CharlieWilliamsFinalProject

A Theory of Fractal Geometry Focusing on

Iterated Function Systems

Charlie Williams

Student Number: 8378184

Supervisor: Dr. Jonathan Fraser

Unit Code: MATH40000

April 29, 2016

Contents

1 Introduction 1

2 Relevant Background Theory 5

2.1 A Brief History of Stefan Banach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Banach’s Contraction Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.3 A Theorem on Iterated Function Systems . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Interesting Fractals and their Generators 13

3.1 The Middle Third Cantor Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.2 Barnsley’s Fern . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.3 The Heighway Dragon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.4 The Sierpinski Tetrahedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

4 Examples 17

5 Measure Theory 22

5.1 What is Measure Theory? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

5.2 The Hausdorff Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5.3 Proof that the Lebesgue Measure of F is 0 . . . . . . . . . . . . . . . . . . . . . . . . 26

6 Proofs Concerning F 28

6.1 |F | = |R| = 2ℵ0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

6.2 F is Totally Disconnected . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

6.3 F is Perfect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

7 What is Dimension? 33

7.1 Calculating an Upper Limit for the Box-counting Dimension of F . . . . . . . . . . . 37

7.2 Calculating a Lower Limit for the Hausdorff Dimension of F . . . . . . . . . . . . . . 37

7.2.1 The Mass Distribution Principle . . . . . . . . . . . . . . . . . . . . . . . . . 38

7.2.2 Potential Theoretic Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

8 An Alternative Approach to Calculating Dimension 45

8.1 Implicit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

8.2 Implicitly Calculating the Dimension of F . . . . . . . . . . . . . . . . . . . . . . . . 50

8.3 Implications of These Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

9 The Hausdorff Content 52

9.1 What is it? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

9.2 Hs∞(F ) = Hs(F ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

References 56

1 Introduction

Figure 1.1: Romanesco Broccoli

No one has ever drawn a perfect circle. What we call cir-

cles are simply approximations to the abstract curve rep-

resenting the set of points in R2 satisfying x2 + y2 = r2,

or any of the infinitely many other definitions. This begs

the question: what use is there in working with this ‘per-

fect’ object if it does not appear in nature? Clearly the

main motivator is that of pure interest, but also it gives

us a better understanding of these approximations which

exist in nature. However, not all objects found in the real

world can be represented by ‘normal’ objects; to illustrate

this consider the examples presented by river deltas, Ro-

manesco broccoli and the internal structure of the human lung. These 3 things cannot be considered

approximations to classical geometric objects so we need to resort to the world of fractals to ad-

equately describe them. So clearly one feature of fractals which sets them apart from other more

traditional disciplines is normally we use mathematics to approximate the real world but here we

have the opposite situation. This reveals another important motivator for the study of fractals: to

gain a deeper understanding of these things which cannot be classified in the conventional manner.

Figure 1.2: A river delta

So what precisely is a fractal? The term was originally

coined by Benoit Mandelbrot and originates from the

Latin ‘fractus’ meaning ‘broken’. In 1982 he published

The Fractal Geometry of Nature [1] in which Mandelbrot

pioneered the field of Fractal Geometry. In it he states

“Fractal geometry is not just a chapter of mathematics,

but one that helps every man to see the world differ-

ently”. Here is one of the most fascinating things about

this topic; its relative newness compared to many other

far more established branches of Mathematics. The Frac-

tal Geometry of Nature propelled the topic into mainstream mathematics and laid the foundations

of modern fractal geometry. This means the field has only really existed in its current form for

around 34 years and resulted in Mandelbrot being widely recognised as The Father of Fractals.

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Figure 1.3: Benoit Mandelbrot

As yet, there still does not exist a precise definition of what it means for an object to be a fractal.

In Fractal Geometry, Mathematical Foundations and Applications, [2] Kenneth Falconer says that

a rigorous definition is the wrong approach. Instead, he draws an analogy with attempting to

define what it means to be alive. There is no solid definition but he identifies these 5 defining

characteristics which are normally exhibited. A fractal will often:

(i) Have a fine structure, i.e. detail on arbitrarily small scales.

(ii) Be too irregular to be described in traditional geometrical language, both locally and globally.

(iii) Have some form of self-similarity, perhaps approximate or statistical.

(iv) Have ‘fractal dimension’ greater than its topological dimension.

(v) Be defined in a very simple way, perhaps recursively.

For my project I shall be focusing on Iterated Function Systems. According to Fractals Everywhere

by Michael Barnsley [8], an I.F.S. is a finite collection of contraction mappings on a complete metric

space. We shall explore what he means by this in detail later. They often lead to a set comprised of

many smaller copies of the original set. This leads me on to the main reason why they interest me;

for the most part, fractals generated in this manner exhibit all of the above qualities. Therefore it

could be argued that I.F.S.s are the purest method of creating fractals.

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Figure 1.4: Constructing the Sierpinski Triangle

To support this claim let us consider a famous fractal which can be generated in this manner, the

Sierpinski Triangle. A method of creating it using an Iterated Function System is you start with a

solid triangle then apply contractions to shrink it by a factor of 12 , create 3 copies then position the

3 resulting smaller triangles in the shape of the original but leaving a gap in the middle. Figure

1.4 shows the result of applying the I.F.S. described 5 times. If we continue and apply it infinitely

many times it yields the set known as the Sierpinski Triangle. If we glance at the list of defining

characteristics we can see (i) and (iii) are true as, no matter how deep we delve, there will still be a

detailed self similar structure. We see that at each stage of construction its area is 34 of the previous.

We iterate infinitely many times to achieve the set itself so, as limk→∞( 34 )k = 0, the resulting set

has ‘area’ 0 (this idea is formalised later when discussing the Lebesgue measure). This means we

have an object containing uncountably many points yet no area so it is clearly too irregular for

traditional geometry and (ii) applies. (iv) is true because its topological dimension is 1 (J. Bell,

[3], p.13) yet its fractal dimension is log 3log 2 ≈ 1.585 (J. Bell, [3], p.14). Finally (v) is clearly the case

from our method of construction.

I will start my project by stating and proving some theorems relevant to the topics being considered

which shall be useful later on. I will then present some interesting examples of fractals which can be

generated with an Iterated Function System, from which I shall move on to some worked examples

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to demonstrate some of the important principles regarding Iterated Function Systems. For the bulk

of the project the overarching theme shall be the fractal generated by applying the contractions

S1(x) = cx, S2(x) = cx + (1 − c) where c ∈ (0, 12 ) to the unit interval. This shall be done in the

context of firstly looking at Measure Theory, secondly proving some interesting characteristics of

this set, then discussing dimension which will act as a starting point to calculate its Hausdorff and

Box Counting Dimensions. Finally I will move on to considering the Hausdorff Content.

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2 Relevant Background Theory

2.1 A Brief History of Stefan Banach

Figure 2.1: Stefan Banach

One of the most influential mathematicians at the turn of the

last century was a man called Stefan Banach. In his 2003 book,

Through a Reporter’s Eyes: The Life of Stefan Banach [4], Ro-

man Kaluza tells his story. He was born in 1892 in Poland

and spent a significant portion of his working life at the Lwow

Polytechnic where he set up The Lwow School of Mathematics.

Tragically, when Germany invaded Poland in World War II, all

universities including his were shut down. After 3 years working

as a lice-feeder, i.e. providing blood for lice which were used to

help develop vaccines, the Russians liberated Lwow. This meant

he could help re-establish the university, however he sadly died

of lung cancer in 1945 before he could make any more major con-

tributions. Arguably his most important result is the Banach-

Tarski paradox, which states that given a solid 3-dimensional

ball, it is possible to take apart then reassemble it to make 2 identical balls through only moving

and rotating pieces, with no shape changing necessary. Banach’s influence is emphasized by the

fact Hugo Steinhaus, who discovered and nurtured him as a young mathematician, said “Banach

was my greatest scientific discovery” (H. Steinhaus, [5], p.121). His main relevance to this project

is that his Contraction Mapping Theorem is an incredibly useful result when studying Iterated

Function Systems. In the this section we shall state and prove it.

2.2 Banach’s Contraction Mapping Theorem

For this theorem we will need the following definitions:

Definition (S. Shirali & H.L. Vasudeva, [6], p.132)

Let (X, d) be a metric space. A continuous map S : X → X is called a contraction if ∃α ∈ (0, 1)

such that ∀x, y ∈ X, d(S(x), S(y)) ≤ αd(x, y).

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A metric space (X, d) is complete if every Cauchy sequence in X has a limit in X. A sequence

{xn}∞n=1 in X is Cauchy if ∀ε > 0, ∃N ∈ N such that ∀n,m ≥ N , d(xn, xm) < ε.

The following theorem and its proof are based on one which can be found on pages 133-134 of

Metric Spaces by S. Shirali & H.L. Vasudeva [6].

Theorem 2.1 (Banach’s Contraction Mapping Theorem) Let S : X → X be a contraction

on the complete metric space (X, d). Then S has a unique fixed point in X.

Proof

Take a complete metric space (X, d) and S a contraction on it. Let x0 ∈ X and define a sequence

{xn}∞n=0 by xn+1 = S(xn) ∀n ≥ 0.

As S is a contraction, ∃α ∈ (0, 1) such that:

d(S(x), S(y)) ≤ αd(x, y) ∀x, y ∈ X

We must show {xn}∞n=0 is Cauchy. So consider, for n ≥ 1, d(xn+1, xn) = d(S(xn), S(xn−1)) ≤αd(xn, xn−1). From this we can deduce:

d(xn+1, xn) ≤ αd(xn, xn−1) ≤ α2d(xn−1, xn−2) ≤ ... ≤ αnd(x1, x0)

Now take 2 distinct positive integers m,n. (X, d) is a metric space so d(xm, xn) = d(xn, xm)

meaning we can assume without loss of generality that m > n. Again using the fact we are working

in a metric space, we can repeatedly apply the triangle inequality to show:

d(xm, xn) ≤ d(xm, xm−1) + d(xm−1, xm−2) + ...+ d(xn+1, xn)

≤ (αm−1 + αm−2 + ...+ αn)d(x1, x0)

= αn(αm−n−1 + ...+ 1)d(x1, x0)

=

(αn

m−n−1∑i=0

αi)d(x1, x0)

<

(αn

∞∑i=0

αi)d(x1, x0)

=αn

1− αd(x1, x0)

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However since n is a positive integer and α ∈ (0, 1), limn→∞ αn = 0 which implies, as m > n,

limn→∞ d(xm, xn) −→ 0. This shows {xn}∞n=0 is Cauchy so therefore, by the definition of a com-

plete metric, there exists an element x ∈ X such that x = limn→∞ xn.

S is a contraction and hence continuous so we can deduce:

S(x) = S(

limn→∞

xn

)= limn→∞

S(xn) = limn→∞

xn+1 = x

So x is a fixed point under the contraction S. Now assume for contradiction that ∃xa, xb ∈ X such

that xa 6= xb with S(xa) = xa and S(xb) = xb, i.e. xa and xb are distinct fixed points in X under S.

In a metric space X, d(x, y) ≥ 0 with equality⇐⇒ x = y, ∀x, y ∈ X, so d(xa, xb) > 0. Futhermore,

as xa, xb are fixed points, we have that:

d(xa, xb) = d(S(xa), S(xb))

≤ αd(xa, xb)

< d(xa, xb)

Which is a contradiction so we must have that xa = xb which means the fixed point of a contraction

on a complete metric space is unique. �

2.3 A Theorem on Iterated Function Systems

For this theorem we will need the following definitions and theorem:


Let (X, d) be a metric space. A subset E is closed if it contains each of its limit points.

Definition (K.J. Falconer, [2], p.4)

In a metric space (X, d), an open ball with centre x and radius r is the set of points within r distance

of x:

B∗r (x) = {y ∈ X | d(x, y) < r}


A subset E of a metric space (X, d) is open if, ∀x ∈ E, ∃r > 0 such that B∗r (x) ⊆ E.

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A space X is compact if ever cover⋃i{Ui} ⊇ X, where each Ui is open, contains a finite subcover.

Definition

Let K([0, 1]) be the set of non-empty, compact subsets of [0, 1].


Given a set X ∈ K([0, 1]), its δ-neighbourhood Xδ is defined to be the superset of X and all points

within δ of it so:

X ⊆ Xδ =⋃x∈X{y ∈ K([0, 1]) | |x− y| ≤ δ}

Theorem 2.2 (Heine-Borel) If a subset X ⊂ Rn is closed and bounded ⇐⇒ it is compact.

