Video 4, chapter 11

Fractal Dimension, Between a Line and an Area

How the similarity dimension log N over log s comes from counting scaled copies, why Sierpinski's triangle has fractal dimension about 1.585, and how logarithms, Hausdorff dimension and topological dimension fit in.

  • Video 4, chapter 11
  • Measuring the unfamiliar

A segment has dimension 1 and a filled square has dimension 2. Sierpinski’s triangle has no area at all, yet the total length of the edges drawn while building it grows without bound, so neither number seems to describe it. What could it mean to say that its dimension is about 1.585? This page builds that number by comparing the size of a shape’s smaller copies with how many of them it takes to rebuild the shape, solves for it with logarithms, and names the kinds of dimension involved and the conditions under which they agree.

Key terms

Fractal dimension

Definition A fractal dimension is a number dd that describes how a set fills space across scales. Every kind of fractal dimension makes precise, in its own way, the idea that covering the set with pieces of a small size ε\varepsilon takes about (1/ε)d(1/\varepsilon)^d pieces as ε\varepsilon shrinks toward 00. The main kinds are the similarity dimension, which counts exact smaller copies of the set, the box-counting dimension, which counts the boxes of a grid that the set meets, and the Hausdorff dimension, which allows covers by sets of any shape. For a segment, a filled square and a solid cube all three give the familiar values 1, 2 and 3, but for other sets they need not be whole numbers. The surprise is that Sierpinski’s triangle, which is made of three half-size copies of itself, has dimension

log⁡3log⁡2≈1.585\frac{\log 3}{\log 2} \approx 1.585

in all three senses. Such a value records how fast the number of pieces grows as the pieces shrink. It does not mean that the set has one and a half directions to move in. For some other sets the different kinds of fractal dimension disagree, so a value should always be given together with its kind.

In plain words Make the pieces half as big and count how many you need. A segment needs 2 times as many, a filled square 4 times and a solid cube 8 times. Sierpinski’s triangle needs 3 times as many, and the exponent that turns 2 into 3 is about 1.585.

Example Sierpinski’s triangle, also called the Sierpinski gasket, is what remains of a filled equilateral triangle after removing the open middle triangle, whose corners are the midpoints of the three sides, then the open middle triangle of each of the three triangles that are left, and so on forever. Its three corner parts are copies of the whole at half the size. Compare it with shapes whose dimension is familiar.

ShapeHalf-size copies neededAs a power of 2
Segment2212^1
Filled square4222^2
Solid cube8232^3
Sierpinski’s triangle32d2^d with d≈1.585d \approx 1.585

A filled triangle of the same outline would need 4 half-size copies, but the gasket has lost its middle and needs only 3. The exponent dd with 2d=32^d = 3 lies between 1 and 2, because 21=22^1 = 2 and 22=42^2 = 4.

Similarity dimension

Definition A similarity with ratio rr is a map of the plane or of space that multiplies every distance by the same number rr, possibly combined with turning, reflecting and shifting. Suppose a set FF is the union of N≥2N \ge 2 pieces, each the image of FF under a similarity with the same ratio rr, where 0<r<10 < r < 1, and let s=1/rs = 1/r be the magnification that brings a piece back to full size. The similarity dimension of this description of FF is the number dd with N=sdN = s^d, which is

d=log⁡Nlog⁡s.d = \frac{\log N}{\log s}.

When the ratios r1,…,rNr_1, \ldots, r_N differ, it is the one number d≥0d \ge 0 with

r1d+⋯+rNd=1.r_1^d + \cdots + r_N^d = 1.

The similarity dimension belongs to the list of maps, not directly to the set. It equals the Hausdorff dimension and the box-counting dimension of FF when the pieces satisfy the open set condition, which asks for a bounded open set VV, not empty, that each map sends into VV, with the images of VV under different maps not overlapping. The condition holds for the Cantor set, the Koch curve, Sierpinski’s triangle and the other examples on this page whose pieces meet at most along edges or at single points. Without the condition, the similarity dimension is still an upper bound for the Hausdorff dimension, but it can be too large.

In plain words Count the copies, find how much each copy must be blown up to match the whole, and ask which power of the magnification gives the count. The answer can be trusted as the dimension of the set only when the copies do not overlap.

Example The Cantor set is what remains of the interval from 0 to 1 after removing the open middle third, then the open middle third of each remaining interval, and so on forever. It is made of 2 copies at one third of the size, so N=2N = 2, s=3s = 3 and

d=log⁡2log⁡3≈0.631.d = \frac{\log 2}{\log 3} \approx 0.631.

The Koch curve is built by replacing a segment with four segments one third as long, the middle two forming a tent, and repeating this on every segment forever. It is made of 4 copies at one third of the size, so its similarity dimension is log⁡4/log⁡3≈1.262\log 4 / \log 3 \approx 1.262. In both sets the pieces meet at most at single points, and the open set condition holds. Now describe the interval from 0 to 1 by the three maps x↦x/2x \mapsto x/2, x↦(x+1)/2x \mapsto (x + 1)/2 and x↦x/2x \mapsto x/2 again. The interval is the union of its three images, so the counting rule would give log⁡3/log⁡2≈1.585\log 3 / \log 2 \approx 1.585. The set has not changed, though, and its dimension is still 1. Two of the three pieces are the same half, so the open set condition fails, and the count of 3 counts that half twice.

