Cantor Sets and Fractal Dimensions
Construction of the Cantor Set
Similarity Dimension of Fractals
The Devil's Staircase
Cantor Set
The Cantor Set is defined iteratively
We start with C 0 the closed interval [0,1]
To get C1 we remove the open middle third of the interval, to obtain the union of intervals [0,1/3] and [2/3,1]
To get C2 we remove the open middle thirds of the intervals of C 1 to obtain the union of intervals [0,1/9], [2/9,1/3],[2/3,7/9], and [8/9,1]
We continue removing the middle thirds in this fashion to get the limiting set C ∞ which we call the Cantor set
Graphs of the First Few Cantor Iterations 0
0
0
0
1/9
1/27
2/27
1/9
2/9
2/9
7/27
1/3
2/3
1/3
2/3
8/27
1/3
2/3
19/27
7/9
20/27
7/9
8/9
8/9
25/27
1
C0
1
C1
1
C2
26/27 1
C3 C4 C5
Ternary Representation of Cantor Se t
Expanding a number in [0,1] into base 3 has a nice geometric representation. Say x=(0.x1x2x3...)3 , then the first digit x 1 tells us which third of the interval x is in. If x1=0 then x is in the left third, x 1=1 then x is in the middle third, x 1=2 then x is in the right third.
The second digit x2 tells us which third x is in with respect to the first digit. So if x=0.02... then it is in the right part of the left third.
Now we can define the Cantor Set to be all numbers in [0,1] whose base 3 expansion only contains zeroes and twos. 0.0...
0.00...
0.01...
0.1...
0.02...
0.2...
0.20...
0.21...
0.22...
Properties of the Cantor Set
C∞ has structure at arbitrarily small scales.
C∞ is made of scaled copies of it self, we call it self similar
C∞ is uncountable.
This is easily proved by an argument similar to the proof for uncountability of the Real Numbers
C∞ has total length 0.
Notice C0 has length 1, C2 has length 2/3, Cn has length (2/3)n so as n→∞ the length of Cn → 0
C∞ is totally disconnected, that is it contains no intervals. However, it contains no isolated points, ie. for every point in the Cantor Set, we can find another arbitrarily close.
The dimension of C is not an integer
Self Similarity Dimension
So how do we determine the dimension of a self similar set? Well consider a square. If we scale the square down by a factor of r, then it will take r 2 of the smaller squares to cover the original square. Similarly for a cube that is scaled down by a factor of r, it will take r 3 of the smaller cubes to reconstruct the original. So suppose we have some self similar set, which when scaled by a factor of r , takes m copies make up the original. Then we define the similarity dimension d , to be the exponent of the relationship m=r d Solving for d, we get d=ln(m)/ln(r)
Similarity Dimension of the Cantor Set
Looking back at our graphs of the Cantor Set, we see that each iterate is composed of two copies of the previous iterate scaled by a factor of 3. So the similarity dimension of the Cantor Set is d=ln(2)/ln(3)=.63093
Cantor Function (Devil's Staircase) Problem 11.2.6
Suppose we pick a point at random from the Cantor set. What is the probability that this point lies to the left of x, where 0≤x≤1 is some fixed number? The answer is given by a function P(x) called the devil's staircase . Right away we should make note that since P(x) is a probability we know that P(0)=0 and P(1)=1. Let us start by building up P(x) from our iterates of the Cantor Set.
11.2.6
Since C0 is the whole interval [0,1], we easily see that if x is some fixed point in [0,1], then x is in C0. So the probability that another point in C0 lies to the left of x is just P 0(x)=x
P1(x) is slightly more interesting. If x is anywhere in [1/3,2/3] we see that half of C 1 lies to the left of [1/3,2/3]. So P 1(x)=1/2 for 1/3≤x≤2/3
11.2.6
P1(x) will be defined piecewise in 3 pieces. 0≤x≤1/3, 1/3≤x≤2/3, 2/3≤x≤1
We already know P1(x) is constant on the middle interval, and we have 2 boundary conditions for each of the left and right intervals. So let us linearly interpolate for both of these intervals.
11.2.6
Assume P1(x)=ax+b for 0≤x≤1/3
P1(0)=0, P1(1/3)=1/2
We get the system of equations: 0=b 1/2=(1/3)*a
So P1(x)=(3/2)x for 0≤x≤1/3
11.2.6
Now assume P1(x)=ax+b for 2/3≤x≤1
P1(2/3)=1/2, P1(1)=1
We get the system of equations: 1/2=(2/3)x+b 1=a+b
Solving for a and b, we get a=3/2, b=-1/2
So P1(x)=(3/2)x-1/2 for 2/3≤x≤1
11.2.6
We can use a similar process to find piecewise functions for P n(x) Here are some plots for n=0,1,2,3,4,5
P0(x)
P1(x)
11.2.6
P2(x)
P3(x)
11.2.6
P4(x)
P5(x)
11.2.6
The limiting function P ∞(x) is the devil's staircase. Strangely enough P ∞(x) can be shown to be continuous.
The derivative of P∞(x) will be zero, nearly everywhere.
