Examensarbete Fractal sets and dimensions Patrik Leifsson

Examensarbete Fractal sets and dimensions Patrik Leifsson
Examensarbete
Fractal sets and dimensions
Patrik Leifsson
LiTH - MAT - EX - - 06 / 06 - - SE
Fractal sets and dimensions
Applied Mathematics, Linkopings Universitet
Patrik Leifsson
LiTH - MAT - EX - - 06 / 06 - - SE
Examensarbete: 20 p
Level: D
Supervisor: Jana Bjorn,
Applied Mathematics, Linkopings Universitet
Examiner: Jana Bjorn,
Applied Mathematics, Linkopings Universitet
Linkoping: May 2006
Sprak
Datum
Division, Department
Date
Matematiska Institutionen

581 83 LINKOPING
SWEDEN
May 2006
Rapporttyp
Report category
Language
x
Avdelning, Institution
Svenska/Swedish
Engelska/English
x
Licentiatavhandling
Examensarbete
ISBN
ISRN
LiTH - MAT - EX - - 06 / 06 - - SE
C-uppsats
Serietitel och serienummer
D-uppsats
Title of series, numbering
ISSN
0348-2960

Ovrig
rapport
URL for elektronisk version
http://www.ep.liu.se/exjobb/mai/2006/tm/006/
Titel
Fractal sets and dimensions
Forfattare
Patrik Leifsson
Title
Author
Sammanfattning
Abstract
Nyckelord
Keyword
Fractal analysis is an important tool when we need to study geometrical objects less
regular than ordinary ones, e.g. a set with a non-integer dimension value. It has
developed intensively over the last 30 years which gives a hint to its young age as a
branch within mathematics.
In this thesis we take a look at some basic measure theory needed to introduce certain
denitions of fractal dimensions, which can be used to measure a set's fractal degree.
Comparisons of these denitions are done and we investigate when they coincide.
With these tools dierent fractals are studied and compared.
A key idea in this thesis has been to sum up dierent names and denitions referring
to similar concepts.
box dimension, Cantor dust, Cantor set, dimension, fractal, Hausdor dimension,
measure, Minkowski dimension, packing dimension, Sierpinski gasket, similarity,
space-lling curve, topological dimension, von Koch curve.
vi
Abstract
Fractal analysis is an important tool when we need to study geometrical objects
less regular than ordinary ones, e.g. a set with a non-integer dimension value. It
has developed intensively over the last 30 years which gives a hint to its young
age as a branch within mathematics.
In this thesis we take a look at some basic measure theory needed to introduce certain denitions of fractal dimensions, which can be used to measure a
set's fractal degree. Comparisons of these denitions are done and we investigate when they coincide. With these tools dierent fractals are studied and
compared.
A key idea in this thesis has been to sum up dierent names and denitions
referring to similar concepts.
Keywords: box dimension, Cantor dust, Cantor set, dimension, fractal, Haus-
dor dimension, measure, Minkowski dimension, packing dimension, Sierpinski gasket, similarity, space-lling curve, topological dimension, von
Koch curve.
Leifsson, 2006.
vii
viii
Acknowledgements
I would like to thank my supervisor and examiner Jana Bjorn for your constant
support, guidance and patience. Your advice has been invaluable.
I would like to thank Nina for your support during this time. My opponent
Daniel Petersson also deserves my thanks. Finally, I would like to thank my
family and friends.
Leifsson, 2006.
ix
x
Preliminaries
In this section we collect some basic notations and denitions.
x y means that there exists c > 0 such that xc < y cx.
I denotes the set of irrational numbers.
N denotes the set of natural numbers.
Q denotes the set of rational numbers.
R denotes the set of real numbers.
Z denotes the set of integers.
Ball For e0 2 Rn and R 3 " > 0, we dene an open ball, B , as
B (e0 ; ") = fe 2 Rn : je e0 j < "g;
and a closed ball, B , as
B (e0 ; ") = fe 2 Rn : je e0 j "g;
where e0 is the center and " the radius of the ball.
Open set, closed set A set A Rn is open if there exists B (e; ") A for all
e 2 A. A set A is closed if Rn n A is open. A set A E is open in E Rn if
for all x 2 A there exist a ball B (x; ") such that
B (x; ") \ E A \ E:
Bounded set The set E bounded, i.e.
Rn
is bounded if the diameter of E , diam E , is
diam E = supfjx
yj : x; y 2 E g < 1:
Compact set A set E Rn is compact if it is both closed and bounded.
Closure of a set For a set E Rn and a point x0 2 Rn we say that
x0 is an accumulation point of E if every ball B (x0 ; ") contains points
from E not equal to x0 ;
r(E ) is the set of all accumulation points of E ;
the closure of E is dened as E = E [ r(E ).
Topological base We say that a collection F of open sets in E Rn is a
topological base if for every open set G in E , there exists a subcollection G F
such that
G=
[
F 2G
F:
Contents
1 Introduction
1.1 Purpose of the thesis . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Structure of the thesis . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 The topological dimension
2.1
2.2
2.3
2.4
The small inductive dimension
The large inductive dimension .
The covering dimension . . . .
The topological dimension . . .
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3 The Hausdor measure and dimension
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3.1 The Hausdor measure . . . . . . . . . . . . . . . . . . . . . . . .
3.2 The Hausdor dimension . . . . . . . . . . . . . . . . . . . . . .
1
1
1
2
3
3
3
4
4
7
7
9
4 Minkowski dimensions
13
5 Fractals and self-similarity
23
6 Cantor sets
27
7 The Sierpinski gasket
33
8 The von Koch snowake
39
9 Space-lling curves
43
10 Conclusions and nal remarks
49
4.1 The packing dimension . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2 Product relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
5.1 Fractals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
5.2 Self-similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
6.1 The ternary Cantor set . . . . . . . . . . . . . . . . . . . . . . . . 27
6.2 Cantor set using ternary numbers . . . . . . . . . . . . . . . . . . 28
7.1 The Sierpinski gasket using the ternary tree . . . . . . . . . . . . 36
7.2 The Sierpinski sieve . . . . . . . . . . . . . . . . . . . . . . . . . 36
8.1 The von Koch curve versus C 2 . . . . . . . . . . . . . . . . . . . 42
9.1 Peano space-lling curve . . . . . . . . . . . . . . . . . . . . . . . 43
9.2 The Heighway dragon . . . . . . . . . . . . . . . . . . . . . . . . 45
Leifsson, 2006.
xi
xii
Contents
Chapter 1
Introduction
A fractal can be described as an object less regular than "ordinary" geometrical
objects. The term fractal came in use as late as 1975, by Mandelbrot who also
gave one mathematical denition of what should be considered fractals. In this
denition the use of fractal dimensions plays a big role and can be used to
measure the fractal degree of a fractal, thereby allowing comparisons between
dierent fractals. Though the denition is of relatively recent date, examples
of sets now known as fractals in the sense of Mandelbrot date back to the late
19th century, e.g. the Weierstrass function
1
X
3
f (x) = ai cos(bi x); 0 < a < 1; ab > 1 + ;
2
i=0
which is continuous everywhere but nowhere dierentiable. Another classical
example from this period is the triadic Cantor set, which will be studied thoroughly later on in this thesis.
The use of fractal analysis is wide. It ranges from probability theory, physical theory and applications, stock-market and to number theory among many
others. Fractal objects and fenomena in nature such as mountains, coastlines
and earthquakes is an area well studied by Mandelbrot. In the theory of fractal
dimensions and fractals there is still much to be explored.
1.1 Purpose of the thesis
The purpose of this thesis is to sum up and investigate dierent theories and
notations within some selected areas of fractal analysis in one comprehensible
and well connected text. A reader with basic knowledge of abstract set theory
and calculus should be able to enjoy most of the contents in this thesis.
1.2 Structure of the thesis
There are nine chapters (besides this introduction). In Section 1.3 conditions
for dimensions that we will require to be fullled are dealt with. Chapters 2 - 4
deal with dierent types of dimensions and measures associated with them. In
Chapter 4 we also extend our dimension algebra with some product relations
Leifsson, 2006.
1
2
Chapter 1. Introduction
which are later on put to the test. In Chapter 5 we look at Mandelbrot's
denition of a fractal set. The concepts of self-similarity and the similarity
dimension are also studied here. These rst chapters cover the basic theory
we need to study and compare dierent examples of fractals, beginning with a
thorough study of the triadic Cantor set and its properties in Chapter 6. Two
additional fractals are then introduced and studied in dierent perspectives in
Chapters 7 - 8. In Chapter 9 we take a closer look at space-lling curves. Finally,
all is rounded o with some conclusions and nal remarks in Chapter 10.
1.3 Dimensions
For a set A Rn we will require the following to be satised, concerning the
dimension dim A of A:
8
< (i) dimfag = 0; where fag is the singleton set.
(ii) dim I 1 = 1; where I 1 is the unit interval.
(I)
: (iii) dim I m = m; where I m is the m-dimensional hypercube.
(II) Monotonicity: If A E then dim A dim E .
n
(III) Countable stability: If fAi g1
i=1 is a sequence of closed subsets of R , then
1 [
dim
Ai = sup dim Ai :
i 1
i=1
(III') Finite stability: If A1 ; A2 ; : : : ; Am are closed subsets of Rn , then
dim
m
[
Ai = max dim Ai :
1im
i=1
(IV) Invariance: If : Rn ! Rn is a homeomorphism, i.e. a continuous
bijection whose inverse is continuous, then dim (A) = dim A.
