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Prof. Dr. Sandra Hayes
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Fractals
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[Chapter 2]
Technische Universität München
[Course Name]
Prof. Dr. Sandra Hayes
hayes@mathematik.tu-muenchen.de
Fractals
2
2 . AT T R A C T O R C O N S T R U C T I O N U S I N G C O N T R A C T I O N S
The Cantor Middle – Thirds Set D can be constructed as the unique compact invariant set of two
contractions.
0
1
S1
Do = I = [ 0, 1 ]
S2
D1 = S(Do)
0
1/3
2/3
1
S1, S2 : I
S1(x) =
D2 = S2(Do)
x
2
x
, S2(x) = + , xI.
3
3
3
Define K(I) := {KI: K  , K closed} and S: K(I) by S(K) = S1(K)S2(K)  K(I).
Then D1 = S(Do), D2 = S(D1) = S2(Do), and Dn = Sn(Do) no. Thus, D =

S
n
(D0 ) .
0
Dn is composed of 2 n disjoint closed subintervals of I each of length 3  n . Thus, D has zero
Lebesque measure  , because  (D)  ( 2/3) n for all n.
Furthermore, D is invariant under S, i.e. S(D) = D, implying that D is a fixed point under S. This
follows from the fact that

D={x =
an
3
n
n 1
,
an  {0,2 } }.
It will be shown later that D has Hausdorff dimension ln2/ln3, which is about 0.6309.
Example: The Cantor Middle Thirds Set D

D
For D : =
n
,
S(D) = D.
0
Proof: “”

S(D)   S ( Dn ) =
0

D
n 1
= D, since D1  Do .
0
“” For x  D  [ 0, 1/3 ], x =

n 2
.a2a3… =


1
an
3
a n 1
, i.e. D  [ 0, 1/3]
3n
n
= .0a2a3…, with an  {0,2 }. Then, x = S1(y) for y =
Technische Universität München
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hayes@mathematik.tu-muenchen.de
Fractals
Analogously, S2 (D) = D[
3
2
, 1]  S(D) = D.
3
Technische Universität München
[Course Name]
Prof. Dr. Sandra Hayes
hayes@mathematik.tu-muenchen.de
Fractals
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Remark
If K(I) is a complete metric space and if S is a contraction, S has exactly one fixed point D, the
Cantor Middle Thirds Set, by the Contraction Mapping Principle. Thus, D is completely
characterized as the unique compact invariant set of the contractions S1 and S2 ..
Definition
Let (X, d) be a metric space. A map
S:X
is a contraction on X if there is a constant 0<c<1:
d(S(x), S(y))  cd(x, y)
x, yX ;
c is called the contractivity. If
x, yX,
d(S(x), S(y)) = cd(x, y)
S is called a similarity.
Remarks
(1)
A contraction is Lipschitz continuous, and therefore uniformly continuous.
(2)
A similarity is injective, and if KX is compact, S induces a map
K → S(K)
which is a homeomorphism. Thus, S transforms K into a geometrically smaller copy of
itself.
(3)
If S(x) = x and S(y) = y for a contraction, then x = y, i.e. there is at most one fixed
point for a contraction.
Examples
(1)
The linear transformation
 x
 x
 c 0  x 
 x
   = c   with 0<c<1
S   := A   = 
 y
 y
0 c  y
 y
is a similarity for the metric induced by the maximum norm
S:2
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|(x, y)|m
:= max (|x|, |y|).
S(x1, y1) – S(x2, y2) = (cx1, cy1) – (cx2, cy2)
Proof :
= c(x1 – x2, y1 – y2)
and
|c(x1 – x2, y1 – y2)|m = c|(x1 – x2, y1 – y2)|m.
(2)
 0 
Let S1 : = S2 +   with S2 as in (1) for c = ½.
1 2 
Q
(3)
S1(Q)
S2(Q)
1 2 
Let S3 : = S2 +   with S2 as in (1) for c = ½.
 0 
S1
Q
S2
S3
(4) Let Q be the closed unit square and
K(Q) := {KQ: K  , K closed}.
Define a map S:= K(Q)
by S(K) := S1(K) S2(K) S3(K) with S1, S2, S3 as above
Technische Universität München
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Fractals
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in (2) and (3).
Q
S(Q) = Q1
S1(Q1)
S2 (Q) = Q2
Let Q1 = S(Q). Then Q1 Q and Q2 := S(Q1) = S2 (Q)  Q1.
Note that the closed unit interval on the x- and the y -axis remain in Sn(Q) for all n.

