and the middle third was taken from the line segment, and then the middle third of the two remaining line segments, etc. My question is: what would the equation for this look like? And more generally, what does "cantor" set and "Julia" set mean? How do we define equations for these? I know for the Seirpenski triangle the sum of the area is $$\frac{3^n}{4^n}\sqrt{\frac{3}{4}}$$ (I think!). Anyway, thanks! I'm not sure if this is a hard or soft question, so I'll err on the side of caution calling it soft. Thanks again!

• This is a perfectly good question. You're making the simple error of assuming that all sets must have equations which describe them. They don't have to. I recommend looking at Wikipedia (don't go too deep!) and then coming back to edit your question more specifically, because you've got several questions in one right now. Aug 30 '17 at 21:33
• Thanks for the feedback! I'm confused how you are supposed to show fractals without an equation or expression of some sort - the answer below defines some ways of expressing the ternary cantor set, so if there are others, please tell me! I agree that I have many questions all at once, but that's how I learned to approach math - if someone tells you that a thing such as a fractional dimension exists, you're going to want to know more, and the more you know the more questions you have! Next time I'll clarify that I was asking if the julia and cantor set were related. Thanks! Aug 31 '17 at 4:16

1. The usual thing that people think about when they say Cantor set is the "usual" ternary Cantor set. This set is a collection of real numbers, which can be described in several ways.

• The ternary Cantor set consists of all real numbers between 0 and 1 (inclusive) that possess a ternary expansion that consists of only 0 and 2. For example, the number $\frac{1}{3}$ can be written as $$0.1_3 = 0.0\overline{2}_3,$$ so $\frac{1}{3}$ is an element of the Cantor set. Similarly $$0.202_3 = 2\cdot 3^{-1} + 0\cdot 3^{-2} + 2\cdot 3^{-3} = \frac{2}{3} + \frac{2}{27} = \frac{20}{27} \approx 0.741_{10}$$ is an element of the Cantor set. On the other hand, $$\frac{1}{2} = 0.\overline{1}_{3}$$ is not an element of the Cantor set, since it does not have a ternary representation consisting of only 0 and 2.

• The ternary Cantor set can also be constructed via the technique you suggest. That is, start with the closed unit interval $[0,1]$, i.e. all of the real numbers between 0 and 1 (inclusive). To fix some notation, call this set $C_0$. Then, remove the open interval $(\frac{1}{3},\frac{2}{3})$. This leaves two closed intervals, namely $[0,\frac{1}{3}]$ and $[\frac{2}{3},1]$. Call the union of these two intervals $C_1$. As was just noted, $C_1$ consists of two closed intervals. Remove the middle third from each of these to get $C_2$, which consists of four closed intervals, each of length $\frac{1}{9}$. Remove the middle third from each of these to get $C_3$, which consists of eight closed intervals, each of length $\frac{1}{27}$. Keep doing this until the end of time (and beyond!). The object that you get in the limit is the Cantor set.

(Technically, we should probably define the Cantor set to be $$\bigcap_{n=0}^{\infty} C_n,$$ but this detail it not really needed to get the right intuition.)

Since this seems related to your computation of the "area" of the Sierpinski gasket, this construction also gives us a way of trying to determine the "size" of the Cantor set. At the $n$-th stage of the construction, we have $2^{n}$ closed intervals, each of which has length $3^{-n}$ (for example, $C_2$ consists of $2^2 = 4$ intervals, each of which is of length $3^{-2} = \frac{1}{9}.$ Thus the "size" of the Cantor set is $$\lim_{n\to\infty} 2^n 3^{-n} = \lim_{n\to\infty} \left( \frac{2}{3} \right)^{n} = 0.$$ In other words, the Cantor set has no "length," which implies that the Cantor set is rather small. On the other hand, it has uncountably many points, so it is also rather large.

• There is at least one more nice construction of the Cantor set (at least, it is the one that I find most useful in my own work). We can define two functions on the real line: $\varphi_1,\varphi_2 : \mathbb{R}\to \mathbb{R}$ by $$\varphi_1(x) = \frac{1}{3} x \qquad\text{and}\qquad \varphi_2(x) = \frac{1}{3}x + \frac{2}{3}.$$ Basically, all this says is that both functions take real numbers as input, give real numbers as output, and behave as the above formulae indicate. That is, $\varphi_1$ multiplies its input by $\frac{1}{3}$, while $\varphi_2$ multiplies its input by $\frac{1}{3}$, then adds $\frac{2}{3}$. Together, these functions form an iterated function system, or IFS. Using some fancy theorems (for those that desire the words: the Banach fixed point theorem, applied to the space of compact subsets of $\mathbb{R}$ with respect to the Hausdorff distance), we can show that there is a unique non-empty set $\mathscr{C}$ of real numbers such that $$\mathscr{C} = \varphi_1(\mathscr{C}) \cup \varphi_2(\mathscr{C}).$$ This set, called the attractor or invariant set of the IFS is the same Cantor set discussed in the two previous constructions. We can think of $\mathscr{C}$ as being the object we get when we apply the two functions to some "sufficiently nice" set infinitely often.

