Fractal Gauges for Hyperspace: One Limit Point
Dissertation
Presented in Partial Fulfillment of the Requirements for the Degree
Doctor of Philosophy
in the Graduate School of The Ohio State University
By
Na Peng, B.S.
Graduate Program in Mathematics
The Ohio State University
2010
Dissertation Committee:
Professor Gerald A. Edgar, Advisor
Professor Neil Falkner
Professor Jeffery McNeal
Copyright by
Na Peng
2010
Abstract
We discuss the problem of finding the exact fractal gauge function for a hyperspace
(the space of nonempty compact subsets) of a metric space. We consider only the
simplest nontrivial example, where the metric space is a sequence with a single limit
point.
ii
Acknowledgements
I am greatly indebted to my advisor Dr. Edgar, for his continual guidance and
constant encouragement and help throughout the period of my graduate studies.
Also, I want to give my thanks to Dr. Pittel, without who I would hardly realize the
beauty of the Probability Theory. Finally, many thanks to Wilson, for the support
and help through these years.
iii
Vita
2005 . . .
B.S., Southeast University, Nanjing, China
2005-10
Graduate Teaching Associate,
Department of Mathematics,
The Ohio State University,
Columbus, Ohio
Fields of Study
Major Field: Mathematics
Specialization: Fractal Geometry
iv
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
Vita . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1 Hyperspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Covering and Packing Measures . . . . . . . . . . . . . . . . . . . . .
1.3 General Setting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
3
4
8
2 Example 2−n + 2−m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1 Constituents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Densities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Covering Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Packing Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
12
13
13
15
3 General 2−n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1 ρ2 (m, n) = 2− min(m,n) . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 ρ3 (m, n) = |2−m − 2−n | . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
19
20
23
4 Distance is in the form of f (m ∧ n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2−x
. . . . . . . . . . . . . . . . . . . . . . . .
4.1 First example: f (x) =
x
4.2 Three-parameter example: fα,β,γ (x) = xβ/α 2−x/α (ln x)γ/α . . . . . . .
4.3 fα (x) = x−α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
−1/α
4.4 fα,s,M (x) = M 1/α ln x − αs ln ln x
. . . . . . . . . . . . . . . . . .
4.5 The choice of ϕ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
v
28
29
30
31
31
5 The
5.1
5.2
5.3
1/n metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Classify K by the idea of missing number . . . . . . . . . . . . . . . .
The Markov Process . . . . . . . . . . . . . . . . . . . . . . . . . . .
Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
35
38
41
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
vi
Chapter 1
Introduction
The notion of dimension arose naturally as humans explored the existing world. But,
as the knowledge grew, sets with non-integer dimensions were introduced as a generalization of this notion. They are called fractal dimensions.
First, Hausdorff introduced the Hausdorff measure Hs , where the corresponding
gauge function (Def 1.1) is ϕ(r) = (2r)s and s ∈ R+ . Later, mathematicians introduced the idea of gauge functions, replacing ϕ(r) by any positive increasing function.
Then, Tricot and Taylor [7][6] introduced the notion of packing measure. This opened
a new window for more calculations, and we are not limited to the subsets of Rn any
more.
Meanwhile, mathematicians started to study the hyperspace equipped with the
Hausdorff metric and to look for the relation between the hyperspace and the original
space. E.Boardman [8] studied the hyperspace of the unit interval [0, 1]. Let fα (t) =
−1
−α
2−αt , gα (t) = 2−t
for α > 0, t > 0, then consider the Hausdorff measure generated
by either functions, we have gα (H[0, 1]) = ∞, f1 (H[0, 1]) = 0. When S is a countable
subset of [0, 1], this offers some insight to the problem I am working on here.
At the same time, when S has the entropy dimension 21 , the guess is the possible
gauge function for the hyperspace here could be ϕ(r) = 2−M r
[2](6.2.3).
1
−1/2
where M > 0
I will work on the hyperspace of a countable set with one limit point, with various
metrics.
Chapter Two begins in an easy special case with a calculation of the covering
measure and the packing measure. First, after defining the iterated function system,
we can see that, except for a measure zero set, the hyperspace is self-similar. This
gives us the fractal gauge. Then density theorems are used in a direct calculation to
find the two desired measures. Then, coverings and packings are found to provide
alternate proofs.
In Chapter Three, we generalize the result of Chapter Two by studying various
metrics satisfying a certain property. Our goal is to find the corresponding fractal
gauge to give the exact covering and packing measures as before. Due to the similarity
of the metrics, the basic structure of the hyperspace is preserved and the argument
in Chapter Two can be used in this case.
Things get more interesting in Chapter Four. With the help of the min function,
we define several generalized types of metrics involving functions with different orders.
The search for the fractal gauge heavily depends on the asymptotic behavior of the
metric functions. The goal here is to make both the covering and packing measures
positive and finite. Again, the basic structure of the hyperspace is the same, and our
work simplifies. Next, in section 4.5, the direction changes: given a function, how can
we find a corresponding metric to make these two measures positive and finite?
Finally, we come to the Chapter Five. The regular Euclidean metric on R is chosen.
Unfortunately, this time the basic structure of the hyperspace is different and other
arguments have to be used. To get an estimate of the size of the constituents, we
study the Markov Process of “no a consecutive heads in flipping n coins”. Then,
we discuss the covering gauge for two different sets, a measure 1 set and the whole
2
space. First, the definition of the fractal measures would naturally give a bound on
the order of the gauge function. Second, we find specific covering for each sets and
direct calculation gives another bound for the order of the gauge function. Last, we
give a lower bound for the packing gauge by constructing a specific packing.
1.1
Hyperspace
We will define the Hausdorff metric and the hyperspace.
Definition 1.1. Let X be a metric space with metric ρ, let A, B ⊆ X, r > 0. Define
Nr (A) = {x ∈ X : there exists y ∈ A such that ρ(x, y) < r},
σ(A, B) = inf{r > 0 : A ⊆ Nr (B) and B ⊆ Nr (A)}.
Then σ is the Hausdorff metric for closed subsets of X.
Definition 1.2. Let X be a complete metric space. Then the hyperspace of X is the
set H(X) of all nonempty compact subsets of X equipped with the Hausdorff metric.
With the Hausdorff metric, the hyperspace shares some topological properties
with X.
Proposition 1.3. If X is a complete metric space, then H(X) is a complete metric
space. If X is separable, then H(X) is separable. If X is compact, then H(X) is
compact.
For proof, see [2, 2.4.1, 2.4.4].
3
1.2
Covering and Packing Measures
We will recall definitions and theorems to be used later. To begin with, let us define
two fractal measures on a metric space, in particular on this hyperspace. First, let us
start with the gauge function. We will follow the definition from [1].
Definition 1.4. A gauge is a function ϕ defined on an interval (0, δ) for some δ > 0
satisfying the following two conditions
• ϕ(r) > 0 for all r > 0.
• if r1 < r2 , then ϕ(r1 ) ≤ ϕ(r2 ).
Definition 1.5. A gauge ϕ is called blanketed iff lim sup
r→0
ϕ(2r)
<∞.
ϕ(r)
We will need this later in the Density Theorems.
Definition 1.6. Let X be a metric space, a ∈ X, r > 0. The open ball ∆(a, r) is
defined as
∆(a, r) = {x ∈ X : ρ(x, a) < r}.
Correspondingly, the closed ball ∆(a, r) is
∆(a, r) = {x ∈ X : ρ(x, a) ≤ r}.
Definition 1.7. A constituent in X is an ordered pair (a, r) where a ∈ X, r > 0.
Definition 1.8. Let X be a metric space, let a ∈ X, let µ be a finite Borel measure
on X, and let ϕ be a gauge. The upper ϕ-density of µ at a is
ϕ
Dµ (a)
µ ∆(a, r)
= lim sup
.
ϕ(r)
r→0
4
The lower ϕ-density of µ at a is
Dϕµ (a)
µ ∆(a, r)
= lim inf
.
r→0
ϕ(r)
Definition 1.9. Let X be a metric space, and let A ⊆ X. A cover or centered cover
of A is a set β of constituents such that
A⊆
[
and x ∈ A for all (x, r) ∈ β
∆(x, r),
(x,r)∈β
If δ > 0, then we say the cover β is a δ-fine cover provided r < δ for all (x, r) ∈ β.
Define
(
Cδϕ (A) = inf
)
X
ϕ(r) : β is a δ-fine cover of A
(x,r)∈β
C0ϕ (A) = sup Cδϕ (A) = lim Cδϕ (A)
δ→0
δ>0
C ϕ (A) = sup{C0ϕ (E) : E ⊆ A}
Here C ϕ is called the ϕ-covering outer measure
Definition 1.10. Let X be a metric space, and let A ⊆ X. A packing of A is a set
π of constituents such that x ∈ A for all (x, r) ∈ π, and ρ(x, x0 ) > r + r0 for all
(x, r), (x0 , r0 ) ∈ π with (x, r) 6= (x0 , r0 ). If δ > 0, then we say the packing π is δ-fine
provided r < δ for all (x, r) ∈ π. Define
(
ϕ
P δ (A)
)
X
= sup
ϕ(r) : π is a δ-fine packing of A
(x,r)∈π
ϕ
ϕ
ϕ
P 0 (A) = inf P δ (A) = lim P δ (A)
δ>0
δ→0
5
ϕ
P (A) = inf
(∞
X
ϕ
P 0 (En )
:A⊆
n=1
∞
[
)
En
n=1
ϕ
Here P is called the ϕ-packing outer measure
Definition 1.11. [5] Suppose that X is a metric space and E ⊂ X is a nonempty
subset, a gauge function is called a covering gauge of E if 0 < C ϕ (E) < ∞, a packing
gauge of E if 0 < P ϕ (E) < ∞.
Definition 1.12. Let X be a metric space. Let µ be a Borel measure on X. Then µ
has the Strong Vitali Property iff, for every Borel set E ⊆ X and every fine cover β
of E, there exists a (countable) centered closed ball packing π ⊆ β such that