Proof

Omitted but details can be found on pages 56-57 of A Taste of Topology by Volker Runde [7].


Fractal Geometry, Mathematical Foundations and Applications, 2nd ed. by K.J. Falconer [2].

Theorem 2.3 Every Iterated Function System on [0, 1] has a unique, non-empty, compact attrac-

tor equal to set found by applying it to [0, 1] infinitely many times.

Consider the Iterated Function System given by the contractions {S1, ..., Sm} on [0, 1] so that,

|Si(x)− Si(y)| ≤ ci|x− y| x, y ∈ [0, 1] ci ∈ (0, 1) (2.1)

Then there exists a unique set F ∈ K([0, 1]) called the attractor such that:

F =

m⋃i=1

Si(F ) (2.2)

Each Si is a contraction map Si : K([0, 1]) → K([0, 1]) so we can define a transformation on a set

E ∈ K([0, 1]) by:

S(E) =

m⋃i=1

Si(E) (2.3)

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For E ∈ K([0, 1]), if we define Sk(E) to be the kth iterate of S on E

i.e. S0(E) = E, Sk(E) = S(Sk−1(E)) for k ≥ 1

Then,

F =

∞⋂k=0

Sk([0, 1]) (2.4)

Proof

Recall how we defined a δ-neighbourhood and take the distance function d for sets A,B ∈ K([0, 1])

to be that of the Hausdorff metric (K.J. Falconer, [2], p.124):

d(A,B) = inf{δ | A ⊂ Bδ, B ⊂ Aδ}

We first must verify that (K([0, 1]), d) satisfies the conditions for a metric, which are:

i) d(x, y) ≥ 0 and d(x, y) = 0 ⇐⇒ x = y

ii) d(x, y) = d(y, x)

iii) d(x, z) ≤ d(x, y) + d(y, z)

(i)

Clearly d(A,B) ≥ 0, ∀A,B ∈ K([0, 1]).

A = B =⇒ B ⊆ A0 and B ⊆ A0

=⇒ inf{δ | A ⊆ Bδ, B ⊆ Aδ} = 0

=⇒ d(A,B) = 0

Now suppose d(A,B) = 0. First observe that A,B ∈ K([0, 1]) =⇒ A,B are compact =⇒ A,B

are closed =⇒ A = A and B = B. Now let b ∈ B, as inf{δ|A ⊂ Bδ, B ⊂ Aδ} = 0 we see that

∀n ∈ N+, b ∈ B ⊆ A 1n

. Therefore ∃a ∈ A such that d(b, a) ≤ 1n . This means that b lies in A = A

by definition and, as b ∈ B is arbitrary, we have B ⊆ A. We can then repeat an identical argument

but with A and B interchanged to demonstrate that A ⊆ B, giving us that A = B. Note that is

argument relies on the compactness of A and B, without which we could not say A = A and B = B

meaning d(A,B) = 0 would not imply A = B.

So we have shown d(A,B) = 0 ⇐⇒ A = B

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(ii)

d(A,B) = inf{δ | A ⊂ Bδ, B ⊂ Aδ} = inf{δ | B ⊂ Aδ, A ⊂ Bδ} = d(B,A)

(iii)

Let d(A,B) = d1 and d(B,C) = d2

Then for δ1 > d1 and δ2 > d2 we have:

A ⊂ Bδ1 , B ⊂ Aδ1 , B ⊂ Cδ2 , C ⊂ Bδ2

=⇒ A ⊂ Bδ1 ⊂ Cδ1+δ2 , C ⊂ Bδ2 ⊂ Aδ1+δ2=⇒ d(A,C) ≤ d1 + d2

Which as δ1 > d1, δ2 > d2

=⇒ d(A,C) ≤ d1 + d2 = d(A,B) + d(B,C)

So we have d(A,C) ≤ d(A,B) + d(B,C) as required. This shows that the distance function defined

above does indeed induce a metric on K([0, 1]).

Consider A,B ∈ K([0, 1]). By 2.3 we see:

d(S(A), S(B)) = d

(m⋃i=1

Si(A),

m⋃i=1

Si(B)

)

Also from the definition of as δ-neighbourhood:

Si(B) ⊂ (Si(A))δ ∀i ⇐⇒m⋃i=1

Si(B) ⊂

(m⋃i=1

Si(A)

)δ

And the same is true with A and B interchanged. This, when combined with the definition of d,

gives:

d

(m⋃i=1

Si(A),

m⋃i=1

Si(B)

)≤ max

1≤i≤md(Si(A), Si(B))

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And by 2.1 we have:

max1≤i≤m

d(Si(A), Si(B)) = (max ci)1≤i≤md(A,B)

And combining all of these gives the inequality

d(S(A), S(B)) ≤ (max ci)1≤i≤md(A,B)

Recall we have a continuous map S : K([0, 1]) → K([0, 1]). Furthermore A and B are arbitrary so

∀A,B ∈ K([0, 1]), d(S(A), S(B)) ≤ αd(A,B), where α = max1≤i≤m ci ∈ (0, 1). Therefore S is a

contraction.

For us to be able to apply Banach’s Contraction Mapping Theorem we need a contraction on

a complete metric space. So let us show (K([0, 1]), d) is complete. Let Xn be an arbitrary Cauchy

sequence in the metric space (K([0, 1]), d). To prove completeness we must show that Xn tends to

a limit in K([0, 1]). Consider:

X = {x ∈ K([0, 1]) | There is a sequence {xn}, such that {xn} → x and, ∀n, xn ∈ Xn}

Note that Xn ∈ K([0, 1]) =⇒ Xn 6= ∅. Therefore there is a convergent sequence {xn} ∈ Xn, ∀n,

such that {xn} → x so, by our definition, X 6= ∅.

Now let α be an arbitrary limit point of X. To show X is closed we must demonstrate that

α ∈ X. If α is a limit point then there is a sequence {αm}, with αm ∈ X ∀m, such that {αm} → α.

Now as each αm ∈ X there must be another sequence {xn} such that {xn} → αm and every element

of this sequence xn ∈ Xn, our arbitrary Cauchy sequence.

Using the fact it is Cauchy, ∃p1 ∈ N such that xp1 ∈ Xp1 and d(xp1 , α1) < 1. Moving 1 along

the sequence we see ∃p2 ∈ N such that xp2 ∈ Xp2 and d(xp2 , α2) < 12 . We can continue this indef-

initely to get a sequence {pm}, with pm ∈ N ∀m, such that d(xpm , αm) < 1m . As we are working

in a metric space we can apply the triangle inequality to see d(xpm , α) ≤ d(xpm , αm) + d(αm, α).

Note:

(i) As d(xpm , αm) < 1m , letting m→∞ means d(xpm , αm)→ 0

(ii) As α is the limit of {αm}, letting m→∞ means d(αm, α)→ 0

Therefore as m → ∞, d(xpm , α) → 0 so α is the limit of {xpm}, and hence {xpm} is Cauchy. Fur-

thermore, by construction, every element of {xpm} belongs to Xpm . This means there is a Cauchy

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sequence {xn} ∈ K([0, 1]) such that every term belongs to Xn with the same limit as {xpm}. By

the construction of X, α ∈ X so by definition X is closed. We have that X = X 6= ∅. Clearly X is

bounded so, by Heine-Borel, X is compact meaning X ∈ K([0, 1]).

Now we must show Xn → X. As previously stated Xn ∈ K([0, 1]) =⇒ Xn 6= ∅ so there is a

sequence {xn} ∈ Xn ∀n. Now Xn ⊂ [0, 1] and Xn is compact so Xn = Xn. Therefore by the com-

pleteness of [0, 1], {xn} → x for some x ∈ [0, 1]. This means that every neighbourhood of x contains

points in Xn meaning x ∈ Xn = Xn so Xn → X. Therefore d induces a complete metric on K([0, 1]).

Therefore we have that S is a contraction on the complete metric space (K([0, 1]), d). Banach’s

Contraction Mapping Theorem applies and ∃!F ∈ K([0, 1]) such that S(F ) = F which proves 2.2.

As Sk([0, 1]) → F as k → ∞ and Si([0, 1]) ⊂ [0, 1] ∀i we have Sk([0, 1]) is a decreasing sequence

of non-empty compact sets containing F . So we have 2.4, i.e. the intersection⋂∞k=0 S

k([0, 1]) = F .

�

Remark

We showed during the preceding proof that d induces a metric on K([0, 1]), but what about P([0, 1])?

P([0, 1]) is the power set of [0, 1] i.e. the set of all subsets of [0, 1], not just the non-empty compact

ones as in the case of K([0, 1]). Consider the 2 sets [0, 1], [0, 1] ∩Q ∈ P([0, 1]).

Clearly [0, 1] ∩Q ⊂ [0, 1] =⇒ [0, 1] ∩Q ⊆ ([0, 1])0. Now ∀x ∈ [0, 1],∀n ∈ N+, x ∈ ([0, 1] ∩Q) 1n

as Q

is dense in R meaning each real number has rationals arbitrarily close.

d([0, 1], [0, 1] ∩ Q) = 0 yet clearly [0, 1] 6= [0, 1] ∩ Q so the first condition which must be satis-

fied for a metric fails meaning d does not induce a metric on P([0, 1]). This example illustrates

a broader idea. Note that although [0, 1] is compact by Heini-Borel as it is closed and bounded

but [0, 1] ∩ Q is not. The implication d(A,B) = 0 =⇒ A = B relies on the fact A = A and

B = B because otherwise d(A,B) = 0 =⇒ B ⊆ A and A ⊆ B, which does not imply A = B.

We can illustrate by returning to the example. First observe that, since every real number is the

limit of a sequence of rational numbers, [0, 1] ∩Q = [0, 1] by definition. So we see that for [0, 1],

[0, 1] ∩Q ∈ P([0, 1]), we have [0, 1] 6= [0, 1] ∩Q, but:

d([0, 1], [0, 1] ∩Q) = 0 =⇒ [0, 1] ⊆ [0, 1] ∩Q = [0, 1] and [0, 1] ∩Q ⊆ [0, 1] = [0, 1]

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3 Interesting Fractals and their Generators

Although later on much of the focus of this project will be one specific fractal this section presents

4 different examples of fractals which can be generated with an Iterated Function System. To give

a glimpse into the variety which can be created using this method, included are Iterated Function

Systems with 1, 2 and 3 dimensional starting sets, which use both the real and complex plane and

which result in both self-similar and self-affine fractals.

3.1 The Middle Third Cantor Set

Despite this set taking his name Georg Cantor did not explicitly define this set. Instead he described

an abstract one with its properties and the precise construction seen here came along later. To

define the Middle Third Cantor Set we start with the unit interval, defining C0 = [0, 1]. Then to

get C1 we remove the open middle 13 to leave 2 closed intervals each of length 1

3 . Then to get C2 we

remove the open middle 19 to leave 4 closed intervals each of length 1

9 . And so to get Ck, k ≥ 1 we

remove the open middle 13k

from each of the 2k−1 closed intervals which constitute Ck−1 to leave

2k intervals each of length 13k

. We then define the Middle Third Cantor set to be C =⋂∞k=0 Ck

(C.D. Aliprantis & O. Burkinshaw, [9], p.41-42). Here is an Iterated Function System which, when

applied to [0, 1], will generate C:

S1(x) =1

3x S2(x) =

1

3x+

2

3

As is illustrated by Figure 3.1 both S1 and S2 will scale down by a factor of 13 , then S2 will also

reposition the result. This set exhibits some fascinating characteristics which have made it the focus

of extensive research. These include it have Lebesgue measure 0 and being uncountable, totally

disconnected and perfect, all of which shall be proved and discussed in more detail later. As a result

we shall be considering this set and variations of it throughout this project, so henceforth we will

be using C to denote the Middle Third Cantor Set.

Figure 3.1: Constructing the Middle Third Cantor Set

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3.2 Barnsley’s Fern

This fractal was first described by Michael Barnsley in his 1993 book ‘Fractals Everywhere’ [8]. It is

incredibly visually striking and was constructed as a demonstration to show it is possible to create

sets such as this with Iterated Function Systems. It differs from the others in this section because

it is an example of a self-affine set, rather than self similar. Self-affine fractals are ones which are

scaled by differing amounts on each axis.

S1(x, y) =

0 0

0 0.16

·xy

S2(x, y) =

0.85 0.04

−0.04 0.85

·xy

+

0

1.6

S3(x, y) =

0.2 −0.26

0.23 0.22

·xy

+

0

1.6

S4(x, y) =

−0.15 0.28

0.26 0.24

·xy

+

0

0.44

Let us break down what each of the Si does; S1 generates the stem, S2 generates progressively

smaller leaves then S3 and S4 generate the largest left and largest right hand leaves respectively.