Logarithm

Definition For a base b>0b > 0 with b≠1b \ne 1 and a number x>0x > 0, the logarithm log⁡bx\log_b x is the exponent yy for which by=xb^y = x. The base 10 and the base e≈2.718e \approx 2.718, whose logarithm is written ln⁡\ln, are the most common. For any two allowed bases bb and cc,

log⁡bx=log⁡cxlog⁡cb.\log_b x = \frac{\log_c x}{\log_c b}.

So for s>0s > 0 with s≠1s \ne 1 and N>0N > 0, the equation sd=Ns^d = N has exactly one solution, d=log⁡N/log⁡sd = \log N / \log s, where both logarithms use the same base and any base gives the same quotient.

In plain words A logarithm answers the question of which power of a given base produces a given number. It is the tool for finding a missing exponent, which is exactly what a dimension is here.

Example For Sierpinski’s triangle, 2d=32^d = 3. With base 10 logarithms,

d=log⁡103log⁡102≈0.47710.3010≈1.585,\begin{aligned} d &= \frac{\log_{10} 3}{\log_{10} 2} \\[0.4em] &\approx \frac{0.4771}{0.3010} \approx 1.585, \end{aligned}

and natural logarithms give 1.0986/0.6931≈1.5851.0986 / 0.6931 \approx 1.585 as well. As a check, 21.585≈3.00012^{1.585} \approx 3.0001. The value is a little closer to 2 than to 1. That is because exponents measure ratios, and 3 is closer to 4 than to 2 as a ratio, since 4/34/3 is smaller than 3/23/2. Indeed d−1=log⁡2(3/2)≈0.585d - 1 = \log_2 (3/2) \approx 0.585 and 2−d=log⁡2(4/3)≈0.4152 - d = \log_2 (4/3) \approx 0.415.

Hausdorff dimension

Definition The diameter of a set UU, written diam⁡U\operatorname{diam} U, is the least upper bound of the distances between pairs of its points. Let FF be a set on a line, in the plane or in space, and let d≥0d \ge 0 and δ>0\delta > 0. Cover FF by finitely or countably many sets U1,U2,…U_1, U_2, \ldots, each of diameter at most δ\delta, and form the sum

∑i(diam⁡Ui)d.\sum_i \bigl(\operatorname{diam} U_i\bigr)^d.

The greatest lower bound of these sums over all such covers is written Hδd(F)\mathcal{H}^d_\delta(F). As δ\delta shrinks, fewer covers are allowed and this number can only grow, and its limit Hd(F)\mathcal{H}^d(F), possibly infinite, is the dd-dimensional Hausdorff measure of FF. For every set FF there is one number D≥0D \ge 0 such that Hd(F)\mathcal{H}^d(F) is infinite for every d<Dd < D and 00 for every d>Dd > D. This number DD is the Hausdorff dimension of FF. At d=Dd = D the measure may be 00, a positive number or infinite. For a segment the measure H1\mathcal{H}^1 is its length, and for a region in the plane H2\mathcal{H}^2 is a fixed multiple of its area, so segments and regions get the dimensions 1 and 2. The Hausdorff dimension is defined for every set, with no exact copies needed. It is named after Felix Hausdorff, who introduced it.

In plain words Cover the set with small pieces and add up their sizes raised to a trial power. If the power is too small the total explodes as the pieces shrink, and if it is too large the total vanishes. The Hausdorff dimension is the power where the switch happens.

Example Take Sierpinski’s triangle with sides of length 1. After kk removal steps, 3k3^k filled triangles with sides 2−k2^{-k} remain, and they cover the gasket. Each has diameter 2−k2^{-k}, so this cover gives the sum

3k(2−k)d=(32d)k.3^k \bigl(2^{-k}\bigr)^d = \Bigl(\frac{3}{2^d}\Bigr)^k.

If d>log⁡3/log⁡2d > \log 3 / \log 2, then 2d>32^d > 3 and the sum tends to 00 as kk grows, so Hd\mathcal{H}^d of the gasket is 00 and its Hausdorff dimension is at most log⁡3/log⁡2\log 3 / \log 2. If dd is smaller, these particular sums grow without bound, but some cleverer cover might be cheaper, so this alone does not prove the matching lower bound. That is what the open set condition provides. The three corner triangles have interiors that do not overlap, so the interior of the whole triangle works as the set VV, and a classical theorem of Moran and Hutchinson then gives Hausdorff dimension exactly log⁡3/log⁡2\log 3 / \log 2.