µ E \
[
∆(x, r) = 0
∆(x,r)∈π
We say that the packing π almost covers the set E.
Theorem 1.13. Vitali Theorem[1] Let X be a metric space and E ⊆ X be a
subset. A fine cover of E is a collection β of constituents such that: x ∈ E for every
(x, r) ∈ β, and for every x ∈ E and every δ > 0, there exists r > 0 such that r < δ
and (x, r) ∈ β. Then there exists either:
1. an infinite (centered closed ball) packing {∆(xi , ri )} ⊆ β such that inf ri > 0, or
2. a countable (possibly finite) centered closed ball packing {∆(xi , ri )} ⊆ β such
that for all n ∈ N,
E\
n
[
∆(xi , ri ) ⊆
i=1
∞
[
i=n+1
6
∆(xi , 3ri )
Theorem 1.14. Density Theorem for the covering measure[1]
Let X be a metric space, let ϕ be a gauge, let µ be a finite Borel measure on X, and
let E ⊆ X be a Borel set.
1. Then
ϕ
µ(E) ≤ C ϕ (E) sup Dµ (x)
x∈E
except when the product is 0 times ∞.
2. Assume ϕ is blanketed. Then
ϕ
C ϕ (E) inf Dµ (x) ≤ µ(E)
x∈E
Theorem 1.15. Density Theorem for the packing measure[1]
Let X be a metric space, let ϕ be a gauge, let µ be a finite Borel measure on X, and
let E ⊆ X be a Borel set.
1. Then
ϕ
P (E) inf Dϕµ (x) ≤ µ(E)
x∈E
2. If µ has the strong Vitali Property, then
ϕ
µ(E) ≤ P (E) sup Dϕµ (x)
x∈E
except when the product is 0 times ∞.
7
1.3
General Setting
In the following, we consider the metric space S = {0, 1, 2, 3, . . . } equipped with
various metrics ρ all satisfying one condition:
0 is the only limit point and all other points are isolated.
Then, under this kind of metric, H(S) = K ∪ F, where
F = {A ⊆ S : 0 ∈
/ A and A is not nempty }
K = {A ⊆ S : 0 ∈ A}
We introduce two notations here, which will be repeatedly used later.
Let An = {A ∈ K : A ⊂ {n, n + 1, n + 2, . . . } ∪ {0}, n ∈ A} and similarly,
let Bn = {B ∈ F : B ⊂ {0, 1, . . . , n}, n ∈ B}.
For each n, symmetrically, there are 2n copies of An , each has the format
{ A \ {n} ∪ Ck : A ∈ An } with Ck varies among the subsets of {1, · · · , n}. Denote
them as {Ak (n)} with A1 (n) = An .
Next, we introduce the the natural measure µ on H(S).
µ(F) = 0,
µ(K) = 1 with µ(Ak (n)) = 2−n for any n ≥ 0, k ∈ {1, · · · , 2n }
For all the ρ in this paper, we will show that µ satisfies the Strong Vitali Property.
This would allow us the use the second statement from Theorem 1.15.
Consider ρ’s where for any δ > 0, there exists N > 0, such that ρ(n, m) > 2δ, if
n, m > N and ρ(n, m) < 2δ, if n, m > N .
ϕ
ϕ
Claim. Under such ρ and the natural measure, we have P (K) = µ(K)P 0 (K)
8
ϕ
Proof. By definition, P (K) = inf
ϕ
(∞
X
ϕ
P 0 (En )
:K⊆
n=1
ϕ
∞
[
)
En . But K ⊆ K, so we
n=1
have P (K) ≤ P 0 (K).
To show the other direction of the inequality, we first look at Ak (n)’s. Fix an
n, for any δ > 0 small enough, we have ρ(n − 1, n) > 2δ, which implies each Ak (n)
is at least 2δ away from any other Ak (n)’s. Let π be an arbitrary δ-fine packing of
A1 (n), then { A \ n ∪ Ck ), r : A ∈ A1 (n)} is an δ-fine packing of Ak (n) as well.
ϕ
By symmetry, we see that P δ (Ak (n)) are the same for each k. Let δ go to 0, we have
ϕ
ϕ
P 0 (K) = 2n P 0 (A1 (n)).
To proceed, we turn to the case where E is only measurable. Then, by the
S
definition of µ, we see that E = ∞
n=1 En , where each En is a symmetric copy of some
Am(n) and they are disjoint. Here En is numerated according to the size of m(n).
Then, given δ > 0, there is some N > 0 such that ρ(N, N + 1) < δ. This implies
ϕ
P 2δ (E) =
X
ϕ
ϕ
P 2δ (En ) + P 2δ (
n<N
[
En )
n≥N
Since each En is disjoint, as δ goes to 0, we have N → ∞ and therefore µ E \
SN
E
→ 0. This implies
i
n=1
ϕ
P 0 (E) =
X
ϕ
P 0 (En ) =
X
n
Now, for any {En } such that K ⊆
ϕ
ϕ
µ(En )P 0 (K) = µ(E)P 0 (K)
n
S
n
En , we group their index into two sets
according based on whether there are measurable or not, demote them as N and M.
9
Note that
S
n∈M
En is measurable, we have
X
ϕ
ϕ
[
P 0 (En ) ≥ P 0
n∈M
[
ϕ
En = µ(
En )P 0 (K)
n∈M
n∈M
Therefore, taking sum with respect to N and M, we get
X
n
ϕ
P 0 (En ) =
X
n∈N
ϕ
P 0 (En ) +
X
ϕ
ϕ
ϕ
P 0 (En ) ≥ µ(K)P 0 (K) = P 0 (K)
n∈M
ϕ
ϕ
Since this works for any {En }, taking infimum, we see that P (K) ≥ P 0 (K).
10
Chapter 2
Example 2−n + 2−m
In this chapter, we give S the metric ρ1 defined by
ρ1 (n, 0) = 2−n ,
ρ1 (n, m) = 2−n + 2−m
We see that under ρ1 , 0 is the only limit point of S.
Let ϕ(r) = 2r, in this chapter, I will find the covering and packing measures for
H(S).
First, let A01 = K \ A1 , F0 = {B ∈ F : 1 ∈
/ B}, F1 = {B ∈ F : 1 ∈ B}, and define
a function f : S → S by f (0) = 0, f (x) = x + 1 for x 6= 0. Then, for any x 6= y in
S, we have ρ(f (x), f (y)) = 21 ρ(x, y); that is, f is a similarity with ratio 21 . Then, we
define two functions g0 , g1 : H(S) → H(S) by:
g0 (E) = f [E] ∪ {1} = {f (x) : x ∈ E} ∪ {1},
g1 (E) = f [E]
Then, g0 maps H(S) onto A01 ∪ F0 and g1 maps H(S) onto A1 ∪ F1 . Also, g1 , g2 are
similarities:
1
σ(g0 (E1 ), g0 (E2 )) = σ(E1 , E2 ),
2
11
1
σ(g1 (E1 ), g0 (E2 )) = σ(E1 , E2 )
2
for any E1 , E2 ∈ H(S). Therefore, the pair (g0 , g1 ) is an iterated function system on
H(S) with ratio (1/2, 1/2) (See [2]). Moreover, the two images A1 ∪ F1 , A01 ∪ F0 are
disjoint, so this IFS satisfies the open set condition [2]. The attractor of the IFS is the
set K, and K is a self-similar Cantor set. Then we see that the Hausdorff dimension
for H(S) is 1. [2]
2.1
Constituents
First, we calculate all the balls.
Given any n and B ∈ Bn , we have
∆(B, 2−` ) = ∆(B, 2−` ) = {B}, for all ` ≥ n
And for any r ∈ (2−` + 2−m , 2−` + 2−(m−1) ] with m > ` > n, we have
∆(B, r) = ∆(B, r) = ∆(B, 2−` + 2−m ) = {B}
Given any n and A ∈ An , we have
∆(A, 2−` ) ∩ K = {E ∈ K : E ∩ {1, . . . , `} = A ∩ {1, . . . , `}}
∆(A, 2−` ) ∩ K = {E ∈ K : E ∩ {1, . . . , ` − 1} = A ∩ {1, . . . , ` − 1}}
And for any r ∈ (2−` + 2−m , 2−` + 2−(m−1) ] with m > ` > n, we have
∆(A, r) ∩ K = ∆(A, r) ∩ K = {E ∈ K : E ∩ {1, . . . , ` − 1} = A ∩ {1, . . . , ` − 1}}
12
2.2
Densities
We can find the densities with respect to the natural measure µ.
Since µ(F) = 0 and according to the previous calculation, for r small enough, ∀ n
and B ∈ Bn ∆(B, r) = ∆(B, r) = {B}, thus
ϕ
Dµ (B) = Dϕµ (B) = 0
Given any m > ` > n, for all A ∈ An and r ∈ (2−` + 2−m , 2−` + 2−(m−1) ],
µ ∆(A, 2−` )
1
= ,
−`
ϕ(2 )
2
µ ∆(A, r)
2−`+1
1
≤
=
−`
−m
ϕ(r)
ϕ(2 + 2 )
1 + 2`−m
µ ∆(A, 2−` )
µ ∆(A, r)
1
2−`+1
1
=
,
≥
=
−`
−`
−(m−1)
ϕ(2 )
2
ϕ(r)
ϕ(2 + 2
)
1 + 2`−m+1
i S
Since 0, 12 = m>` (2−` + 2−m , 2−` + 2−(m−1) ], we conclude that
ϕ
Dµ (A)
µ ∆(A, r)
1
= lim
=1
= lim sup
2
m=` ,`→∞ 1 + 2`−m
ϕ(r)
r→0
∆(A,
r)
µ
1
1
Dϕµ (A) = lim inf
=
lim
=
`−m+1
r→0
m=`+1,`→∞ 1 + 2
ϕ(r)
2
2.3
Covering Measure
In this section, we will prove C ϕ (H(S)) = 1 using two different methods.
First of all, let us do it using the densities.
13
On the one hand, for K,
C ϕ (K) ≤
µ(K)
=1
ϕ
inf A∈K Dµ (A)
C ϕ (K) ≥
µ(K)
=1
ϕ
supA∈K Dµ (A)
Therefore, noting that ϕ is blanketed, C ϕ (K) = 1.
On the other hand, for F, given any m > ` > n, let δ = 2−` and
κm = {∆(B, 2−m ) : B ∈ Bn }, then κm is a δ-fine cover of Bn and
X
Cδϕ (Bn ) ≤
ϕ(2−m ) = 2n−1 · 2−m+1 = 2n−m
(x,2−m )∈κm
Let m → ∞, we have Cδϕ = 0 and C0ϕ (B) = limδ→0 Cδϕ = 0.
But κm is a δ-fine cover for any subset of Bn , so we conclude that C ϕ (Bn ) = 0.
P
ϕ
ϕ
ϕ
ϕ
Therefore, C ϕ (F) ≤ ∞
n=1 C (Bn ) = 0 and C (H(S)) ≤ C (F) + C (K) = 1.
µ(H(S))
Also, C ϕ (H(S)) ≥
= 1.
ϕ
supA∈H(S) Dµ (A)
Consequently, C ϕ (H(S)) = 1
Secondly, let us do it by the original definition, i.e, by the coverings.
Given any ` > n and A ∈ An , we have
µ ∆(A, r) =