(IV') Lipschitz invariance: If g is a bi-Lipschitz transformation, i.e.
L1 jx yj jg(x) g(y)j L2 jx yj
for all x; y 2 A and some 0 < L1 L2 < 1, then
dim g (A) = dim A:
Remark 1.1. A function g that fullls condition (IV') is also known as a lipeomorphism.
Chapter 2
The topological dimension
There are three dierent denitions of the topological dimension: ind, Ind and
Cov. The rst two are inductively dened. The covering dimension, Cov, is also
known as the Lebesgue covering dimension or the topological dimension.
All of these dimensions coincide in a separable metric space and since we will
restrict ourselves to subsets of Rn , we may consider any of the three dimensions
as the topological dimension. We will denote the topological dimension of a set
E by dimT E .
2.1 The small inductive dimension
Denition 2.1. The small inductive dimension of a set E inductively as follows:
Rn
is dened
ind ; = 1, where
; is the empty set.
For an integer k 0, we have ind E k if and only if there exists a
topological base U for the open sets in E such that ind @U k 1 for all
U 2 U.
We say that ind E = k if and only if ind E k and ind E k 1.
If ind E k for all k 0, then ind E = 1.
2.2 The large inductive dimension
First we need the concept of separated sets.
Denition 2.2. For sets A; E M we say that the set S Rn separates A
and E in M if there exist disjoint open sets V and W in Rn such that A V ,
E W , and S = M n (V [ W ).
Denition 2.3. The large inductive dimension of a set E inductively as follows:
Rn
is dened
Ind ; = 1.
Leifsson, 2006.
3
4
Chapter 2. The topological dimension
For an integer k 0, we have Ind E k if and only if two disjoint closed
sets in E can be separated in E by a set C such that Ind C k 1.
We say that Ind E = k if and only if Ind E k and Ind E k
1.
If Ind E k for all k 0, then Ind E = 1.
2.3 The covering dimension
As with the large inductive dimension we need to introduce a few terms before
we dene the covering dimension.
Denition
2.4. A family F of subsets of
S
E
F.
Rn
is a cover of a set E
Rn
if
F 2F
Denition 2.5. ([7] p. 95) If E and F are two covers of a metric space G and
we have that for every F 2 F there is an E 2 E with F E , then F is a
renement of E .
Denition 2.6. For a family F of sets, the order, ord F , of F is less than or
equal to k if and only if we have an empty intersection for any k + 2 of the sets.
The order of F is equal to k if and only if ord F k and ord F k 1.
Example 2.7. The family F = f(i 1; i + 1); i 2 Zg constitutes a cover of R
by open intervals and ord F = 1.
Denition 2.8. The covering dimension of a set E Rn is dened as follows:
Cov ; = 1.
For an integer k 0, we have Cov E k if and only if all nite open
covers of E have an open renement with order less than or equal to k.
We say that Cov E = k if and only if Cov E k and Cov E k
1.
If Cov E k for all k 0, then Cov E = 1.
2.4 The topological dimension
Denition 2.9. We say that
E Rn is dense in M Rn if E = M .
M is separable if there exists a countable set E such that E = M .
Proposition 2.10. Rn is separable.
Proof. We rst show that if A = fa1 ; a2 ; : : : g and B = fb1 ; b2 ; : : : g are countable
then
A B = f(a; b) : a 2 A; b 2 B g
is countable. This is easily proved e.g. by the Cantor diagonalization method.
Simply arrange the numbers ai in a horizontal list versus the numbers bi in a
vertical list and then, starting at (a1 ; b1 ) move successively through the list of
2.4. The topological dimension
5
numbers (ai ; bj ), i.e. from (a1 ; b1 ) we move to (a2 ; b1 ), then to (a1 ; b2 ), from
this point we move to (a1 ; b3 ) and then to (a2 ; b2 ) and so on. In this manner
we can count all the numbers (ai ; bj ) in the list and thus conclude that A B
above is countable.
Now, since Z is countable and there is an injection from Q to Z Z, it
follows that Q is countable (in accordance with the Schroder-Bernstein theorem
[3] p. 100), Q2 = Q Q is countable, and by induction, Qn is countable for all
n. Thus Rn is separable since Rn = Qn for all n.
Now, what is interesting for us is that for a separable set ind, Ind and Cov
coincide. Since we restrict ourselves to subsets of Rn matters are now simplied
a bit.
Theorem 2.11. (Theorem 8.10 in [1]) For every separable set E ,
ind E = Ind E = Cov E:
Corollary 2.12. If E Rn , then
ind E = Ind E = Cov E:
Thus for a subset E of Rn , we need only think of the topological dimension
as one dimension, and we dene the topological dimension of E as
dimT E = ind E:
Denition 2.13. The set E is said to be totally disconnected if for any
e1 ; e2 2 E , e1 6= e2 , we have that e1 and e2 can be separated by the empty set.
Proposition 2.14. ([6], Theorem A.4.13) A compact set E Rn has
dimT E = 0 if and only if E is totally disconnected.
Proposition 2.15. The topological dimension, dimT , fullls the dimension requirements stated in Section 1.3.
Proof. (I)(i) For the singleton set fag we have dimT fag = 0 by Proposition 2.14.
(I)(ii) The open intervals
Ix;" = (x "; x + "); x 2 I 1 = I; " > 0;
constitute a topological base for the interval I . Now since
dimT @Ix;" = dimT fx
"; x + "g = 0
by (i), we have that dimT I 1 and the fact that I is connected gives us
dimT I 6= 0 and thus dimT I = 1.
(I)(iii) That dimT I m = m can be shown by an argument similar to that for
(I)(ii).
(II) Let A E Rn and dimT E = k. Thus we have a topological base U
of open sets for E with dimT @U k 1 for all U 2 U . Thus the collection
fU \ A : U 2 Ug
forms a topological base for A. Since
dimT (@ (A \ U ) \ A) < dimT @U
k
1;
6
Chapter 2. The topological dimension
we obtain dimT A k = dimT E .
(III) See e.g. Theorem 3.7 in [1].
(IV) (See e.g. Theorem 3.1.6 in [7]). This can intuitively be understood by
studying a topological base U for E . Since f and f 1 are continuous, it follows
that
U 0 = ff (U ) : U 2 Ug
is a topological base for f (E ). Shortly, the base is preserved under the mapping
f and the inversion f 1 . From this it can be shown that
dimT f (E ) dimT E;
and similarly for the opposite inequality.
Remark 2.16. In condition (III) the fact that the Ai 's are closed sets is important. For sets Ai that are not closed, the condition is not true in general.
Let us illustrate Remark 2.16 with an example.
Example 2.17. Let I and Q be denoted as earlier. For q1 and q2 both in Q such
that q1 < q2 , the empty set separates them in Q. Thus Q is totally disconnected
by Denition 2.13 and hence
dimT Q = 0
by Proposition 2.14. In an analogous manner with the same references it follows
that
dimT I = 0:
So we have that
but
dimT I = dimT Q = 0;
dimT (I [ Q) = dimT R = 1 6= maxfdimT Q; dimT Ig:
Proposition 2.18. (Theorem 3.2.10 in [7]) If A; E Rn , then
dimT (A [ E ) 1 + dimT A + dimT E:
Example 2.17 shows that this connection cannot be improved.
Proposition 2.19. (Theorem A.4.14 in [6]) dimT Rn = n.
Remark 2.20. In this chapter we have seen examples of dierent methods to
calculate a set's topological dimension. The key idea to calculate dimT E for a
set E is to study @E as we have seen, e.g. a line I has topological dimension 1
since @I consist of the endpoints of the line which have topological dimensions
0. A square can be enclosed by a closed curve with topological dimension 1 and
thus the square has topological dimension 2 and so on.
Chapter 3
The Hausdor measure and
dimension
3.1 The Hausdor measure
1
S
Denition 3.1. We say that fEi g1
Ei and
i=1 is a -cover of a set E if E i=1
0 diam Ei , for all i. For a set E Rn , s 0, and > 0, dene
Hs (E ) = inf
1
X
i=1
(diam Ei )s ;
where the inmum are taken over all (countable) -covers fEi g1
i=1 of E . Also
dene the s-dimensional Hausdor measure by
Hs (E ) = lim
Hs (E ) = sup Hs (E ):
!0 >0
To put it in words, the Hausdor measure approximates a sets lenght, area
or volume through covers with diameters less than or equal to . The letter s
denotes what is approximated, i.e. lenght, area or volume. The approximation
gets better the smaller sets we use in the covering which makes it natural to let
! 0 in the denition.
Denition 3.2. An outer measure, , on
subsets of Rn that satises:
Rn
is a positive set function on all
(;) = 0;
monotonicity: (A) (E ) if A E Rn ;
countable sub-additivity:
for all Ei Rn .
1
[
i=1
Ei
1
X
i=1
(Ei )
Proposition 3.3. Hs is an outer measure.
Leifsson, 2006.
7
8
Chapter 3. The Hausdor measure and dimension
Proof. With the above denition of outer measure we have
Hs (;) = 0 since diam ; = 0 .