The nonempty compact set
S
n
(Q) is called the Sierpinski Right Triangle.
0
Remark: A decreasing sequence Kn Kn +1 of nonempty compact sets in a topological space has
a nonempty compact intersection K .
Proof: K K1 and K is closed, thus compact. Because the finite intersection of the Kn is
nonempty, so is K by the finite intersection property of compact sets.
Example
In the following 4 sketches, each sketch is composed of line segments. Each segment defines a
similarity by mapping the 2 big dots onto the end points of the segment.
For example, the second figure means reduction by 1/3 in the x direction, i.e.
Technische Universität München
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 x   x 3
S1   =   . The third figure represents reduction by 1/3 in the x direction then translating by
 y  y 
(1/3,0)
and rotating by  /2 counterclockwise if (1/3,0) is taken as the origin. Note that a
rotation by θ counterclockwise is the linear transformation
 x
 x
T   = A   ,
 y
 y
 cos
A := 
 sin 
 sin  
,
cos 
Q
θ
For θ =
 0  1

Α = 
1 0 

,
2
. Then
1
 x
 y
Τ   = (0,1) and T   =   .
 0
 y
 x 
T(Q)
A rotation by

counterclockwise is an isometry, i.e. |T(x,y)|m = |(x,y)|m.
2
Note that a rotation by -

clockwise is given by A =
2
 0 1

 :
  1 0
Technische Universität München
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Definition
Let X be a metric space. A finite set of contractions Sj:X, 1 j  n, is called an iterated
function scheme, abbreviated IFS ( which is a notation coined by Barnsley and Demo in 1985
[B/D] ). Let
K(X) := {KX: K  , K compact},
and define
n
S:= K(X)
by
S(K):=
S
j
(K ) .
j 1
Then K  K(X) is invariant if S(K) = K.
The fundamental theorems concerning finitely many contractions were proved by J. Hutchinson in
1981 in the paper “Fractals and Similarities”, Indiana Univ. J.Math. 30, 713-747. In this paper
Hutchinson developed the mathematical background for Mandelbrot’s visionary book “The Fractal
Geometry of Nature” , Freeman, 1977.
Theorem 2.1 (Hutchinson, 1981 (3.1))
Let X be a complete metric space. There is a unique nonempty compact invariant set AX
associated to finitely many contractions Sj: X, 1  j  n; A is called the attractor for S.
For any BK(X)
A = lim Sk(B)
k 
with Sk the k-fold composition of S with itself. The convergence is with respect to the Hausdorff
metric on K(X).
Technische Universität München
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If S(B)B, then

A=
S
k
(B ) .
1
Remarks
1) In the first part of 2.1 the set B can even be a singleton, i.e. a point in X.
2) The proof is based on the fact that there is a natural metric making K(X) a complete metric space
and S a contraction; the Contraction Mapping Principle then implies the first part of 2.1. This metric
will be introduced now.
Hausdorff metric on K(X), X a metric space
Recall that a metric d on a metric space X can be used to define the distance between subsets A and
B of X.
Definition
If BX is nonempty and xX, define
d(x, B) := inf d(x , y)  0.
yB
Hausdorff metric on K(X), X a metric space
Recall that a metric d on a metric space (X, d) can be used to define the distance between subsets A,
B X.
Let (X, d) be a metric space.
Definition
If BX is nonempty and xX, define
d(x, B) := inf d(x , y)  0.
yB
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If A is also nonempty, define
d(A, B) := inf d(x , B)  0, where the infimum is taken over all x in A.
Remarks
(1)
d(x, B)  [0,), since {d(x, y): yB}  .
(2)
If B is compact, there is a closest point in B to any point xX:
d(x, B) = d(x, b) for at least one point bB.
Proof:
d: X  X →  is continuous
 d(x, ) : X → , y→d(x, y), is continuous.
Since BX is compact, d(x,  ) has a minimum in B,
i.e.  b B:
(3)
d(x, b) = min d(x, y) = d(x, B).
yB
d(x, B) = 0  x  B . This implies that d: X X →  does not induce a metric on the set
P(X) of all subsets of X.
Proof: x B  x = lim b n
Technische Universität München
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Example
X = , x = 0, B = (0, 1). Then
d(0, (0,1)) = 0 but {0} (0, 1) = .
B
1
x
0
Definition
Let (X, d) be a metric space and ε>0. The (open) ε – neighborhood of AX is
Aε : = {xX:  a  A with d(x, a)<ε },
i.e. A is thickened by ε.
A
ε
Examples
(1)
X = , A = (0, 1), ε>0  Aε = (-є, 1 + ε).
-ε
0
A
1
1+ε
(2) Χ = 2, Α = Β1(0), ε>0  Aε is the open disc with center 0 and radius 1+ε
(3) Χ = 2, Α = (-1, 1), ε>0  Aε is the open rectangle of length 2 + 2ε and height 2ε with
d = | |m.
Technische Universität München
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-1
1
Aε
Remarks
(1)
Aε =
 B (a)  A,
where B  (a) is the ball around a point a in A with radius  .
a A
(2)
A  B  Aε  Βε.
(3)
r<s
(4)
B=
 Ar  As.
 B
,
 0
“” is trivial, and if xB1/n n, then for every n there is a
1
bn  B with d(bn, x) <
 bn → x  xB , because B is closed.
n
Definition
If A, B  X, define
D(A, B):= inf{ε>0: ABε, BAε}  [0, )
with inf  := .
Examples
(1)
X = , A = , B = {0}

