An Introduction to Fractals and Hausdorff Measures by David C Seal A Senior Honors Thesis Submitted to the Faculty of The University of Utah In Partial Fulfillment of the Requirements for the Honors Degree of Bachelor of Science In Mathematics Approved: ————————————– ————————————– Davar Khoshnevisan Aaron Bertram Supervisor Chair, Mathematics ————————————– ————————————– Gordan Savin Martha Bradley Departmental Honors Advisor Director, Honors Program
Transcript
An Introduction to Fractals and Hausdorff Measures
by
David C Seal
A Senior Honors Thesis Submitted to the Faculty ofThe University of Utah
In Partial Fulfillment of the Requirements for the
————————————– ————————————–Gordan Savin Martha BradleyDepartmental Honors Advisor Director, Honors Program
Abstract. It is possible for one to define fractals as those sets whichhave non-integral Hausdorff Dimension. This paper defines HausdorffDimension and rigorously introduces the necessary theory to prove thatone such set is indeed a fractal. The first chapter includes a proof ofBanach’s Fixed Point Theorem. The Hausdorff Metric is defined andit is mentioned how one produces ‘generators’ for creating self-similiarfractals. The second chapter defines Hausdorff Dimension and provessome characteristics regarding it, including its invariance under differentLp metrics. Frostman’s lemma is introduced as a tool for calculating theHausdorff Dimension of sets, and examples of fractals are introduced inorder to illustrate how one proceeds in calculating the dimension offractals.
Contents
1
CHAPTER 1
Classical Fractals
This chapter outlines basic theory needed to construct certain ClassicalFractals. An application of a specialized version of Banach’s Fixed PointTheorem will be used to construct most if not all the classical fractals en-countered in this paper.
1. Background Theory
If X and Y a metric spaces, we say a function f : X → Y is a lipschitzfunction (or contraction mapping) if there exists a constant 1 > c ≥ 0 suchthat
(1.1) dY ((f(x), f(y)) ≤ cdX(x, y) ∀x, y ∈ X.
If f is a lipschitz function, then we immediately have that f is a continuousfunction.
Theorem 1.1. Banach’s Fixed Point Theorem. If (X, d) is a completemetric space, and a function f : X → X is a lipschitz function, then f hasa unique fixed point.
Proof. Pick any x0 ∈ X. Define a sequence xnn∈N, by xn+1 = f(xn).If x1 = f(x0) = x0, we have already found a fixed point, (uniqueness will beshown shortly), so we may assume x1 6= x0. We have the following estimateon any two consecutive terms in the sequence:
d(xn+1, xn) = d(f(xn), f(xn−1))
≤ cd(xn, xn−1) ≤ . . . ≤ cnd(x0, x1).
In fact, xnn∈N is a cauchy sequence. Let n, k ∈ N.
there exists an N ∈ N, for every n ≥ N , we have cn < ǫ 1−cd(x0,x1)
. The
previous inequalities show that d(xn+k+1, xn) < ǫ. Therefore, xnn∈N is acauchy sequence, and since (X, d) is a complete metric space, we have thatlimn→∞ xn = x∗ converges. Then, the following recursive relation:
xn+1 = f(xn),
allows us to take limits of both sides. Hence,
limn→∞
xn+1 = limn→∞
f(xn).
Since f is continuous, we can take the limit inside. This implies
x∗ = limn→∞
xn+1 = limn→∞
f(xn) = f(
limn→∞
xn
)
= f(x∗),
which is the desired property of a fixed point. To show uniqueness, supposef(x1) = x1 and f(x2) = x2 for some x1, x2 ∈ X. Then,
d(f(x1), f(x2)) = d(x1, x2) ≤ cd(x1, x2).
This shows that
0 ≤ (c − 1)d(x1, x2).
Since c < 1 and d(x1, x2) ≥ 0, we must have that d(x1, x2) = 0. This verifiesthe uniqueness of x∗, since X is Hausdorff.
For the purposes of this paper, we will take X = C, the set of all compactsubsets in R
n. But first, we must define a metric on C. For any ǫ > 0, theǫ-extension of A is defined as:
Aǫ := x : |x − y| < ǫ, y ∈ A =⋃
y∈A
Bǫ(y).
Obviously, Aǫ ⊇ A for any set A. For two compact sets A,B ∈ C, we definethe Hausdorff Metric as:
d(A,B) := infǫ : A ⊆ Bǫ, B ⊆ Aǫ.
We must first show this is a metric:
(1) d(A,B) ≥ 0, and d(A,B) = 0 ⇐⇒ A = B for any A,B ∈ C.
Proof. d(A,B) ≥ 0 follows immediately from the definitionsof Aǫ and Bǫ. Suppose d(A,B) = 0. Let a ∈ A. For every ǫ >0, a ∈ Bǫ since d(A,B) = 0. If a /∈ B, then a ∈ Bc ⇒ ∃ ǫ >0, Bǫ(a) ⊂ Bc, so inf|a−b| : b ∈ B ≥ ǫ. Therefore a ∈ A ⇒ a ∈ Bif d(A,B) = 0, so we have A ⊆ B. The identical argument showsB ⊆ A. Therefore, A = B.
(2) d(A,B) = d(B,A) for any A,B ∈ C.
Proof. This is immediate, since order does not matter for the‘and’ statement.
4 1. CLASSICAL FRACTALS
(3) d(A,C) ≤ d(A,B) + d(B,C) for any A,B,C ∈ C. We will showthis with the following lemma:
Lemma 1.2. For any set A ∈ C, and every ǫ, δ > 0, we havethat (Aδ)ǫ ⊆ Aδ+ǫ.