Let A E Rn and " > 0.
that
1
X
i=1
Then there is a -cover fEi g1
1=1 of E such
(diam Ei )s Hs (E ) + ":
But fEi g1
1=1 is also a valid -cover of A and thus
1
X
s
H (A) (diam Ei )s Hs (E ) + ":
i=1
Now, letting " ! 0 gives us
Hs (A) Hs (E ):
1 s
P
H (Ei ) < 1 we have for an arbitrary " > 0 that for each
i=1
i 1 there exists a -cover fAij g1
j =1 of Ei such that
Assuming
1
X
j =1
(diam Aij )s < Hs (Ei ) + 2 i ":
Thus fAij gi;j 1 constitutes a valid -cover of
X
(diam Aij )s i;j 1
1
S
i=1
Ei and
1
1
X
X
Hs (Ei ) + 2 i " = Hs (Ei ) + ":
i=1
i=1
Letting " ! 0 proves the claim.
Proposition 3.4. Hs is a measure, i.e. if fAi g1
i=1 is a pairwise disjoint countable collection of measurable sets, then
1 X
1
[
Ai = (Ai ):
i=1
i=1
(3.1)
Proof. See for example Theorem 4.2 in [16].
Remark 3.5. Hs is an outer measure though not a measure.
Let us illustrate the truth of Remark 3.5.
Example 3.6. We will consider the set E = (Q \ [0; 1]) [0; 1]. If we let
Ai = fqi g [0; 1], for each qi in Q \ [0; 1], we have
1
X
H1 (Ai ) = inf (diam Eij )1 = 1;
(3.2)
j =1
3.2. The Hausdor dimension
9
where Eij are -covers of Ai with diam Eij
sian products fqi g [0; 1] we have
. Since we are considering Carte-
1
X
(diam Eij ) 1
j =1
and thus (3.2) is fullled. Then using balls, Bi , of radii to cover E we need
approximately c 12 of these, where c 2 R. Thus
H
1
1
[
i=1
Ai = inf
1
X
i=1
(diam Bi ) c 1
2
1
<
1
X
i=1
H1 (Ai ) = 1:
Hence the necessary condition (3.1) is not fullled in order for
measure.
Hs
to be a
Denition 3.7. The Lebesgue measure Ln on Rn is dened as follows. For A
of the form
dene
A = f(x1 ; : : : ; xn ) 2 Rn : ai xi bi g
(3.3)
Ln (A) = (b1 a1 )(b2 a2 ) (bn an );
and extend Ln to general subsets of Rn by
1
1 o
nX
[
Ln (Ai ) : A Ai for all Ai of the form (3:3):
Ln (A) = inf
i=1
i=1
Remark 3.8. It can be proved that when s = n on Rn , then Hn is the Lebesgue
measure (within a constant multiple). (See for example Theorem 30 in [19]).
3.2 The Hausdor dimension
Turning our attention once again towards dimensions, we can now with the aid
of the s-dimensional Hausdor measure dene the Hausdor dimension.
Denition 3.9. The Hausdor dimension of a set E Rn is
dimH (A) = supfs : Hs (E ) > 0g = supfs : Hs (E ) = 1g
= inf ft : Ht (E ) < 1g = inf ft : Ht (E ) = 0g:
Remark 3.10. The previous denition can also be expressed as
if s< dimH (E );
Hs (E ) = 1
0 if s> dim (E );
H
and Hs (E ) can attain any value in [0; 1] for s = dimH (E ).
In other words, dimH E is the critical value where the s-dimensional Hausdor measure of the set E so to speak jumps from innity to zero.
Proposition 3.11. The Hausdor dimension, dimH , satises the conditions
for dimensions outlined in Section 1.3.
10
Chapter 3. The Hausdor measure and dimension
Proof. We again treat each condition separately.
(I)(i) We have dimH fag = 0 for the singleton set fag, since diamfag = 0 gives us
Hs (fag) (diamfag)s = 0
for all s > 0, and thus
dimH fag = inf fs : Hs (fag) = 0g = 0:
(I)(ii) We have 0 < H1 (I 1 ) < 1 and hence dimH I 1 = 1, using Remark 3.10.
(I)(iii) Let I m be an m-dimensional hypercube in Rn , 1 m n. We have
Hm (I m ) = cLm (I m ) 2 (0; 1):
Hence dimH I m = m, using Remark 3.10.
(II) Monotonicity: Let A E Rn . Since Hs (A) Hs (E ) when
A E Rn , it follows from Denition 3.9 that dimH (A) dimH (E ).
(III) Countable stability: From the monotonicity we have for all j ,
dimH
1
[
i=1
Ei
dimH Ej
and hence, taking supremum over all j ,
dimH
1
[
The inequality
dimH
j =1
1
[
i=1
Ej
Ei
sup dimH Ej :
j 1
sup dimH Ej
j 1
follows from the fact that if s > dimH Ei for all i, then Hs (Ei ) = 0
S
1 for all i, which gives us Hs
Ei = 0 and hence
i=1
dimH
1
[
i=1
Ei
s:
(IV') Lipschitz invariance: See for example Chapter 2 in [9].
Remark 3.12. It can be shown that the von Koch curve (see Chapter 8) which
4
has Hausdor dimension log
is homeomorphic to [0,1] with Hausdor dimension
log 3
1 and thus we can directly see that invariance is not fullled for the Hausdor
dimension, (see e.g. [17]).
Proposition 3.13. If E is a nite or countable set, then dimH (E ) = 0.
3.2. The Hausdor dimension
11
Proof. This is justied by the fact that for a singleton set Ei we have
H0 (Ei ) = 1:
Thus
dimH (Ei ) = 0:
Hence, by the countable stability
dimH
[1
E = 0:
i=1 i
This is a usefull proposition which we will take advantage of later on.
12
Chapter 3. The Hausdor measure and dimension
Chapter 4
Minkowski dimensions
Other commonly used names for the Minkowski dimension are e.g. box-counting
dimension, box dimension, fractal dimension, metric dimension, capacity dimension, entropy dimension, logarithmic density and information dimension. The
logic in these names can often be seen through their context. We will however
in favor of simplicity only use Minkowski dimension to mean any of these.
The Minkowski dimension of a non-empty bounded subset of Rn is dened
through an upper and lower dimension, which need not coincide.
Denition 4.1. For a non-empty bounded subset E of Rn we dene the upper
Minkowski dimension as
dimM E = inf fs : lim sup N (E; ")"s = 0g;
"!0+
where 0 < " < 1 and N (E; ") is the least number of balls with radius " needed
to cover E . In a similar manner we dene the lower Minkowski dimension as
dimM E = inf fs : lim inf N (E; ")"s = 0g:
"!0+
Instead of using the covering numbers N (E; ") one can also use the packing
numbers
P (E; ") = maxfk : 9 disjoint balls B (xi ; "); i = 1; : : : ; k; with xi 2 E g: (4.1)
Proposition 4.2. For all E Rn the following holds
N (E; 2") P (E; ") N (E; "=2):
Proof. To convince ourselves of the validity of
N (E; 2") P (E; ");
let k = P (E; ") and consider the disjoint balls B (xi ; "), xi 2 E , i = 1; : : : ; k.
S
Now, if there exists an x in E n ki=1 B (xi ; 2") then the balls
B (x1 ; "); : : : ; B (xk ; "); B (x; ") are pairwise disjoint and thus
k + 1 P (E; ") = k
Leifsson, 2006.
13
14
Chapter 4. Minkowski dimensions
which is a contradiction. So the balls B (xi ; 2") cover E and thereby
N (E; 2") k = P (E; "):
For the validity of
P (E; ") N (E; "=2);
let k1 = N (E; "=2) and k2 = P (E; ") and let x1 ; : : : ; xk1 2 Rn and
y1 ; : : : ; yk2 2 E be such that
E
k1
[
i=1
B (xi ; "=2)
and the balls B (yl ; "), l = 1; : : : ; k2 , are disjoint. This results in that all of the
yl 's are in some B (xi ; "=2) and no B (xi ; "=2) have more than one point yl (since
the balls B (yl ; ") are disjoint). This gives us k2 k1 and thus
P (E; ") N (E; "=2):
It follows from Denition 4.1 and Proposition 4.2 that
dimM E = lim sup
"!0+
log P (E; ")
log N (E; ")
= lim sup
log(1=")
"!0+ log(1=")
(4.2)
dimM E = lim inf
log P (E; ")
log N (E; ")
= lim inf
:
"!0+ log(1=")
log(1=")
(4.3)
and
"!0+
Let us show the rst equality in (4.2). The second equality in (4.2) follows from
Proposition 4.2. Suppose s > dimM E . Then for all there exists an "0 > 0
such that
N (E; ")"s < ; for all " < "0 :
Then
so
log N (E; ")"s < log s>
log log N (E; ")
;
log "
and letting " ! 0+ and taking inmum over all s > dimM E , we get
dimM E lim sup
"!0+
log N (E; ")
:
log(1=")
Now suppose
log N (E; ")
< s < dimM E:
log(1=")
Then there exist "j ! 0 and 0 > 0 such that
lim sup
"!0+
N (E; "j )"sj > 0 > 0
15
which implies
log N (E; "j ) + s log "j > log 0 :
But then
s lim sup
"j !0+
log N (E; "j ) log 0
log(1="j )
N (E; ")
lim sup loglog(1
;
=")
"!0+
which is a contradiction. Thus
dimM E = lim sup
"!0+
log N (E; ")
:
log(1=")
When (4.2) and (4.3) are equal, the common value is called the Minkowski
dimension of E , and we write
log P (E; ")
log N (E; ")
= lim
:
"!0 log(1=")
"!0 log(1=")
dimM E = lim
(4.4)
In this denition of the Minkowski dimension, the number N (E; ") can be replaced by any of the following numbers:
the smallest number of closed balls of radius " needed to cover E ;
the smallest number of cubes of side " needed to cover E ;
the number of "-mesh cubes that intersect E , (where an "-mesh cube is
a cube of the form [e1 "; (e1 + 1)"] [en "; (en + 1)"], e1 ; : : : ; en are
integers);
the smallest number of sets with diameter at most " covering E .