D(, {0}) = .
Proof:
A =   Bε  ε and Aε =   ε. However, for every ε , B  Aε = , since B  .
In general, if B is a subset of X and if B  , then D(, B) = , i.e. D can be used
Technische Universität München
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to define a metric at most on the set {AX: A  } P(X) .
(2)
Χ = , Α = {0}, Β = [0, ) 
D(A, B) = .
Proof:
A  B  A  Bε  ε but there is no ε>0 with BAε. Therefore,
D can be used to define a metric at most on the set {AX: AØ, A bounded}  P(X).
(3)
Χ = , Α = (0, 1), Β = [0, 1]
 D(A, B) = 0
(but A  B).
Proof:
A  B  A  Bε  ε.
Β  Aε ε and
inf{ε>0} = 0.
Remark: D defines a metric at most on the set
B(X) := {A  X: A  , A closed and bounded}
which contains
K(X).
Theorem 2.2
If X is a metric space, then
dH: K(X)  K(X) → ,
(A, B) → dH(A, B) : = inf{ε>0: ABε, BAε}
defines a metric on K(X):={AX: A, A compact} called the Hausdorff metric .
Proof:
(1)
dH is well-defined:
dH(A, B)  [0, ), since A and B are compact and
dH(A, B) = max {d(x, B) :
(2)
x  A }.
Symmetry:
dH(A, B) = dH(B, A) is obvious.
(3)
dH(A, B) = 0  A = B :
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Proof:
“” Follows from
  >0, AB  , since dH(A, B) = 0.
“” “”
A 
inf{ε>0} = 0.
 B
= B by a preceding remark.
 0
Therefore, A  B .
“” similarly.
(4)
Triangle Inequality: d(A, B)  d(A, C) + d(C, B) :
Proof:
Let α > dH(A, C), β>dH(C, B).
It will be shown that
α + β  dH(A, B) :
A  Cα, C  Bβ  A  Bα+β, because if aA , there is a
cC with d(a, c) < α and for c there is an element bB with d(c, b) < β .
The Triangle Inequality for d implies that
d(a, b) < α + β
 ABα+β.
Similarly, B  Aα+ß  α + β 
With
α = dH(A,C) +
dH(A, B).
1
, ß = dH(C, B) + 1/n,
n
(4) follows.
Remarks
iff  ε > 0  N:
dH(Kn, K) < ε n .
(1)
Kn → K in
(2)
dH(A, B) = max{inf{ε>0: B  Aε}, inf {ε>0: B  Aε}}.
Examples
(K(X), dH)
Technische Universität München
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Fractals
(1)
15
Χ = , Α = [0, 1], Β = {0}  dH(A, B) = 1.
Proof:
-ε
0=B ε
1
B  A  Aε  ε
Α  Βε = (-ε, ε)  ε>1.
dH(A, B) = inf{ε>1} = 1.
Note: d(0, [0, 1]) = 0, i.e. the Euclidean distance is 0, but the Hausdorff distance is 1.
(2)
.
B
(3)
d H (A,B)
.
B
A
d H (A,B)
Remark: If X is a metric space, K  X is compact if and only if every sequence in K contains a
subsequence which converges in K (Bolzano - Weierstraß Property).
Technische Universität München
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Theorem 2.3 (Hutchinson)
If (X, d) is a complete metric space, then (K(X), dH) is complete.
Proof:
Let (Kn)n be a Cauchy sequence in K(X).
Define K:={xX:  xm  Km, for every m and xm → x}.
(1)
Kn → K:
Let ε> 0. Since (Kn)n is Cauchy,  N:
(*)
dH(Kk, Kn) <

2
 k, n  N .
It will be shown that K (Kn)ε and Kn  Kε for n  N, implying dH(Kn, K)  ε
 n  N.
Let n  N.
a) K  (Kn)ε :
For xK  xm  Km m so that xm → x   M:
d(xm, x) <

2
mM.

by (*) .
2
 d(x, y)  d(x, xk) + d(xk, y) < ε  k  max{N, M}
If k  N,
 y  Kn with
d(xk, y) <
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 K  (Kn)ε.
b) Kn  Kε
Let yKn.
For j  , set
εj :=

2 j2
.
Since (Kn)n is a Cauchy sequence,  Nj  :
(**)
d(Kk, Kl) < εj
 l, k  Nj
A sequence n < N1 < N2 … and a sequence xNj  KNj satisfying
d(xNj, xNj+1) < εj
and
d(y, xNj) < (3/4)ε  j
will be constructed. The first inequality insures that (xNj) j is a Cauchy sequence
in X , and since X is complete, it has a limit x in X. After the Extension Lemma
for Cauchy Sequences, there is a
Cauchy sequence ( ~
x n )n in X with ~
x n Kn  n such that
~
xN j = xN j
j.
Then x  K follows. The second inequality above implies
d(y, x) = d(y, lim xNj) = lim d(y, xNj)  (3/4)ε < ε
j 
 y  Kε
and
j 
Kn  Kε.
Let N1 > n satisfy (**) for j =1.
By (*), Kn  (KN 1 ) 
  xΝ 1  KΝ 1 : d(y, xN 1 ) <
2