In the natural world many things including all the examples mentioned in the introduction (river

deltas, Romanesco broccoli and the internal structure of human lungs) are structured in a similar

manner to a fern. What is meant by this is there a main stem with progressively smaller parts

which to some degree resemble the whole branching off. Therefore this fern can be used as a starting

point to model many natural phenomena. This provides evidence to substantiate the claim in the

introduction that fractals are a fantastic way to model the natural world.

Figure 3.2: Barnsley’s Fern

14 of 57

3.3 The Heighway Dragon

In the introduction it was said that Iterated Function Systems can created fractals with incredibly

fine detail in a simple manner, and this is a perfect example of that. The Heighway dragon is

defined by 2 very simple formulae which generate this beautifully intricate set. This has led to its

inclusion in popular culture: in Jurassic Park there are 7 depictions of it after varying amounts

of iterations displayed in between chapters. Fictional mathematician Ian Malcolm claims that his

modelling of Jurassic Park resulted in this fractal, and hence it was too chaotic to ever work (J.M.

Crichton, [10]). This has resulted in it being known to many as ‘The Jurassic Park Dragon’.

Figure 3.3: The Heighway Dragon

Below we see 2 pairs of contractions {S1, S2} and {S′1, S′2}. An interesting thing to note in this

example is that the resulting attractor generated by applying {S1, S2} to [0, 1] ⊂ R2 is identical to

the one which results from applying {S′1, S′2} to [0 + 0i, 1 + 0i] ⊂ C (A. Scott, [11]). This shows

that Iterated Function Functions Systems are equally applicable to the complex plane, although in

the project we shall be focusing on subsets of Rn.

S1(x, y) =

1/2 −1/2

1/2 1/2

·xy

S2(x, y) =

−1/2 −1/2

1/2 −1/2

·xy

+

1

0

S′1(z) =

(1 + i)z

2S′2(z) = 1− (1− i)z

2

Both of these will rotate and scale down by a factor of√22 , S1 rotates by −π4 , S2 rotates by π

4 and

translates, as can be seen in Figure 3.4.

Figure 3.4: Constructing the Heighway Dragon

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3.4 The Sierpinski Tetrahedron

Here we have an example of a Iterated Function System which can be applied to a 3-dimensional

starting set. Recall how we defined the Sierpinski Triangle in the introduction. There is a 3 dimen-

sional analogue to this called the Sierpinski Tetrahedron. It can be made by starting with a solid

tetrahedron then apply a contraction to shrink it by a factor of 12 , create 4 copies and positioning

them resulting smaller tetrahedrons in the shape of the original with the vertices touching but

leaving a gap in the middle. Consider this set of contractions:

S1(x, y, z) =

1/2 0 0

0 1/2 0

0 0 1/2

·x

y

z

S2(x, y, z) =

1/2 0 0

0 1/2 0

0 0 1/2

·x

y

z

+

1/2

0

0

S3(x, y, z) =

1/2 0 0

0 1/2 0

0 0 1/2

·x

y

z

+

1/4

0√

3/4

S4(x, y, z) =

1/2 0 0

0 1/2 0

0 0 1/2

·x

y

z

+

1/4√

3/12

1/√

6

This I.F.S. will generate the Sierpinski Tetrahedron when applied to the tetrahedron with vertices

at a = (0, 0, 0), b = (1, 0, 0), c = ( 12 , 0,

√32 ) and d = ( 1

2 ,√36 ,√63 ). Figure 3.5a illustrates this. As is

clear from how all Si multiply by 12I3×3, where I3×3 is the 3× 3 identity matrix, all 4 shrink by a

factor of a half and then the vectors added to each of the last 3 have the job of repositioning the

resulting object in the correct place, so as to replicated the structure of the original. For a better

indication of what the Sierpinski Tetrahedron looks like itself see Figure 3.5b.

Figure 3.5: The Sierpinski Tetrahedron

(a) 1st stage of construction (b) After several more iterations

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4 Examples

Problem 1

Find a pair of similarity transformations on R for which [0, 1] is the attractor, are there infinitely

many such pairs?

Consider:

S1(x) = 12x S2(x) = 1

2x+ 12

S1([0, 1]) =

[0,

1

2

]S2([0, 1]) =

[1

2, 1

]So S1([0, 1]) ∪ S2([0, 1]) = [0, 1] which shows [0, 1] is a fixed point under the S1, S2 defined above.

Therefore, by the Theorem 2.3, [0, 1] is the unique attractor.

In the more general case of:

S1(x) = cx S2(x) = (1− c)x+ c

S1([0, 1]) = [0, c] S2([0, 1]) = [c, 1]

So S1([0, 1])∪S2([0, 1]) = [0, c]∪[c, 1] = [0, 1]. Therefore, by the same reasoning as above, [0, 1] is the

unique attractor. As c ∈ (0, 1) this gives us uncountably many pairs of similarity transformations

on R for which [0, 1] is the unique attractor.

Problem 2

Find a set of 4 similarity transformations for which the Middle Third Cantor Set is the attractor.

Is it possible to find a set of 3 similarity transformations which will also generate it?

As mentioned in the previous section to generate the C with 2 similarity transformations we apply

this to the unit interval:

Sa(x) =1

3x, Sb(x) =

1

3x+

2

3

17 of 57

So to find 4 similarity transformations which generate it when applied to [0, 1] consider:

Sa(Sa(x)) =1

3

(1

3x)

=1

9x

Sa(Sb(x)) =1

3

(1

3x+

2

3

)=

1

9x+

2

9

Sb(Sa(x)) =1

3

(1

3x)

+2

3=

1

9x+

2

3

Sb(Sb(x)) =1

3

(1

3x+

2

3

)+

2

3=

1

9x+

8

9

So we have that the attractor of this set of 4 similarity transformations, when applied to [0, 1], will

generate C:

S1(x) =1

9x, S2(x) =

1

9x+

2

9, S3(x) =

1

9x+

2

3, S4(x) =

1

9x+

8

9

As we can see from the method of construction, by using 4 transformations instead of 2 essentially

we are essentially skipping a stage of construction, as is illustrated in Figure 4.1.

Figure 4.1: 4 Similarity Transformations that Generate the Middle-Third Cantor Set

In much the same manner it is possible to generate C with 3 similarity transformations. In terms

of Sa and Sb as defined above, we can calculate Sa(Sa(x)), Sa(Sb(x)) and combine it with Sb(x)(or alternatively using Sa(x), Sb(Sa(x)) and Sb(Sb(x))

)to get a system with an identical attractor.

Therefore this set of similarity transformations generate C when applied to [0, 1]:

S1(x) =1

9x, S2(x) =

1

9x+

2

9, S3(x) =

1

3x+

2

3

Figure 4.2 illustrates this. We said using 4 transformations to generate C is tantamount to skipping

a stage of construction. When using 3 things become more asymmetric, so let us prove that C is

still the unique attractor of this system as claimed. Note:

S1(Ck) =1

9Ck S2(Ck) =

1

9Ck +

2

9S3(Ck) =

1

3Ck +

2

3

18 of 57

Furthermore we can define Ck in the construction of C by C0 = [0, 1] and, for k ≥ 0, Ck+1 =

13Ck ∪ ( 1

3Ck + 23 )

=⇒ Ck+2 =1

3Ck+1 ∪

(1

3Ck+1 +

2

3

)=

1

3

(1

3Ck ∪

(1

3Ck +

2

3

))∪(1

3Ck+1 +

2

3

)=

1

9Ck ∪

(1

9Ck +

2

9

)∪(1

3Ck+1 +

2

3

)= S1(Ck) ∪ S2(Ck) ∪ S3(Ck+1)

Now recall C =⋂∞k=0 Ck, and note the fact that k1 > k2 =⇒ Ck1 ⊂ Ck2 . Therefore if we have a

fixed n ∈ N,⋂∞k=0 Ck =

⋂∞k=0 Ck+n. Also, as all Si are contractions and hence continuous so ∀k,

Si(⋂∞k=0 Ck) =

⋂∞k=0 Si(Ck). Therefore we have that:

S1(C) ∪ S2(C) ∪ S3(C) = S1

( ∞⋂k=0

Ck)∪ S2

( ∞⋂k=0

Ck)∪ S3

( ∞⋂k=0

Ck)

=

( ∞⋂k=0

S1(Ck)

)∪( ∞⋂k=0

S2(Ck)

)∪( ∞⋂k=0

S3(Ck+1)

)

=

∞⋂k=0

S1(Ck) ∪ S2(Ck) ∪ S3(Ck+1)

=

∞⋂k=0

Ck+2

=

∞⋂k=0

Ck

= C

Showing C is a fixed point under the system {S1, S2, S3}. Therefore, by Theorem 2.3, C is the

unique attractor.

Figure 4.2: 3 Similarity Transformations that Generate the Middle-Third Cantor Set

19 of 57

Problem 3

The Middle Third Cantor Set is named as such because of one of its methods of construction. We

remove the middle third open interval from the unit interval, then removing the open middle third

of the two remaining intervals and repeat ad infinitum. One can also construct variations of this

by removing the open middle quarter or any other amount less than 1 at every stage. So this begs

the question, is it possible to construct the Middle-λ Cantor set, for λ ∈ (0, 1), through similarity

transformations?

At the first stage we remove λ from [0, 1], leaving [0, 12 (1 − λ)] and [ 12 (1 + λ), 1] i.e. 2 intervals

of length 12 (1− λ). Next we remove 1

2λ(1− λ) (i.e. λ multiplied by the length of the interval) from

the 2 remaining intervals, leaving 4 intervals of length 12

(12 (1− λ)− 1

2λ(1− λ))

= 14 (1− 2λ+ λ2).

To get the 3rd stage we remove 14λ(1− 2λ+ λ2) from each to leave 8 remaining intervals of length

12

(14 (1− 2λ+ λ2)− 1

4λ(1− 2λ+ λ2))

= 18 (1− 3λ+ 3λ2 − λ3).

We are starting to see a trend emerging here; let us define lk to be the interval length and rk

the amount removed in the kth stage of construction. Clearly r0 = 0, l0 = 1 and we see rk = λlk−1

and lk = 12 (lk−1 − rk−1) = lk−1( 1−λ

2 ).

This gives us that lk = ( 1−λ2 )k, showing us that the contraction constant in this case is ( 1−λ

2 ) ∈ (0, 12 )

since λ ∈ (0, 1). Note also that this explains why the first few stages which we explicitly calcu-

lated take the form of a 12k

multiplied by the binomial expansion of (1 − λ)k. From here it is

straightforward to deduce that the pair of similarity transformations:

S′1(x) =

(1− λ

2

)x, S′2(x) =

(1− λ

2

)x+

(1 + λ

2

)Generate the Middle-λ Cantor set when applied to [0, 1]. A different way of regarding this result is

that the pair of similarity transformations:

S1(x) = cx S2(x) = cx+ (1− c)

Where c ∈ (0, 12 ) generate the Middle-(1 − 2c) Cantor set when applied to [0, 1]. Let us take F to

be the unique attractor generated by applying S1 and S2 to the unit interval. In the later sections

we shall consider this Fractal F in much greater detail.

20 of 57

Problem 4

Is it possible to define a pair of similarity transformations for which the Smith-Volterra-Cantor Set

is the attractor?

A variation on the more well known Middle Third Cantor set is the Smith-Volterra-Cantor set.

Again we start with the unit interval, defining S0 = [0, 1]. Then to get S1 we remove the open

middle 122·1 = 1

4 to leave 21 = 2 intervals each of length 12 (1− 1

22·1 ) = 38 . Then to get S2 we remove

the open middle 122·2 = 1

16 to leave 22 = 4 intervals each of length 12 ( 3

8 −1

22·2 ) = 532 . So for Sk we

remove the open middle 122k

from each of the 2k−1 closed intervals which constitute Sk−1 to leave

2k intervals each of length 12 (lk−1 − 1

22k), where lk is the length of each closed interval in Sk. We

then define the Smith-Volterra-Cantor set to be S =⋂∞k=0 Sk (T. Beatty, [12]).