Arguments similar to the proof of Proposition 4.2 show that this leads to the
same denotation.
The Minkowski dimension can also be dened by means of the n-dimensional
volume of an "-neighbourhood of E Rn . The "-neighbourhood, E" , of E is
dened as
E" = fx 2 Rn : jx yj " for some y 2 E g:
(4.5)
Then, using the above dened Lebesgue measure, we have the following equivalent formulas for the Minkowski dimension.
Proposition 4.3. Let E Rn . Then
dimM E = n + lim sup
"!0+
log Ln (E" )
;
log(1=")
(4.6)
dimM E = n + lim inf
log Ln (E" )
;
log(1=")
(4.7)
"!0+
and
if it exists.
log Ln (E" )
"!0+ log(1=")
dimM E = n + lim
(4.8)
16
Chapter 4. Minkowski dimensions
Proof. First we prove that the inequalities
c"n P (E; ") Ln (E" ) c(2")n N (E; ")
(4.9)
hold, where c is the volume of the unit ball in Rn . If we have a cover of E by
N (E; ") balls with radii ", then we have that E" can be covered by balls with
radii 2". Thus
Ln (E" ) c(2")n N (E; "):
To understand that
c"n P (E; ") Ln (E" );
simply notice that the space that the P (E; ") disjoint balls ll is covered by the
"-neighbourhood of E and the n-dimensional Lebesgue measure of E" exceeds
or equals c"n P (E; ") thereby.
Now, using (4.9), we have
log(c 1 " n Ln (E" ))
log P (E; ")
lim sup
log(1=")
"!0+
"!0+ log(1=")
log Ln (E" ) log c n log "
= lim sup
log "
"!0+
log Ln (E" )
= lim sup
+n
log "
"!0+
log Ln (E" )
= n lim inf
:
log "
"!0+
dimM E = lim sup
This was attained from c"n P (E; ") Ln (E" ). Using the last inequality in (4.9)
we have
log N (E; ")
log(c 1 (2") n Ln (E" ))
lim sup
log(1=")
"!0+ log(1=")
"!0 +
n
log L (E" ) log c n log(2")
= lim sup
log "
"!0+
n
log L (E" )
= lim sup
+n
log "
"!0+
log Ln (E" )
= n lim inf
:
log "
"!0+
dimM E = lim sup
Thus we conclude that
dimM E = n
lim inf
+
"!0
log Ln (E" )
;
log "
which proves (4.6). Now, (4.7) follows in an manner analogous to the above.
Finally,
log Ln (E" )
dimM E = n + lim+
"!0 log(1=")
follows from (4.6) and (4.7).
17
From the denition of the Hausdor measure we can deduce the useful relation Hs (E ) N (E; ) s . In other words, if fAi g1
i=1 is a -cover of E , then as
diam Ai for all i, we have
1
X
i=1
s
s
(diam Ai )s |s + s +
{z + } = N (E; ) :
N (E;)
Theorem 4.4. For all E Rn ,
dimH E dimM E dimM E:
(4.10)
Proof. If s < dimH E , then
0 < Hs (E ) = lim
Hs (E ) !
lim N (E; ) s
!0+ 0+
and thus log N (E; ) + s log > log Hs (E ) 1, for small enough
this we have
s lim inf
!0+
> 0. From
log N (E; ) log Hs (E )
= dimM E dimM E:
log(1= )
Taking supremum over all s gives us the desired inequality.
That the inequalities in Theorem 4.4 may be strict is best shown with an
example.
Example 4.5. A (compact) set needs not have its Hausdor dimension equal
to its Minkowski dimension. The set E = f0; 1; 21 ; 13 ; 14 ; : : :g has Hausdor dimension 0 since it is countable. However, dimM E = 12 . To show this, let " > 0
and k be the smallest integer such that
1
k 1
1
k
=
1
k(k 1)
< ":
A rst-order approximation of " in terms of k is " k12 . The number of balls of
radius " it takes to cover the points 1; 12 ; 13 ; 14 ; : : : ; k 1 1 equals k 1 p1" . And
to cover the points which lie in E \ [0; k1 ] by balls of radius ", it takes about
1
2p1 " balls. Hence the number of balls needed to cover E is essentially
2k"
N (E; ") 1
1
1
p
+p p
2 "
"
"
from which we obtain
dimM E =
log N (E; ")
=
"!0+
log "
lim
1
log( p1" )
log " 1
= lim 2
= :
"!0+ log "
"!0+ log "
2
lim
A question that still remains is whether the Minkowski dimension fullls the
dimension requirements from Section 1.3.
Proposition 4.6. The upper Minkowski dimension, dimM , satises conditions
(I), (II), (III') and (IV') from Section 1.3. The lower Minkowski dimension,
dimM E , fullls conditions (I), (II) and (IV') of these.
18
Chapter 4. Minkowski dimensions
Proof. (I)(i) We have
dimM fag = inf fs : lim sup N (fag; ")"s = 0g:
"!0+
Since N (fag; ") = 1 for all " > 0, we have for all s > 0 that
lim sup N (fag; ")"s = 0
"!0+
and hence 0 dimM fag dimM fag = 0.
(I)(ii)-(iii)We have N (I m ; ") " m and hence
log N (I m ; ")
= m:
"!0+ log(1=")
dimM I m = dimM I m = dimM I m = lim
(II) If A E then the number of balls with radius " needed to cover A at
most equals the number of balls with radius " needed to cover E , i.e.
N (A; ") N (E; "):
Thus we have
dimM A dimM E and dimM A dimM E
and so dimM A dimM E by (4.2){ (4.4).
(III') Let fEij g1
j =1 be a -cover of Ai for i = 1; : : : ; n. Then
Sn
fEij ; i = 1; : : : ; n; j = 1; : : : ; 1g
is a -cover of i=1 Ai , so
N
n
[
i=1
Ai ; n
X
i=1
N (Ai ; ):
If s > dimM Ai for all i, then we have in accordance with Denition 4.1 that
lim sup N (Ai ; ) s = 0:
!0+
Adding over i = 1; : : : ; n gives us
lim sup N
!0+
n
[
i=1
Ai ; s = 0:
S
Thus from Denition 4.1 we have dimM ( ni=1 Ai ) s. Hence
dimM (
n
[
i=1
Ai ) i=1
max dimM Ai :
;:::;n
S
Now, suppose that dimM ( ni=1 Ai ) < max dimM Ai , i.e. there is an i such
i=1;:::;n
that
n
[
dimM Ai > dimM
Ai :
i=1
4.1. The packing dimension
19
But this contradicts the monotonicity. Thus
dimM
n
[
Ai = i=1
max dimM Ai :
;:::;n
i=1
(IV') The Lipschitz invariance is fullled due to the fact that if
jg(x) g(y)j Ljx yj
and the set E can be covered by N (E; ") sets with diameter less than or equal
to ", then the images of these N (E; ") sets form a cover of g (E ) by sets with
the diameter less than or equal to L", so that N (g (E ); L") N (E; ").
This shows that
dimM g (E ) dimM E and dimM g (E ) dimM E;
using (4.2) and (4.3). Applying this same argument to g
opposite inequality, ([9], p. 44).
1
instead, gives us the
Example 4.7. Let us now consider yet another set and calculate its Hausdor
and Minkowski dimensions. The set to consider is F = f0; 1; 21 ; 14 ; 18 ; : : :g. Since
F E , where E is the set considered in Example 4.5, we immediately know
that dimM F dimM E = 21 from the monotonicity condition. We also know
that dimH F = 0 since F is countable as well. Now, with 0 < = 2 n we we
get a valid -cover of F by N (F; 2 n ) = n + 2 balls. Since
log N (F; 2 n )
n!1
log 2n
lim
we have that
log n
log n
nlim
= lim
= 0;
n
!1 log 2
n!1 n log 2
dimM F = dimM F = 0:
Now consider the mapping
:x!
1 x1
2
1
;
from E to F for each nonzero element x in E . Since
1 x1 1
= 0;
lim
x!0 2
we have (0) = 0 and thus the mapping is continuous.
Remark 4.8. Examples 4.5 and 4.7 shows that the countable stability (III) and
the invariance (IV) fail for the Minkowski dimension.
4.1 The packing dimension
As we have seen before (Example 4.5), the Minkowski dimension does not satisfy
the countable stability criterion for dimensions. Neither is the nite stability
criterion fullled for dimM . However, these complications can be overcome if
we introduce the packing dimension for a set E Rn .
20
Chapter 4. Minkowski dimensions
Denition 4.9. The lower and upper packing dimensions are
n
o
dimP E = inf sup dimM Ei
and
i1
n
o
dimP E = inf sup dimM Ei ;
i1
where the inma are taken over all countable covers fEi g1
i=1 of E
bounded sets.