Proof. Let ǫ, δ > 0 and A ∈ C. Fix any a ∈ (Aδ)ǫ. Thereexists an a0 ∈ Aδ such that |a0−a| < ǫ. Since a0 ∈ Aδ, there existsan a1 ∈ A such that |a1 − a0| < δ. The triangle inequality gives:
|a − a1| ≤ |a − a0| + |a0 − a1| < ǫ + δ,
which shows that a ∈ Aǫ+δ.
Proof of Metric Property (3). Fix three sets, A,B,C ∈C. Pick any δ1 > d(B,C), δ2 > d(A,C). We have the followinginclusions:
B ⊆ Cδ1 ;(1.2)
C ⊆ Bδ1 ;(1.3)
A ⊆ Cδ2 ;(1.4)
C ⊆ Aδ2 .(1.5)
Equation (??) shows that
(1.6) Cδ2 ⊆ (Bδ1)δ2 .
Applying lemma (??) and equation (??) to the previous inclusion,we have:
(1.7) A ⊆ Bδ1+δ2 .
Similiarly, (??) shows that
(1.8) Cδ1 ⊆ (Aδ2)δ1 .
Applying lemma (??) and equation (??) to the previous inclusion,we have that
(1.9) B ⊆ Aδ1+δ2 .
Since d(A,B) is the infimum over all such extensions, we have that
(1.10) d(A,B) ≤ δ1 + δ2
for every δ1 > d(B,C) and δ2 > d(A,C). Letting δ1 ↓ d(B,C)followed by δ2 ↓ d(A,C), we have that d(A,B) ≤ d(B,C)+d(C,A).
It remains to show that this space of compact subsets of Rn is complete
under the Hausdorff Metric.
Theorem 1.3. Suppose a sequence of non-empty sets, An ⊆ C is acauchy sequence under the Hausdorff Metric. Then limn→∞ An = A exists,and is non-empty.
1. BACKGROUND THEORY 5
We will define the set
(1.11) A := x : ∀x ∃ xn ∈ An, limn→∞
xn = x,
and start by proving that this set is non-empty.
Lemma 1.4. The set A as defined in (??) is non-empty.
Proof. The trick to this proof is to produce a cauchy sequence of pointsand apply the completeness of Euclidean space. First we start by applyingthe fact that An is cauchy. There exist N1 < N2 < N3 < . . . such that
AN1⊂ (AN1+k)2−1(1.12)
AN2⊂ (AN2+k)2−2
AN3⊂ (AN3+k)2−3
...
for every k ∈ N. To have a complete sequence starting at 1 rather than N1,we may pick any points a1, . . . , aN1−1 in each corresponding A1, . . . , AN1−1.(From now on it will be assumed that each ai ∈ Ai). Start with any aN1
, andchoose aN1+1, aN1+2 . . . aN2
so that |aN1−aj1| < 1
2 for N1 ≤ j1 ≤ N2. Giventhis aN2
and applying the fact that AN2⊂ (AN2+k)2−2 for any k, we may
select aN2+1, aN2+2 . . . aN3so that |aN2
− aj2 | < 122 for each N2 ≤ j2 ≤ N3.
Continuing in this manner, we construct a sequence of points, an∞n=1, with
the following properties:
|aN1− aj1 | < 1
2 for N1 ≤ j1 ≤ N2,(1.13)
|aN2− aj2 | < 1
22 forN2 ≤ j2 ≤ N3,
|aN3− aj3 | < 1
23 for N3 ≤ j3 ≤ N4,
...
|aNi− aji
| < 12i for Ni ≤ ji ≤ Ni+1
...
It remains to show that this construction produces a Cauchy sequence. Letǫ > 0 and k ∈ N. There exists an r ∈ N such that 1
2r < ǫ. Set N := Nr,and suppose n ≥ Nr. We must have that Nr ≤ Nj ≤ n < Nj+1 and thatNs ≤ n + k < Ns+1 for some j, s ∈ N. (Note that r < j). The following
6 1. CLASSICAL FRACTALS
estimate will be useful:
|aNj− an+k| ≤ |aNj
− aNj+1| + |aNj+1
− aNj+2| + . . .(1.14)
. . . + |aNs−1− aNs | + |aNs − an+k|
<1
2j+
1
2j+1+ . . . +
1
2r
<1
2j
(
1 +1
2+
1
4+ . . .
)
=1
2j
(
1
1 − 12
)
=1
2j(2).
The finite sum had been passed to the infinite geometric series, and theinequalities came from repeated use of equations (??) together with thetriangle inequality. Then the following inequality shows our sequence iscauchy, after applying an inequality from (??) one more time:
|an − an+k| ≤ |an − aNj| + |aNj
− an+k|(1.15)
≤1
2j+ 2
(
1
2j
)
< 4
(
1
2j
)
≤ 4
(
1
2r
)
< 4ǫ.
Since ǫ was arbitrary, we have shown that A is non-empty, since there is aconvergent sequence of points, an ∈ An.
Lemma 1.5. The sequence An converges to the set A as defined above.
Proof. Let ǫ > 0. There exists an N ∈ N, for every n ≥ N and everyk ∈ N, we have:
(1.16) d(An, An+k) < ǫ/2.