It was in part developed by all 3 of the people after whom it is named; in 1875 the British mathe-

matician Henry Smith investigated sets obtained by iteratively removing parts of the unit interval,

in 1881 Volterra used the idea to construct the function that takes his name and in 1883 Cantor

rediscovered the idea (T. Beatty, [12]). Perhaps is most interesting application is in Volterra’s

aforementioned function. We shall not go into details but it is essentially constructed by defining a

function on every interval removed in the construction of the Smith-Volterra-Cantor set. Its deriva-

tive exists and is bounded everywhere yet its derivative is not Riemann-integrable. This means

it provided a counterexample to the second version of Fundamental Theorem of Calculus, namely

if F is differentiable and its derivative F ′ exists and is bounded then∫ baF ′(x)dx = F (b) − F (a),

meaning mathematicians had to re-evaluate (W.W. Dunham, [13], p.171-172).

Coming back to the problem at hand, note that limk→∞1

22k= 0 so, as k → ∞, we see that

lk ≈ 12 lk−1. This is a problem because to define a pair of similarity transformations we need a

constant contraction factor. The contraction factor from S0 to S1 is 38 , S1 to S2 is 5

32 ÷38 = 5

12

and as k increases it tends to 12 . The precise numbers are irrelevant, but what is important is that

these values are not constant (as is the case with the Middle-λ Cantor set) so it is not possible to

write down 2 similarity transformations in the form S1(x) = a1x + b1 and S2(x) = a2x + b2 with

a1, a2, b1, b2 ∈ R which we can apply to [0, 1] to generate this set.

21 of 57

5 Measure Theory

5.1 What is Measure Theory?

To define a measure we must first define a σ-algebra.

Definition (V.I. Bogachev, [16], p.4)

If we have a set X, a collection of subsets Σ is a σ-algebra if:

i) ∅ ∈ Σ

ii) Σ is closed under compliments i.e. E ∈ Σ =⇒ Ec = X\E ∈ Σ

iii) Σ is closed under countable unions i.e. if we have a countable collection {Ei}∞i=1 ⊂ Σ, then⋃∞i=1Ei ∈ Σ

Definition (V.I. Bogachev, [16])

If we have a set X and Σ a σ-algebra over X a, function µ : Σ→ R ∪ {+∞} is called a measure if,

∀E ∈ Σ, it satisfies:

i) Non-negativity: µ(E) ≥ 0

ii) Empty set condition: µ(∅) = 0

iii) Monotonicity: E1 ⊂ E2 =⇒ µ(E1) ≤ µ(E2)

iv) Countable sub-additivity: For a countable collection of subsets Ei

µ

( ∞⋃i=1

Ei

)≤∞∑i=1

µ(Ei)

Measures are essentially a way of assigning what can be intuitively interpreted as size to abstract

sets. A very useful application of Measure Theory is to make integration possible on a much

wider variety of sets; with a measure called the Lebesgue Measure we can integrate many functions

for which it is impossible to calculate the Riemann integral. The 1-Dimensional Lebesgue Outer

Measure is defined to be:

L1(E) = inf

{ ∞∑k=1

l(Ik)∣∣ every Ik open and E ⊆

∞⋃k=1

Ik

}

Where, if Ik = (a, b), then l(Ik) = b − a (V.I. Bogachev, [16], p.26). It assigns length to subsets

of R. One of its noteworthy characteristics is, ∀E ⊂ R, ∀x ∈ R, L1(x + E) = L1(E) where

22 of 57

x + E = {x + y | y ∈ E} (V.I. Bogachev, [16], p.80). This is known as translation invariance and

it essentially says that the Lebesgue measure of a subset of R will not change irrespective of how

far you shift it up and down the real line, which fits with our existing idea of length. Furthermore,

from this we can define a null set. L1(E) = 0⇐⇒ E is a null set. An interesting point to note here

is that all countable sets are null i.e. E countable =⇒ L1(E) = 0. To prove this we must first note

that, as E is countable, it must have an enumeration {en}∞n=1. Now let ε > 0 be arbitrary. Clearly:

E ⊂∞⋃n=1

([en −

ε

2n+2, en +

ε

2n+2

])From which we can deduce that:

L1(E) ≤ L1

( ∞⋃n=1

([en −

ε

2n+2, en +

ε

2n+2

]))(5.1)

≤∞∑n=1

L1([en −

ε

2n+2, en +

ε

2n+2

])(5.2)

=

∞∑n=1

2ε

2n+2(5.3)

=ε

2

∞∑n=1

1

2n(5.4)

=ε

2(5.5)

< ε (5.6)

Noting that to obtain 5.1 we have used the monotonicity of the Lebesgue measure, 5.2 we have

used the sub-additivity of the Lebesgue measure and 5.3 we have used that, if [a, b] is an interval,

L1([a, b]) = l([a, b]) = b−a. Therefore, as ε > 0 is arbitrary, we have that E countable =⇒ L1(E) =

0. �

23 of 57

5.2 The Hausdorff Measure

A measure closely related to the topics later in this project is the Hausdorff measure. To define

this we first need the following definitions.


Given metric space (X, d), we define the diameter of a set E ⊂ X to be:

|E| = diam(E) = sup{d(x, y) | x, y ∈ E}


Given a set E, a countable collection of sets {Ui}∞i=1 is a δ-cover for E if E ⊆⋃∞i=1 Ui and

∀i, |Ui| ≤ δ.

Now with this in mind let:

Hsδ(E) = inf

{ ∞∑i=1

|Ui|s∣∣∣{Ui}∞i=1is a δ-cover of E

}

Then we define the s-dimensional Hausdorff measure of E to be Hs(E) = limδ→0Hsδ(E) (K.J.

Falconer, [2], p.27). As we let δ → 0 the amount of permissible δ-covers will decrease meaning the

infinium can only increase. Therefore Hsδ(E) ≤ Hs(E) and limδ→0Hsδ(E) will always exist as we

are working in R≥∪{∞}. Now let’s prove the s-dimensional Hausdorff measure is indeed a measure.

Firstly note if we have a cover {Ui}∞i=1 for a set E, for each constituent set of the cover |Ui| ≥0 =⇒

∑∞i=1 |Ui|s ≥ 0 =⇒ inf{

∑∞i=1 |Ui|s} ≥ 0 =⇒ Hs(E) ≥ 0.

Next see ∅ ⊆ ∅ so ∅ is a cover for ∅ =⇒ Hsδ(∅) = 0 =⇒ Hs(∅) = 0.

To demonstrate monotonicity consider E1 ⊂ E2 and let {Ui} be a δ-cover for E1 and {Vj} be

a δ-cover for E2. This means E1 ⊂⋃j Vj too, and:

Hsδ(E1) = inf

{ ∞∑i=1

|Ui|s}≤ inf

{ ∞∑j=1

|Vj |s}

= Hsδ(E2)

And letting δ → 0 gives us Hs(E1) ≤ Hs(E2).

Now we move on to sub-additivity. Let {Ei}∞i=1 be a countable collection of sets. If Hs(Ei) = ∞

24 of 57

for some i then the result is immediate so assume, ∀i, Hs(Ei) <∞ =⇒ Hsδ(Ei) <∞ ∀δ > 0.

Take ε > 0. For each Ei there is at least 1 countable δ-cover {U (i)j }∞j=1 whose sum achieves

the infinium in the above definition so that, as Hsδ(E) is defined to be the infinium, we can always

make it an increment larger, giving us:

∞∑j=1

|U (i)j |

s ≤ Hsδ(Ei) +ε

2i

And⋃∞i=1{U

(i)j }∞j=1 is a countable δ-cover for

⋃∞i=1Ei so:

Hsδ

( ∞⋃i=1

Ei

)≤

∞∑i=1

∞∑j=1

|U (i)j |

s

≤∞∑i=1

(Hsδ(Ei) +

ε

2i

)

=

∞∑i=1

Hsδ(Ei) + ε

This is true ∀ε > 0 so letting δ → 0 gives us the desired result:

Hs( ∞⋃i=1

Ei

)≤∞∑i=1

Hs(Ei)

So we have that Hs(E) is a measure. �

We shall see some applications of the Hausdorff Measure later when discussing dimension and

the Hausdorff Content.

25 of 57

5.3 Proof that the Lebesgue Measure of F is 0

Recall how we defined F to be the fractal generated by applying the contractions S1(x) = cx, S2(x) =

cx+(1−c) where c ∈ (0, 12 ) to [0, 1] infinitely many times. Let Fk be the union of 2k closed intervals

of length ck that remain after k iterations of S1 and S2, so that:

F =

∞⋂k=0

Fk

We will show by induction that L1(Fk) = (2c)k

L1(F0) = 1

L1(F1) = L1([0, c] ∪ [1− c, 1]) = 2c

L1(F2) = 4c2

Assume L1(Fn) = (2c)n and let us consider L1(Fn+1). Note that, as S1(Fn) and S2(Fn) are disjoint,

L1(S1(Fn) ∪ S2(Fn)) = 2L1(S1(Fn)) so we have that:

L1(Fn+1) = L1(S1(Fn) ∪ S2(Fn))

= 2L1(S1(Fn))

= 2cL1(Fn)

= (2c)(2c)n

= (2c)n+1

Therefore, by induction, L1(Fk) = (2c)k

c ∈(

0,1

2

)=⇒ 2c ∈ (0, 1)

=⇒ limk→∞

(2c)k = 0

=⇒ limk→∞

L1(Fk) = 0

F =⋂∞k=0 Fk so we have the desired result L1(F ) = 0. �

Remark

It would be tempting to say that it is obvious L1(F ) = 0 as at every stage of construction we reduce

the length of the sum of the constituent intervals. We repeat ad infinitum so surely this amount

will reduce to 0. This is not true; recall the Smith-Volterra-Cantor set as defined in the previous

26 of 57

section. Note that to achieve the kth stage of construction of S we remove 2k−1 intervals of length

2−2k. Therefore we can calculate the Lebesgue measure of the resulting set as follows:

L1(S) = 1−∞∑k=1

2k−1

22k

= 1− 1

2

∞∑k=1

1

2k

= 1− 1

2

(12

1− 12

)= 1

2

> 0

If you look at Figure 5.1, a rough illustration of this set, we see that the amount being removed

diminishes rapidly at each stage. In the S the amount removed for the kth stage is 122k

which

decreases exponentially and independently of the length of the interval length in the preceding

stage. When we compare this to the Middle Third Cantor Set, we see rk = λlk−1 so is therefore

proportional to the preceding stage’s interval length. Consequently it will decrease a lot more slowly

so it makes sense that we will end up with a positive Lebesgue measure. So here we have a set

which at a glance will look fairly similar to F and is constructed in a similar manner, yet has a

positive Lebesgue measure, showing that the previous result is far from trivial.

Figure 5.1: The Smith-Volterra-Cantor Set

27 of 57

6 Proofs Concerning F

6.1 |F | = |R| = 2ℵ0

Consider c ∈ (0, 12 ), x ∈ [0, 1], S1(x) = cx, S2(x) = cx+ (1− c)and let F be the attractor.

If d is the distance function for the Hausdorff metric, for a set E ⊆ [0, 1],

d(S(E), F ) = d(S(E), S(F )) ≤ cd(E,F ) =⇒ d(Sk(E), F ) ≤ ckd(E,F )

Because c < 1, ck → 0 as k →∞ so the sequence of iterates Sk([0, 1]) converges to the attractor F

i.e.

limk→∞

d(Sk([0, 1]), F ) = 0

So for each k,

Sk([0, 1]) =⋃Ik

Si1 ◦ · · · ◦ Sik([0, 1]) =⋃Ik

Si1(Si2(· · · (Sik) · · · ))

Where Ik is the set of all k-term sequences with ij ∈ {1, 2}. These sequences provide a natural

coding for elements x ∈ F with,

x = xi1xi2 · · · =∞⋂k=1

Si1 ◦ · · · ◦ Sik([0, 1])

This natural coding can be used to show there is a bijection between F and {1, 2}N. It is injective

because each element of {1, 2}N is uniquely determined by an infinite sequence of contractions. It

is surjective because, by the coding method described, every element of {1, 2}N corresponds to an

infinite sequence of contractions.

So we have |F | = |{1, 2}N|

Suppose for contradiction that {1, 2}N is countable i.e. |F | = ℵ0. Then we can enumerate it

by saying:

{1, 2}N = {y(n) | n ∈ N}

But each y(n) is a sequence so:

y(n) = (y(n)m )∞m=1, y(n)m ∈ {1, 2}

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If we define a sequence y = (ym)∞m=1 by the rule:

yn =

1 y(m)m = 2

2 y(m)m = 1

Then y ∈ {1, 2}N but differs from y(m) at the mth place ∀m ∈ N. This is a contradiction and

hence |{1, 2}N| > ℵ0 so, assuming the Continuum Hypothesis as detailed below, |{1, 2}N| = 2ℵ0 .