r,
r = 2−`
.


2−`+1 ≤ 2r, r ∈ (2−` , 2−(`−1) )
In other words, µ ∆(A, r) ≤ 2r.
P
P
P
Hence, ∀ δ ∈ [2−` , 2−`+1 ), ∆(A,r)∈κ ϕ(r) = 2 ∆(A,r) r ≥ ∆(A,r) µ ∆(A, r) ≥ 1,
where κ is any δ-fine cover of K. Thus, C ϕ (H(S)) ≥ C0ϕ (K) ≥ 1.
14
At the same time, let us denote 2{1,2,...,`−1} as {G(i)}, then ∀ δ ∈ [2−` , 2−`+1 ), and
n
o
∀ m > `, consider the δ-fine cover κ =
G(i) ∪ {0}, 2−` + 2−m , then
X
ϕ(r) = 2 · 2`−1 · (2−` + 2−m ) → 1, as m → ∞
κ
Now, given any E ⊆ H(S), we can rewrite E =
S2n−1
i=1
Ei .
Here Ei = {E ∈ E : E ∩ {1, 2, . . . , n − 1} is the same}.
Case I: If Ei ∩ K 6= ∅, then Ei can be covered by 2`−n copies of {(G(i) ∪ {0}, 2−` +
2−m )};
Case II: If Ei ∩ K = ∅, then Ei can be covered by 2`−n copies of {(G(i), 2−` + 2−m )}.
Therefore, we can find a δ-fine cover of E consisting of either (G(i), 2−` + 2−m ) or
G(i) according to the above 2 cases. Then, C0ϕ (E) = limδ→0 Cδϕ (E) ≤ 1 and C ϕ (H(S)) ≤
C0ϕ (E) = 1.
Combine the two, we conclude that C ϕ (E) = 1.
2.4
Packing Measure
ϕ
Similarly, we will prove that P (H(S)) = 2, again using two methods.
Firstly, we will do it using densities. But before that
Claim. Under ρ1 , µ satisfies the Strong Vitali Property.
Proof. Due to the structure of the balls in ρ1 , we see that given any 0 < r < 1, there
exists n such that 2−n−1 ≤ r < 2−n . This implies that there are at most 2n+1 disjoint
balls with radius more than r. Therefore, according to the Vitali Theorem, given any
subset E of H(S) and any fine cover β of E, we can find a countable centered closed
15
ball packing {∆(xi , ri )} ⊆ β such that, for all n,
E\
n
[
∆(xi , ri ) ⊆
i=1
∞
[
∆(xr , 3ri )
i=n+1
Note that all these ∆(xi , ri ) are disjoint, we have
P µ ∆(xi , ri ) ≤ 1. On other
hand, due to the metric, to cover a ball with radius 3ri , we just need 4 balls of radius
P P ri . Therefore,
µ ∆(xi , 3ri ) ≤ 4 µ ∆(xi , 3ri ) ≤ 4. Take n to infinity, we have
S
µ E\ ∞
∆(x
,
r
)
= 0.
i
i
i=1
Consider the set Dn = K ∪ Bn , since Bn % F, we have Dn % K ∪ F = H(S).
ϕ
For P (K), due to the above claim, now we can apply Theorem 1.14 and it shows
that
2=
µ(K)
µ(K)
ϕ
≤ P (K) ≤
=2
ϕ
supA∈K Dµ (A)
inf A∈K Dϕµ (A)
ϕ
Moreover, note that dist(K, Bn ) ≥ 2−n , since P is a metric outer measure and Bn is
a finite set thus has packing measure 0, we see that
ϕ
ϕ
ϕ
P (Dn ) = P (K) + P (Bn ) = 2
ϕ
ϕ
Therefore, P (H(S)) = limn→∞ P (Dn ) = 2
Secondly, we will do it using the packings.
ϕ
ϕ
According to above, we have already proved that P (H(S)) = P (K), so it is
ϕ
ϕ
ϕ
enough to show P 0 (K) = 2. The claim from 1.3
 will guarantee P (K) = µ(K)P 0 (K).