In particular, we have that
(1.17) d(AN , An) < ǫ/2
for every n ≥ N . To show convergence, we need to show:
A ⊆ (AN )ǫ;(1.18)
AN ⊆ (A)ǫ.(1.19)
We will first show (??). Let x ∈ A. There exists a sequence xnn∈N
where each xn ∈ An and limn→∞ xn = x. There exists an nx ≥ N (maydepend on x) such that |xnx − x| < ǫ/2. Because of (??) we can find ayN ∈ AN where |yN − xnx | < ǫ/2. Since
we have that A ⊆ (AN )ǫ.For the second inclusion, pick any xN ∈ AN . We will use an argu-
ment almost identical to the one given to show A is non-empty, although inthis case, we already have a start due to equation (??). Again the points
2. APPLICATIONS OF THE HAUSDORFF METRIC 7
x1, x2, . . . , xN−1 can be arbitrary, we will only concern ourselves with thetail end of this sequence. There exist N = N1 < N2 < N3 < . . . such that
AN1⊂ (AN1+k)ǫ2−1(1.21)
AN2⊂ (AN2+k)ǫ2−2
AN3⊂ (AN3+k)ǫ2−3
...
for every k ∈ N. We can again construct a similar sequence, this time withthe distances multiplied by a factor of ǫ. This sequence, xn
∞n=1, has the
following properties:
|xN − xj1 | < ǫ2 for N1 ≤ j1 ≤ N2,
|xN2− xj2 | < ǫ
22 for N2 ≤ j2 ≤ N3,
|xN3− xj3 | < ǫ
23 for N3 ≤ j3 ≤ N4,
...
|xNi− xji
| < ǫ2i for Ni ≤ ji ≤ Ni+1
...
This sequence also doesn’t wander very far from xN , since if k ∈ N, thenNs ≤ n + k < Ns+1 for some s and
|xN − xN+k| ≤ |xN − xN2| + |xN2
− xN3| + . . .(1.22)
. . . + |xNs − xN+k|
=ǫ
2+
ǫ
22+ . . . +
ǫ
2s
< ǫ.
This sequence converges to some x∗ ∈ A, and this point x∗ is near xN , since|xN −x∗| ≤ |xN −xm|+ |xm −x∗|. Since limm→∞ xm = x∗, the second termcan become arbitrarily small, and the first term is small from the previousinequality. Hence, |xN − x∗| < ǫ, so xN ∈ A, the desired result.
2. Applications of the Hausdorff Metric
With these preliminaries out of the way, we may begin with definingsome fractals. By creating interesting contraction mappings, we may applyour specialized version of Banach’s Fixed Point theorem in order to createthese objects.
Suppose a finite set of functions, f1, f2, . . . , fr are contraction map-pings defined on R
n. We declare a new function F : C → C,
(2.1) F (A) :=
r⋃
i=1
(fi(A))
8 1. CLASSICAL FRACTALS
where fi(A) := y : ∃ x ∈ A, fi(x) = y is the image of a compact set Aunder fi. It will be shown that F : C → C is a contraction mapping, and byapplying Banach’s Fixed point theorem we may ascertain the existence of a“fixed point”. This fixed point will oftentimes be a fractal itself. However,we wish to discard of trivial examples that do not create any interesting“fixed points”. (Consider the function that breaks the square into fourdisjoint parts, whose union is equal to the original square).
Lemma 1.6. If F and fi are as given in (??), then F is a lipschitzfunction.
Proof. For each fi, there exists a ci < 1 such that |fi(x) − fi(y)| ≤ci|x − y| for every x, y. Set c := maxc1, c2, . . . , cr < 1. It is sufficient toshow that
(1) F (B) ⊆ F (A)cd(A,B),(2) F (A) ⊆ F (B)cd(A,B)
for every A,B ∈ C, since d (F (A), F (B)) is the infimum over all such exten-sions. We proceed by fixing two sets, A,B ∈ C.
Pick any point z ∈⋃
i fi(B). There exists an x ∈ B, and a fixed i suchthat z = fi(x). Consider any ǫ > d(A,B). Then x ∈ Aǫ since ǫ > d(A,B)and B ⊆ Aǫ. There exists a y ∈ A such that |y − x| < ǫ. We have:
(2.2) |fi(x) − fi(y)| ≤ ci|x − y| ≤ c|x − y| < cǫ
Which shows that
(2.3) z = fi(x) ∈ fi(A)cǫ ⊆⋃
i
fi(A)cǫ.
Therefore, F (B) ⊆ F (A)cǫ. After we take ǫ ↓ d(A,B), we have that F (B) ⊆F (A)cd(A,B). Inclusion number (2) follows if this argument is repeated byreplacing each A with a B.
The set of functions f1, f2, . . . , fn are called the generator functionsfor a fractal. For example, the middle thirds Cantor set, as discussed inthe next chapter, can be generated with two functions, f1(x) = 1
3x and
f2(x) = 13x + 2
3 . The cross product of the Cantor set can be generated withthe four functions:
(1) f1(x, y) = (13x, 1
3y);
(2) f2(x, y) = (23 + 1
3x, 13y);
(3) f3(x, y) = (23 + 1
3x, 23 + 1
3y);
(4) f4(x, y) = (13x, 2
3 + 13y).
Many examples of fractals can be constructed by inventing a set of gen-erator functions, and applying them a given compact set. The next chapterproceeds with a precise definition of ‘fractional’ Hausdorff dimension, andwe will compute the dimension for a few simple examples.
CHAPTER 2
Hausdorff Measure, Dimension and Examples
1. Basic Measure Theory on Euclidean Space
We say µ is a measure on elements in the σ−algebra of subests of Ω ifµ satisfies the following three properties:
(1) µ(∅) = 0;(2) µ(A) ≤ µ(B) if A ⊆ B, and A,B ∈ σ−algebra;(3) If A1, A2, . . . is a countable sequence of sets, then
µ
(
∞⋃
i=1
Ai
)
≤∞∑
i=1
µ(Ai)
and
µ
(
∞⋃
i=1
Ai
)
=
∞∑
i=1
µ(Ai)
if Ai are disjoint Borel sets.