=:εo.
2
Choose N2 > N1 satisfying (**) for j:= 2. By (**) for j = 1 (k = N1, l = N2)
K N 1  ( K N 2 ) 1
  xN 2  K N 2 with d( xN 1 , xN 2 ) < ε1
 d(y, xN 2 )  d(y, xN 1 ) + d(xN 1 , xN 2 )  ε0 + ε1 =


+ 3
2
2
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By induction, sequences n < N1 < N2 < … and xNj  KNj are constructed so that
(xNj)j is a Cauchy sequence and
d(y, xNj) < ε0 + ε1 + ... εj-1 = ε (
=ε(
(2)
1
1
1
1
1
+ 3 + 4 + ..+ j 1 ) < ε ( +
2
2
2
2
2

1
2
k 3
k
)=
1
1
3
+ )=
ε.
2
4
4
K  K(X) :
K   , since  Cauchy sequence (xn), xn Kn, constructed as above.
K closed: Let aj  K, j   : aj → a X. It will be shown that aK.
By definition of K, for every j there is a sequence (xj,k ) in Kk  k with
aj = lim xj,k .
(*)
k 
Since aj → a,
1
.
j
Due to (*), there is a subsequence of integers m1 < m2 < ... with
 N1 < N2 < … such that d( aNj, a) <
d( xN j ,m j , aN j ) <
 d(xN j ,m j, a) 
1
j
.
2
.
j
Let ym j := xNj,mj  Km j  ym j → a. By the Extension Lemma for Cauchy
Sequences, a is in K.
K compact:
It will be shown that K is totally bounded, i.e. for every  > 0
there are finitely many open balls with radius  and centers in K covering K.
Assume the contrary. Then there is a positive  such that there is no covering of K by finitely
many balls of radius  with centers in K. Now a sequence (x i ) in K can be constructed with
(+)
d(x i , xj)   for i  j.
Choose any point x 1 in K. Since B  (x 1 ) doesn’t cover K , there is an x 2 in K with
Technische Universität München
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d(x 2 , x1)   . If x1, …, x n are in K and satisfy (+), choose x n 1 in K with d(x n 1 , x j )  
for 1  j  n. Using part (a) of the proof, for every i there exists a point y i in K N with
However, K N
d(x i , y i ) <  /3.
is compact, and there must be a convergent subsequence (y i j ) of (y i ) which is
Cauchy. Thus, an M exists with
d(y i j , y ik ) <  /3
for i,k  M, implying that
d(x i j , x ik ) < 
for i,k  M, a contradiction to (+).
Note : Theorem 2.3 holds when K(X) is replaced by B(X).
Extension Lemma for Cauchy Sequences
If (X, d) is a metric space, let (Kn)n be a Cauchy sequence in K(X). Let
0 < n1 < n2 < …
be a subsequence of integers. If ( x n j )j is a Cauchy sequence in X with x n j  K n j  j,
there is another Cauchy sequence ( ~
x )n in X with ~
x Kn  n such that
n
n
~
x n j = xn j
j.
Proof:
Contractions on (K(X), dH) for a metric space (X, d)
Theorem 2.4
If (X, d) is a metric space, then every contraction S: X with contractivity c defines a
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contraction S: K(X) on (K(X), dH) with the same contractivity c by
S(A) := { S(x): xA}
for A  K(X) := {AX: A  , A compact }.
Proof:
(1)
S: K(X)  is well defined:
S(A)  K(X), because S is continuous and A is compact.
(2)
dH(S(A), S(B))  c dH(A, B),
A, B  K(X):
Let ε > dH(A, B). It will be shown that
dH(S(A), S(B))  cε.
A  Bε  S(A)  S(Bε).
S(Bε)  (S(B))cε , since  xBε  bB: d(x, b) < ε
 d(S(x), S(b))  cd(x, b) < cε
 S(A)  (S(B))cε.
Similarly, S(B)  (S(A))cε.
 dH(S(A), S(B))  cε
If εn := dH(A, B) +
ε > dH(A, B).
1
c
, then dH(S(A), S(B))  cdH(A, B) +
 n.
n
n
Taking lim gives (2).
n
Remark
Aε  Bε  (A  B)ε for Α, Β  Κ(Χ).
Proof:
A, B  A  B  Aε, Bε  (A  B)ε.
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Corollary 2.5 ( Hutchinson, 1981 (3.2))
Finitely many contractions Sj: X with contractivity cj, 1  j  n, define a contraction S: K(X)
with contractivity c := max1jn cj by
n
S(A) :=
S
j
( A) .
j 1
Proof:
(1)
S: K(X)  well defined:
Sj(A)  K(X), 1  j  n  S(A)  K(X).
Notice that the hypothesis “finitely” is used.
(2)
dH(S(A), S(B))  max1jn dH(Sj(A), Sj(B)) =: M:
Note that the hypothesis “finitely” is also used here to insure the existence of such an M
 .
Let ε > M