r = 2−`
2r,
For any ` > n and A ∈ An , µ ∆(A, r) =


2−`+1 > r, r ∈ (2−` , 2−(`−1) )
Thus, r < µ ∆(A, r) .
16
On the one hand, ∀ δ > 0, ∃ `, s.t 2−` ≤ δ < 2−(`−1) .
Then, given any δ-fine packing Π of K,
X
X
=
∆(A,r)∈Π
2r < 2
∆(A,r)∈Π
X
µ ∆(A, r) ≤ 2µ(K) = 2
∆(A,r)∈Π
Moreover, for the δ same as above, let rm = 2−`−1 + 2−`−2 +, . . . , +2−m , m > `2 ,
then rm % 2−` . To show the equality holds, we consider the packing
Πm = {(B ∪ {0}, rm ) : B ∈ B` }, then there are only 2` disjoint balls there and
Pδϕ (K) ≥
X
ϕ(rm ) = 2rm ∗ 2` = 1 +
Πm
ϕ
Therefore, we proved that P 0 (K) = 2.
17
1
+ . . . + 2`+1−m → 2
2
Chapter 3
General 2−n
We will move our focus to other metrics, with ρ(n, o) = 2−n , and our goal is to find
ϕ
the corresponding ϕ(r) for each metric such that the result Cµϕ = 1 and P µ = 2 held.
ϕ
For the discussion below, we will reach this goal by showing D (A) = 1,
Dϕ (A) =
1
2
a.e.
Before introducing those metrics, let us spend some time studying the general
structure of the elements of A, which would give some insight for later calculation.
For all A ∈ K satisfying:
There is a N such that A ∩ {N, N + 1, · · · } = ∅ or {N, N + 1, · · · } ⊆ A
We see that there are only countably many of them.
Therefore, almost everywhere with respect to µ, for any A ∈ An , there is a
subsequence of natural numbers, say `k ’s, such that `k % ∞ and {`k } ⊆ A. Also,
almost everywhere with respect to µ, for any A ∈ K, there is another subsequence
of natural numbers, say `0k , such that `k % ∞ and {`0k , · · · , `0k + 21 ln(2`0k )} ∩ A = ∅.
Even more, this implies the existence of some {`00k } such that `00k % ∞, {`00k } ∩ A = ∅
and {1 + `00k } ⊆ A.
Also, by slightly changing the proof, we can see that µ satisfies the Strong Vitali
Property under any such metrics. Therefore, we can apply the full version of the
Theorem 1.14.
18
One more thing to notice, for each of the following cases, by an argument similar
to the beginning of chapter 2, we see that the corresponding Hausdorff dimension is
ϕ
still 1. But, to get C ϕ = 1 and P = 2, we might need some changes on ϕ.
3.1
ρ2 (m, n) = 2− min(m,n)
We introduce ρ2 (m, n) = 2− min(m,n) , m 6= n 6= 0;
ρ2 (m, 0) = 2−m . Then, under ρ2 ,
0 is still the only limit point for S.
First of all, let us find all the constituents. Given ` > n, for any A ∈ An and for
any r ∈ (2−` , 2−`+1 ), we have
K ∩ ∆(A, 2−` ) = {E : E ∩ {1, · · · , `} = A ∩ {1, · · · , `}}
K∩∆(A, r) = K∩∆(A, 2−` ) = K∩∆(A, r) = {E : E∩{1, · · · , `−1} = A∩{1, · · · , `−1}}
Since they have the some structure as the constituents from chapter 1, we see that
under ρ2 , we have the same densities as ρ1 .
Naturally, let us generalize this metric by introducing
for all α > 0, ρ(α) (m, n) = (ρ2 (m, n))1/α , ∀ m ≤ 0, n ≤ 0
This time, we let ϕ(r) = 2 · 2α . Then, follow the same calculation as above,
for ` > n and ∀ A ∈ An , r ∈ 2−`/α , 2−(`−1)/α ,we have
K ∩ ∆(A, 2−`/α ) = {E : E ∩ {1, · · · , `} = A ∩ {1, · · · , `}}
K∩∆(A, r) = K∩∆(A, r) = K∩∆(A, 2−`/α ) = {E : E∩{1, · · · , `−1} = A∩{1, · · · , `−1}}
19
µ ∆(A, r)
2−(`−1)
Note that
<
= 1, and it approaches this supremum when
ϕ(r)
ϕ(2−`/α )
r → 2−(`+1)/α +. Therefore, we have
ϕ
Dµ (K ∩ A) = lim sup
r→0
µ(K ∩ ∆(A, r))
µ(K ∩ ∆(A, 2−`/α +))
= lim
=1
`→∞
ϕ(r)
ϕ(2−`/α +)
µ ∆(A, r)
2−(`−1)
1
Similarly, note that
>
= , and it approaches this infi−(`−1)/α
ϕ(r)
ϕ(2
)
2
−(`−1)/α
mum when r → 2
−. Therefore, we have
−(`−1)/α
µ
K
∩
µ
K
∩
∆(A,
r)
∆(A,
2
−)
1
= lim
=
Dϕµ (K ∩ A) = lim inf
−(`−1)/α
r→0
`→∞
ϕ(r)
ϕ(2
−)
2
3.2
ρ3 (m, n) = |2−m − 2−n |
Now, let us define another metric
ρ3 (m, n) = |2−m − 2−n |, m 6= n 6= 0;
ρ3 (m, 0) = 2−m
Again, 0 is the only limit point under ρ3 .
Moreover, let ϕ̃(r) = 2 · 2−` , for all r ∈ (2−(`+1) , 2−` ].
Again, we try to find the structure of the constituents first.
This time, the structure of the constituents does depend of the structure of A.
Given any n + 1 < ` < m − 1, for all A ∈ An , ∀ r ∈ (2−` − 2−(m−1) , 2−` − 2−m ), we
20
start with the constituents first.
{E : E ∩ {1, · · · , ` − 1} = A ∩ {1, · · · , ` − 1}}




6 ∅}, ` ∈ A

{E : E ∩ {`, · · · } =
−`

K ∩ ∆(A, 2 ) = T 
K,
`∈
/ A, A ∩ {` + 1, · · · } =
6 ∅






{E : ` ∈
/ E},
A ∩ {`, · · · } = ∅
Therefore, µ K ∩ ∆(A, 2−` ) =




2−`+1 , ` ∈ A




2−`+1 , ` ∈
/ A, A ∩ {` + 1, · · · } =
6 ∅.






2−` ,
A ∩ {`, · · · } = ∅
{E : E ∩ {1, · · · , ` − 2} = A ∩ {1, · · · , ` − 2}}




{E : E ∩ {` − 1, `} =
6 ∅}, ` − 1 ∈ A



−`

K ∩ ∆(A, 2 ) = T
K,
`−1∈
/ A, ` ∈ A






{E : ` − 1 ∈
/ E},
`−1∈
/ A, ` ∈
/A
21
Therefore, µ K ∩ ∆(A, 2−` ) =




3 · 2−` , ` − 1 ∈ A




2−`+2 ,






2−`+1 ,
`−1∈
/ A, ` ∈ A
`−1∈
/ A, ` ∈
/A
{E : E ∩ {1, · · · , ` − 1} = A ∩ {1, · · · , ` − 1}}


K ∩ ∆(A, 2−` − 2−m )


{E : E ∩ {`, · · · , m − 1} =
6 ∅}, ` ∈ A




= T
K ∩ ∆(A, r)
`∈
/ A, A ∩ {`, · · · , m − 1} =
6 ∅
K,

K ∩ ∆(A, r)




{E : ` ∈
/ E},
A ∩ {`, · · · , m − 1} = ∅



−`
−m

, `∈A
2
·
2
−
2




µ K ∩ ∆(A, 2−` − 2−m )
= 2−`+1 ,

µ K ∩ ∆(A, r) = µ K ∩ ∆(A, r)