Furthermore, we say µ is a probability measure if µ(Ω) = 1. For thepurposes of this paper, we will always consider Ω ⊆ R
n.
2. Hausdorff Dimension
If U is any non-empty subset of n-dimensional Euclidean space, Rn, the
diamater of U is defined as |U | := sup|x − y| : x, y ∈ U. For now, we willuse the usual Euclidean metric: |x−y| := ((x1−y1)
2+(x2−y2)2+. . .+(xn−
yn)2)1/2. However, as will be shown shortly, we may use any Lp metric. IfF ⊆ R
n, and a collection Uii∈I satisfies the following conditions:
(1) |Ui| ≤ δ ∀i ∈ I(2) F ⊆
⋃
i∈I Ui,
then we say the collection Uii∈I is a δ-cover of F. For the purposes of thispaper, we may assume the collection is always countable.
If F ⊆ Rn and s, δ > 0, we define:
(2.1) Hsδ(F ) = inf
∞∑
i=1
|Ui|s,
where the infimum is taken over all δ-covers of F .As δ becomes smaller, the number of available covers becomes reduced.
For 0 < δ1 < δ2 < 1, we have Hsδ1
(F ) ≥ Hsδ2
(F ) since any δ1 cover of F is
9
10 2. HAUSDORFF MEASURE, DIMENSION AND EXAMPLES
also a δ2 cover of F . i.e. Hsδ(F ) is increasing as δ decreases. We define the
s-dimensional Hausdorff measure as:
(2.2) Hs(F ) := limδ→0
Hsδ(F ).
The following equation will become very useful. For any δ-cover, Uiof F , and for any t > s, we have:
(2.3) Htδ(F ) ≤ δt−sHs
δ(F ).
Proof. Consider any δ-cover Ui of F . Then each |Ui|t−s ≤ δt−s since
|Ui| ≤ δ, so:|Ui|
t = |Ui|t−s|Ui|
s ≤ δt−s|Ui|s.
Summing over all Ui’s we obtain∑
i |Ui|t ≤ δt−s
∑
i |Ui|s. The result follows
after taking the infimum.
Two applications of equation (??) should be noted:
(1) If Hs(F ) < ∞, and if t > s, then Ht(F ) = 0.
Proof. Equation (??) shows that Htδ(F ) ≤ δt−sHs
δ(F ) for anypositive δ. The result follows after taking limits, since Hs(F ) <∞.
(2) If Hs(F ) > 0, and if t < s, then Ht(F ) = ∞.
Proof. Equation (??) shows that 1δs−tH
sδ(F ) ≤ Ht
δ(F ) for ev-
ery positive δ. After taking limits, we see that Ht(F ) = ∞, sincelimδ→0
1δs−t = ∞ and limδ→0 H
sδ(F ) = Hs(F ) > 0.
One immediate consequence of these observations is that Hs(F ) = 0 or∞ everywhere except at a unique value s, where this value may be finite. Asa function of s, Hs(F ) is decreasing function. Therefore, the graph of Hs(F )will have a unique value where it jumps from infinity to 0. Thus, if we definethe Hausdorff dimension of a set F as dim
H(F ) := sup s : Hs(F ) = ∞ =
inf s : Hs(F ) = 0, the graph of Hs(F ) looks like
(2.4) Hs(F ) =
∞ if s < dimH
(F );
0 if s > dimH
(F );
undetermined s = dimH
(F ).
Remark 2.1. It should be noted that we obtain the same HausdorffDimension if we use open or closed covers.
Consider dimH
(F ), the dimension we obtain when we use closed coverscompared to dim
H(F ), the dimension we obtain if we use open covers. If
we take any open δ-cover Ui of F , then the closure of each of these, Uicreates a closed δ-cover of F . This immediately shows that Hs
δ(F ) ≤ Hsδ(F ).
After taking lim δ → 0, we have that dimH
(F ) ≤ dimH
(F ). On the otherhand, if we consider a closed δ-cover Ci of F , we can take an open ǫ − δextension of each of these, |Ui| = |Ci|+ (ǫ− δ) ≤ ǫ, we obtain an open cover
2. HAUSDORFF DIMENSION 11
Ui where each Ci ⊂ Ui and |Ui| ≤ ǫ for each i. This shows that for everyǫ > δ, we have Hs
δ(F ) ≥ Hsǫ(F ). After taking limǫ↓δ, followed by limδ→0 we
have that Hs(F ) ≥ Hs(F ), which finishes the proof.Because of this jump discontinuity given in the graph of (??), we may
show that Hausdorff Measure is invariant under Lp metrics.
2.1. Invariance of Hausdorff Dimension under Lp metrics. Forp ≥ 1 and x = (x1, x2, . . . , xn) ∈ R
n we define the Lp norm:
(2.5) ‖x‖p := (|x1|p + |x2|
p + . . . + |xn|p)1/p.
For p = ∞, we define:
(2.6) ‖x‖∞ := max|x1|, |x2|, . . . , |xn|.
Since max|xi|ni=1 ∈ |xi|
ni=1 and a > 0, 0 ≤ b ≤ c ⇒ 0 ≤ ba ≤ ca, (xa is
a non-decreasing function since axa−1 > 0 for positive x), we immediatelyhave ‖x‖p
∞ ≤∑
i |xi|p ⇒ ‖x‖∞ ≤ ‖x‖p after taking the 1/p root of both
sides.Similiarly, we have
∑
i |xi|p ≤
∑
i max |xi|p = n max |xi|
p ⇒ ‖x‖p =
(∑
i |xi|p)1/p ≤ (n max |xi|
p)1/p = n1/p max |xi| = n1/p‖x‖∞. Together,these inequalities show:
(2.7) ‖x‖∞ ≤ ‖x‖p ≤ n1/p‖x‖∞.