Topological dimension

Definition The topological dimension of a set is a whole number defined step by step. The empty set has topological dimension −1-1. A set has topological dimension at most nn if every point of it has arbitrarily small neighborhoods within the set whose boundaries, taken within the set, have topological dimension at most n−1n - 1. The topological dimension is the smallest such nn. It does not change when a set is bent or stretched by a one-to-one continuous map whose inverse is also continuous, so it describes how a set is connected up rather than how its detail scales. For every set in the plane or in space, the topological dimension is at most the Hausdorff dimension. Benoit Mandelbrot, who coined the word fractal, proposed calling a set a fractal when its Hausdorff dimension is larger than its topological dimension, although no single definition of fractal is used by everyone.

In plain words Topological dimension asks how many dimensions it takes to cut a small piece out of the set. A piece of a curve is cut out by removing a few points, a piece of a surface by removing a curve, and each step down counts one dimension.

Example The Cantor set has topological dimension 0. Every point of it lies in arbitrarily small parts of the set, the parts inside the short intervals of the construction, and each such part is cut off from the rest of the set, so its boundary within the set is empty. The Koch curve is a bent and stretched segment, so its topological dimension is 1. Sierpinski’s triangle splits into small parts that touch the rest only at a few corner points, so its dimension is at most 1, and it contains whole segments, its edges, so its dimension is exactly 1. The four-corner dust is built by keeping the four corner squares of side 14\tfrac{1}{4} of a unit square and repeating this in every kept square forever. Its four pieces are separate, so it is a dust of points with topological dimension 0, yet its Hausdorff dimension is log⁡4/log⁡4=1\log 4 / \log 4 = 1.

SetTopological dimensionHausdorff dimension
Segment11
Filled square22
Cantor set0log⁡2/log⁡3≈0.631\log 2 / \log 3 \approx 0.631
Koch curve1log⁡4/log⁡3≈1.262\log 4 / \log 3 \approx 1.262
Sierpinski’s triangle1log⁡3/log⁡2≈1.585\log 3 / \log 2 \approx 1.585
Four-corner dust01

The last four rows are fractals in Mandelbrot’s sense, including the dust, whose Hausdorff dimension is a whole number.

Check yourself

  1. Change a number. The Sierpinski carpet is made by dividing a filled square into 9 equal smaller squares, removing the open middle one, and repeating this in each of the 8 remaining squares forever. Find its similarity dimension, and compare it with the filled square.
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The carpet is made of 8 copies at one third of the size, so N=8N = 8, s=3s = 3 and

d=log⁡8log⁡3≈1.893.d = \frac{\log 8}{\log 3} \approx 1.893.

A filled square is made of 9 copies at one third of the size, which gives log⁡9/log⁡3=2\log 9 / \log 3 = 2. Removing one ninth at every step lowers the dimension a little below 2. The 8 pieces meet only along edges, and the open unit square works as the set VV of the open set condition, since its 8 images are separate open squares inside it. So the Hausdorff dimension of the carpet is also about 1.893.

  1. A classmate says that the Koch snowflake has dimension about 1.262. The snowflake is built by growing a Koch curve outward on each side of an equilateral triangle. When is the claim right?
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It is right for the boundary curve and wrong for the filled snowflake. The boundary is the union of three Koch curves, each of Hausdorff dimension log⁡4/log⁡3≈1.262\log 4 / \log 3 \approx 1.262, and a union of finitely many sets has the largest of their dimensions, so the boundary has dimension about 1.262 too. The filled snowflake contains the original triangle together with its inside, a region of positive area, so its measure H2\mathcal{H}^2 is positive and its Hausdorff dimension is at least 2. No set in the plane has a larger dimension, so it is exactly 2. Its topological dimension is 2 as well, while the boundary curve has topological dimension 1. A dimension should always be attached to the right set, the curve or the region it encloses.

  1. Extend the pattern. Build a set from the interval from 0 to 1 by keeping the first quarter, from 0 to 14\tfrac{1}{4}, and the second half, from 12\tfrac{1}{2} to 1, and repeating this in every kept interval forever. The two pieces now have different sizes. Find the similarity dimension.
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The rule N=sdN = s^d needs pieces of one size, so use the equation for different ratios, here 14\tfrac{1}{4} and 12\tfrac{1}{2},

(14)d+(12)d=1.\Bigl(\frac{1}{4}\Bigr)^d + \Bigl(\frac{1}{2}\Bigr)^d = 1.

With x=(1/2)dx = (1/2)^d, the first term is x2x^2, so x2+x−1=0x^2 + x - 1 = 0 and the positive root is x=(5−1)/2≈0.618x = (\sqrt{5} - 1)/2 \approx 0.618. Then

d=log⁡((1+5)/2)log⁡2≈0.694,\begin{aligned} d &= \frac{\log\bigl((1 + \sqrt{5})/2\bigr)}{\log 2} \\[0.4em] &\approx 0.694, \end{aligned}

since 1/x=(1+5)/21/x = (1 + \sqrt{5})/2, the golden ratio. The answer lies between log⁡2/log⁡4=0.5\log 2 / \log 4 = 0.5, the value if both pieces were quarters, and 11, the value if both were halves. The two kept intervals do not overlap, so the open interval from 0 to 1 works as VV, and the Hausdorff dimension is also about 0.694.

How this lesson connects