2−` ,
`∈
/ A, A ∩ {`, · · · , m − 1} =
6 ∅
A ∩ {`, · · · , m − 1} = ∅
{E : E ∩ {1, · · · , ` − 1} = A ∩ {1, · · · , ` − 1}}




6 ∅}, ` ∈ A

{E : E ∩ {`, · · · , m} =
−`
−m

K ∩ ∆(A, 2 − 2 ) = T 
K,
`∈
/ A, A ∩ {`, · · · , m} =
6 ∅






{E : ` ∈
/ E},
A ∩ {`, · · · , m} = ∅
Then, we have µ K∩∆(A, 2−` −2−m ) =




2−(`−1) − 2−m , ` ∈ A




2−`+1 ,






2−` ,
22
`∈
/ A, A ∩ {`, · · · , m} =
6 ∅.
A ∩ {`, · · · , m} =
6 ∅
According to the discussion at the beginning of this chapter, except for a measure
zero set, for any A ∈ K, there exist sequences of indices `0k and {`00k }.
ϕ̃
First, let us use the {`00k } to find Dµ (A). Judging by the size of µ ∆(A, r) corresponding to `, we see that such maximum would occur
in the interval
00
µ ∆(A, r)
2−`k +1
00
00
r ∈ (2−(`k +1) , 2−`k ]. But, in this interval
= −`00 +1 = 1.
ϕ̃(r)
2 k
µ ∆(A, r)
ϕ̃
Therefore, we conclude that Dµ (A) = lim sup
= 1.
ϕ̃(r)
r→0
Second, we will use {`0k } to find Dϕ̃µ (A). Again, judging by the size of µ ∆(A, r)
corresponding to `, we see that such minimum would occur
in the interval
0
µ ∆(A, r)
2−`k
1
−(`0k +1) −`0k
≥ −`0 +1 = , where the
r ∈ (2
, 2 ]. But, in this interval
ϕ̃(r)
2
2 k
−`0k
equality holds when r 6= 2 .
∆(A,
r)
µ
1
Therefore, we have Dϕ̃µ (A) = lim inf
= .
r→0
ϕ̃(r)
2
3.3
General Case
In this section, we will find some generalized results.
Given any metric ρ with ρ(n, 0), let ϕ(r) be the corresponding gauge such that
ϕ
Cµϕ = 1 and P µ = 2.
Claim. If there is N > 0 big enough such that ρ(m, n) ≥ 2−m∧n = ρ2 (m, n),
∀m 6= n > N , then ϕ(r) = 2r.
Proof. Thanks to the Triangle Inequality, we have ρ2 ≤ ρ ≤ ρ1 besides finite many
integers. Then, for any r ∈ (0, 1) and any A ∈ K, we have ∆1 (A, r) ⊆ ∆(A, r) ⊆
∆2 (A, r), same for the closed balls. Here ∆i (A, r) stand for ∆(A, r) under the metric
ρi . Note that we chose the same ϕ for both ρ1 and ρ2 , then stick to the same choice,
23
ϕ
we will still get Dµ (K ∩ A) = 1, Dϕµ (K ∩ A) = 0.5. Therefore, ϕ(r) = 2r is the desired
gauge.
Next, we turn to another kind of ρ where ρ3 (m, n) < ρ(m, n) < ρ2 (m, n)
if m 6= n > N for some N big enough.
Claim. In this case, the desire gauge is ϕ̃(r) = 2 · 2−` , for all r ∈ (2−(`+1) , 2−` ], as in
the previous section.
Proof. Thanks to the Triangle Inequality, we see that
ρ(n + 1, 0) ≤ ρ(n, n + 1) ≤ ρ(n, n + 2) ≤ · · · < ρ(n, 0), ∀ n > N
So, given any n + 1 < ` < m, for all A ∈ An and r ∈ ρ(`, m), ρ(`, m + 1)
{E : E ∩ {1, · · · , ` − 1} = A ∩ {1, · · · , ` − 1}}




6 ∅}, ` ∈ A

{E : E ∩ {`, · · · } =
−`

K ∩ ∆(A, 2 ) = T 
K,
`∈
/ A, A ∩ {` + 1, · · · } =
6 ∅






{E : ` ∈
/ E},
A ∩ {`, · · · } = ∅




2−`+1 , ` ∈ A




−`
Therefore, µ K ∩ ∆(A, 2 ) = 2−`+1 , ` ∈
/ A, A ∩ {` + 1, · · · } =
6 ∅.






2−` ,
A ∩ {`, · · · } = ∅
Since ρ(` − 1, `) > ρ3 (` − 1, `) = 2−` ,
K ∩ ∆(A, 2−` ) = {E : E ∩ {1, · · · , ` − 1} = A ∩ {1, · · · , ` − 1}}
24
Therefore, µ K ∩ ∆(A, 2−` ) = 2−`+1 .
Moreover, since ρ(` + 1, `) > ρ(` + 1, 0), for any r ∈ ρ(` + 1, 0), ρ(`, ` + 1)
K∩∆ A, ρ(`, `+1) = K∩∆(A, r) = K∩∆(A, r) = {E : E∩{1, · · · , `} = A∩{1, · · · , `}}
So, µ ∆ A, ρ(`, ` + 1) = µ ∆(A, r) = µ ∆(A, r) = 2−` .
K ∩ ∆ A, ρ(`, m + 1)
K ∩ ∆(A, r)
K ∩ ∆(A, r)
{E : E ∩ {1, · · · , ` − 1} = A ∩ {1, · · · , ` − 1}}




{E : E ∩ {`, · · · , m} =
6 ∅}, ` ∈ A




= T
K,
`∈
/ A, A ∩ {`, · · · , m} =
6 ∅






{E : ` ∈
/ E},
A ∩ {`, · · · , m} = ∅



2−`+1 − 2−m , ` ∈ A




µ K ∩ ∆ A, ρ(`, m + 1)
= 2−`+1 ,

µ K ∩ ∆(A, r) = µ K ∩ ∆(A, r)




2−` ,
`∈
/ A, A ∩ {`, · · · , m} =
6 ∅
A ∩ {`, · · · , m} = ∅
{E : E ∩ {1, · · · , ` − 1} = A ∩ {1, · · · , ` − 1}}