Equation (??) and the unique jump discontinuity of dimH
(F ) from equa-tion (??) can be used to show the invariance of Hausdorff Dimension underany Lp metric. Consider
Hsδ,∞(F ) := inf
∞∑
i=1
|Ui|s∞;(2.8)
Hsδ,p(F ) := inf
∞∑
i=1
|Ui|sp;(2.9)
where each Ui is a δ-cover of F under their respective metrics. We define
Hs∞(F ) := lim
δ→0Hs
δ,∞(F );(2.10)
Hsp(F ) := lim
δ→0Hs
δ,p(F ).(2.11)
We will show the equivalence of the dimension obtained by the Lp metric tothe L∞ metric.
Theorem 2.2. The Hausdorff dimension, dimH
(F ) = sups : Hsp =
∞ = infs : Hsp = 0 = sups : Hs
∞ = ∞ = infs : Hs∞ = 0, is invariant
under any Lp metric.
Proof. Take any δp-cover of F : (i) F ⊆⋃
i Ui, (ii) |Ui|p ≤ δ for alli. Then, with the aid of equation (??), in addition to the extra obvious
12 2. HAUSDORFF MEASURE, DIMENSION AND EXAMPLES
definitions:
|Ui|p := sup|x − y|p : x, y ∈ Ui,(2.12)
|Ui|∞ := sup|x − y|∞ : x, y ∈ Ui,(2.13)
we have
(2.14) |Ui|∞ ≤ |Ui|p ≤ n1/p|Ui|∞.
Since F ⊆⋃
i Ui under the L∞ metric as well, (Observe that open sets,Ui = x : ∃ r > 0, Br(x) ⊆ Ui are invariant under any Lp metric), we havethat
(2.15) Hsδ,∞ ≤ Hs
δ,p.
After taking limδ→0,
(2.16) Hs∞(F ) ≤ Hs
p(F ).
Conversely, suppose Ui is a δ-cover under the L∞ metric: (i) F ⊆⋃
i Ui, (ii) |Ui|∞ ≤ δ for all i. Equation (??) shows that np|Ui|p ≤ |Ui|∞.Which shows that
(2.17) np · Hsδ,p ≤ Hs
δ,∞.
After taking limits, we have that
(2.18) np · Hs∞ ≤ Hs
p.
Therefore we have a similiar inequality for the Hsp and Hs
∞:
(2.19) Hsp ≤ Hs
∞ ≤ n1/pHsp.
This relationship, together with the jump discontinuity given by(??), showsthat Hs
p = Hs∞ everywhere either one is either 0 or ∞.
It should be noted that if either value Hsp or Hs
∞ is positive and finite,Hs
p 6= Hs∞ in general. What has been proven is that the value s, at which this
jump occurs is unique. We will be concerned with computing the HausdorffDimension s, rather than the s-dimensional Hausdorff measure which is whywe need not be concerned that Hs
p 6= Hs∞, only that either one is positive
and finite.Without further ado, we will present an example of a set with non-
integral Hausdorff dimension.
2. HAUSDORFF DIMENSION 13
2.2. Ternary Cantor Set. The construction of the Ternary CantorSet proceeds as follows:
Where in each stage of construction, the middle third is removed from eachinterval in the previous set. In general, for the nth stage of the construction,we have 2n intervals, each of length 3−n. The Ternary Cantor Set is definedas
C :=∞⋂
n=1
Cn.
Since Cn+1 ⊂ Cn ∀n ∈ N, and since each Cn is compact, we know thatC is non-empty. For example, a little observation will reveal that the set0, 1/3, 1/9, 1/27, . . . ⊂ C. In fact, without proof we can say: |C| = |R|. Inorder to find an upper bound for dim
H(C), the most obvious cover will work.
Take Ei to be a closed cover, where ∪2n
i=1Ei = Cn And each |Ei| = 3−n isthe ith interval of Cn. It follows that
(2.21)
2n∑
i=1
|Ei|s =
2n∑
i=1
(3−n)s = 2n3−ns.
When s = log3 2 equation (??) becomes
(2.22) 2n3−ns = 2n3−n log3 2 = 2n3log3 2−n
= 2n2−n = 1.
We can conclude that dimH
(C) ≤ log3(2), since for every δ > 0, this covercan become arbitrarily small. I.e., Hs
δ(C) ≤ 1 ∀ δ > 0. To show that log3(2)is indeed the Hausdorff Dimension of C, we will need the following lemma.
Lemma 2.3 (Frostman’s Lemma). Suppose F is a measurable subset ofR
n, and suppose ∃ s, a > 0 and a probability measure µ on F such that
(2.23) lim supr→0
supx∈F
µ(Br(x))
rs≤ a.
Then, Hs(F ) ≥ 1a·2s . (In particular, dim
H(F ) ≥ s.)
Proof. Applying (??), we can make supx∈Fµ(Br(x))
rs sufficiently closeto a for all b > a. Formally: if b > a, there exists ǫ > 0 such that uniformlyfor all x ∈ F , µ(Br(x)) ≤ brs whenever r ≤ 2ǫ.
14 2. HAUSDORFF MEASURE, DIMENSION AND EXAMPLES
There exist open sets, U1, U2, . . . ⊆ F, |Ui| = ri
2 ≤ ǫ such that:
(1) F ⊆ ∪∞i=1Ui, and
(2) Hs(F ) + ǫ ≥∑∞
i=1(ri/2)s.