We have established a bijection between F and {1, 2}N and proved the later to be uncountable so

we have the required result: |F | = 2ℵ0 . �

Remark

This result is interesting because in the previous section we showed that L1(F ) = 0 so F is an un-

countable set with Lebesgue measure 0. We proved previously that all countable sets have Lebesgue

measure 0 whereas the uncountable ones such as [a, b] ⊂ R for which L1([a, b]) = L1((a, b)) = b−a >0 do not. Therefore one may be inclined to assume a set E is countable ⇐⇒ L1(E) > 0. However

this is not true generally and F provides a clear counter example.

A Note on Cantor and the Continuum Hypothesis

The part of the previous proof which establishes that |{1, 2}N| > ℵ0 is more commonly known

as Cantor’s diagonalisation argument. It was published by Georg Cantor in 1892, 18 years after

his first article to include a proof of the uncountability of R. However, it was the first to make

use of infinite tables, an incredibly useful technique used in many things including Godel’s first

incompleteness proof. Unfortunately, the mathematical community of the time did not take well

to Cantor’s ideas. Leopold Kronecker, who went so far as to say “I don’t know what predominates

Cantor’s theory - philosophy or theology, but I am sure that there is no mathematics there”, sum-

marises the general consensus on Cantor’s theorems well (J.W. Dunham, [14]).

This bad reception was mainly due to the fact that the concept of infinity had largely been avoided

throughout the mathematical world up until Cantor’s time. The Cult of Pythagoras associated

good and evil with the finite and the infinite respectively, and accordingly omitted the latter from

their mathematics. Although infinity existed in some forms such as in irrational numbers (π, e, etc.)

it was avoided for the most part. During the 19th century its use became more prevalent, initially

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through Cauchy’s use of it to define integrals in terms of limits. It was not until the advent of Set

Theory that it was possible to view infinity as an object it own right. It was therefore from this

academic heritage that Cantor’s revolutionary theorems were viewed (G.D. Allen, [15], p.15-19).

It should be noted at this point that in modern mathematics Cantor’s proofs are widely accepted.

Those who still do not accept infinity and refuse to use it in their mathematics are in a stark minority.

One person who recognised Cantor’s genius at the time was David Hilbert whom, in 1900, set

out a list of what he considered the biggest 23 unsolved problems in Mathematics. The first of

these was the Continuum Hypothesis. It states that there is no set whose cardinality is strictly in

between that of the integers and the real numbers i.e. There is no set X such that ℵ0 < |X| < 2ℵ0.

Here we see from where its name is derived; the continuum is another name for the real numbers.

However it was later proved to be undecidable using Godel’s incompleteness proof (G.D. Allen,

[15], p.15-19), meaning it is impossible to either prove or disprove. Therefore, in the context of the

previous proof, assuming it to be true gives us that |{1, 2}N| > ℵ0 =⇒ |{1, 2}N| = 2ℵ0 . Therefore

we have the desired result that the cardinality of F is equal to the cardinality of the continuum.

6.2 F is Totally Disconnected

In 2 dimensions, a totally disconnected set is one which contains no intervals, so that is what we

shall prove is the case for F .

Let Fk = Sk([0, 1]), Fk is the set of 2k closed intervals of length ck.

Let x, y ∈ F and we can assume x < y without loss of generality that ∀k ≥ 0:

x, y ∈ F =⇒ x, y ∈ Fk (6.1)

Then ∃k ∈ N such that |x− y| > ck

As each interval has length ck, x and y must be in different intervals. This means that, as every

interval in F is disjoint, there must be an element z, with x < z < y, such that z /∈ Fk =⇒ z /∈ F ,

by taking the contrapositive of 6.1.

As this is true for arbitrary x, y ∈ F , F must be totally disconnected. �

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6.3 F is Perfect

For a set to be perfect it must be closed and contain no isolated points. F is obtained by taking

compliments of opens sets in [0, 1] so it must be closed. To show F has no isolated points we must

demonstrate that:

∀ε > 0, ∀x ∈ F, (Bε(x)\{x}) ∩ F 6= ∅

Take x ∈ F and ε > 0. As F =⋂∞k=0 Fk, x ∈ Fk ∀k ≥ 0

If we take large enough k ∈ N, we have ck < ε

Fk is the finite union of 2k disjoint closed intervals of length ck so is also closed. Let [a, b] be

one of these intervals containing x, so that [a, b] ⊆ Bε(x).

Now consider Fk+1. There are 2 intervals [a, a′], [b′, b] ⊂ [a, b], one of which must contain x.

We can say without loss of generality that x ∈ [a, a′].

By construction F ∩ [b′, b] 6= ∅ so take y ∈ [b′, b]. This means y ∈ (Bε(x)\{x}) ∩ F and, as x

is arbitrary and we have taken any ε > 0, F is perfect. �

Remark

We have shown above that F is both perfect, meaning it is closed and has no isolated points, and

totally disconnected disconnected meaning it contains no intervals. This seems to be a contradiction

- how can a closed set have no connected components, yet contain no isolated points? Here is an

analogy to help visualise how this is possible, consider the set:

{0} ∪{ 1

n

∣∣∣ n ∈ N+}

The set is clearly totally disconnected but {0}, unlike the rest of the set, is not an isolated point

as ∀ε > 0, ∃N ∈ N+ such that n > N =⇒ 1n < ε, so we have that Bε(0)\{0} 6= ∅. Due to its

layered construction, every element of F is like {0} in this example and is not connected to any

other element yet has other elements arbitrarily close. This is how it manages to be both perfect

and totally disconnected. This is a good example of why fractals are a good source of mathematical

interest; they sometimes exhibit qualities which appear paradoxical at first glance such as these.

Interestingly it turns out that, up to homeomorphism, the middle-third Cantor set is the only one

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which exhibits this quality of being both totally disconnected and perfect, meaning that all sets

sets such as these are topologically equivalent to C.

A consequence of this is that the S, the Smith-Volterra-Cantor set, is also totally disconnected

and perfect, in fact you can see with minimal effort we can alter the 2 proofs for F to apply for S as

well. Intuitively the fact S is perfect is not particularity surprising; C is and L1(S) = 12 > 0 = L1(C)

so it makes sense that if C has no isolated points then neither will S. However the fact S is also

totally disconnected is far more surprising: here we have a set with positive measure whilst at the

same time containing no intervals. We proved previously that a set E countable =⇒ L1(E) = 0

in an argument that relies on the fact E is made up singleton points from which we constructed a

cover. One may be inclined to believe that a set being made up of singleton points is synonymous

with a set containing no intervals but S provides a clear counter example.

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7 What is Dimension?

One of the characteristics which Falconer says fractals tend to exhibit is that their fractal dimension

is greater than their topological dimension. What does this mean? If you have a 0-dimensional dot

and scale it by a factor of 2 its total ‘size’ remains the same i.e. increases by a factor of 20 = 1. If

you take a 1-dimensional line and scale it by a factor of 2 its total ‘size’, interpreted here as length,

increases by a factor of 21 = 2. If you take a 2-dimensional square and scale it by a factor of 2 its

‘size’, interpreted here as area, increases by a factor of 22 = 4. If you take a 3-dimensional cube

and scale it by a factor of 2 its ‘size’, interpreted here as volume, increases by a factor of 23 = 8. So

here we have an inkling of what the ‘dimension’ of an object means; it is a measure of how much

the total space it takes up scales when you scale its components. Following this line of thought we

can deduce the following. If an object’s dimension is s then rs = N where r is the amount we scale

by and N is the resulting amount of constituent unit ‘blocks’, Figure 7.1 illustrates this. We can

then rearrange this to get s = logNlog r . This is not rigorous, merely a demonstration to give an idea

of what dimension is.

Figure 7.1: Illustrating Dimenion

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Dimension is not a particularly interesting concept in the real world as it is 3-d. Accordingly

mathematics concerned with the study of real world objects in general deals exclusively with those

of integer dimension. One of the fascinating things about fractals, and in fact one of the key char-

acteristics used to classify them, is they normally have a non-integer dimension. This concept is

hard to comprehend in the normal world. So what is dimension and how is it possible to rigorously

calculate it?

The 2 most popular definitions of dimension are the Hausdorff and the Box-counting and they

are the ones we shall be working with here. Recall how we defined the Hausdorff Measure Hs(E)

in Section 5.2. From that starting point we then define the Hausdorff dimension to be:

dimH E = inf{s ≥ 0 | Hs(E) = 0} = sup{s | Hs(E) =∞}

Figure 7.2: A graph showing the ‘jump’ in the Hausdorff measure

Given an arbitrary set E, Figure 7.2 shows a graph with s on the x-axis and Hs(E) on the y-axis.

We see that at the red dotted line Hs(E) ‘jumps’ from∞ to 0, and it is this value of the ‘jump’ that

we take as the value for dimH E (K.J. Falconer, [2], p.31). This is what motivates the definition

and explains the equality of inf{s ≥ 0 | Hs(E) = 0} and sup{s | Hs(E) =∞} in the definition. An

alternative way of viewing this diagram is shown below.

Hs(E) =

∞ 0 ≤ s < dimH E

0 s > dimH E

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It should be noted that at dimH E, Hs(E) ∈ [0,∞]. The Hausdorff dimension is the most widely

used and exhibits all of the properties which are considered ‘desirable’ in a dimension, which will

be discussed shortly. However, as mentioned previously, it is not the only one we shall be working

with, the other dimension we shall consider with is the Box-counting dimension. Let E 6= ∅ be a

bounded subset of Rn and define Nδ(E) to be the smallest number of sets of diameter less than or

equal to δ such that E is contained in their union. Then we define the lower and upper box-counting

dimensions to be (K.J. Falconer, [2], p.41):

dimBE = limδ→0

logNδ(E)

− log δ

dimBE = limδ→0logNδ(E)

− log δ

A benefit of these 2 is, provided the set under consideration is bounded, they are always guaranteed

to exist. This is because a bounded set will always have an infinium and supremum, and taking

this limit is essentially taking a limit of elements in such a set. Furthermore if dimBE = dimBE

then we say that the Box-counting dimension of E is (K.J. Falconer, [2], p.41):

dimB E = limδ→0

logNδ(E)

− log δ

Let us compare the two. If, given a set E, we can cover it by Nδ(E) sets of diameter δ, i.e. there

is a cover {Ui}Nδ(E)i=1 such that |Ui| ≤ δ, ∀i. As Hsδ(E) = inf{

∑∞i=1 |Ui|s

∣∣∣{Ui}∞i=1is a δ-cover of E}:

Hsδ(E) ≤Nδ(E)∑i=1

|Ui|s ≤ δsNδ(E)

So Hsδ(E) ≤ δsNδ(E). Clearly Hsδ(E) ≥ 0 so we see δsNδ(E) ≥ 0 =⇒ logNδ(E) + s log δ ≥ 0.

Therefore s ≤ − logNδ(E)log δ . Then by letting δ → 0 we see:

s ≤ limδ→0

− logNδ(E)

log δ= dimBE ≤ limδ→0

− logNδ(E)

log δ= dimBE

Now dimBE < ∞ =⇒ s < ∞ =⇒ s ≥ dimH E. Therefore we can relate the Hausdorff and

Box-counting dimensions with the inequality:

dimH E ≤ dimBE ≤ dimBE (7.1)

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So what are the characteristics of these differing definitions of dimension? Below you see some

properties which are considered ‘desirable’ in a dimension (K.J. Falconer, [2], p.41):

Monotonicity : E1 ⊂ E2 ⇒ dimE1 ≤ dimE2

Countable Stability : {Ei} countable⇒ dim(⋃∞i=1Ei) = sup1≤i<∞Ei

Lipschitz Invariance : f a bi-Lipschitz transformation ⇒ dimE = dim f(E)

Countable Sets : E finite or countable⇒ dimE = 0

Open Subsets : E an open subset of Rn ⇒ dimE = n

Smooth Manifolds : E a smooth m-dimensional manifold ⇒ dimE = m

The last 2 properties, Open Subsets and Smooth Manifolds, ensure that whatever dimension we

are working with is consistent with the usual definition of dimension. Consequently they could

be considered the most important; there seems little use in a rigorous definition of dimension if it

fails to coincide with what we already know as dimension. Below we see a table showing how the

Hausdorff and Box-counting dimensions compare:

dimH E dimBE dimBE

Monotonicity X X X

Countable Stability X × ×Lipschitz Invariance X X X

Countable Sets X × ×Open Sets X X X

Smooth Manifolds X X X

As we can see, Hausdorff dimension has all properties (K.J. Falconer, [2], p.32) and the Box-

counting has all except 2 (K.J. Falconer, [2], p.48). To see why the Box-counting dimensions

fails where it does we first must note that, if E denotes the closure of E, dimBE = dimBE and

dimBE = dimBE. This is because if we let B1, ...,Bn be a finite collection of Nδ(E) closed balls

diameter δ, then E ⊆⋃ni=1 Bi =⇒ E ⊆

⋃ni=1 Bi and therefore:

dimBE = limδ→0

logNδ(E)

− log δ= limδ→0

logNδ(E)

− log δ= dimBE

And the upper equivalent statement for the upper Box-counting limit can be demonstrated in

the same manner. Now consider the set E = [0, 1] ∩ Q. As Q is everywhere dense in R we see

that E = [0, 1] by definition, so dimB E = dimB E = dimB([0, 1]) = 1. Therefore, as [0, 1] ∩ Q

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is countable but has a Box-counting dimension of 1, dimB does not satisfy the Countable Sets

condition. Furthermore E is the union of countably many singleton sets all of which have dimension

0, but the union of which has dimension 1 so Countable Stability does not hold generally for the

Box-counting dimension either.