6 ∅}, ` ∈ A

{E : E ∩ {`, · · · , m} =

K ∩ ∆ A, ρ(`, m) = T 
K,
`∈
/ A, A ∩ {`, · · · , m} =
6 ∅






{E : ` ∈
/ E},
A ∩ {`, · · · , m} = ∅
25
Then, we have µ K∩∆ A, ρ(`, m) =




2−(`−1) − 2−m , ` ∈ A




2−`+1 ,






2−` ,
`∈
/ A, A ∩ {`, · · · , m} =
6 ∅.
A ∩ {`, · · · , m} = ∅
Let us use ϕ̃. Since the structure of the constituents in this case is quite similar
as in the previous case, by a similar argument, we see that
ϕ̃
Dµ (A)
µ ∆(A, r)
= lim sup
= 1,
ϕ̃(r)
r→0
µ
∆(A,
r)
1
Dϕ̃µ (A) = lim inf
=
r→0
ϕ̃(r)
2
Then, what happen to other metrics ρ where no such N could be found? Out of
these two gauges, can we just settle with one? The answer is yes.
Claim. The gauge ϕ̃(r) works here as well, if and only if, there are `000
k % ∞ such
000
000
that ρ(`000
k , m) < ρ2 (`k , m) for any m > `k and any k > 0.
Proof. By the argument in the beginning of this chapter, we see that, almost every000
where on K, {`000
k , `k + 1, . . . } ∩ A 6= ∅ for any k > 0 and A ∈ K. Therefore, for any
000
−`000
k +1 . Therefore, to
,
m
+
1)
where m > `000
,
m),
(ρ
r ∈ ρ(`000
k , we have µ(A, r) = 2
k
k
ϕ
ensure Dµ (A) = 1, we have to choose ϕ(r).
e
For Dϕµ (A), since ϕ̃(r) ≥ 2r, and 0.5 can
000
000 be reaches in 2−`k −1 , 2−`k , we will not have any issues here.
At the same time, if not such `000
k can be found, then for n big enough, we have
ρ(n, m) ≥ ρ2 (n, m) for all m > n. This falls into the category of the first claim of
this section.
26
Chapter 4
Distance is in the form of f (m ∧ n)
In this chapter, we will introduce a set of metrics with the following properties:
ρ(m, n) = f (m ∧ n) and ρ(m, 0) = f (m) for m 6= n > N
where ∀ x > N, 0 < f (x) ≤ 21 , f 0 (x) < 0.
ϕ
Now, our goal is to find the corresponding ϕ(r) such that both Cµϕ and P µ are
positive and finite.
First of all, let us show the validity of the Triangle Inequality. On the one hand,
note that f is decreasing, then given any k, n, m > N , if k < n ∧ m, ρ(n, m) =
f (n ∧ m) < f (k) = ρ(k, n) < ρ(k, n) + ρ(k, m). Also, if m ∧ n < k, then ρ(n, m) =
f (n ∧ m) = f (n ∧ k) ∨ f (m ∧ k) < ρ(n, k) + ρ(m, k). On the other hand, say k < m ∧ n
and k < N , then ρ(m, n) ≤
1
2
= ρ(k, m) < ρ(k, m) + ρ(k, n). Also, of m ∧ n < k and
m ∧ n < N , then ρ(m, n) =
1
2
= ρ(m, k) ∨ ρ(n, k) < ρ(m, k) + ρ(n, k).
Secondly, one thing to notice, given any such metric ρ, for any n > 0 and any
A ∈ An , for ` > n ∨ N and any r ∈ f (`), f (` − 1) we have
K ∩ ∆(A, f (`)) = {E : E ∩ {1, · · · , `} = A ∩ {1, · · · , `}}
K∩∆(A, r) = K∩∆(A, r) = K∩∆(A, f (`)) = {E : E∩{1, · · · , `−1} = A∩{1, · · · , `−1}}
µ ∆(A, r)
µ ∆(A, r)
2−(`−1)
2−(`−1)
≤
,
as
well
as
≥
.
Therefore,
ϕ(r)
ϕ(f (`)+ )
ϕ(r)
ϕ(f (` − 1)− )
27
4.1
First example: f (x) =
2−x
x
Because f (x) satisfies all the prerequisites with N = 1, we define ρ5 (m, n) =
2−(m∧n)
,
m∧n
2−m
. According to the discussion in the beginning of this
m
chapter, we see that ρ is indeed a metric.
2
· r| ln r|, then ϕ(r) is an increasing function with ϕ(r) → 0
Then, let ϕ(r) =
ln 2
as r → 0. So, it satisfies the prerequisite for a gauge function. Furthermore,
with ρ5 (m, 0) = f (m) =
ϕ(2r)
ln 2 lim
= lim 2 · 1 +
=2
r→0 ϕ(r)
r→0
ln r So, ϕ is blanketed and Density Theorem could be used for both cases.
According to the discussion at the beginning of this chapter, we have
ϕ
Dµ (K
Dϕµ (K
µ K ∩ ∆(A, f (`)+ )
` ln 2
∩ A) = lim
= lim
=1
+
`→∞
`→∞ ` ln 2 + ln `
ϕ(f (`) )
µ K ∩ ∆(A, f (` − 1)− )
(` − 1) ln 2
1
∩ An ) = lim
= lim
=
−
`→∞
`→∞ 2 · [(` − 1) ln 2 + ln(` − 1)]
ϕ(f (` − 1) )
2
ϕ
Then, using the same arguments as in 2.3 and 2.4, we have both Cµϕ and P µ are
positive and finite.
Let us generalize this metric a little bit to a two-parameter example. Pick any
α, β > 0 and let
fα,β (x) = 2−x/α x−β/α , here fα,β (x) is decreasing with limx→∞ = 0. Then we define a
metric by ρ(m, n) = fα,β (m ∧ n), ∀ m 6= n 6= 0 and f (m, 0) = fα,β (m).
Next, let ϕ(r) = 2αβ (ln 2)−β rα | ln r|β . Here ϕ(r) is an increasing function with
ϕ(r) → 0 as r → 0. So, it satisfies the prerequisite for a gauge function as well. The
28
argument above shows ϕ is blanketed. So, the Density Theorem shows both measures
are positive and finite.
Similarly, we have
ϕ
Dµ (A)
Dϕµ (A)
4.2
−β
µ K ∩ ∆(A, f (`)+
β ln `
= lim
= lim 1 +
=1
`→∞
`→∞
ϕ(f (`)+ )
` ln 2
−β
µ K ∩ ∆(A, f (` − 1)− )
1
β ln(` − 1)
1
= lim
1+
=
= lim
−
`→∞ 2
`→∞
ϕ(f (` − 1) )
(` − 1) ln 2
2
Three-parameter example: fα,β,γ (x) = xβ/α 2−x/α (ln x)γ/α
This time, for any α > 0, β, γ ≥ 0, we introduce fα,β,γ (x) as stated. Note that
0
fα,β,γ
(x) = xβ/α−1 2−x/α (ln x)γ/α−1 α−1 ln x(β − x ln 2) + γ
0
(x) < 0 if x ≥ N 0 . Therefore, there is some
Then for some N 0 large enough, fα,β,γ
N > N 0 such that f (x) ≤ 12 .
Now, let us introduce the metric ρ(m, n) = fα,β,γ (m ∧ n) if m ∧ n > M , and
ρ(m, 0) = fα,β,γ (m). Also, for any m, and n ≤ M , we let ρ(m, n) = 21 .
1
1
Next, let us introduce ϕ(r) = 2(ln 2)β α−β rα (ln )−β (ln ln )−γ , then ϕ(r) is an
r
r
increasing function on (0, 1) with ϕ(r) → 0 as r → 0. Furthermore,
−β −γ
ϕ(2r)
ln 2 ln | ln 2r|
α lim
= lim 2 · 1 +
= 2α
r→0 ϕ(r)
r→0
ln r ln | ln r|
So, ϕ is blanketed and Density Theorem could be used for both cases.
29
Then, again, we have
ϕ
Dµ (A)
−β 1 γ
µ K ∩ ∆(A, f (`)+ )
β ln ` γ ln ln `
= lim
= lim 1 −
−
1+O
=1
`→∞
`→∞
ϕ(f (`)+ )
` ln 2
` ln 2
ln `
Dϕµ (A)
4.3
µ K ∩ ∆(A, f (` − 1)− )
= lim
`→∞
ϕ(f (` − 1)− )
−β γ
β ln(` − 1) γ ln ln(` − 1)
1
1
1−
−
1+O
= lim
`→∞ 2
(` − 1) ln 2
(` − 1) ln 2
ln(` − 1)
1
=
2
fα (x) = x−α
In the previous two sections, f (x) = O(ax ) as x → ∞ for some 0 < a < 1. But, here
we do not need f to go to zero exponentially due to the fact ρ(m, n) = f (m ∧ n).
This time, let us introduce ρ(m, n) = fα (m ∧ n) = (m ∧ n)−α , then ρ is a metric.
Next, let ϕ(r) = 2−r
−1/α
. Note that ϕ(r) is increasing with ϕ(r) → 0, it satisfies
the prerequisite of the gauge. But
ϕ(2r)
ϕ(r)
→ ∞ as r → 0, which leads to the situation
that ϕ(r) is not blanketed. Then, different from the previous sections, when it comes
to C ϕ (K), the density theorem can only give an lower bound. But, the upper bound
part could be done by referring to last part of 2.3, where we use a special covering κ.
ϕ
Dµ (A)
Dϕµ (A)
µ K ∩ ∆(A, f (`)+ )
= lim
= lim 2−(`−1)+` = 2
+
`→∞
`→∞
ϕ(f (`) )
µ K ∩ ∆(A, f (` − 1)− )
= lim
= lim 2−(`−1)+`−1 = 1
`→∞
`→∞
ϕ(f (` − 1)− )
30
4.4
fα,s,M (x) = M 1/α ln x −
s
α
ln ln x
−1/α
This time for any α > 0, M ≥ 0, s, we introduce gα,s = ln x − αs ln ln x and let
−1/α
0
fα,s,M (x) = M 1/α gα,s . Note that gα,s
= x1 1 − α lns x , then for some N large enough,
0
0
(x) > 0, which gives fα,s,M
(x) < 0. Therefore,
if x > N , then both gα,s (x) > 0 and gα,x
there is some N 0 > N such that fα,s,M (x) ≤ 12 .
Next, let us introduce metric ρ(m, n) = fα,s,M (m∧n) if m∧n > N 0 , and ρ(m, 0) =
fα,s,M (m). Also, for any m, and n ≤ N 0 , we let ρ(m, n) = 12 . The argument for the
Triangle Inequality is the same as in the previous case.
Next, let us introduce ϕ(r) = 21 − M
s/α −s
r
exp(M r−α ) , then ϕ(r) is an increasing
function on (0, 1) with ϕ(r) → 0 as r → 0. But ϕ(r) is not blanketed either.
Thus,
+
µ K ∩ ∆(A, f (`) )
ϕ
Dµ (A) = lim
`→∞
Dϕµ (A) = lim
ϕ(f (`)+ )
µ K ∩ ∆(A, f (` − 1)− )
`→∞
ϕ(f (` −
1)− )
s/α
= lim 2−` + `(1 − (s ln ln `)/α ln `) = 1
`→∞
s/α
1
= lim 2−` + (` − 1)[1 − O(ln ln `/ ln `)] =
`→∞
2
ϕ
Same as in the previous section, we are still able to show that both Cµϕ and P µ are
positive and finite by 2.3 and 2.4
4.5
The choice of ϕ
At this stage, a question occurs. What are the reasonable conditions so that ϕ(r) is
a gauge function of H(S) with nontrivial measure? Also, if it is, what is the choice
of the metric ρ?
31
What we want to do is to define ρ as in the beginning of this section, so when it
comes to the choice of f (x), we only need to know what are {f (`)}` .
First of all, if ϕ(r) is strictly monotone in [0, δ) for some δ > 0 and ϕ(0) = 0, taking
the absolute value if necessary, we get an positive increasing function, still call it ϕ.
Thus, ϕ−1 (r) exists and is also positive increasing in [0, δ). Define f (x) = ϕ−1 (2−x ),