(We can always find an open cover within any ǫ tolerance of Hs(F ), sinceHs(F ) is the greatest lower bound of the sum of such covers.) For eachi ∈ N, take any point xi ∈ Ui and note that Bri
(xi) ⊇ Ui. Then,
1 = µ
(
∞⋃
i=1
Ui
)
≤∞∑
i=1
µ(Ui) ≤∞∑
i=1
µ(Bri(xi))(2.24)
≤ b∞∑
i=1
(ri)s = 2sb
∞∑
i=1
(ri/2)s ≤ 2sb (ǫ + Hs(F )) .
This implies:
(2.25)1
2sb≤ ǫ + Hs(F ).
Let ǫ ↓ 0, then b ↓ a and we have the desired result.
To show that dimH
(C) = log3 2 it suffices to prove that dimH
(C) ≥log3 2. If we take the base-3 expansions for all real numbers x ∈ [0, 1], wecan write x =
∑∞i=1 3−ixi where each xi ∈ 0, 1, 2. If we consider
(2.26) C∗ :=
∞∑
i=1
3−ixi : xi ∈ 0, 2
,
then we obtain a subset of C, and the set difference, C \ C∗ = x ∈ C :x /∈ C∗ is a countable set. Consider a sequence of i.i.d. (independent,identically distributed) random variables, Xi
∞i=1, where
(2.27) PX1 = 2 = PX1 = 0 = 1/2.
For a fixed n, and by independence, we have PX1 = x1,X2 = x2, . . . ,Xn =xn = PX1 = x1PX2 = x2 · · ·PXn = xn = 2−n if each xi ∈ 0, 2.If we set X =
∑∞i=1 3−iXi, we can define a probability measure µ on C∗ by
µ(A) = PX ∈ A for measurable subsets A ⊆ [0, 1]. For a fixed x ∈ C∗,if y ∈ B3−n(x) then y1 = x1, . . . , yn = xn ⇒ µ(B3−n(x)) ≤ 2−n = 3−ns
where s = log3 2. To finish the estimate, let 1 > ǫ > 0. There exists ann ∈ N, 3−(n+1) ≤ ǫ ≤ 3−n. In particular, 3−n3−1 ≤ ǫ ⇒
(2.28) 3−n ≤ 3ǫ;
also,
(2.29) µ(Bǫ(x)) ≤ µ(B3−n(x)),
since Bǫ(x) ⊆ B3−n(x). Together, equations (??) and (??) imply
Then this probability measure immediately satisfies
(2.31) supx∈F
µ(Bǫ(x)) ≤ 2ǫs.
Appealing to Frostman’s Lemma, we have dimH
(C) ≥ log3 2. Another in-teresting variation of the Ternary Cantor set allows for a different size to beremoved in each step of the construction of C.
2.3. Middle-α Cantor Set. The Middle-α Cantor Set (0 < α < 1) isconstructed in a manner identical to the Ternary Cantor Set. In each stepof the construction, we remove the middle-α from each interval.
When α = 1/3 the situation reduces to the normal Ternary Cantor Set.It may be possible that a number-theoretic argument with numbers ofthe form:
∑∞
i=1 ((1+α2 )i + (1−α
2 )i)xi : xi ∈ 0, 1
and an argument iden-tical to that given for the middle third Cantor Set may be used to showdim
H(C(α)) = log 2/ log 2
1−α . However, we will use a more general argu-ment by defining a topology and distance function on ternary trees. By thenature of construction for the middle-alpha set, this generalization easilylends itself for an application to this problem.
2.4. Topology on Trees. Consider a 3-adic tree, T for which anyelement in the tree can be described with an infinite sequence of 0’s, 1’sand 2’s. This may be thought of a ‘tree’, with roots and each step one maychoose which of the 3 branches to take.
Set T := x : x = (x1, x2, . . .), xi ∈ 0, 1, 2 for i ∈ N. Fix a finitesequence (x1, x2, . . . xn), xi ∈ 0, 1, 2 and set
x|n := (x1, x2, . . . , xn),(2.33)
B(x|n) := y : yi = xi, i ≤ n, yi ∈ 0, 1, 2 for i ∈ N.(2.34)
Then the set
(2.35) B := ⋃
x∈T
B(x|n) : n ∈ N ∪ ∅
16 2. HAUSDORFF MEASURE, DIMENSION AND EXAMPLES
defines a basis for a topology on T . B is indeed a basis for a topology onT , for it satisfies the following two conditions:
(1) If x ∈ T , then B(x1) ∈ B and x ∈ B(x1).(2) Suppose x ∈ B1∩B2 6= ∅. Say B1 = B(y1,...yn1
) and B2 = B(z1,...zn2)
for some z and y. Then, y1 = z1 since the intersection is non-empty.Set n3 = minn1, n2 ≥ 1. Since x ∈ B1 ∩ B2, we have x|n1
= y|n1
and x|n2= z|n2
. In particular, we have z|n3= x|n3
= y|n3so we can
set B3 = B(y1,...yn3) = B(z1,...yn3
). In particular, we have B3 ⊆ B1
and B3 ⊆ B2 so x ∈ B3 ⊆ B1 ∩ B2.
Since B satisfies the previous two conditions, we may define a topologygenerated by B. We say a subset U ⊆ T is open if ∀x ∈ U,∃B ∈ B suchthat x ∈ B and B ⊆ U . If we set τ =
⋃
U for all such U , then one can seethat τ satisfies the three criterion for making τ a topology:
(1) ∅, T ∈ τ ;(2) If Uii∈I is a collection of subsets of τ , then
⋃
i∈I ;Ui ∈ τ ;(3) If Ui is a finite collection of sets in τ , then
⋂ni=1 Ui ∈ τ .