Rn
by
With this denition, we get dimP E = dimP E = 0 when E is countable, and
the countable stability criterion for the packing dimension is fullled. The other
dimension requirements valid for the Minkowski dimension are still valid.
Denition 4.10. Let 0 < " < 1 and 0 s < 1. For a set E Rn we dene
P"s (E ) = sup
1
X
i=1
(diam Bi )s and P s (E ) = lim P"s (E ) = inf P"s (E ):
"!0+
">0
The supremum is taken over all collections of disjoint balls fBi g of radius less or
equal to " and centers in E . We then dene the s-dimensional packing measure
of a set E Rn as
1
1 o
nX
[
s
s
P (Ei ) : E = Ei :
P (E ) = inf
i=1
Proposition 4.11.
i=1
P s (E ) is a measure on Rn .
Proof. See for example [16] p. 82.
We now use the s-dimensional packing measure to dene the packing dimension.
Theorem 4.12. For a set E Rn ,
dimP E = inf fs : P s (E ) = 0g = inf fs : P s (E ) < 1g
= supfs : P s (E ) > 0g = supfs : P s (E ) = 1g:
Proof. See Theorem 5.11 in [16].
Denition 4.13. For E Rn , we dene the packing dimension as
dimP E = dimP E:
Proposition 4.14. If E Rn then
dimT E dimH E dimP E dimM E:
Proof. The rst inequality follows from Proposition 5.1. The second inequality,
i.e. dimH E dimP E , follows from the fact that
Hs (E ) P s (E ) for all E Rn ;
(see Theorem 5.12 in [16]), and thus we have
dimH E dimP E
4.2. Product relations
21
due to Denition 3.9 and Theorem 4.12. The following proof of the last inequality can be found on p. 46 in [9].
For arbitrary t and s such that t < s < dimP E we have that
P s (E ) P s (E ) = 1:
So for 0 < 1 there are disjoint balls fBi g1
i=1 with centers in E and radii at
most equal to such that
1
X
1 < (diam Bi )s :
i=1
Next we assume that for all k we have that nk of these balls satisfy
2 k
1
< diam Bi 2 k :
Then
1<
1
X
k=0
nk 2 ks
(4.11)
is also satised. Now, unless we want (4.11) contradicted there need to be some
k with
nk > 2kt (1 2t s )
as we also sum over k. The nk balls all have centers in E and we can shrink
them to have radii 2 k 1 < .
Hence P (E; 2 k 1 ) nk and
(2 k 1 )t P (E; 2 k 1 ) nk (2 k 1 )t > 2 t (1
where 2 k
1
2t s ) ;
< . Thus
lim sup P (E; ) t 2 t (1
!0
2t s ) > 0
so dimM E t, for all t < dimP E and the claim follows thereby.
The following proposition gives a sucient condition for when the packing
and Minkowski dimensions coincide.
Proposition 4.15. (Corollary 3.9 in [9]) If E Rn is a compact set such that
dimM (E \ G) = dimM E;
for all open sets G intersecting E , then
dimP E = dimM E:
4.2 Product relations
There are some valuable formulas which can reduce the amount of eort needed
to calculate the dimension. We shall now consider some of them.
22
Chapter 4. Minkowski dimensions
Proposition 4.16. For sets A; E Rn we have
dimH (A E ) dimH A + dimH E ;
dimH (A E ) dimH A + dimM E ;
dimM (A E ) dimM A + dimM E ;
(4.12)
(4.13)
(4.14)
Proof. See Chapter 7.1 in [9].
There is also a product formula concerning the topological dimension of sets.
Proposition 4.17. (Theorem 3.9 in [1]) Let A Rn and E Rn be two sets,
not both empty. Then
dimT (A E ) dimT A + dimT E:
Chapter 5
Fractals and self-similarity
5.1 Fractals
Proposition 5.1. ([12], p. 3) For any set E we have dimT E dimH E .
A complete proof will not be given. Some things can be noted however. For
E = ; we obviously have dimT E < dimH E since dimT ; = 1 and dimH E 0.
Remembering Proposition 3.13 we know that for a countable set E we have
dimH E = 0. Referring to Proposition 2.14, which says that a compact totally
disconnected set E Rn has dimT E = 0, we have these cases covered too. The
cases left to study we leave unproven. (See e.g. p. 104 in [13]).
Denition 5.2. We say that a set E Rn is fractal if dimT E < dimH E . The
fractal degree of the set E is (E ) = dimH E dimT E .
Proposition 5.3. A set E is fractal if the value of dimH (E ) is non-integer.
Proof. The proposition follows from the fact that dimT (E ) only takes on integer
values, so if dimH (E ) is not integer then neither is (E ).
5.2 Self-similarity
A self-similar set is loosely speaking a set consisting of scaled copies of itself.
Denition 5.4. For a closed set E Rn , the mapping T : E
contraction on E if there is a c 2 (0; 1) such that
jT (x) T (y)j cjx yj
! E is called a
(5.1)
for all x; y 2 E .
The smallest c satisfying (5.1) is called the contraction ratio of T . Moreover,
a contraction is a continuous mapping.
Denition 5.5. A xed point of a mapping T : E ! E is a point x 2 E that
remains unchanged under the mapping, i.e. T x = x.
The following proposition is proved in [15] p. 323.
Leifsson, 2006.
23
24
Chapter 5. Fractals and self-similarity
Proposition 5.6. Let E 6= ; be a closed set with the contraction T : E ! E
dened on it. Then T has precisely one xed point.
When we have equality in (5.1), then T preserves the geometrical similarity,
and we call T a similarity or simlitude. For the smallest c fullling (5.1) we
call T a similar contraction.
Denition 5.7. For a family T = fT1 ; T2 ; : : : ; Tm g of similarities with contraction ratios c1 ; c2 ; : : : ; cm , m 2, we say that a nonempty compact set E is
invariant under T if
E=
m
[
i=1
Ti (E ):
Proposition 5.8. For any T as in Denition 5.7 there is a unique invariant
set.
Proof. See [12] p. 19.
An invariant set under a family of similarities T is called a self-similar set.
Denition 5.9. We say that the contractions T1 ; T2 ; : : : ; Tm fulll the open set
condition if there is a nonempty bounded open set O such that
m
[
i=1
Ti (O) O
with the Ti (O)'s pairwise disjoint.
With the prerequisites thus far gained, we introduce yet another dimension
concept.
Denition 5.10. Let E be a self-similar set such that
E = T1 (E ) [ T2 (E ) [ [ Tm (E );
where Ti , i = 1; : : : ; m, are similarities with contraction ratios ci 2 (0; 1), and
the Ti (E )'s are disjoint. The similarity dimension of E , dimS E , is the unique
solution s to the Moran equation
cs1 + cs2 + + csm = 1:
(5.2)
In the special case when
c1 = c2 = = cm = c
we have that
Thus
mcs = 1 and hence log m + s log c = 0:
log m
:
(5.3)
log(1=c)
Despite its name, the similarity dimension does not satisfy the dimension
conditions from Section 1.3 in general. However, under certain circumstances it
coincides with the other dimensions, thus justifying its notion as a dimension.
dimS E = s =
5.2. Self-similarity
25
Proposition 5.11. (Theorem 2.7 in [10], Theorem 4.14 in [16]) Let Ti be similarities on Rn satisfying the open set condition with contraction ratios ci ,
i = 1; 2; : : : ; m. If E is the invariant set of fTi gm
i=1 , then
dimH E = dimP E = dimM E = dimS E ;
0 < Hs (E ) < 1 and P s (E ) < 1, where s = dimS E ;
There exist e1 ; e2 2 (0; 1) such that for s = dimS E ,
e1 rs Hs (E \ B (x; r)) e2 rs
for all x 2 E and 0 < r 1.
Remark 5.12. If the open set condition is not fullled in Proposition 5.11, then
we instead get the relation
dimH E = dimP E = dimM E dimS E:
(See e.g. [12], Theorem 2.3).
Proposition 5.13. (Proposition 9.6 in [9]). For contractions T1 ; T2 ; : : : ; Tm
with contraction ratios ci < 1; i = 1; : : : ; m, on a closed invariant set E Rn
we have that
where (5.2) is fullled.
dimH E s and dimM E s;
26
Chapter 5. Fractals and self-similarity
Chapter 6
Cantor sets
Generally, for 0 < < 12 , we dene Cantor sets on R as the limit set
C () =
1 [
2i
\
i=0 j =1
Ei;j
where E0;1 = [0; 1]; E1;1 = [0; ]; E1;2 = [1 ; 1], and for dened intervals
Ei 1;1 ; : : : ; Ei 1;2i 1 , the intervals Ei;1 ; : : : ; Ei;2i are dened through removing
intervals of length (1 2) diam Ei 1;j = (1 2)i 1 from the middle of each
interval Ei 1;j . Thus each Ei;j has length i .
The following proposition will be veried in the following sections.
Proposition 6.1. Some important properties of C () are the following:
It is uncountable, compact and totally disconnected.
L(C ()) = 0.
log 2
dimH C () = log(1
=) .
Hs (C ()) = 1, where s = dimH C ().
