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Dynamical -Homework #1 - Solved
1 Notebook Question
Problem 1.
Let G = {g0,...,gk−1} be a contracting iterated function system with contracting similarity ratio c < 1. Fix i ∈ {0,...,k −1}. We have d(gi(x),gi(y)) = cd(x,y) for all x,y ∈ Rn. Problem 2.
The limit set given by Definition 1 is defined as n where Λ0 = K and Λn for n ≥ 0 is defined as Λn+1 = Ski=0−1 gi(Λn).
The limit set for Definition 2 is
{z ∈ Rn : for any neighborhood N of z there are infinitely many finite words w
such that the image Φ(w) lies in N}
Problem 3.
Section 1.2 of Edgar states that the Sierpinski gasket can be constructed using three dilation with ratio 1/2 and centers at the three vertices of the triangle S0. Let one of these vertices be at the origin, (0,0). We are given that S0 has side length 1, so the other two vertices must be at (1,0) and (1/2,1/2). Hence, the three dilations are given by,
f1(x,y) = f2(x,y) = f3(x,y) =
2 Questions from Silva
2.1 Section 2.1
Problem 3.
Suppose we have some interval from the definition of outer measure. That is, an interval Ij such that Ij is open and bounded. Then we can write Ij = (a,b) where a,b ∈ R. Let us fix k ∈ Z such that k/2j ≤ a. We will construct a sequence of dyadic intervals [k/2j,(k + 1)/2j),[(k + 1)/2j,(k + 2)/2j,...[(k + `)/2j,(k + ` + 1)/2j such that (a,b) ⊂ S`i=0[(k + i)/2j,(k + i + 1)/2j).
From Lemma 2.1.2, we have,
` `
b − a ≤ X|[(k + i)/2j,(k + i + 1)/2j)| = X1/2j
i=0 i=0
= (` + 1)/2j
Problem 4.
Let A ⊂ R and let t ∈ R. Let us define A + t = {a + t : a ∈ A}. Suppose that C = SIj is an open cover for A.
Now suppose that C + t = {c + t : c ∈ C} is not an open cover for A + t. Then there exists some a ∈ A such that a ∈ C but a + t 6∈ C + t. Note that since C is a union of open intervals, ∃ja ∈ N such that a ∈ Ija = (xja,yja). By our assumption that a + t ∈6 C + t, we must have that a + t ∈6 (xja + t,yja + t). However, we know that
xja < a < yja ⇐⇒ xja + t < a + t < yja + t
Thus, we have a contradiction and so C + t is an open cover for A + t.
Now take some countable collection D = SIj of open intervals. Suppose D + t is an open cover of A + t but D is not an open cover of A. Then there exists some a ∈ A such that a + t ∈ D + t but a 6∈ D. Note that since D is a union of open intervals, ∃ka ∈ N such that a + t ∈ Ika + t = (cka + t,dka + t). By our assumption that a 6∈ D, we must have that a 6∈ (cka,dka). However, we know that
cka + t < a + t < dka + t ⇐⇒ cja < a < dja
Thus, we have a contradiction and so D is an open cover for A.
From the above, we have shown that the unions of countable open intervals which cover A and cover A + t are precisely the same covers, up to a shift by t. Now note that for I = (b,a), I + t = (a + t,b + t), we have
|I| = |b − a|
and
|I + t| = |b + t − (a + t)| = |b − a|
Hence, the intervals that make up each countable union retain their length when shifted by t. Thus,
∞ ∞
X [
|Ij| : A ⊂ Ij, where Ij are bounded open intervals =
j=1 j=1
∞ ∞
X [
|Ij + t| : A + t ⊂ Ij + t, where Ij are open bounded intervals
j=1 j=1
As a result, the above two sets must have the same infimum and so λ∗(A) = λ∗(A + t).
Problem 5.
Suppose N is a null set. Then λ∗(N) = 0 by definition. Now fix some set A ⊂ R. By countable subadditivity, we have that
λ∗(A ∪ N) ≤ λ∗(A) + λ∗(N) = λ∗(A)
Now observe that A ⊂ A ∪ N. Hence, by Proposition 2.1.1 (3), we have that λ∗(A) ≤ λ∗(N ∪ A). Thus, by both of the above statements, we have
λ∗(A ∪ N) = λ∗(A)
Problem 7.
Suppose we have countably many null sets N1,N2,.... By countable subadditivity, we have that,
∞ ! ∞
λ∗ [ Nk ≤ X λ∗(Nk) = 0 k=1 k=1
In addition, we know that is bounded below by 0 because we are taking the infimum of a sum of interval lengths, where length is defined to be a non-negative real number. Hence, and thus is a null set, as required.
Problem 8.
Let A ⊂ R and t ∈ R. Define tA = {ta : a ∈ A}. If t = 0, then tA = {0} and so
λ∗(tA) = λ∗({0}) = 0 = |t|λ∗(A)
as required.
Now suppose t 6= 0. Suppose that C = SIj is an open cover for A.
Now suppose that tC = {tc : c ∈ C} is not an open cover for tA. Then there exists some a ∈ A such that a ∈ C but ta 6∈ tC. Note that since C is a union of open intervals, ∃ja ∈ N such that a ∈ Ija = (xja,yja). By our assumption that ta 6∈ tC, we must have that ta 6∈ (txja,tyja). However, we know that when t is positive
xja < a < yja ⇐⇒ txja < ta < tyja
and when t is negative
xja < a < yja ⇐⇒ txja > ta > tyja
Thus, we have a contradiction and so tC is an open cover for tA.
Now take some countable collection D = SIj of open intervals. Suppose tD is an open cover of tA but D is not an open cover of A. Then there exists some a ∈ A such that ta ∈ tD but a 6∈ D. Note that since D is a union of open intervals, ∃ka ∈ N such that ta ∈ tIka = (tcka,tdka). By our assumption that a 6∈ D, we must have that a ∈6 (cka,dka). However, we know that when t is positive
tcka < a < tdka ⇐⇒ cja < a < dja
and when t is negative
tcka < a < tdka ⇐⇒ cja > a > dja
Thus, we have a contradiction and so D is an open cover for A.
From the above, we have shown that the unions of countable open intervals which cover A and cover tA are precisely the same covers, up to a scaling by t. Now note that for I = (b,a), tI = (ta,tb), we have
|I| = |b − a|
and
|tI| = |tb − ta| = |t| · |b − a|
Hence, the intervals that make up each cover for A have their lengths scaled by |t|. Thus, P|tIj| = P|t||Ij| = |t|P|Ij|. Thus, from the reasoning above about the length of each interval and the fact that the covers for tA are precisely the covers for A scaled by t, we have
∞ ∞
λ∗(tA) = inf X|tIj| : tA ⊂ [ tIj, where Ij are open bounded intervals
j=1 j=1
∞ ∞
X [
= inf |t| · |Ij| : A ⊂ Ij, where Ij are open bounded intervals
j=1 j=1
∞ ∞
X [
= |t|inf |Ij| : A ⊂ Ij, where Ij are open bounded intervals
j=1 j=1
= |t| · λ∗(A)
as required.
2.2 Section 2.2 Problem 5.
Problem 6.
Note that K ⊂ [0,1], and for any x,y ∈ [0,1], we have that 0 ≤ x + y ≤ 2. Thus, K + K ⊂ [0,2]. Now we need to show that [0,2] ⊂ K + K