Sj(A)  (Sj(B))ε j . Therefore,
S(A) 
n
 (S
j 1
and it follows from the remark above that S(A)  (S(B))ε.
Similarly, S(B)  S(A)ε.
 dH(S(A), S(B))  ε
For εn := M +
1
,
n
ε > M.
dH(S(A), S(B))  M +
1
n
 n.
Taking lim gives (2).
n
(3) For j{1, …, n} with M = dH(Sj(A), Sj(B)),
dH(S(A), S(B))  dH(Sj(A), Sj(B))  cjdH(A, B)  cdH(A, B).
Remarks
(1) If A, B, C  K(X) with A  B  C, then dH(A, C)  dH(A, B)
Proof:
j
( B)) 
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Let ε > dH(A, C)  C  Aε  B  Aε  ε  dH(A, B).
(2)
If X is a metric space, a subset K of X is compact if and only if every sequence in K
contains a subsequence which converges in K ( the Bolzano-Weierstrass Property).
Lemma 2.6
Let X be a metric space, and let
Kn  K(X), n, be decreasing, i.e.
Kn  Kn+1, n  .
If K  K(X) with K  Kn, n, then

K =
K
n

dH(Kn, K)  0.
lim Kn = K, i.e.
n
1
n
Proof: “”:
K

K
 Kn . After the remark above, 0  dH(K,
n
K
1
 dH(K,

n
)  dH(K, Kn)  n
1

 Kn ) = 0
 K=
1

K
n
.
1
“” :

K :=
K
n
 Kn+1  Kn.
1
Let rn : = dH(Kn, K) = inf {  > 0 : Kn
 K  } for n . By the remark above,
rn  rn+1  0, n   r  0: rn → r.
If r > 0, let ε : = r/2. Because rn  r >  , it follows that
and for every n there is an x n
 Kn
such that x n
Kn  K  ,
 K .
Since Kn  K 1 for every n, the sequence (x n ) has a convergent subsequence (xk j
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and thus a limit x in K 1 . Then  N:
d(xk j , x) < ε  j  N.
If x  K, a contradiction to xk j  Kε follows, and the proof is finished.
Fix m   and consider j > m. Then k j > k m and Kk j  Kk m

xk j  Kk m  j > m .

x Kk m
for all m   implying x  K .
Proof of Theorem 2.1: After 2.3, 2.5 and 2.6, Theorem 2.1 follows.
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Example: The Sierpinski Right Triangle
The attractor for the 3 similarities in 2 given by
(see Example (4) ) is the Sierpinski Right Triangle defined recursively as follows using infinite
removals.
Let Do be a right triangle with side lengths 1 (hypotenuse
2 ). Divide Do into 4 right triangles by
1
joining the midpoints of each side. Each of the 4 smaller triangles has side length
and one of
2
these triangles is inverted.
D0
D1
D2
Define D1 by removing the open inverted triangle, i.e. remove its interior but not its boundary.
Proceed recursively: Define Dn+1 by removing the open inverted triangle from each of the 3n right
1
triangles in Dn which have side lengths n .
2

D :=
D
n
is the Sierpinski Right Triangle (W.Sierpinski, 1915); D is nonempty and compact,
1
Because Dn  Dn+1 for every n , and Dn is nonempty and compact.
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Remark
The area of D is 0 , i.e. the Sierpinski Triangle has zero Lebesque measure l(D) in 2.
Step
Area
1
=a
2
3
a
4
32
a
42
…
Do
D1
D2
…
Number of triangles
Side lengths
1
1
3
32
…
n
 3
  a
 4
Dn
3n
1
2
1
22
…
1
2n
n
 3
l(D)    a
 4
n  l(D) =0.
Remark
The boundaries of each triangle in Dn remain in Dn+1 and therefore, D contains all of these line
segments. At step n, 3n new segments arise, each with a length of at least 2 –n. The length of all the
segments in Dn is at least 3n2-n → , implying that the Sierpinski Triangle has infinite length, i.e.
infinite Lebesque measure in .
Later we will see that it has a dimension of about 1.58.
Remark
The Sierpinski Right Triangle is the attractor for the three contracting similarities
 x 
 x
 x 1


 
  
x
x
 x




S1   =  2  ,
S2   =  2  ,
S3   =  2 2  .
 y  1 
 y 
 y 
 y
 y
 y
 2 2
2
 2 
Proof: The claim is that S(D) = D. It is immediate that Dn+1 = S(Dn) for every n, implying

S(D) = S(
D
1
n
)

 S (D
n
) = D.
1
To prove that D  S(D) , let p be in D. Then p  D1 = S(Do) = S1(Do)  S2(Do)  S3(Do) .
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It is no restriction to assume that p belongs to only one of the three subtriangles Sj(D0), since
otherwise p is either (0,½), (½, ½), or (½,0) , which is S1(q), S2(q) or S3(q) for q = (0,1), (1,0) or
(0,0). Therefore, assume that p is in S1(D0) but not in S2(D0) or in S3(D0).
 n, p  Dn+1 = S(Dn) = S1(Dn)  S2(Dn)  S3(Dn).
Dn  Do 
S2(Dn)  S2(Do) and S3(Dn)  S3(Do)
 n  qn  Dn : p = S1(qn) =

pS1 ( Do )
p  S1(Dn) n.
qn
 qn = 2p  n  qn = qn+1 =: q  D and
2
p = S1(q)  S1(D)  S(D ) = S1(D)  S2(D)  S3(D).
Remarks
(1) For the closed unit square Q in Example (4), above, the Sierpinski Right Triangle