6.1 The ternary Cantor set
Among the dierent choices of for C (), = 1=3 is the most frequently used
one. We shall therefore show some of the general properties of Cantor sets for
this one to make it less abstract. The general case can be treated in an analogous
manner. The set
C( ) =
1
3
1 [
2i
\
i=0 j =1
Ei;j ; where E0;1 = [0; 1]; E1;1 = [0; 13 ]; E1;2 = [ 23 ; 1]
and so on is called the ternary or triadic Cantor set or simply the Cantor dust.
Following the procedure recently described we start with the unit interval. If
we denote C (1=3) with just C , we have that the unit interval is our set C0 . To
receive C1 we remove the open middle third interval from C0 , i.e.
C1 = C0 n ( 13 ; 23 ) = [0; 13 ] [ [ 32 ; 1]:
Leifsson, 2006.
27
28
Chapter 6. Cantor sets
From each of these two intervals we then remove the open middle third intervals
of length
1 1 2 1 1
1 2
= :
3 3
9
Thus
C2 = C1 n (( 91 ; 92 ) [ ( 79 ; 89 )) = [0; 19 ] [ [ 29 ; 31 ] [ [ 23 ; 79 ] [ [ 89 ; 1]:
Continuing like this in innitely many steps gives us our limit set
C=
1
\
i=0
Ci :
This set is obviously quite porous containing no intervals of positive length,
hence its name Cantor dust. A simple illustration of the rst dierent generations Ci of C follows below.
Figure 6.1: The rst generations of the triadic Cantor set.
6.2 Cantor set using ternary numbers
An alternative view of the triadic Cantor set is with base three, i.e. ternary,
expansions of each number in [0; 1]. First of all, any number can be written with
a dierent expansion than the one it already has. Consider for example the base
two expansion of the number seven (written in base ten), i.e. 7 = 22 + 21 + 20
so we have 710 = 1112 . Another example is 810 = 23 = 10002 . In an analogous
manner we can rewrite any fraction written in base n into another base m 6= n.
E.g.
45 1 1 1
1
= + + + = 0:1011012 :
64 2 8 16 64
What we need to know in our further investigation is how base three expansion
works. This is accomplished in a similar fashion as converting base ten into base
two. For instance,
710 = 2 31 + 30 = 213 and 810 = 2 31 + 2 30 = 223 :
Likewise we convert the base ten fraction 4=7 as 0:120102120 : : :3 = 0:1201023 ,
where the underline indicates a repeating decimal expansion of the underlined
digits. Now, looking at the Cantor dust from a base three expansion point of
view gives us what is illustrated below.
Thus, the ternary Cantor set consists of all numbers between zero and one with
ternary expansions in zeros and twos only. Some numbers have two dierent
expansions, e.g.
1
= 0:1000:::3 = 0:0222:::3 ;
3
6.2. Cantor set using ternary numbers
29
Figure 6.2: Base three expansion of the triadic Cantor set.
but in these cases it is most important whether the number can be written with
only zeros and twos, because then it counts as a member of the set even if it
also has its expansions using zeros and ones. The points 0 = 0:03 and 1 = 0:23
evidently count as well. Let us verify some of C 's properties mentioned earlier.
Proposition 6.2. The Cantor set C is uncountable.
Proof. Consider the second generation C1 of C and the binary sequences (xi )1
i=1
with xi 2 f0; 1g. Now, for every c 2 C we let x1 = 0 if c belongs to the left
segment of C1 and x1 = 1 if c is found in the right segment of C1 . After this
step is done we now need to consider in which of the two possible segments
of C2 's four parts c is in. Letting this procedure continue further yields a
binary sequence (x1 ; x2 ; : : :) for each c 2 C . Similarly each of those sequences
corresponds to a c in C . Thus we have a bijection between C and the binary
sequences (xi )1
i=1 . Since the set of binary sequences is uncountable, so is C .
Proposition 6.3. The (triadic) Cantor set has dimT = 0.
Proof. The Cantor Tset is compact since it is both closed and bounded. It is
closed since in C = 1
i=0 Cn each Cn consists of a nite union of closed intervals
and using the fact that the union of a nite collection of closed sets is closed
according to De Morgan's laws and that any intersection of closed sets is closed.
To see that C is totally disconnected, assume that c1 ; c2 2 C and c1 < c2 . Let
= c2 c1 . Each interval Cn 2 C is of length 3 n . Choosing n such that
3 n < places c1 and c2 in dierent intervals. Supposing I = [a; b] is the last
interval in the construction with c1 ; c2 2 I gives us that
a+b
a+b
c1 <
< c2 and
2= C:
2
2
Thus there is a c = a+2 b 2= C such that c1 < c < c2 . Hence
A = C \ [0; c) and E = C \ (c; 1]
are nonempty separated sets with A [ E = C . Thus C is totally disconnected
and hence dimT C = 0 by Proposition 2.14.
The Cantor set is an invariant set. It has the similar contractions
T1 =
i.e.
x
3
and T2 = 1
x
C = T1 (C ) [ T2 (C ):
3
;
30
Chapter 6. Cantor sets
From (5.3) we thus have
log 2
:
log 3
The open set condition is fullled with O = (0; 1). Hence we have
dimS C =
dimH C = dimP C = dimM C = dimS C =
log 2
log 3
due to Proposition 5.11. From Proposition 5.3 we can now also verify that C
is a fractal set. As a further exercise we calculate dimM C = log 2= log 3, thus
verifying the equality dimM C = dimS C .
Example 6.4. Looking at Figure 6.1, our starting interval C0 = [0; 1] only
needs one box of diameter 1 to cover it. Thus N (C0 ; 1) = 1. Next, we see
that C1 in its turn needs N (C1 ; 1=3) = 2 boxes to be covered, where 1=3 is the
scaling or similarity ratio. Following the pattern we have
N (C2 ; 1=9) = N (C2 ; 1=32 ) = 4 = 22 ; N (C3 ; 1=27) = N (C3 ; 1=33 ) = 8 = 23
and in general
N (Cn ; (1=3)n ) = 2n :
Thus
log N (Cn ; (1=3)n )
n!1 log(1=(1=3)n )
log(2n )
n log 2 log 2
= lim
= lim
=
:
n!1 log(3n ) n!1 n log 3
log 3
dimM C = lim
Proposition 6.5. If C is the ternary Cantor set we have
(C ) = dimH C
dimT C =
log 2
>0
log 3
and C is a fractal set.
Proof. The proposition follows from Proposition 5.3 since dimH C = log 2= log 3
is non-integer.
We nish this chapter with an interesting theorem by Hausdor and then
an interesting example of another Cantor set.
Proposition 6.6. (Theorem 6.6 in [21]) Every compact set is a continuous
image of the Cantor set.
Proof. See p. 100 in [21].
Example 6.7. Let us now consider the Cartesian product of the set C (1=4)
with itself, i.e. C (1=4) C (1=4). Let us call it C 02 . From Proposition 6.1 we
have that
an thus
dimH C 0 =
log 2 1
= ;
log 4 2
dimH C 02 = 2 1
2
6.2. Cantor set using ternary numbers
31
by Proposition 4.16. Now, C 02 is obviously totally disconnected and thus
dimT C 02 = 0, which also can be seen by using Proposition 4.17 which gives
us
dimT C 02 dimT C 0 + dimT C 0 = 0;
since C 0 is totally disconnected as well. And since C 02 is not the empty set we
know that dimT C 02 0. Hence the fractal degree of C 02 is
(C 02 ) = dimH C 02 dimT C 02 = 1;
and thus a fractal set need not have an integer fractal degree value which one
could have thought.
32
Chapter 6. Cantor sets
Chapter 7
The Sierpinski gasket
In this chapter we will start exploring the formulas in Section 4.2. We will
however rst calculate the Minkowski dimension of a fractal set known as the
Sierpinski gasket, S , with the aid of triangle shaped coverings. Consider a closed
equilateral triangle, S0 , of unit side. To obtain the Sierpinski gasket, start with
dividing S0 into four equally big triangles by joining the midpoint of each side
with one another. Now, remove the middle one of these, i.e. the open triangle
containing the center of S0 . The remaining set, S1 , consists of three smaller
copies of the original one, now with the side length 1=2. Continuing with the
same procedure with each of these three triangles leaves us with nine smaller
equally big triangles and after that we have 27 smaller triangles and so on.
Iterating further in innitely many steps nally gives us the limit set
S=
1
\
n=0
Sn ;
known as the Sierpinski gasket as illustrated on next page.
A rst covering of S would of course be with a triangle of unit size. The
next step is to cover S with three triangles of side "1 = 1=2, thus giving us
N (S; "1 ) = N (S; 1=2) = 3. Next we cover S with triangles of side "2 = "21 = 14
giving us a covering of S with nine triangles of side 41 . Following the pattern
we have that S then needs to be covered with 27 triangles of side "3 = "22 = 18 .
So in numbers we have:
N (S; "0 ) = N (S; 1) = 1
N (S; "1 ) = N (S; 1=2) = 3
N (S; "2 ) = N (S; 1=4) = N (S; 1=22 ) = 9 = 32
N (S; "3 ) = N (S; 1=8) = N (S; 1=23 ) = 27 = 33
and in general
Thus
Leifsson, 2006.
N (S; "n ) = N (S; (1=2)n ) = 3n :
log N (S; "n )
log N (S; "n )
= lim
n!1 log(1="n )
"n !0+ log(1="n )
dimM S = lim
33
34
Chapter 7. The Sierpinski gasket
Figure 7.1: The Sierpinski gasket
log N (S; (1=2)n )
log(3n )
n log 3 log 3
=
lim
= lim
=
;
n!1 log(1=(1=2)n )
n!1 log(2n ) n!1 n log 2
log 2
using (4.4).