In fact, we may define a metric on this tree with the following twodefinitions:
ρ(x, y) := supn ∈ N : yi = xi, i ≤ n(2.36)
d(x, y) := 2−ρ(x,y).
The number ρ(x, y) counts how many branches the numbers x and y have incommon with each other. The number d(x, y) forms a metric for elementsin τ . We will list the properties that qualify d as a metric, and prove themin turn.
(1) d(x, y) ≥ 0 and d(x, y) = 0 iff x = y.
Proof. d(x, y) ≥ 0 is immediate since 2p ≥ 0 for every p ∈N ∪ ∞. If x = y, then d(x, y) = 2−ρ(x,y) = 0 since p(x, y) =∞. If x 6= y then there is an N for which xN 6= yN . The setn ∈ N : yi = xi, i ≤ n is bounded above by N , so ρ(x, y) ≤ N <
∞. Then d(x, y) = 2−ρ(x,y) ≥ 2−N > 0, which shows d(x, y) =0 ⇒ x = y. (If the reader is uncomfortable with saying 2−∞ = 0,we may have defined d(x, y) as usual if x 6= y and d(x, y) = 0 ifx = y).
(2) d(x, y) = d(y, x).
Proof. ρ(x, y) = ρ(y, x) is obvious, and proves the symmetricproperty.
(3) If x, y ∈ T , then d(x, y) ≤ d(x, z) + d(z, y) for any z ∈ T .
Proof. Fix some x, y and z in T . The case where x = y isclear, because d(x, y) = 0 ≤ d(x, z) + d(z, y) follows from item (1)for metrics. If x 6= y, then we have the following two cases:
2. HAUSDORFF DIMENSION 17
(a) Case 1: ρ(x, z) ≤ ρ(y, x). Then, 2−ρ(y,x) ≤ 2−ρ(x,z), whichimplies d(y, x) ≤ d(x, z), which implies d(y, x) ≤ d(x, z) +d(z, y), and we are done.
(b) Case 2: ρ(x, z) > ρ(y, x). From the definition of ρ, we knowthat zi = xi for all i ≤ ρ(x, z). In particular, since ρ(x, z) >ρ(y, x), we have that xi = yi = zi for all i ≤ ρ(y, x) soρ(y, z) ≥ ρ(y, x). (y agrees with z at least until ρ(y, x)).Since zρ(y,x)+1 = xρ(y,x)+1 6= yρ(y,x)+1, we are forced to con-clude that ρ(y, z) ≤ ρ(y, x) which shows that ρ(y, z) = ρ(y, x).
This shows the desired result: d(y, x) = 2−ρ(y,x) = 2−ρ(y,z) =d(y, z) ≤ d(y, z) + d(z, x).
2.5. Hausdorff Dimension of Middle-α Cantor Set.
2.5.1. Lower bound for C(α). Every element in [0, 1] can be uniquelydescribed, with the exception of a countable set, by elements in a ternarytree, whose digits in 0, 1, 2 describe which third of the Cn(α), includingthe removed middle third, the element resides in for each n. If we considerC∗(α) ⊂ C(α), where the elements in C∗(α) are those which do not contain a1 in their decimal expansion, we can define a probability measure on C∗(α).
Furthermore, we have an estimate for the “size”, or how many pointsreside in small areas around any one given point. This is exactly what weneed in order to define a probability measure on Borel subsets of C∗(α).
In the nth stage of construction, we have 2n intervals, each of length(1−α
2 )n. If x ∈ C∗(α), then x|n = (x1, . . . , xn) for some xi ∈ 0, 2. Thesmoothest probability measure is the same that was used for the TernaryCantor Set, and the most obvious choice. Take a sequence of iid randomvariables Xi
∞i=1, PXi = 0 = PXi = 2 = 1/2. For Borel subsets
A ⊆ [0, 1], set
(2.37) µ(A) = P(X1,X2, . . .) ∈ A.
Where the infinite sequence X := (X1,X2, . . .) denotes where X resides inthe unit interval [0, 1]. For a fixed x ∈ C∗(α),
(2.38) µ(B( 1−α2
)n(x)) ≤ 2−n =
(
2
1 − α
)−ns
where s = log(2)/ log( 21−α). Now, let ǫ > 0. Since 1−α
2 < 1, there exists ann ∈ N, such that
(
1 − α
2
)n+1
≤ ǫ ≤
(
1 − α
2
)n
.
In particular,(
21−α
)−n (2
1−α
)−1≤ ǫ which implies
(2.39)
(
2
1 − α
)−n
≤
(
2
1 − α
)
ǫ;
18 2. HAUSDORFF MEASURE, DIMENSION AND EXAMPLES
also,
(2.40) µ(Bǫ(x)) ≤ µ
(
B( 2
1−α)−n(x)
)
,
since Bǫ(x) ⊆ B( 2
1−α )−n(x). Together, equations (??) and (??) imply
µ(Bǫ(x)) ≤ µ(B( 2
1−α )−n(x)) ≤
(
2
1 − α
)−ns
≤(2.41)
((
2
1 − α
)
ǫ
)s
=
(
2
1 − α
)s
ǫs = 2ǫs.
Then this probability measure immediately satisfies
(2.42) supx∈C∗(α)
µ(Bǫ(x)) ≤ 2ǫs.
Appealing to Frostman’s Lemma, we have dimH
(C) ≥ log( 2
1−α ) 2, which
yields a lower bound.2.5.2. Upper bound for C(α). To obtain an upper bound, we use the
most obvious cover. Fix an n ∈ N. Take Ei to be the closed set that covers
the ith interval of Cn(α). |Ei| =(
21−α
)−nfor each i. Then,
(2.43)n∑
i=1
|Ei|s = 2n
(
2
1 − α
)−ns
= 1
when s = log( 2
1−α) 2. This yields the proper upper bound for C(α) since
this cover can become arbitrarily small. I.e. dimH
(C(α)) ≤ log( 2
1−α) 2.