7.1 Calculating an Upper Limit for the Box-counting Dimension of F

As previously mentioned, if we can cover F with Nδ(F ) sets of diameter less than or equal to δ,

then the upper limit of the box counting dimension for F is:

dimBF = limδ→0− logNδ(F )

log δ

So let us fix δ ∈ (0, 1) and then we can choose a k ∈ N so that ck < δ ≤ ck−1. Now recall how

we defined F =⋂∞k=0 Fk =⇒ ∀k ≥ 0, F ⊂ Fk and each Fk is a ck-cover of F so, as ck ≤ δ, Fk is

a δ-cover of F . Fk consists of 2k intervals so we could take Nδ(F ) = 2k but noting that ck ≤ δ,

we see that it may be possible to reduce this if the difference between ck and δ allows us to create

a cover out of fewer intervals so Nδ(F ) ≤ 2k. Note also that by letting δ → 0, ck → 0 which, as

c ∈ (0, 12 ), implies k →∞. From this we see that:

dimBF = lim supδ→0

logNδ(F )

− log δ

≤ lim supk→∞

log 2k

− log ck−1

= lim supk→∞

k log 2

−(k − 1) log c

= lim supk→∞

log 2

− log c+ log ck

= lim supk→∞

log 2

− log c+ log ck

=log 2

− log c

So we have: dimBF ≤ log 2− log c , i.e. our upper bound for the upper box counting dimension of F is

equal to log 2− log c .

7.2 Calculating a Lower Limit for the Hausdorff Dimension of F

As dimH F ≤ dimBF ≤ dimBF , if we can demonstrate that the lower limit of the Hausdorff

dimension for F to be equal to the upper limit of its Box-counting dimension then it shows F ’s

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Hausdorff and Box-counting dimensions are equal to that value. Calculating and upper bound for

dimBF was relatively straight forward as we simply considered a specific cover and evaluated what

happened as we decreased the size of the constituent covering sets. A similarly direct calculation

for a lower bound for dimH F we would far harder, as we would need to consider how all potential

covers behaved. A way to work around this is to demonstrate no constituent set Ui of a δ-cover

can cover too much of the set which is the focus of the calculation when compared to its weighted

‘size’ |Ui|s. This means, as {Ui} would cover the set,∑i |Ui|s cannot be too small, which we can

then relate to the Hausdorff dimension by taking infima and letting δ → 0. The following method

makes use of a mass distribution to achieve this.

7.2.1 The Mass Distribution Principle

We shall first need the following definitions:

Definition (V.I. Bogachev, [16], p.6)

Recall how we defined a σ-algebra in Section 5. A subset E ⊂ Rn is Borel if it belongs to the

smallest σ-algebra containing all open sets.

Therefore a Borel set is one which can be formed by taking compliments and countable unions

of open sets. It is worth noting that this does not preclude closed sets from being Borel; the com-

pliment of an open set is closed.


The support of a measure µ on R the smallest closed set E ⊂ R such that µ(R\E) = 0. It is denoted

supp(µ).


We define a measure µ on a bounded subset of E ⊂ Rn to be a mass distribution if supp(µ)⊆ E

and 0 < µ(E) <∞.


Techniques in Fractal Geometry by K.J. Falconer [17].

Theorem 7.1 (The Mass Distribution Principle) Given a mass distribution µ on a set E,

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suppose that for some s, ∃λ, ε > 0 such that µ(U) ≤ λ|U |s, for all Borel sets U ⊂ E such that

|U | ≤ ε, then:

dimH E ≥ s

Proof

{Ui} an arbitrary δ-cover for E implies:

0 < µ(E) ≤ µ(⋃

i

Ui

)≤∑i

µ(Ui) ≤ λ∑i

|Ui|s

(Using non-negativity, monotonicity, sub-additivity, original assumption)

By taking infima we see Hsδ(E) ≥ µ(E)λ then letting δ → 0 gives Hs(E) ≥ µ(E)

λ . By defining µ

to be a mass distribution can say 0 < µ(E) < ∞ and λ > 0 so we see µ(E)λ > 0 =⇒ Hs(E) > 0.

Now recall:

dimH E = inf{s ≥ 0 | Hs(E) = 0} = sup{s | Hs(E) =∞} =⇒ dimH E ≥ s

Which is the desired result. �

Let Fk be defined as previously. We can define a mass distribution by repeated subdivision. Let

µ(F0) = µ(Fk) = 1. Fk is made up of 2k intervals {I(i)k }2k

i=1 each of length ck. So we have

µ(Ik) = 2−k and l(Ik) = ck → 0 as k →∞.

So essentially we are starting a mass of 1 then dividing evenly between each of the 2k intervals

at the kth stage so, for for each k:2k∑i=1

µ(I(i)k ) = 1

Let U ⊆ [0, 1] be Borel. Then we can choose some k ∈ N+ such that ck+1 ≤ |U | < ck. This means

U can intersect no more than 2 of the intervals in Fk so we have:

µ(U) ≤ 2µ(Ik)

= 2 · 2−k

= 2(ck)x (for some x ∈ R)

= 2(c−1 · ck+1)x

= 2c−x(ck+1)x

≤ 2c−x|U |x

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Figure 7.3

So we have that µ(U) ≤ 2c−x|U |x. Therefore if we set λ = 2c−x > 0 as c ∈ (0, 1) the conditions for

the Mass Distribution Principle are satisfied and so we have that dimH F ≥ x. So what is x? From

the previous calculation we see:

2 · 2−k = 2(ck)x =⇒ log 2(1−k) = log(ck)x

=⇒ (1− k) log 2 = kx log c

=⇒ x =(1− k) log 2

k log c

=⇒ x =log 2k − log 2

log c−→ − log 2

log cas k −→∞

So we have that s = log 2− log c =⇒ dimH F ≥ log 2

− log c .

7.2.2 Potential Theoretic Method

Before we can start talking about this method we need a definition and 2 theorems, whose relevance

shall become clear later on.


When working in Rn with the usual notion of distance, ∀x, y ∈ Rn, d(x, y) = |x − y|, then closed

ball with centre x and radius r is the set of points within r distance of x:

Br(x) = {y ∈ Rn | |x− y| ≤ r}

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Theorem 7.2 (Covering Lemma) Let {Bi | i ∈ I} be a countable collection of balls in a bounded

region of Rn. Then there is a countable disjoint sub-collection {Bj | j ∈ J} where J ⊂ I such that:⋃i

Bi ⊂⋃j

4Bj

Meaning every ball in {Bj} is concentric with one in {Bi} and the {Bi} is contained in the union

of all balls with centres coinciding with each Bj and 4 times the radius.

Proof

Omitted but details can be found on pages 66-67 of Fractal Geometry, Mathematical Foundations

and Applications, 2nd ed. by Kenneth Falconer [2].

The following theorem and its proof are based on one which can be found on page 67 of Frac-

tal Geometry, Mathematical Foundations and Applications, 2nd ed. by K.J. Falconer [2].

Theorem 7.3 Let µ be a mass distribution on Rn, E ⊂ Rn be Borel and γ ∈ (0,∞) be constant.

if ∀x ∈ E:

limr→0µ(Br(x))

rs< γ

Then Hs(E) ≥ µ(E)γ

Proof

Given δ > 0 we define:

Eδ = {x ∈ E | µ(Br(x)) < γrs, ∀r ∈ (0, δ]}

Let {Ui} be a δ-cover of E and as Eδ ⊆ E it will also be a δ-cover of Eδ. For each Ui containing

an element x ∈ Eδ, clearly Ui ⊂ B|Ui|(x) so by the monotonicity of a measure we see µ(Ui) ≤µ(B|Ui|(x)). Then by taking r to be |Ui| in our definition of Eδ we get:

µ(Ui) ≤ γ|Ui|s

From which we can, as Eδ ⊂ {Ui} and again using the monotonicity of a measure, deduce:

µ(Eδ) ≤∑i

{µ(Ui) | Ui ∩ Eδ 6= ∅} ≤ γ∑i

|Ui|s

Going back to the definition of Eδ:

x ∈ Eδ =⇒ µ(Br(x))

rs< γ, r ∈ (0, δ]

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This means, as δ → 0, r → 0 so by the statement of the theorem, d(E,Eδ) → 0. Therefore, by

taking infima at letting δ → 0, we get µ(Eδ) = µ(E) so:

µ(E) ≤ γHs(E)

Now γ ∈ (0,∞) so we can divide by it giving us Hs(E) ≥ µ(E)γ , the desired result. �

We shall see the relevance of this result when applying it later. Now moving on, for s ≥ 0 we

define the s-potential at a point x ∈ Rn due to a mass distribution µ on Rn to be:

φs(x) =

∫dµ(y)

|x− y|s

And from here we define the s-energy of a mass distribution µ to be:

Is(µ) =

∫φs(x)dµ(x) =

∫∫dµ(x)dµ(y)

|x− y|s

So this is essentially the total s-potential of the whole system. To get an inkling of how to interpret

these things physically consider this example: In R3 with s = 1 and taking µ to be the standard

mass measure we see:

φ1(x) =

∫R3

dm(y)

|x− y|Which is the Newtonian gravitational potential. This makes sense physically; a system of greater

total mass will give a larger value for gravitational potential but, the further x is from the rest of

the system, the greater |x−y| will be which in turn leads to lower values for gravitational potential.

This appears unrelated to the study of dimension at first glance but this is not the case. We are

going to prove the following result, which shows s-energy is a incredibly useful when trying to find

a lower bound for the Hausdorff dimension.

The following theorem and its proof are based on one which can be found on pages 70-71 of Fractal

Geometry, Mathematical Foundations and Applications, 2nd ed. by K.J. Falconer [2].

Theorem 7.4 For E ⊂ Rn, if there is a mass distribution µ on E such that Is(µ) < ∞ then

dimH E ≥ s.

Proof

Let E ⊂ Rn. Suppose there is a mass distribution µ such that Is(µ) <∞ and supp(µ) ⊂ E. Define:

E1 =

{x ∈ E

∣∣∣ limr→0

µ(Br(x))

rs> 0

}

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Due to our construction, ∀x ∈ E1, we can find ε > 0 and a sequence {ri} such that {ri} → 0 and

for each ri:µ(Bri(x))

ris> ε =⇒ µ(Bri(x)) ≥ εrsi > 0

If µ({x}) > 0 then Is(µ) = ∞, due to the infinitesimal nature of integration, which contradicts

our original assumption that Is(µ) < ∞, so we must have that µ({x}) = 0, i.e. the measure of

every singleton set is 0. µ is a measure and hence continuous so let us take qi ∈ (0, ri) and consider

the Annulus Ai = Bri(x)\Bqi(x). We can take subsequences if necessary so assume without loss of

generality that ri+1 < qi, ∀i so that all the Ai are disjoint and centred at x. It follows from the

Covering Lemma that µ(Ai) ≥ 14εr

si .