then f is positive, decreasing, with f (x) → 0 as x → ∞.
Then, we define ρ(m, n) = f (m ∧ n) and ρ(m, 0) = f (m), by argument similar to
section 4.1, we see that ρ is indeed a metric.
If in this case, ϕ is continuous, then we have ∀`, ϕ(f (`)− ) = ϕ(f (`)) = ϕ(f (`)+ ),
and
µ K ∩ ∆(A, f (`)+ )
ϕ
Dµ (A) = lim
ϕ(f (`)+ )
`→∞
Dϕµ (A) = lim
`→∞
= lim 2−(`−1)+` = 2
µ K ∩ ∆(A, f (` − 1)− )
ϕ(f (` − 1)− )
`→
= lim 2−(`−1)+`−1 = 1
`→
Moreover, given any fe(x) defined on [1, ∞), if fe is positive, decreasing with
0 <= a = lim inf
x→∞
fe(x)
fe(x)
≤ lim sup
=b<∞
f (x)
x→∞ f (x)
Then, fe could replace f to give nontrivial measures in both cases if ϕ is blanketed.
We could check, for sections from this chapter, the choices of f all fell into this case.
If, ϕ is not continuous, then ϕ(f (`)− ) ≤ ϕ(f (`)) ≤ ϕ(f (`)+ ), but the equality
does not necessarily hold for each `. In this case, f still gives nontrivial measures.
And, when we are looking for the alternative fe for f , the above argument works if ϕ
is again blanketed.
Secondly, if ϕ(r) is monotone but not strictly monotone, then let (ak , bk ] k be
32
the collection of intervals contained in (0, δ) where ϕ(r) = ϕ(ak ), ∀ r ∈ (ak , bk ],
ak+1 < bk+1 < ak < bk and ϕ(r1 ) 6= ϕ(r2 ) if r1 6= r2 ∈
/ (ak , bk ] for any k. Then, we can
define ϕ−1 and the image of it does not fell into any such (ak , bk ]. Everything else
follows the previous discussion.
Thirdly, if ϕ(r) is not monotone in any [0, δ), but ϕ(0) = 0, then we are looking
ϕ
ϕ2
for ϕ1 and ϕ2 such that C ϕ = C ϕ1 and P = P .
On the one hand, let ϕ1 be nondecreasing and ϕ1 ≤ ϕ by replacing all the bumping
intervals of ϕ inside [0, 1) by a horizontal line connecting the endpoints. Bumping
interval is defined as the interval where function changes from decreasing to increasing
at the right endpoints and the left end point is picked such that both endpoints
correspond to the same ϕ value. One thing to notice, they are infinitely many such
intervals. Moreover, we see that ϕ1 (r) ≤ ϕ(r) where the equality holds when r does
not belong to any bumping intervals.
Let (ak , bk ) be the collection of all bumping intervals, with ak+1 < bk+1 <
ak < bk . Let δk = bk + εk , where εk is chosen to be small and ϕ is increasing in
(bk , bk + εk ). Given any A ⊆ H(S) and βm , any δm −fine covering of A, if for some
∆(E, r) ∈ βm , r ∈ (ak , bk ) for some k , then let r0 = bk , otherwise, r0 = r. Things
0
= {∆(E, r0 ) : ∆(E, r) ∈ βm } is another δm −fine
to notice, ϕ1 (r) = ϕ(r0 ) and βm
covering of A. Therefore
X
∆(E,r)∈βm
ϕ(r) ≥
X
ϕ1 (r) =
∆(E,r)∈βm
X
ϕ(r0 )
0
∆(E,r0 )∈βm
By taking infimum, we see that Cδϕm (A) = Cδϕm1 (A). Since A is arbitrarily chosen, the
conclusion follows.
33
Similarly, let ϕ2 be nondecreasing and ϕ ≤ ϕ2 by replacing all the left invisible
intervals of ϕ by a horizontal line connecting the endpoints. Then, we see that ϕ ≤ ϕ2 .
Given any set A and π, a δ−fine parking of A, if for some ∆(E, r) ∈ π, r ∈ (ar , br ),
where (ar , br ) is a maximal left invisible interval, which means not contained in any
other left invisible interval, then we see that ∆(x, ar ) ⊂ ∆(x, r). Let r0 = ar in this
case, otherwise, r0 = r. Define π 0 = {∆(x, r0 ) : ∆(x, r) ∈ π}, we get
X
∆(x,r)∈π
X
ϕ(r) ≤
ϕ2 (r) =
∆(x,r)∈π
ϕ
X
ϕ(r0 )
∆(x,r0 )∈π 0
ϕ2
By taking supreme, we see that P δ = P δ . Therefore, the conclusion follows.
Since both ϕ1 and ϕ2 are monotone, we can use the method mentioned before to
find the desired metric.
34
Chapter 5
The 1/n metric
In this chapter, we are dealing with the following metric
1
1 ρ(m, n) = − , m 6= n 6= 0,
m n
We will show that the covering gauge is ϕ(r) = 2−αr
ρ(m, 0) =
−1/2 +g(r)
1
m
with g(r) = o(r−1/2 ), α <
2.4.
Pick any A ∈ K, then for E ∈ ∆(A, r), let ξ(r) =
A
\
{0, 1, · · · , ξ(r)} = E
\
hq
1
4
+ 1r −
1
2
i
. We have
{0, 1, · · · , ξ(r)}
and no requirement for E ∩ {[r−1 ] + 1, [r−1 ] + 2, · · · }. So the only portion we are
interested in, with respect to a given r, is r−1/2 , r−1 ∩ E.
5.1
Classify K by the idea of missing number
It will be useful to represent H(S) as a subspace of the infinite binary sequences.
Let’s introduce the idea of “missing number”. For A = {an } ∈ K, we say A misses
the number m if ρ(m, n) ≥ r for all n such that n ∈ A, i.e, an = 1. We will explore
∆(A, r) first.
35
Due to the structure of the metric ρ, r is not easy to handle in this case. So, first
let us go back to the space S. Let
r
αk = r
−1/2
1 2
k
k r + k − , βk = r−1/2
4
2
r
1 2
k
k r+k+
4
2
For any n, if n ∈ (αk , αk+1 ), then ρ(n, n − k) < r < ρ(n, n − k − 1). Similarly, if
n ∈ (βk , βk+1 ), then ρ(n, n + k) < r < ρ(n, n + k + 1). For r small enough and
k ≤ r−1/3 /2, we would always have αk < βk < αk+1 . Let Ik = [βk , αk+1 ], then
|Ik | ∼
1
√
r−1/2 ,
2 k
and for any n ∈ Ik , we have {m : ρ(m, n) < r} = [n − k, n + k].
Thus, instead of looking at r, we only need to watch the pattern of missing numbers
in each Ik . One more thing to notice, between Ik−1 and Ik , there are only k slots.
Next, let us proceed to H(S). Fix certain k as in the above, I want to classify all
A’s by solely looking at an ’s with n ∈ Ik . Let Nk = |Ik |, and to simplify writing, we
will omit all k subindex below.
First of all, since we are in I, then to miss certain number, say m1 , we need an = 0
for |n − m1 | ≤ k. Otherwise, m1 ∈ ∆(n, r), contradiction.
Secondly, given m1 < m2 , if A misses only these two numbers out of the interval
[m1 − 1, m2 + 1], then m2 − m1 ≥ 2k + 2 with am1 −k−1 = am1 +k+1 = am2 −k−1 =
am2 +k+1 = 1.
Thirdly, given m1 < m2 and m2 − m1 ≤ 2k + 1, if A misses both, then A misses
all the numbers on [m1 , m2 ].
Now, let us look inside I. There are many closed subintervals of missed numbers,
disjoint in the sense that no left endpoint is the immediate successor of some right
endpoint. Then, for any two such consecutive intervals, if there is only one number
in between, say n, then n ∈ A. If there are more than 1 number in between, then we
36
only need the immediate follower of the first interval and the immediate predecessor
of the second interval to be included in A. Let ` count how many numbers are missed
P
by A, then we have j jϑ(j) = `, where ϑ(j) is the number of such subintervals with
P
exactly j integers inside. Denote β = j ϑ(j) and a = 2k + 1. Note that to miss j
consecutive numbers, we need at least j + 2k an ’s be equal to 0. So, the number of
slots where freedom is allowed in the interval I is less than
N−
X
X
(2k + j)ϑ(j) − (
ϑ(j) − 1) = N − aβ − ` + 1
j
j
and more than
N−
X
j
(2k + j)ϑ(j) − 2(
X
ϑ(j) − 1) = N − (a + 1)β − ` + 2
j
Also among these slots the length of consecutive 00 s is at most 2k as well.
One thing to notice, between Ik−1 and Ik , there are k free slots. So, when we
consider the numbers of different coverings, we should extend each Ik to [αk , αk+1 ]
to include these free slots and N + k will replace N in the calculation. But, since
k = o(N ) for r small enough, this would not change the later result in section 5.4.
Given A ∈ K, let {n1 , n2 , . . . , n` } be all the numbers missed by A, then
∆(A, r) ∩ K = {E ∈ K : ni ∈
/ E, i = 1, 2, · · · , ` and E does not miss n if n ∈ A}
Remark 5.1. For A 6= B with same missing numbers, denoted as {n1 , n2 , . . . , n` }.
1. Since σ(A, B) < R, we have A ∈ ∆(B, r) as well as B ∈ ∆(A, r).
37
2. ∆(A, r) 6= ∆(B, r). To see this, pick any m ∈ A∆B, then there is some E ∈ K
such that E misses m. This implies E ∈
/ ∆(A, r) ∩ ∆(B, r).
5.2
The Markov Process
To find µ ∆(A, r) for any A, r, we will need to count on the probability of those
intervals where some numbers are missing. Note that, to miss a number n in some Ik ,
A ∩ {n − k, · · · , n, · · · , n + k} = ∅. Therefore, if A does not miss any number from Ik ,
there are no 2k + 1 consecutive an ’s equal to 0. This corresponds to the probability
of flipping a coin m times without a consecutive heads. Therefore, let us detour a
little bit by looking at this Markov Process.
Fix a, and let p(m) be the above probability.

