Dimensions of some generalized Cantor sets

by Slav and Victor Sharapov

Here we will discuss dimensions of Cantor-type fractals in the segment [0,1], and in general, in an n-dimensional unit cube. First, recall the classic Cantor set is obtained from the unit segment [0,1]/NOBR> by removing the open interval (1/3,2/3), then from the remaining intervals [0,1/3] and [2/3,1], removing the middle thirds (1/9,2/9) and (7/9,8/9), and so on. In other words, the initiator of the Cantor set is [0,1], and the generator is the union [0,1/3] U [2/3,1]. The dimension of the cantor set is found from the relation

2*(1/3)^d = 1,

hence d = Log(2)/Log(3), approximately 0.63093.

Now, instead of (1/3,2/3), we remove the interval (x, 1-x), where x lies between 0 and 1/2, and build a self-similar fractal by removing the central part of length (1 - 2x)L from every remaining interval of length L. The dimension d of this fractal is given by

2*xd = 1,

(1)

Proposition 1 For any real r, 0 < r < 1, there is a fractal with dimension d = r.

Proof: From equation (1) we see d = -Log(2)/Log(x). The function f(x) = -Log(2)/Log(x) has positive derivative for x in (0,1/2), so f(x) is strictly increasing. Observe lim_x->0+f(x) = 0 and lim_x->(1/2)-f(x) = 1. Noting f(x) is continuous on (0,1/2) completes the proof.

For a Cartesian product of copies of the same Cantor set, the dimension of the product is the sum of the dimensions. That is,

d(C₁ x C₂) = d(C₁) + d(C₂)

(The requirement that C₁ = C₂ guarantees the product is self-similar, so we avoid the many problems of computing the dimension of self-affine fractals.)

As a consequence of this observation and the previous proposition, we see

Proposition 2 For any natural number n, and for any real number r, 0 < r < n, there is an n^th Cartesian power of a Cantor set with the dimension d of the product satisfying d = r.

In particular, for any natural number m, there is a fractal of dimension m.

Proof:By Proposition 1, there is a fractal of dimension r/n. Take the n^th Cartesian power.

We can conclude that for every real number r there is a countable collection of fractals each having dimension r. Can you see how to do this? We will upgrade this result shortly.

Now we consider a more general case, where instead of the central interval (x, 1-x), we remove a shifted interval (x+h, 1-x+h), where x is beween 0 and 1/2, and h is between -x and x. That is, the initiator and generator are

The dimension d of the fractal generated in this fashion satisfies the equation

(x+h)d + (x-h)d = 1,

(2)

Clearly, it is enough to consider nonnegative h because negative values produce mirror images of positive values.

Lemma The dimension d of equation (2) increases with h,

Proof: Consider equation (2) as an implicit function F(h,d) = 1, where d depends on h. Then taking the derivative with respect to h of both sides,

D_dF(h,d)*d' + D_hF(h,d) = 0

where d' denotes the derivative of d with respect to h. Solving for d'

d' = -D_hF(h,d)/D_dF(h,d)

= -(d(x+h)^d-1 - d(x-h)^d-1)/((x+h)^dln(x+h) + (x-h)^dln(x-h))

= (d((x+h)^d-1 - (x-h)^d-1d(-ln(x+h)) + (x-h)^d(-ln(x-h)))

The numerator is positive because x+h is greater than x-h; the denominator is positive because both x+h and x-h are less than 1, so their logarithms are negative. Therefore, the derivative of d with respect to h is positive, so d increases with h.

Proposition 3For any real r, 0 < r < 1, there is a continuum of fractals with dimension d = r.

Proof: By Proposition 1, we can create a Cantor set fractal with any given dimension r. Then, in its generator we can shift the removed middle part by h, 0 < h < D, for some D > 0. By the Lemma, the dimension of hte corresponding fractal increases. However, we can compensate for this increase by increasing hte removed interval. This is possible because of the continuity of equations (1) and (2). Therefore, we obtain a set of fractals of dimension r equivalent to the set of points in the interval (0, D), that is, having the cardinality of the continuum.

In the beginning, we saw that generators of smaller measure produced fractals of smaller dimension. However, there is an interesting construction that allows us to build a fractal of arbitrarly large dimension with a generator of arbitrarily small measure.

To see this, we will consider another generalization of the Cantor set. Here the initiator is the unit interval, and the generator consisit of n disjoint segments, each of length L. (It is also possible to consider segments of different lengths.)

The dimension d of this fractal satisfies the equation

n*Ld = 1,

(3)

Proposition 4 For and s > 0 and for any t > 0, there is a fractal with the measure n*L of its generator less than s, and its fractal dimension d > 1 - t.

Proof: For any given s, form the generator by arbitrarily scattering n disjoint segments of length L = s/(2n) in the unit interval [0, 1]. In this case, equation (3) becomes n*(s/(2n))^d = 1, or

d = (Log(n))/(Log(n) - Log(s/2)).

This has limit 1 as n -> infinity.

Note that two different generalizations of the Cantor set may have the same initial step, though they may turn out to have different dimensions. For example, if we remove the interval (.7, .8) to form a generator, the dimension of the corresponding fractal satisfies the equaution .7^d + .2^d = 1, so d is about 0.83978. On the other hand, let us partition [0, 1] into ten equal intervals. Each interval consists of numbers with the same first digit after their decimal point. We reomve all the numbers that start with a 7, and partition each of the nine remaining intervals into ten intervals and remove all the numbers that have second digit 7. Continue. Note the first stage of both constructions are the same, while the fractals produced are quite different. The dimension of this second fractal satisfies the equation 9*(1/10)^d = 1, so d is about 0.95424.

These fractals ahve the same dimension, but are obviously not the same. The difference lies in the placement and size of the gaps, a subtle topic called lacunarity.

Here are some examples of fractals of dimension 1 in the unit square.

Here the dimension d can be found by solving 4*(1/4)^d = 1, so d = 1.

Here the dimension d can be found by solving 9*(1/9)^d = 1, so d = 1.

In general, if we divide the segment [0, 1] into k² equal intervals and choose k of them to form the generator, then the Cartesian square built from this generator gives a fractal of dimension 1. Taking an arbitrary even Cartesian power, say 2m, gives a fractal with dimension m.