Note also that |x− y|s ≥ r−si on Ai. Therefore ∀x ∈ E1:

φs(x) =

∫dµ

|x− y|s≥∞∑i=1

∫Ai

dµ(y)

|x− y|s≥∞∑i=1

14εr

si r−si =∞

However by assumption of the theorem Is(µ) =∫φs(x)dµ < ∞ so we must have that φs(x) <

∞, ∀x ∈ supp(µ), from which we can deduce that µ(E1) = 0. Note that by the monotonicity

of the Hausdorff measure, E\E1 ⊂ E =⇒ Hs(E\E1) ≤ Hs(E). Furthermore as for x ∈ E\E1,

limr→0µ(Bri (x))

rs = 0 we can apply Theorem 7.3 to E\E1, giving us that for some γ > 0:

Hs(E) ≥ Hs(E\E1) ≥ µ(E\E1)

γ≥ (µ(E)− µ(E1))

γ=µ(E)

γ

If we let γ → 0 then we see Hs(E) → ∞ giving us Hs(E) = ∞. Now recall that dimH E =

sup{s | Hs(E) =∞} so therefore dimH F ≥ s. �

With this in mind let us consider F and take µ to be the mass distribution defined in the pre-

vious section. Consider 2 distinct elements x, y ∈ F then, for some k ≥ 1, x and y are in the same

interval of Fk−1 but in different ones of Fk, so we can say without loss of generality that x ∈ I and

y ∈ J with I, J ⊂ Fk and I ∩ J = ∅. The integral we wish to bound is:

Is(µ) =

∫F

∫F

dµ(x)dµ(y)

|x− y|s

Now recall I, J are 2 of the constituent intervals of Fk so, excluding the situation when I = J , are

disjoint. Therefore we can find an upper bound for the s-energy of Fk by summing all the integrals

for the kth level, excluding the situation where I = J to avoid double counting, so we see:

Is(µ) ≤∑I,JI 6=J

∫x∈I

∫y∈J

dµ(x)dµ(y)

|x− y|s

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But we want to bound the s-energy of F not Fk so recall F =⋂∞k=0 Fk. Therefore the s-energy of

F will be bounded by the sum of all the upper bounds for all of the Fk, for k ≥ 1, giving us:

Is(µ) ≤∞∑k=1

∑I,JI 6=J

∫x∈I

∫y∈J

dµ(x)dµ(y)

|x− y|s

To achieve Fk from Fk−1 we remove the middle (1− 2c) from each interval. This means in the Fk,

if x and y are in different intervals as is the case here, ck−1 ≥ |x−y| ≥ ck−1(1−2c). By considering

the latter part of this inequality in conjunction with fact Fk has 2k intervals each with assigned

mass 2−k, we see:

Is(µ) ≤∞∑k=1

2k2−k · 2−k(

ck−1(1− 2c))s

=

∞∑k=1

1

2k · cks · c−s · (1− 2c)s

=cs

1− 2c

∞∑k=1

(1

2cs

)kAs cs

1−2c is independent of k. Futhermore, c ∈ (0, 1) =⇒ cs

1−2c is positive finite constant, so this sum

will converge provided:

1

2cs< 1 =⇒ 2cs > 1

=⇒ log 2 + s log c > 0

=⇒ s <log 2

− log c

Therefore, by Theorem 7.4, dimH F ≥ log 2− log c .

44 of 57

8 An Alternative Approach to Calculating Dimension

We have just calculated an upper bound for dimBF and seen 2 methods of directly calculating a

lower bound for dimH F . As you can see they are fairly cumbersome and the calculations required to

find the value at which s ‘jumps’ from 0 to∞ are complex , requiring ideas which appear unrelated

at first glance resulting in a rather indirect solution. An alternative approach one can use is to

set out conditions which, if fulfilled, guarantee that 0 < Hs(F ) < ∞ at s = dimH F . Then, as we

do not need to calculate the precise value of s first, it can often be found far more easily with the

benefit of hindsight.

8.1 Implicit Theorems

The following theorem and its proof are based on one which can be found on pages 42-43 of Tech-

niques in Fractal Geometry by K.J. Falconer [17].

Theorem 8.1 Consider E ∈ K(Rn) and let a > 0 and δ0 > 0. If for every open set U with

U ∩ E 6= ∅ and |U | < δ0 there is a map g : U ∩ E → E which, for elements x, y ∈ U ∩ E, satisfies:

a|U |−1|x− y| ≤ |g(x)− g(y)| (8.1)

Then, putting s = dimH E we have that Hs(E) ≥ as > 0 and dimBE = dimBE = s.

Proof

Let us assume that Hs(E) < as, which we shall show leads to a contradiction.

E is compact and hence bounded. Due to its boundedness we may assume that it has an open

cover which, as E is compact, must have a finite sub-cover, which we shall denote {Ui}ni=1. By also

noting that Hs(E) < as we see, ∀ 1 ≤ i ≤ n, this cover satisfies:

i) Ui ∩ E 6= ∅

ii) |Ui| < min{a2 , δ0}

iii) E ⊂⋃ni=1 Ui

iv)∑ni=1 |Ui|s < as

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Let us rearrange (iv), then we can find t ∈ (0, s) such that:

a−tn∑i=1

|Ui|t < 1 (8.2)

As a > 0 we can rearrange 8.1 to see that for each set Ui, there is a mapping gi : Ui ∩E → E such

that for x, y ∈ Ui ∩ E:

|x− y| ≤ a−1|Ui||gi(x)− gi(y)| (8.3)

These gi map a small fraction of E on to the whole so if we take their inverses and apply them

to E we see they behave similarly to iterated function systems. Let Ik be the set of all k-term

sequences {ij}kj=1 where ij ∈ {1, ..., n}. Define I =⋃∞k=0 Ik and for each {ij} ∈ Ik define

U{ij} = g−1i1 (...(g−1ik (E))...). Note that this bears a very striking resemblance to how we coded

F when proving |F | = 2ℵ0 but in this case we are applying a set of n ‘contractions’ rather than just

2. This emphasises how the g−1i act like an iterated function system.

Consider 2 elements x, y ∈ U{ij}. Now if we apply 8.3 k times to x, y we see that:

|x− y| ≤ a−k|Ui1 | · · · |Uik ||gi1 ◦ ... ◦ gik(x)− gi1 ◦ ... ◦ gik(y)|

Which implies |U{ij}| ≤ a−k|Ui1 | · · · |Uik ||E|. If we define b = a−1 min1≤i≤n |Ui| then given δ < |E|,for every x ∈ E there is a sequence {ij} ∈ I such that x ∈ U{ij} and bδ ≤ a−k|Ui1 | · · · |Uik ||E| < δ

because we remove at least b to achieve the kth interval. Again we see similarities with Iterated

Function Systems here; this is analogous to we fixed the size of covering sets between ck and ck−1

in all the previous sections. Now recall how we defined Nδ(E) and in this case we see:

Nδ(E) ≤ #{{ij} ∈ I | bδ ≤ a−k|Ui1 | · · · |Uik ||E|

}≤

∑{ij}∈I

(bδ)−t(a−k|Ui1 | · · · |Uik ||E|)t

≤ |E|tb−tδ−t∞∑k=0

a−kt∑{ij}∈I

(|Ui1 | · · · |Uik |)t

= |E|tb−tδ−t∞∑k=0

(a−t

n∑i=1

|Ui|t)k

< αδ−t by 8.2

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For some constant α ∈ (0,∞). By definition:

dimBE = lim supδ→0

logNδ(E)

− log δ

< lim supδ→0

logα− t log δ

− log δ

= lim supδ→0

(t− logα

log δ

)= t

Note that the last step uses the fact that as δ → 0+, log δ → −∞. Therefore we have that

dimBE < t < s. We can apply the Inequality 7.1 to get dimH(E) < s which contradicts that

s = dimH E. This means the initial assumption must be false, proving that Hs(E) ≥ as > 0 as

required. Furthermore taking t arbitrarily close to dimH E = s gives dimBE ≤ s which, again using

the Inequality 7.1, implies s = dimH E = dimBE. �

Therefore, if a set E satisfies the conditions of Theorem 8.1 then we have that Hs(E) > 0 and

dimH E = dimBE = dimBE. However that is not all we need. This next theorem again sets out

conditions which must be fulfilled which if satisfied also guarantee dimH E = dimBE = dimBE but

also that Hs(E) < ∞. Therefore if both are satisfied we know 0 < Hs(E) < ∞ and can therefore

calculate s far more.


Techniques in Fractal Geometry by K.J. Falconer [17].

Theorem 8.2 Consider E ∈ K(Rn) and let a > 0 and δ0 > 0. If every closed ball B with centre in

E and radius δ < δ0 has a mapping f : E → E ∩ B which, for elements x, y ∈ E, satisfies:

aδd(x, y) ≤ d(f(x), f(y)) (8.4)

Then putting s = dimH E, we have that Hs(E) ≤ 4sa−s <∞ and dimBE = dimBE = s.

Proof

Let Nδ(E) be the maximum number of disjoint closed balls B with centres in E of radius δ. Now

let δ < min{a−1, δ0} we have:

δsNδ(E) > a−s (8.5)

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Which we shall show this leads to a contradiction. Choose a t > s such that:

δtNδ(E) > a−t (8.6)

So we have Nδ(E) disjoint closed balls B1, ...,BNδ(E) of radius δ with centres in E. From the

conditions set out in the statement of the theorem, there exist mappings fi : E → E ∩ Bi, where i

runs from 1 to Nδ(E), such that ∀x, y ∈ E:

aδd(x, y) ≤ d(fi(x)), (fi(y)) (8.7)

As with the g−1i in the previous proof, {f1, ..., fNδ(E)} acts like an Iterated Function System map-

ping E onto disjoint similar subsets contained in each Bi, whose attractor is a subset of E, with a

contraction factor of (aδ). Our next step is to find a lower bound for the dimension of this attrac-

tor, for which we shall use the Mass Distribution Principle from the previous section (Theorem 7.1).

Define Bi,Bj to be the result of applying a sequence of k contractions to the original ball con-

taining all of E corresponding to the sequences i, j ∈ {1, ..., Nδ(E)}k by the coding method defined

previously. Let:

d = inf{d(x, y) | x ∈ Bi, y ∈ Bj , i 6= j}

p = inf{p ∈ N | ip 6= jp}

So Bip−1∩ Bjp−1

6= ∅ but Bip ∩ Bjp = ∅ so are disjoint and d(Bip ,Bjp) ≥ d. Recall, as we are

treating the fi as an Iterated Function System, the contraction factor is aδ so with this in mind, p

applications of 8.7 yields:

d(fi1 ◦ · · · ◦ fip(E), fj1 ◦ · · · ◦ fjp(E)) ≥ (aδ)pd(Bip ,Bjp) ≥ (aδ)pd (8.8)

The first inequality is due to the fact that the images of E under different sequences of various f are

contained in each ball, so the minimum distance between them must be greater than the minimum

distance between the balls containing them.

Now note δ < min{a−1, δ0} =⇒ δ < a−1 and recall how we defined a mass distribution on F by re-

peated sub-division. Let us use the same idea starting here; set µ(E) = 1 then µ(fi1 ◦· · ·◦fik(E)) =

(Nδ(E))−k. As we did previously, let U ⊂ Rn be such that U ∩ E 6= ∅ and |U | < d, and let:

k = inf{x ∈ N | (aδ)k+1d ≤ |U | ≤ (aδ)kd} (8.9)

48 of 57

All we have here is a more generalised version of we did when using the Mass Distribution Principle

to calculating the dimension of the attractor F ; pin the size of this set U between to covering sets

of the attractor, so we can find an upper bound for µ(U) in terms of its diameter. The bounds for

|U | are derived from this.

By 8.8 we can say that U intersects fi1 ◦ · · · ◦ fik(E) for at most 1 of the sets obtained by ap-

plying {f1, ..., fNδ(E)} to E corresponding to the k-term sequence i ∈ {1, ..., Nδ(E)}k so by using

8.6 and 8.9 we see that:

(Nδ(E))−(k+1) ≤ µ(U) ≤ (Nδ(E))−k < (aδ)kt < (daδ)−t|U |t (8.10)

From which, by setting λ = (daδ)−t, we can apply the Mass Distribution Principle to see that

dimH E ≥ t > s. However this is a contradiction since dimH E = s so the initial assumption must

be false and therefore Nδ(E) ≤ a−sδ−s. Now:

dimBE = lim supδ→0

logNδ(E)

− log δ

≤ lim supδ→0

log(a−sδ−s)

− log δ

= lim supδ→0

(s

log a

log δ+ s)

= s

Therefore, we have that dimBE ≤ s = dimH E which we can combine with the Inequality 7.1 to

show that all 3 dimensions are equal. Furthermore, recall that all the B1, ...,BNδ(E) are all disjoint.