D=
S
k
(Q ) ,
k 0
(2) The Sierpinski Triangle is not a topological Cantor set ,i.e. not a compact, totally
disconnected set in 2 without isolated points.
Proof:
D is connected , because all triangle boundaries of Dn belong to D.
Attractor for contractions Sj, 1  j  n, and the iteration of the Sj’s.
The attractor for finitely many contractions Sj: X on a complete metric space X is also the closure
of the fixed points of all possible finite compositions of the Sj’s. To prove this the following
definition will be needed.
Definition
Let (X, d) be a metric space. Then the diameter of B  X, B   is
|B| := sup{d(x, y): x, y B}.
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Remarks
(1)
|B| = 0  card B = 1, i.e. B is a singleton.
(2) A  B
 | A|  |B|.
(3) The intersection  B k of a decreasing sequence Bk  Bk+1 of nonempty compact sets in
a metric space X is a singleton x if lim  Bk  = 0.
Furthermore, if xk  Bk for all k , then xk  x.
Proof:
 Bk  K(X), and
| Bk |  | Bk| for all k. Thus, card ( Bk) = 1.
Let x = Bk and  > 0. There is an N so that | Bk | <  for N  k
Since x  Bk for every k,
d(xk, x )  |Bk| <  for N  k, and therefore,
(4) If S: X is a contraction with contractivity c, then
xn  x.
n
|S(B)|  c|B|.
(5) Let S1, S2: X be contractions and define S(B) := S1(B)  S2(B) for BX.
Then S2(B) := S(S(B)) = S(S1(B))  S(S2(B))
= S1(S1(B))  S2 (S1(B)) S1(S2(B))  S2(S2(B)) .
Theorem 2.7 (Hutchinson,1981 (3.1))
Let X be a complete metric space and let Sj: X be a contraction with contractivity cj, 1  j n.
For every map L:  → {1, …, n}, L(k):= Lk, and for every BK(X) with Sj(B)B, 1  j  n, denote
S L1 ... Lk := S L1  S L2  ...  S Lk .
Then

S
k 1
L1 ... Lk
( B)
is a singleton, denoted by xL, and xL is in the attractor A of the Sj’s. Furthermore, xL is independent
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of B and defines a surjective map
φ: {1, …, n} → A,
i.e.
L→ xL,


S L1 ... Lk ( A)  .
L  
 k 1

A =
Proof:
(1)
S L1 ... Lk , Lk 1 ( B)  S L1 ... Lk ( B) k   is a decreasing sequence of nonempty compact sets and
| S L1 ... Lk ( B) |  c L1 | S L2 ... Lk ( B) |  …  c L1  c Lk |B| < ck|B|
for c: = max1jn cj < 1 .
After the third remark above,

S
k 1
L1 ... Lk
( B) =: {xL}.

(2)
But {xL} =
 S L1 ... Lk ( A) , due to A =
k 1


S
k 1
L1 ... Lk
S
k 1
Therefore, S L1 ... Lk ( A)  S L1 ... Lk ( B) for all k, implying that
However,
n
 S k ( B)  S(B) :=

S
k 1
j
( B)  B and
j 1
L1 ... Lk
( A)  {xL}.
( A)   , as a deceasing sequence of sets in K(X), and consequently,

S
k 1
L1 ... Lk
( A) = {xL}.
It follows that xL is independent of B and is in the attractor, because
S(A) = A implies Sj(A)  A  j .
(3)
φ is surjective:
n
Α = S(Α) =
S
j
( A) .
j 1
n
If xA,  L1  {1, …, n} with x  S L1 ( A) = S L1 ( S j ( A)) =
j 1
n
S
j 1
  L2  {1, …, n} with x  S L1 L2 ( A) .
By induction,  k  Lk  {1, …, n} with x  S L1 ... Lk ( A) 
L1 j
( A) 
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x

S
k 1
L1 ... Lk
 φ is surjective.
( A)
Corollary 2.8
For L: → {1, …, n} and k , let
x k : x k
with
τk :  → {1, …, n}
the periodic sequence
L1 … Lk L1 … Lk L1 … Lk….
Then
lim xk = xL
k 
and
S L1 ... Lk ( x k ) = xk,
i.e. xkA is the unique fixed point of the contraction S L1 ... Lk .
Proof:

(1)
{xL} =
S
k 1
 k,
L1 ... Lk
( A)  S L1 ... Lk ( A)

{xk } =
 S
l 1
Set l := k.  xk  S L1 ... Lk ( A)
(2)
 q  and
k (1 ) ... k ( l )
k
 k.
( A)  S k (1) ... k ( l ) ( A)
 l.
 xk → xL after the remark above.
j1, …, jq  {1, …, n},

S j 1 …j q (xk) = S j 1 …j q (  S k (1) ... k ( l ) ( A) ) =
l 1
Now set q = k and jl = Ll for 1  l  k.