Let us look at S from another point of view. Letting fc1 ; c2 ; c3 g be the
vertices of S0 , we dene the contractions Ti : Rn ! Rn by
1
Ti (x) = (x ci ) + ci ; i = 1; 2; 3:
2
Thus
S = T1 (S ) [ T2 (S ) [ T3 (S );
with the contraction ratio 1/2 for each Ti . Now, the Sierpinski gasket satises
the open set condition if we let O be the interior of the starting triangle, S0 .
Hence, by (5.3), we have
log 3
dimS S = s =
:
log 2
Thus with reference to Proposition 5.11 we have
= lim
dimH S = dimP S = dimM S = dimS S =
log 3
:
log 2
Without thorough calculations, we also have the following from Proposition 4.16:
dimH (S I ) = dimH S + dimH I =
since
log 3
+ 1:
log 2
dimH I = dimM I = 1:
Proposition 7.1. The topological dimension of the Sierpinski gasket is
dimT S = 1:
35
Proof. To narrow it down we use Proposition 5.1 and conclude that dimT S 1
since dimH S = 1:584 : : :. We also know that S 6= ; so dimT S 0. Now, since
S is connected in its perimeter we have that dimT S 6= 0 by Proposition 2.14
and thus dimT S = 1.
Let us now sum this up with a proposition.
Proposition 7.2. If S is the Sierpinski gasket then
(S ) = dimH S
dimT S =
log 3
log 2
1 = 0:584 : : : ;
and thus S is a fractal set.
Proof. The proposition follows from Proposition 5.3 since dimH S is non-integer.
Example 7.3. Let us now develop further from what we have learned thus
far in this chapter and verify the nite stability criterion for the Minkowski
dimension. Let E = S [ I , where S is the Sierpinski gasket and I is a horizontal
line segment of length 1, (it is helpful to think of I lying horizontal next to the
base of S ). We want to calculate dimM E .
Our rst covering of E will consist of two squares of side 1, giving us
N (E; 1) = 2 = 1 + 1:
Continuing like before we get:
N (E; 1=2) = 5 = 31 + 21 ;
i.e. three squares to cover S and two squares to cover the line segment.
N (E; 1=4) = 13 = 9 + 4 = 32 + 22 ;
and in general:
and thus
N (E; (1=2)n ) = 3n + 2n
log N (E; (1=2)n )
n!1 log(1=(1=2)n )
log(3n + 2n )
= lim
n!1 log(2n )
log(3n (1 + ( 23 )n )
= lim
n!1
log(2n )
log(3n ) + log(1 + ( 23 )n )
= lim
n!1
log(2n )
n
log(3 )
= lim
n!1 log(2n )
log 3
=
log 2
log 3
= maxf
; 1g
log 2
= maxfdimM S; dimM I g;
dimM E = lim
36
Chapter 7. The Sierpinski gasket
in accordance with (III') in Section 1.3.
7.1 The Sierpinski gasket using the ternary tree
With the knowledge of how the ternary number system works from before, we
can now consider a simple procedure which iterates to the Sierpinski gasket,
S . What we will consider is a so called ternary tree. Starting at a point given
the value 0, we draw three outgoing lines of equal length from this point and
with an angular distance of 120 . One line is drawn to the right, one line
to the left upwards and one line to the left downwards. Thereafter we label
them with 0, 1 and 2 respectively. Then, if we continue to draw lines in the
same manner outwards from the end of each of these lines and adding the
number corresponding to its direction in the end of its value gives us all of the
ternary numbers in [0; 1]. Continuing like this indenitely, we receive a structure
agreeing well with our Sierpinski gasket. Actually, the labeling of numbers in
this case is not so important as long as we make homogeneous iterations all the
time. This is best shown with an illustration.
Figure 7.2: The ternary tree approximation of the Sierpinski gasket.
7.2 The Sierpinski sieve
Let us look at yet another approach to construct the Sierpinski gasket. We now
construct the Sierpinski gasket using modulo 2 arithmetic on Pascal's triangle.
An illustration of the rst rows of Pascal's triangle follows below. Each number
inside Pascal's triangle is the sum of the two numbers above it, i.e 6 = 3 + 3,
15 = 5 + 10 and so on. One can now choose between many dierent approaches
7.2. The Sierpinski sieve
37
Figure 7.3: The rst rows of Pascal's triangle.
to construct the Sierpinski gasket from this triangle. We will use modulo 2
arithmetic, i.e. considering even and odd numbers inside Pascal's triangle. Now
apply modulo 2 arithmetic on Pascal's triangle in the sense that we color each
position in the triangle white if it consists of an even number, and each position
black if it contains an odd number. Iterating this process throughout Pascal's
triangle leaves a good approximation to the Sierpinski gasket known as the
Sierpinski sieve, illustrated below.
Figure 7.4: The rst rows of the Sierpinski sieve.
Looking at the rst two rows of the colored Pascal's triangle we see that this
corresponds to the rst approximation, S1 , of the Sierpinski gasket. Adding
another two rows we notice that these four rows correspond to S2 . Generalizing
this we have the rst 2k colored rows correspond to Sk .
38
Chapter 7. The Sierpinski gasket
Chapter 8
The von Koch snowake
We will now consider another fractal set called the von Koch snowake, illustrated below.
Figure 8.1: The von Koch snowake.
The von Koch snowake consists of three congruent fractals K , called von
Koch curves. It is therefore enough to only study K in order to derive the
properties of the von Koch snowake.
Leifsson, 2006.
39
40
Chapter 8. The von Koch snowake
To construct K , start with a line segment and call this K0 . Now, we remove
the middle third of K0 and replace it with the upright sides of an equilateral
triangle, so each of these four line segments are of equal length. Call this curve
K1 . Next, we repeat this procedure on each of the four segments of K1 , giving
us K2 consisting of 42 = 16 line segments of equal length. Thus K3 consists of
43 = 64 line segments and so on. K is now dened as the limit set obtained
after innitely many iterations, i.e.
K = nlim
!1 Kn :
We now look at one of the properties of K and hence the von Koch snowake
thereby.
Proposition 8.1. The von Koch curve is of innite length.
Proof. Let K0 have length 1. Then K1 has length 4=3, K2 has length 42 =32 and
so on. Thus Kn constructed after n steps is of length 4n =3n . Hence intuitively
K should have length
4n
lim n = 1:
n!1
3
For a rigorous proof, we take an advance look at Proposition 8.3 below and
notice that dimH K > 1. Thus using Remark 3.10 we have that K 's length is
in fact innite.
Corollary 8.2. The von Koch snowake has innite length.
The von Koch curve is obviously self-similar, while the von Koch snowake
is not since it has no copies of itself in its structure. The open set condition is
fullled with O being the open equilateral triangle with side length equal to K0 .
K can be dened through the following contractions using complex numbers:
p
1
3i
T1 = ( +
)z
2
6
and
p
3i
1
)(z 1) + 1:
T2 = (
2
6
Thus K is an invariant set under these contractions. We
p can now
p conclude that
the Moran equation (5.2) is satised with c = j 12 + 63i j = 1= 3 and m = 2,
giving us
log 4
dimS K =
;
log 3
i.e. twice as big as the similarity dimension of the Cantor dust.
Proposition 8.3. Let K be the von Koch curve. Then
dimH K = dimP K = dimM K = dimS K =
log 4
:
log 3
Proof. The proposition follows from our recent arguing above and from Proposition 5.11.
Let us conrm dimM K = dimS K with an example.
41
Figure 8.2: The rst two coverings of K .
Example 8.4. We will calculate the Minkowski dimension of K using square
box coverings. In our rst cover we use three boxes to cover K . Thus we
have "1 = 1=3 and N (K; "1 ) = 3. In the next step we use 12 covers and
"2 = "1 =3 = 1=9. This is illustrated in a simple gure below.
All in all we have
N (K; "1 ) = N (K; 1=3) = 3,
N (K; "2 ) = N (K; 1=9) = N (K; 1=32 ) = 12 = 3 4,
N (K; "3 ) = N (K; 1=27) = N (K; 1=33 ) = 48 = 3 42 ,
and in general N (K; "n ) = N (K; 1=3n ) = 3 4n 1 .
Hence
log N (K; "n )
log N (K; "n )
dimM K = lim
= lim
n!1 log(1="n )
"n !0 log(1="n )
log N (K; 1=3n )
log(3 4n 1 )
= lim
=
lim
n!1 log(1=(1=3)n )
n!1 log(3n )
(n 1) log 4 + log 3 log 4
=
:
= lim
n!1
n log 3
log 3
Proposition 8.5. If K is the von Koch curve then dimT K = 1.
Proof. First of all, since dimH K = 1:261 : : : we know that dimT K 1 by
Proposition 5.1. And since K is not the empty set we can also conclude that
dimT K 0. From Proposition 2.14 we have that dimT K 6= 0 because K is
connected.
Proposition 8.6. The von Koch curve K is a fractal set with the fractal degree
(K ) =
log 4
log 3
1:
42
Chapter 8. The von Koch snowake
Proof. From Proposition 5.3 we have that K is a fractal since dimH K is noninteger and remembering Denition 5.2 we have that the fractal degree of K
is
log 4
(K ) = dimH K dimT K =
1 = 0:2618 : : : :
log 3
8.1 The von Koch curve versus C 2
We now briey compare the von Koch curve, K , with the Cantor product C 2 =
C C , where C is the middle third Cantor set.