This inequality, together with the previous section yields dimH
(C(α)) =log( 2
1−α )(2).
2.6. Hausdorff Dimension for Cross Product of Ternary Cantor
Set.
2.6.1. Lower Bound for C×C. Consider the set C×C = (x, y) : x, y ∈C where C is the usual Ternary Cantor Set. C×C =
⋂∞n=0(Cn×Cn) where
each Cn is given by (??). Consider C∗ × C∗ ⊆ C × C, (x, y) ∈ C∗ × C∗
where (x, y) = (∑
i 3−ixi,∑
i 3−iyi) and each xi, yi ∈ 0, 2 gives the usual
base-3 expansion for points in the unit square. Again, the set difference,(C × C) \ (C∗ × C∗) is countable. Set a sequence of i.i.d. random variablesXi and Yi:
where s = log3 4. Now, let ǫ > 0. There exists an n ∈ N, such that3−(n+1) ≤ ǫ ≤ 3−n. In particular, 3−n3−1 ≤ ǫ ⇒
(2.48) 3−n ≤ 3ǫ.
Then,
(2.49) µ(Bǫ(x, y)) ≤ µ (B3−n(x, y)) ,
since Bǫ(x, y) ⊆ B3−n(x, y). Together these equations imply:
µ(Bǫ(x, y)) ≤ µ (B3−n(x, y)) ≤(2.50)
4−n = 3−ns = (3−n)s ≤ (3ǫ)s = 3sǫs = 4ǫs.
Then this probability measure immediately satisfies
(2.51) sup(x,y)∈C∗×C∗
µ (Bǫ(x, y)) ≤ 4ǫs.
Applying Frostman’s Lemma, we have dimH
(C) ≥ log3 4.2.6.2. Upper Bound for C × C. In the nth stage of the construction of
C × C, we have 4n boxes of diameter 3−n. (Here the ’diameter’ is takenover closed balls with the L∞ metric). The most obvious cover will work.Take Ei to be the closed “ball” covering the ith box in the nth stage of theconstruction for C × C. Then |Ei| = 3−n and
(2.52)∑
i
|Ei|s =
4n∑
i=i
|Ei|s = 4n|Ei|
s = 4n3−ns = 1,
where s = log3 4. Since the mesh of this cover can become arbitrarily small,for this fixed s, we have Hs(C ×C) ≤ 1 ⇒ dim
H(C ×C) ≤ log3 4. Together,
this section and the previous one imply dimH
(C × C) = log3 4.
3. A simple proof that |C × C| = |R|
The cross product of the Cantor Set can be used as a proof that equatesthe cardinality of these two sets: |C × C| = |R|. To start, we define thealgebraic set difference as:
(3.1) C − C := x − y : x, y ∈ C.
20 2. HAUSDORFF MEASURE, DIMENSION AND EXAMPLES
Since C ⊂ [0, 1], it is not too surprising that C − C ⊆ [−1, 1]. However, thesurprising result is that the algebraic set difference fills in the entire interval:C − C = [−1, 1].
To show this result, we will present a rather ingenious proof originallygiven by Cantor.
Proof. If we consider some b ∈ [−1, 1], then we can construct a 45degree line through b. In the xy-plane, this line is described as y = x + b.Once we show that this line intersects C × C for some x∗, y∗ ∈ C × C, wehave that b = y∗−x∗ for some x∗, y∗ ∈ C, which shows that C−C ⊇ [−1, 1].
To begin, let b ∈ [−1, 1], and consider the line y = x + b. The lineobviously intersects C0 × C0 either at (0, b) or at (|b|, 0), depending on ifb > 0 or if b < 0. To show that this line intersects C1 × C1, we will break itup into cases:
(1) If 23 ≤ b ≤ 1, then the line intersects C1 × C1 at the point (0, b).
(2) If 13 ≤ b < 2
3 , then the line intersects C1×C1 at the point (23 − b, 2
3)
(3) If 0 ≤ b < 13 , then the line intersects C1 × C1 at the point (0, b).
(4) If −13 ≤ b < 0, then the line intersects C1 ×C1 at the point (|b|, 0).
(5) If −23 ≤ b < −1
3 , then the line intersects C1 × C1 at the point
(23 , b+ 2
3). (note that since −23 ≤ b < −1
3 , then 0 ≤ b+ 23 < 1
3 ∈ C1,
after adding 23 to both sides).
(6) If −1 ≤ b < −23 , then the line intersects C1×C1 at the point (|b|, 0).
The argument proceeds by induction. However, the argument shownhere says that if a 45 degree line intersects any box in the Cantor set, thenit intersects one of the boxes in the next step of the construction for theCantor set. Hence, this exact same argument can essentially be repeatedto construct a sequence of points (xn, yn)∞n=0 where each (xn, yn) ∈ Cn ×Cn. Since this sequence is bounded, we can apply the Bolzano-Weirstrasstheorem to obtain a convergent subsequence (xnk
, ynk) → (x∗, y∗). Since
Cn × Cn is a closed set, then we have that (x∗, y∗) ∈ Cn × Cn, the desiredresult.
Bibliography
[1] Falconer, Kenneth, “Fractal Geometry: Mathematical Foundations and Applica-tions,” John Wiley & Sons Ltd., 1990.
[2] Khoshnevisan, Davar & Xiao, Yimin, “Fractal Geometry of Random Processes,”(Pre-print 2006).