S
l 1
j1 ... j q k (1) ... k ( l )
( A) = x j1 ... jq L1 ... Lk Lk 1 ..
Then S L1 ... Lk ( x k ) = xk, and
xk is a fixed point .
Remark
Every fixed point of every contraction S L1 ... Lk is in A for all k and all L:  → {1, …, n}.
Corollary 2.9
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A is the closure of the set of fixed points of the S L1 ... Lk  L: → {1, …, n} and
 k.
Proof:
“”  x  A  L  {1, …, n}:
x = xL = lim xk
by 2.7 and 2.8.
“” Every fixed point xk of S L1 ... Lk is in A, and if x = lim xk, then xA, since A is closed.
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Image Data Compression
Pictures may be worth a thousand words, but they require a lot of computer memory to store which
is expensive. Images are stored in computers as bits (a bit is a binary information unit, either 0 or 1,
on or off). In 1994 the storage per bit cost a half a millionth of a dollar,
1 bit
=
1
= 0.510-6 dollars.
6
2  10
Human eyes can process around 8 millions bits so that a moderate quality photo costs about $4 to
store and a photo album about $1000 for 250 images.
Data compression reduces costs and also increases the speed of data transmition, saving time and
money.
The Sierpinski Triangle and the Cantor Middle Third Set show how a small number of contractions
can determine an intricate structure .In this way, finitely many contractions can be used for data
compression.
One of the most popular methods of data compression, the so-called JPEG standard (Joint
Photographic Experts Group ) uses a different approach, namely Fourier Transforms (in particular,
the Discrete Cosine Transform) which filters out the high-frequency Fourier coefficients.
Image compression using contractions was commercialized by Michael Barnsley, the author of
“Fractals Everywhere”, 1988, Academic Press ( 2nd edition 2000, Morgan Kaufmann )who
founded the company Iterated Systems Inc. in Atlanta, Georgia.
Data Compression Problems
(1)
What compact sets are attractors A or can be approximated by attractors A for finitely
many contractions?
(2)
How can finitely many contractions Sj approximating a compact set be found? (The
encoding problem).
The attractor A is approximated by every iterate Sk(B) for BK(X), since Sk(B) → A, and the more
iterates the better the approximation. The next theorem, which is a straightforward corollary of the
classical Banach Fixed Point Theorem, gives an error estimate:
Collage Theorem 2.10
Let X be a complete metric space and let Sj:X be contractions with contractivity cj, 1  j  n.
If c = max1jncj, then for any BK(X) and any k  0
dH(Sk(B), A) 
ck
dH(B, S(B)).
1 c
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Proof: (By induction)
k=0:
dH(B, A)  dH(B, S(B)) + dH(S(B), A)
[Triangle Inequality]
 dH(B, S(B)) + cdH(B, A),
from 2.5 using A = S(A).
k  k + 1:
dH(S
k+1
ck
ck
2
(B), A) = dH(S (S(B),A) 
dH(S (B), S(B)) 
cdH(S(B), B), by 2.5.
1 c
1 c
k
Remarks
(1)
In the visual arts a collage is a collection of pieces cut out from various materials and then
n
pasted together. In this sense, S(B) =
S
j
( B)
is a collage.
j 1
(2)
The smaller dH(B, S(B)) is, the better the approximation dH(B, A) of B to A is.
(3)
The closer c is to 0, the better the approximation of B to A. (
(4)
For encoding (i.e. given BK(X), find n and find n contractions Sj:X whose
1
>1).
1 c
n
attractor A resembles B), find contractions Sj so that S(B) =
S
j
( B) is close to B in the
j 1
Hausdorff metric. If Sj are similarities, Sj(A) is a homeomorphic copy of A.
Examples
1) Sierpinski Right Triangle
Goal:
Find the contractions Sj:2 whose attractor is the Sierpinski Right Triangle.
Method:
How many reduced affine images are needed to cover the original triangle? Obviously, the answer
is three. Find these affine maps.
An affine map  2  is uniquely determined by any 3 points in 2 and their images, provided
such points are not collinear.
Technische Universität München
[Course Name]
Prof. Dr. Sandra Hayes
hayes@mathematik.tu-muenchen.de
Fractals
33
 x
 x
Find S1   := M   + p, p2 and M is a 2x2-matrix, so that S1(0, 0) = (0, 0) ( implying p = 0),
 y
 y
1
1
S1(1, 0) = ( , 0), and S1(0, 1) = (0, ).
2
2
1
1
C
C'
1
A
1
A
B
B'
S1
 x   a b   x   ax  by 
   = 

M   = 
 y   c d   y   cx  dy 
ax  by   

cx  dy   
4 unknowns a, b, c, d in 6 linear equations (2 equations for each point together with its image)
(0, 0) → (0, 0) = (α, β):
(1, 0) → (
1
, 0) = (α, β):
2
(0, 1) → (0,
1
) = (α, β):
2
1