Figure 8.3: The Cantor product C 2 .
With the dimension formulas in Section (4.2) we directly have
dimH (C n ) = dimP (C n ) = dimM (C n ) = dimS (C n ) = n dimH C = n
where
log 2
log 3
(8.1)
n
Cn = C
| C {z C} R :
n times
Now, from (8.1) we have that
dimH C 2 = 2
log 2
;
log 3
which leads us to the rst interesting observation that
dimH C 2 = dimH K:
Also, noticing that C 2 is totally disconnected whilst K is not we know that
their topological dimensions dier, and thus their fractal degrees do as well.
Chapter 9
Space-lling curves
Before digging into any examples of space-lling curves we shall dene its concept.
Denition 9.1. A continuous function f : [0; 1] ! Rn , n 2, is called a spacelling curve if the n-dimensional Lebesgue measure, Ln , of its direct image,
f = f ([0; 1]), is strictly positive, i.e. Ln (f ) > 0.
Remark 9.2. The mapping under a space-lling curve, S , from [0; 1] to the space
that S lls is surjective, however not injective. (See Netto's theorem in e.g. [21]).
9.1 Peano space-lling curve
We shall now consider a fractal curve called the Peano space-lling curve, P ,
which maps the closed unit interval onto the closed unit square i.e. it lls the
unit square. It was Giuseppe Peano who rst discovered a curve of this type.
His approach was purely analytic. The geometric approach we will use here
was deduced by David Hilbert one year later, therefore this Peano curve is also
known as the Hilbert curve. In fact all space-lling curves are called Peano
curves.
Our starting set consists of the closed unit square [0; 1]2 . To generate the
rst approximation P1 of P we think of the unit square as a partition into four
connected squares of side 1/2. Now, starting at t = 0 a line is drawn into the
center of the rst square then leading on to the center of the next square so that
the centers of all four squares are visited only once, and the endpoint of the line
is in t = 1. This is illustrated below. The thinner lines are not part of the curve.
To generate P2 we now think of the unit square as 16 = 42 connected squares
of side 1/4. The procedure is now the same, we want to pass through every
center of the squares once, starting at t = 0 and nishing at t = 1. Following
the pattern we have that Pn can be generated if we think of the unit square as
4n squares of side 1=2n through which we want to draw a polygon curve similar
to the one before, i.e. passing through the center of each subsquare. A simple
illustration of P2 to P5 follows below.
The length of Pn is obviously
2n
Leifsson, 2006.
1
;
2n
43
44
Chapter 9. Space-lling curves
Figure 9.1: The rst iteration, P1 , of the Hilbert curve.
thus
P = nlim
!1 Pn
intuitively has innite length. Referring to Proposition 9.3 and Remark 3.10 as
we have done before, we can conrm that this is true. Let us show that it really
is a mapping from [0; 1] to [0; 1]2 . Looking at the gures we can with a simple
reasoning see that
jPn+1 (t) Pn (t)j Thus for m > n we have that
sup jPm (t)
t 1
1 p2
X
0
i=n
p
2
for t 2 [0; 1]:
2n
m
X1
jPi+1 (t) Pi (t)j
Pn (t)j sup
0t1
i=n
p
2
=2 n
2i
2
! 0;
as m; n ! 1:
Thus (Pn )1
n=1 is uniformly convergent according to the Cauchy criterion for
uniform convergence, which states that a sequence, (Pn )1
n=1 , of functions dened
on a set E Rn is uniformly convergent if and only if
sup jPm (t)
t2 E
Pn (t)j ! 0 as m; n ! 1:
Now, P is continuous on [0; 1] due to the fact that if a sequence of continuous
functions converges uniformly towards a function P on an interval, then P is
continuous (see 2.1.8. in [18]). We also have that [0; 1] is compact, and so
P ([0; 1]) is compact. Thus, since every point in [0; 1]2 is an accumulation point
of P ([0; 1]), we have that
P ([0; 1]) = [0; 1]2 :
Proposition 9.3. The Hilbert curve, P , is a space-lling curve.
Proof. From our calculations above we get that the direct image of P ([0; 1]) is
[0; 1]2 . Since L2 ([0; 1]2 ) = 1 > 0 the claim thus follows.
From (5.3), with m = 4 and c = 1=2, we get
dimS P =
log 4
= 2:
log 2
9.2. The Heighway dragon
45
Figure 9.2: The second to fth iterations of the Hilbert curve.
Remembering Remarks 3.8 and 3.10 we also notice that dimH P = 2 and so
dimH P = dimP P = dimM P = dimS P = 2;
in accordance with Remark 5.12 and Proposition 4.14. The topological dimension of the Hilbert curve, or Peano curve, is 2 since it lls the plane. Hence the
fractal degree of P is
(P ) = dimH P
dimT P = 2
2 = 0:
Thus our denition of a fractal set tells us that The Hilbert curve is not a fractal
set though we intuitively know it is. Fractals with this property are sometimes
referred to as borderline fractals.
9.2 The Heighway dragon
Let us take look at another space-lling curve called the Heighway dragon after
its founder John E. Heighway. It can be constructed in many dierent ways.
We will use line segments to construct it. Our starting set, D0 , consists of the
unit interval.
p To create D1 we replace D0 with two line segments each of length
d1 = 1= 2 joined at a right angle. Next we receive D2 by replacing each line
segment in D1 with two line segments each of length
d2 =
p1 d1 = 12 ;
2
46
Chapter 9. Space-lling curves
again joined at a right angle. If we let this procedure continue ad inmum we
in each iteration Dn , n = 0; 1; 2; : : :, have that every line segment is of length
1
dn = ( p )n
2
and the Heighway dragon, D, is the limit set received, i.e.
D = nlim
!1 Dn :
A simple illustration of the rst ve iterations is provided below, starting with
D0 in the upper left corner.
Figure 9.3: The rst ve iterations of the Heighway dragon curve.
Proposition 9.4. None of the approximations Dn , n = 0; 1; 2; : : :, of the Heighway dragon curve crosses itself.
Proof. See e.g. [7] p. 21.
An intuitive understanding of the validity of Proposition 9.4 can be gained
by rounding the right angles o a bit. An example where this has been done to
D9 is depicted below.
It can also be shown in a similar manner as in the case of the Peano curve
that the Heighway dragon is a space-lling curve, (see Proposition 2.4.3 in [7]).
Proposition 9.5. If D is the Heighway dragon curve we have that
dimH D = dimP D = dimM D = dimS D = 2:
p
Proof. The reduction ratio in every step is c = 1= 2 and in each step every line
segment is replaced by m = 2 new ones. Thus by (5.3) we have
log 2
p = 2:
log 2
Remarks 3.8 and 3.10 imply dimH D = 2 and thus the claim follows thereafter
from Proposition 4.14 and Remark 5.12.
dimS D =
9.2. The Heighway dragon
47
Figure 9.4: The D9 approximation of the Heighway dragon with the right angles
rounded o.
One can easily create dierent variants of the Heighway dragon by changing
the angles. One among many interesting manipulations of the Heighway dragon
leads to an approximation of the Sierpinski gasket. In the following example
we will change the angles in the Heighway dragon curve which will give us an
altered Heighway dragon curve, D0 . Now, joining three of these in an equilateral
triangle results in a set known as the fudgeake (from fudging the von Koch
snowake), which is illustrated below.
Figure 9.5: The Fudgeake, consisting of three copies of D0 put together.
Example 9.6. Changing the angles in the dragon curve from 90 degrees to 120
and using the same starting set, i.e. the unit line, gives us that D10 has side
lengths 12 sin 60 = p13 . Each of the 2n line segments in the n:th step has length
( p13 )n and thus intuitively D0 's total length is innite since
2 n
p = 1:
n!1 3
lim
48
However,
Chapter 9. Space-lling curves
log 2
2
p = 2 log
> 1;
log 3
log 3
thus by Remark 3.10 it has innite length. A noticeable fact here is also that
dimS D0 is twice as big as dimS C and equal to dimS K .
dimS D0 = dimH D0 =
Chapter 10
Conclusions and nal
remarks
Although this thesis is restricted to some basic parts of fractal analysis I think it
is enough to emphasize the need of fractal analysis as an important complement
when our classical calculus does not apply. The thesis could easily have been
broadened by e.g. including other metric spaces besides Rn . However, to allow
more readers with the adequate prerequisites and to get a more connected text,
this has been avoided. There are also many other dimension concepts and
measures not mentioned which could have been included.
To sum it up, I think the purpose of collecting similar concepts has succeeded
and connections between dimensions and certain fractals have been claried. It
would have been interesting to extend the thesis with some areas applicable to
fractal analysis such as number theory, however as mentioned in the introduction
the area of fractal analysis is wide but still much is unexplored within it.
Overall, I think fractal analysis is an enjoyable and rewarding area to study.
Even though the theory in the rst chapters is occasionally quite heavy it is
interesting to see when the dierent dimensions coincide and under which conditions. Then applying the theory to various fractal sets and to investigate how
they are connected and dier makes it well worthwhile I think.
Leifsson, 2006.
49
50
Chapter 10. Conclusions and nal remarks
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