MTH6107 Chaos & Fractals
Exercises 4
Exercise 1. Show that the notion of topological conjugacy defines an equivalence relation on the set of self-maps of [−1, 1].
Recall that f and g are said to be topologically conjugate if there exists a homeomorphism h : [−1, 1] → [−1, 1] such that h ◦ f = g ◦ h
Clearly any f is topologically conjugate to itself: just take h to be the identity map.
The relation is symmetric: if h ◦ f = g ◦ h then H ◦ g = f ◦ H where H = h−1 .
The relation is transitive: if h ◦ f1 = f2 ◦ h and h0 ◦ f2 = f3 ◦ h0 , then setting
H = h0 ◦ h we see that
H ◦ f1 = h0 ◦ h ◦ f1 = h0 ◦ f2 ◦ h = f3 ◦ h0 ◦ h = f3 ◦ H .
Therefore topological conjugacy is an equivalence relation.
Exercise 2. Use the map h(x) = sin(πx/2) to show that the map f : [−1, 1] → [−1, 1]
defined by f (x) = 1 − 2|x| is topologically conjugate to the Ulam map g : [−1, 1] →
[−1, 1] given by g(x) = 1 − 2x2 .
First observe that h : [−1, 1] → [−1, 1] defined by h(x) = sin(πx/2) is indeed a
homeomorphism.
We will show that h ◦ f = g ◦ h.
Firstly, if x ∈ [−1, 0] then h(f (x)) = sin((2x + 1)π/2) = sin(π/2 + πx) = cos(πx),
and if x ∈ [0, 1] then h(f (x)) = sin((1 − 2x)π/2) = sin(π/2 − πx) = cos(πx).
Secondly, g(h(x)) = 1 − 2 sin2 (πx/2) = cos πx.
So g(h(x)) = h(f (x)), as required.
Exercise 3. Determine whether the map F : [−1, 1] → [−1, 1] given by F (x) = 1 − |x|
is topologically conjugate to the map G : [−1, 1] → [−1, 1] given by G(x) = 1 − x2 .
The two maps are not topologically conjugate.
Justification: Every point in [0, 1] has minimal period 2 under F , whereas G only
has a single orbit of minimal period 2 (namely {0, 1}), therefore the maps cannot be
topologically conjugate.
Exercise 4. Let C0 = [0, 1]. In the standard construction of the Cantor ternary set
C = ∩∞
k=0 Ck , describe briefly how the sets Ck are defined, and explicitly write down the
sets C1 and C2 . If Ck is expressed as a disjoint union of Nk closed intervals, compute
the number Nk , and determine the common length of each of the Nk closed intervals.
Use this to show that, assuming the box dimension of the ternary Cantor set C exists,
then it must equal log 2/ log 3
The set Ck−1 is a disjoint union ∪i Ii of closed intervals. If from each of these closed
intervals Ii we remove the ‘open middle third’, we are left with a pair of closed intervals
Ii− and Ii+ , each of length a third the length of I. The union ∪i (Ii− ∪ Ii+ ) of these
intervals is then defined to be the set Ck .
C1 = [0, 1/3] ∪ [2/3, 1], and C2 = [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪ [8/9, 1].
Nk = 2k because N0 = 1 and the recursive procedure doubles the number of intervals
at each step.
The length of each interval in Ck is 1/3k , because the length of the closed intervals
decreases by a factor of 3 at each step, and the length of C0 = [0, 1] is 1.
If εk = 1/3k then N (εk ) = 2k , and so the box dimension equals
k log 2
log 2
log N (εk )
= lim
=
.
k→∞ k log 3
k→∞ − log εk
log 3
lim
Exercise 5. Let H denote the collection of compact subsets of R, let Φ : H → H be the
iterated function system defined by the two maps φ1 (x) = x/10 and φ2 (x) = (x+3)/10,
and let Ck denote Φk ([0, 1]) for k ≥ 0. Write down the sets C1 and C2 , and compute
the Hausdorff distance h(C1 , C2 ) between them. If Ck is expressed as a disjoint union
of Nk closed intervals, compute the number Nk , and determine the common length of
each of the Nk closed intervals whose disjoint union equals Ck . Show that if the box
dimension of C = ∩∞
k=0 Ck exists then it must equal log 2/ log 10.
C1 = [0, 1/10] ∪ [3/10, 4/10], and
C2 = [0, 1/100] ∪ [3/100, 4/100] ∪ [3/10, 31/100] ∪ [33/100, 34/100].
If A = C1 , B = C2 then hBA = 0 since B ⊂ A, whilst
hAB = max %(x, C2 ) = %(1/10, C2 ) = %(1/10, 4/100) = 6/100 = 3/50 ,
x∈C1
so the Hausdorff distance between C1 and C2 is
h(C1 , C2 ) = max(3/50, 0) = 3/50 .
Nk = 2k because N0 = 1 and the recursive procedure doubles the number of intervals
at each step.
The length of each interval is 1/10k , because the length of the closed intervals
decreases by a factor of 10 at each step, and the length of C0 = [0, 1] is 1.
If εk = 1/10k then N (εk ) = 2k , and so the box dimension equals
k log 2
log 2
log N (εk )
= lim
=
.
k→∞ k log 10
k→∞ − log εk
log 10
lim
Exercise 6. Let H denote the collection of compact subsets of R, let Φ : H → H be the
iterated function system defined by the two maps φ1 (x) = x/10 and φ2 (x) = (x+9)/10,
and let Ck denote Φk ([0, 1]) for k ≥ 0. If Ck is expressed as a disjoint union of Nk closed
intervals, compute the number Nk , and determine the common length of each of the
Nk closed intervals whose disjoint union equals Ck . Show that if the box dimension of
C = ∩∞
k=0 Ck exists then it must equal log 2/ log 10.
Nk = 2k because N0 = 1 and the recursive procedure doubles the number of intervals
at each step.
The length of each interval is 1/10k , because the length of the closed intervals
decreases by a factor of 10 at each step, and the length of C0 = [0, 1] is 1.
If εk = 1/10k then N (εk ) = 2k , and so the box dimension equals
k log 2
log 2
log N (εk )
= lim
=
.
k→∞ k log 10
k→∞ − log εk
log 10
lim
Exercise 7. For C as in Exercise 6, give a description of the members of C in terms of
the digits of their decimal expansion. If f : C → C is defined by f (x) = 10x (mod 1)
then find a point x ∈ C which has minimal period 2 under f .
C consists of those numbers between 0 and 1 which have a decimal expansion whose
digits all equal either 0 or 9.
There are two points of minimal period 2, namely
1/11 = 0.090909 . . .
and 10/11 = 0.909090 . . .
Exercise 8. Describe the construction of the Sierpinski triangle P , and show that if the
box dimension of P exists then it must equal log 3/ log 2
Begin with a solid equilateral triangle, then sub-divide it into 4 congruent equilateral
triangles, then remove the central triangle, leaving 3 solid equilateral triangles. Repeat
the above step with each of the remaining 3 triangles, and continue the process ad
infinitum.
Assuming (without loss of generality) that the initial equilateral triangle has side
length 1, we see that N (1/2) = 3, and more generally N (1/2k ) = 3k , so existence of
the box dimension D means that
log N (1/2k )
log 3k
log 3
log N (ε)
= lim
=
lim
=
.
−k
ε→0 − log ε
k→∞ − log 2
k→∞ k log 2
log 2
D = lim