S1(x, y) =  2
 0

α=β=0
a=
1
,
2
b = 0,
c=0
d=
1
2

0   x
 x
   = M   .
1   y
 y

2
S1 is a contraction if ||M||<1 with
||M|| := sup
x0
Mx
=
x
1

M t M = M2 =  4
 0

max ,
λmax is the maximum eigenvalue of MtM . In our case,

0
1
 , λmax = ,
1
4

4
max =
1
.
2
Technische Universität München
[Course Name]
Prof. Dr. Sandra Hayes
hayes@mathematik.tu-muenchen.de
Fractals
34
2) If X = 2 , n = 4, and c = 0.6, then for any B in 2 with dH(B, S(B)) < 0.02,
dH(B, A) < 0.05.
dH(B, A)  2.5.
Note that if dH(B,S(B)) < 1.0, then
Proof:
dH(B, A) 
(1/ 1 – 0.6) dH(B, S(B)) < 0.02/0.4 = 0.05 in the first case,and in the
second case
dH(B, A)  1/1-0.6 = 1/0.4 = 2.5.
Any compact set in d can be approximated arbitrarily closely by a self-similar set, i.e. by an
attractor given by contracting similarities, as will be shown next. Beforehand recall that an
invertible matrix M is orthogonal if M t = M 1 and that the only linear isometries are given by
orthogonal matrices ( see for example Fischer, Lineare Algebra ).
Theorem 2.11
S: n is a similarity if and only if there is an r>0 , a vector a  n , and an orthogonal nxn-matrix
M  O(n) so that
S(x) = r M(x) + a,
i.e. S is a combination of a translation by a vector a, a scaling by r, and an orthogonal linear
transformation ; in particular, S is affine.
Proof: Denote the Euclidean norm of x  n by | x |. Then | x | 2 = <x, x> for the Euclidean
inner product.
“” :
Let |S(x) – S(y)| = r |x – y|  x, yn with r > 0.
Define T(x) :=
1
(S(x) – S(0))
r
 T(0) = 0, and |T(x) – T(y)| =
1
|S(x) – S(y)| = |x – y|,
r
i.e. T is an isometry fixing 0. In particular, T is norm preserving , i.e. |T(x)| = |x| .
Technische Universität München
[Course Name]
Prof. Dr. Sandra Hayes
hayes@mathematik.tu-muenchen.de
Fractals
35
T preserves inner products:
x, y
2 <x, y> = |x|2 + |y|2 – |x – y|2
since by the bilinearity of the inner product
(*)
<x-y, x-y> = <x, y> +
<y, y> - 2 <x, y> .
Therefore, |x – y|2 = |T(x) – T(y)|2 ,since T is an isometry , and by (*)
|T(x) – T(y)|2 = |T(x)|2 + |T(y)|2 – 2 <T(x), T(y)>
= |x|2 + |y|2 – 2 <T(x), T(y)> ,

since T preserves the norm.
<T(x), T(y)> = <x, y> using (*) again.
(+)
T is linear:
Let {ej : 1  j  n } be an orthogonal basis for  ,
i.e. <ei, ej> = δij.
 {Τ(ej): 1  j  n } is also an orthogonal basis, because of (+).
 T(x) =
n
  T ( x), T (e )  T (e )
j
j 1
n
=
  x, e
j 1
j
j
Fourier series
 T (e j )
by (+).
 T is linear .
T is orthogonal, since it is an isometry .
 T(x) = M(x) with M  O(n)
 r T(x) = r M(x) = S(x) – S(0)
 S(x) = r M(x) + S(0).
“”:
Let S(x) = r M(x) + a.
Then S is a similarity, since |S(x) – S(y)| = r |M(x -y)| = r |x – y|,
because orthogonal matrices preserve inner products and thus also norms.
Technische Universität München
[Course Name]
Prof. Dr. Sandra Hayes
hayes@mathematik.tu-muenchen.de
Fractals
36
Examples (Orthogonal Transformations)
 cos 
M:= 
 sin 
(1)
Rotation counterclockwise by 
 cos 
MTM:= 
  sin 
(2)
 sin  

cos  
sin    cos 

cos    sin 
 sin    1 0 
=
.
cos    0 1 
Reflection on the y-axis
  1 0

M = 
 0 1
(x, y) → (-x, y),
(3)
  1 0   1 0  1 0
 
 = 
 .
MM = 
 0 1  0 1  0 1
MT = M,
Reflection on the x-axis
1 0 

M = 
0
1


Theorem 2.12
Let B be a nonempty compact subset of d . For every ε>0 there are finitely many contracting
similarities S1, … Sn: d  for which the attractor A satisfies
dH(B, A) < ε .
B, p98
1
10
5
(1) =
= = 2.5.
1  .6
4
2
(a)
d(B, S(B) < 1.0, c =.6  dH(B, A) 
(b)
B  2, S1, …, Sn: (n = 4) cont. of S is .6
dH(B, S(B) < .02, c =0.6  dH(B, A) < .05,
since dH(B, A) 
0.20
1
d(B, S(B)) <
= 0.05.
4
1  .6
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Technische Universität München
[Course Name]
Fractals
Prof. Dr. Sandra Hayes
hayes@mathematik.tu-muenchen.de
37