2. We may also be interested in the dimension of a fractal set (perhaps this is what you mean by "equation"?). The basic intuition is as follows: we want to know how the "measure" of a set changes when we scale it. For example:

• The "measure" of a line segment is its length. Suppose that we start with a line segment of length 1. If we scale that line segment by some factor $C$, then we get a new line segment of length $C$.

• The "measure" of a square is its area, say length times width. Suppose that we have a square that measure 1 unit on a side (thus it has area 1), and we scale it by a factor of $C$. We get a new square that measure $C$ on each side, for an area of $C^2$.

• The "measure" of a cube is its volume. If we have a cube of unit volume (i.e. it has volume 1), and we scale it by a factor of $C$, then we get a cube of volume $C^3$.

Note the pattern: if we scale a 1-dimensional line segment by $C$, its measure goes like $C^1$; if we scale a 2-dimensional square by $C$, its measure goes like $C^2$; and if we scale a 3-dimensional cube by $C$, it's measure goes like $C^3$. The exponent gives us the dimension. That is, if the measure of our original set $S$ is denoted by $m(S)$, then the measure of the scaled set $CS$ is given by $m(CS) = C^d m(S)$, where the exponent $d$ is the dimension.

So, what does this have to do with the Cantor set? Suppose that we have a way of measuring the Cantor set, and we get an actual nonzero number out of it, say $m(\mathscr{C})$ (we showed above that the "usual" way of measuring 1-dimensional objections doesn't work---it gives us zero---but let's just pretend that we have a better tool for measuring the Cantor set). If we scale the Cantor set by a factor of 3, we end up with exactly 2 copies of the original Cantor set. Thus the measure of the scaled set is exactly twice the measure of the original set. Thus $m(3\mathscr{C}) = 2m(\mathcal{C})$. So we want to find an exponent $d$ such that $2 = 3^d$ (the new measure is equal to the scaling factor raised to some power).

With a little bit of manipulation, we get $$2 = 3^d \implies \log(2) = d \log(3) \implies d = \frac{\log(2)}{\log(3)} = \log_3(2) \approx 0.6309.$$ Thus we might reasonable say that the Cantor set is approximately $0.6309$-dimensional---this could be called the "fractal dimension" of the Cantor set. Indeed, all of this can be made rigorous for various related (but slightly different) notions of dimension. In every case (that I know of), we get that the Cantor set is $\log_3(2)$-dimensional.

3. A Julia set is a very different kind of monster. In the last construction of the Cantor set, we defined two functions, then applied them over and over and over again. Doing this infinitely often gave us the Cantor set.

A similar kind of game can be played with single functions. Very roughly, suppose that we have a function $\varphi:\mathbb{C}\to\mathbb{C}$. If we apply $\varphi$ to a particular point infinitely often, there are two general behaviours that we might see: the point might be sent to a point very far from where it started (and also very far from where its nearby neighbors started), or the point (and all of its neighbors) can remain close together, and not get too far away from where they started. The collection of points that don't go too far is a (filled-in) Julia set. This is not my area of expertise, so I am going to direct you to WolframMathworld and the Wikipedia page on the topic.

• Thanks! For another question, what would a non ternary cantor set be? Okay, enough questions, let me clarify so I can understand: the ternary cantor set has no length for the same reason the seirpenski triangle has no area - it approaches zero length (or area in the case of the triangle) - it approaches 0 length as n approaches infinity. This makes sense because if you keep dividing a line by two thirds, youre going to end up with a set of points. The one thing I'm confused about is how or why you write 1/3 as 0.1sub3 = 0.02 repeating sub 3. Other than that, thanks! Aug 31 '17 at 4:04
• @user463013, for the same reason that 0.9999... = 1 in decimal. math.stackexchange.com/q/11/276406 Aug 31 '17 at 4:26
• what does the sub3 mean? I understand why 0.9999...=1, I don't see how 1/3 is able to be represented using only 0 and 2. Sorry, but I don't understand how this representation was developed, or where it came from. Thanks! Aug 31 '17 at 4:35
• @user463013, it means it's in base 3. Aug 31 '17 at 4:42
• ohhhhhh that changes things. Thanks! Aug 31 '17 at 4:43