Every x ∈ E it must be less than δ away from one of these balls, because otherwise it would fall

into a different one. Therefore we can construct a (4δ)-cover of E with N ′2δ(E) not necessarily

disjoint balls of radius 2δ, with centres at the same points as the original cover. Clearly we see

N ′2δ(E) ≤ Nδ(E) ≤ a−sδ−s. The reason it is a 4δ-cover is each constituent ball has radius 2δ and

hence diameter 4δ, so we see:

Hs4δ(E) = inf

{ ∞∑i=1

|Ui|s∣∣∣{Ui}∞i=1is a 4δ-cover of E

}

≤a−sδ−s∑i=1

(4δ)s

= a−sδ−s(4δ)s

= 4sa−s

And from here letting δ → 0 shows Hs(E) ≤ 4sa−s <∞, the desired result. �

49 of 57

8.2 Implicitly Calculating the Dimension of F

So let us apply this idea to F . Take an arbitrary open set U ⊂ [0, 1] with U ∩ F 6= ∅. Then,

for some k ∈ Z+, ck+1 ≤ |U | < ck. Due to the self similar nature of F the is clearly a mapping

g : U ∩ F → F which scales up with a factor of c−k so, in the context of the Theorem 8.1, taking

a = c the conditions are satisfied so, assuming s = dimH F , dimBF = dimBF = s and 0 < Hs(F ).

Now consider an arbitrary closed ball Bδ(x), where x ∈ F . And let f : F → F ∩ Bδ(x). Now

∃k ∈ N such that 2 · ck ≤ |Bδ(x)| = 2δ < 2 · ck−1 =⇒ ck ≤ δ < ck−1 for which there is a contraction

ratio of ck from F → F ∩Bδ(x). So in the context of Theorem 8.2, setting a = c the conditions are

satisfied and assuming s = dimH F , dimBF = dimBF = s and Hs(F ) <∞.

Now combining these 2 we see 0 < Hs(F ) < ∞ so we can calculate a value for s heuristically.

Let us consider the effect to Hs(F ) of applying our contractions S1, S2 to F :

Hs(F ) = Hs(S1(F )) +Hs(S2(F ))

If {Ui} is a δ-cover of F then clearly {S(Ui)} is a cδ-cover of S(F ) so:

∞∑i=1

|S(Ui)|s =

∞∑i=1

cs|Ui|s = cs∞∑i=1

|Ui|s

By takings infima we can deduce that Hscδ(S(F )) = csHsδ(F ) and letting δ → 0 shows Hs(S(F )) =

csHs(F ). This is called the scaling property of the Hausdorff measure, and if we follow an very

similar proof, it is possible to demonstrate that for E ⊆ Rn, Hs(λE) = λsHs(E) holds ∀λ ∈ (0,∞).

Now returning to the calculation above we see:

Hs(F ) = Hs(S1(F )) +Hs(S2(F )) = csHs(F ) + csHs(F ) = 2csHs(F )

=⇒ 1 = 2cs =⇒ s =log 2

− log c

Note that for the last step of this calculation it is essential that 0 < Hs(F ) <∞ because this allows

us to divide by it in Hs(F ) = 2csHs(F ).

50 of 57

8.3 Implications of These Theorems

Theorems 8.1 and 8.2 are based on ones from Techniques in Fractal Geometry [17] and although the

latter was first proved by its Kenneth Falconer in 1989 [19] 8.1 was first proved by John Mclaughlin

in 1987 [18]. Between them essentially state that, given a set E, if it is possible to map every

non-empty subset onto the whole and map the whole onto smaller balls, then at 0 < Hs(E) <∞ at

s = dimH E = dimBE = dimBE. This has the remarkable consequence that, for sets which exhibit

self similarity everywhere, then dimH E = dimBE = dimBE. This is especially pertinent to this

project because as has been previously mentioned fractals generated by Iterated Function Systems

often have this quality and the connection is emphasized by how mechanisms which essentially work

like I.F.S.s are employed in both proofs. In addition, this means that calculating the dimension of

sets such as these is much more straightforward. For example, if we know that 0 < Hs(E) < ∞we can use the identity

∑ni=1 c

si = 1 where the ci are the contraction constants for {S1, ..., Sn}. To

demonstrate this recall the purely self similar sets from Section 3; the Middle Third Cantor Set

(denoted C), the Heighway Dragon (denoted D) and the Sierpinski Tetrahedron (denoted T ). To

demonstrate how much easier it is to calculate dimension, assuming Theorems 8.1 and 8.2 hold, we

see that:

For C:c1 = c2 =

1

3=⇒ 2

(1

3

)s= 1 =⇒ dimH C = dimB C = s =

log 2

log 3

For D:

c1 = c2 =

√2

2=⇒ 2

(√2

2

)s= 1 =⇒ dimH D = dimB D = s = 2

For T :

c1 = c2 = c3 = c4 =1

2=⇒ 2

(1

2

)s= 1 =⇒ dimH T = dimB T = s = 2

Clearly this is not rigorous but it gives an indication of how much straightforward calculating

dimension is if the conditions for these theorems can be fulfilled. This shows, when applicable, it

is generally much easier to use these Theorems than the methods requiring arduous calculations

outlined in the previous section.

51 of 57

9 The Hausdorff Content

9.1 What is it?

Recall how for a set E we defined:

Hsδ(E) = inf

{ ∞∑i=1

|Ui|s∣∣∣{Ui}∞i=1is a δ-cover of E

}

And that from that we defined the s-dimensional Hausdorff measure to be Hs(F ) = limδ→0Hsδ(E).

We have obtained this by letting δ → 0 in the original definition, so what happens if we go in the

opposite direction and consider arbitrary sets with no restriction on diameter? We call this the

Hausdorff content Hs∞(E), and it is defined to be (A. Farkas & J.M. Fraser, [20], p.405):

Hs∞(E) = inf

{ ∞∑i=1

|Ui|s∣∣∣{Ui}∞i=1 is a countable cover of E by arbitrary sets

}An interesting characteristic of the Hausdorff content is that we can use it to calculate dimension,

using dimH E = inf{s | Hs∞(E) = 0}. Here we see one of the major benefits of using it over the

Hausdorff measure; it still can be used to calculate dimension yet when calculating we have no

constraints on the size of our covering sets which makes life far easier. Let’s see how this compares

to the Hausdorff Measure. Note that as δ decreases the amount of acceptable covers can only

decrease so the infinium Hsδ(E) increases. On the other hand as we limit δ less and less the number

of permissible countable covers can only increase so the infinium Hsδ(E) will decrease, leading to

the inequality:

Hs∞(E) ≤ Hsδ(E) ≤ Hs(E)

Figure 9.1: A graph showing how Hs∞(E) varies as s increase

52 of 57

Figure 9.1 presents a rough indication of how Hs∞(E) varies as s increase. It is rough because we

do not know how Hs∞(E) behaves between 0 and dimH E. However, for all sets E 6= ∅ to avoid

the problem of 00, {E} is a cover so H0∞(E) ≤

∑1i=1 |E|0 = 1 and α ≤ 1. The fact that we

have illustrated this region with a straight line with negative gradient is simply because Hs∞(E) is

decreasing (although not strictly decreasing) here and, for t < dimH E, 0 ≤ Ht∞ ≤ Ht∞ ≤ ∞.

Is Hs∞ a measure? Consider 2 unit intervals positioned at the top and bottom of a unit square, call

them L1 and L2. |L1| = |L2| = 1 and s = 1 so we are simply dealing with the sum of the lengths of

any potential cover. Therefore it is not possible to decrease the sum by decreasing the length of the

covering sets so H1∞(L1) = H1

∞(L2) = 1. Now let us consider L1 ∪ L2. We can cover this union by

the unit square itself, [0, 1]×[0, 1], which has a diameter of√

12 + 12 =√

2 by Pythagoras’ Theorem.

H1∞ is the infinium of this sum so we can obtain an upper bound for it by calculating

∑∞i=1 |Ui| for

any cover {Ui}∞i=1 which covers L1 ∪ L2 with arbitrary sets. Therefore we see that:

H1∞(L1 ∪ L2) = inf

{ ∞∑i=1

|Ui|s∣∣∣{Ui}∞i=1 is a countable cover of L1 ∪ L2 by arbitrary sets

}≤

∣∣[0, 1]× [0, 1]∣∣

=√

2

So H1∞(L1 ∪ L2) ≤

√2 < 2 =⇒ H1

∞(L1 ∪ L2) < H1∞(L1) +H1

∞(L2) meaning Hs∞ is not countably

additive. For a function f : Σ→ R∪{∞}, f countably sub-additive =⇒ f countably additive so, if

we take the contrapositive of this statement we see that, as Hs∞ is not countably additive, it is not

countably sub-additive and hence not a measure. Essentially this is simply a counter example to

show how, when working with the Hausdorff Content, it is possible to use disproportionately large

covering sets to give values smaller than a true measure would.

Figure 9.2: L1 and L2

53 of 57

9.2 Hs∞(F ) = Hs(F )

For an arbitrary set E, if Hs∞(E) = 0 then there is a countable cover {Ui} with no restriction on

the diameter of these covering sets such that:

∞∑i=1

|Ui|s = 0

This implies that there is another countable cover {Vj} such that E ⊆ {Vj} ⊆ {Ui} and, ∀j and

arbitrary δ > 0, |Vj | ≤ δ. As {Vj} ⊆ {Ui} we see that:

∞∑j=1

|Vj | ≤∞∑i=1

|Ui| = 0 =⇒ |Vj | = 0, ∀j

=⇒ |Vj |s = 0, ∀j

=⇒∞∑j=1

|Vj |s = 0

=⇒ inf

{ ∞∑j=1

|Vj |s∣∣ {Vj} is a δ-cover for E

}= 0

Then by letting δ → 0 we see Hs(E) = 0. Also, as previously mentioned, Hs∞(E) ≤ Hs(E) always

holds true so Hs(E) = 0 =⇒ Hs∞(E) = 0 and we have that for any set E:

Hs∞(E) = 0⇐⇒ Hs(E) = 0

Therefore Hs∞(F ) = Hs(F ) if s > dimH F . But what about elsewhere? Let {Ui}∞i=1 be a countable

cover of F by arbitrary sets i.e. there is no limit on |Ui|. We wish to create a δ-cover from this, so

fix δ > 0. Let α = max{|Ui|}, which may be assumed to exist since F is bounded. Then ∀i we see

that:

|S(Ui)| ≤ cα

|S ◦ S(Ui)| ≤ c2α

And as δ > 0 is fixed, ∃n ∈ N such that:

|S ◦ · · · ◦ S(Ui)︸︷︷︸S applied n times

| ≤ cnα < δ

Now let J = {1, 2}n so J is the set of all n-term sequence consisting entirely of 1s and 2s. Let Sj(Ui)

denote the set obtained by applying S1 and S2 to Ui n times corresponding to the coding method de-

scribed in the proof that |F | = 2ℵ0 i.e. if J 3 j = j1j2 · · · jn then Sj(Ui) = Sj1(Sj2(· · ·Sjn(Ui) · · · )).

54 of 57

Therefore {Sj(Ui)}j∈J where i = 1, ...,∞ is a δ-cover of F and we see that:

Hsδ(F ) ≤∑j∈J

∞∑i=1

|Sj(Ui)|s (9.1)

≤∑j∈J

∞∑i=1

cns|Ui|s (9.2)

=

(∑j∈J

cns)( ∞∑

i=1

|Ui|s)

(9.3)

= (2ncns)

( ∞∑i=1

|Ui|s)

(9.4)

=

∞∑i=1

|Ui|s (9.5)

Note that in the previous calculation we can split up the sum∑j∈J

∑∞i=1 c

ns|Ui|s, because cns is

independent of i and |Ui|s is independent of j, to obtain 9.3. |{1, 2}n| = 2n =⇒∑j∈J c

ns = 2ncns

giving us 9.4. Finally recall s = dimH F = − log 2log c so 2ncns = (2cs)n = 1n = 1 to get to 9.5.

As we have that Hsδ(F ) ≤∑∞i=1 |Ui|s, by taking infima we can deduce that Hsδ(F ) ≤ Hs∞(F )

and then letting δ → 0 we see that Hs(F ) ≤ Hs∞(F ). It was demonstrated earlier that for an

arbitrary set E, Hs∞(E) ≤ Hs(E) so we have the desired result:

Hs∞(F ) = Hs(F )

55 of 57

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