.5


.5

.
.
Let M be the transition matrix, then M = 
.


.5

0
.5 0 · · ·
0 0



0 .5 · · · 0 0 


..

.



0 0 · · · 0 .5

0 0 ··· 0 1
.
(a+1)×(a+1)
Here p(m) = 1−Mm (1, a+1), where M i (1, a+1) = P (there is consecutive 0’s with length a).Let
P (λ) be the characteristic polynomial of M , we have
i
(λ − 1) h
a+1
a
P (λ) =
(2λ)
− 2(2λ) + 1 ,
2λ − 1
z a+1 − 2z a + 1 z
let f (z) =
−1
z−1
2
We want to know the distribution of roots of f .
Claim. If a ≥ 3, then f has a − 1 roots inside |z| < 1 + a2 .
38
za − 1
z
− 1 (z − 2)
, then f (z) − g(z) = − 1.
2
2
a z − 1
2
a−2
Moreover, since both 1 +
are increasing on [3, ∞), for a ≥ 5,
and
a
2(a + 1)
Proof. Let g(z) =
z
|g|
|z a − 1|
a−2
= |z − 2|
≥
|f − g|
|z − 1|
2(a + 1)
2
1+
a
a
2
− 1 > 1, on |z| = 1 +
a
Then, by Rouche’s Theorem, in |z| < 1 + a2 , both f and g have a − 1 roots.
For the case a = 4, we can use direct calculation to prove the statement.
Claim. f has a real root inside the interval (2 − 2−a , 2 − 2−a−1 ).
Proof. Since z = 2 is another root of f , the only root left must be real. Denote it as
c. First of all, by the classical result on the stochastic matrix, we see that |c| < 2.
Secondly, since ca+1 −2ca +1 = 0, we have 2−c = c−a < 2−a . Thirdly, if c ≥ 2−2−a−1 ,
then ca+1 − 2ca + 1 = − (2 − c)ca − 1 ≥ − (2−a−1 2−a − 1) ≥ 12 , contradiction.
One more thing to notice is that, if we let P1 (λ) = (2λ)a+1 − 2(2λ)a + 1, we see
a
that P10 (λ) = 0 has only two roots: 0,
and neither of these is a root of P1 (λ)
a+1
itself. Moreover, P1 (1) 6= 0. Therefore, P (λ) has a + 1 distinct roots, which implies
that the matrix M is diagonalizable.
Denote all the

.5


.5

.
.
Let A = 
.


.5

.5
eigenvalues as λi with |λ1 | < |λ2 | < · · · < |λa | < λa+1 = 1.

 
.5 0 · · · 0
0

 





0
0 .5 · · · 0 

 

.
A u
.. ..

 .. .
then
M
=
where
u
=
.
.



 

 
0
1

 
0 0 · · · .5
0

 
0 0
0
.5
a×a
Direct calculation shows that {λ1 , λ1 , · · · , λa } are eigenvalues of A. Thus, A is also
T
diagonalizable and we can find an orthonormal J such that JAJ = D, where D is
39
a diagonal matrix with λi as the i-th diagonal entry. The i-th column of J, denoted
as ui , is the corresponding
eigenvector of λi .


J 1 
Let JM = 
 where 1 is a a × 1 matrix with all the entries are 1, while 0
0 1
is a 1 × x matrix with all the entries are 0. Not surprisingly, JM is

 the eigenvector
T
matrix of M with determinant 1. Direct calculation gives
ν(j) = −
P
i
J−1
M
J
=
0
ν
, where
1
uj (i) and
X
|ν(j)| ≤
|uj (i)|
sX
X
=
|uj (i)|2 +
|uj (i1 )||uj (i2 )|
i
i1 6=i2
s
i
X
Xh
≤
|uj (i)|2 +
|uj (i1 )|2 + |uj (i2 )|2
i
i1 <i2
√
= a
But since λa is very close to 1, more delicate calculation is need here. We use the
fact that µa is the eigenvector of A with respect to eigenvalue λa . First, we have
ν(a) =
ua (j) =
1−a
u (1).
2λa −1 a
1
2λa
On the other hand, we also could get
ua (1) 1
+ · · · + (2λa )n−j+1 ua (1) and thus 2 < |ua (j)| ≤ |ua (1)| for any j 6=
1. Since k ua k= 1, combining the previous two results, we have −ua (1)ν(a) ∈ [1, 2].
40


D
Note that M = JM 
 −1
 JM .
1
p(m) = 1 − Mm (1, a + 1) = −
X
ui (1)λm
i ν(i)
1≤i≤a
=
λm
a
!
X λi m
ui (1)ν(i)
ua (1)ν(a) +
λa 1≤i<a

m


≤ 1 − 2−a−2
2 + a3/2 2−m
m


≥ 1 − 2−a−1
2 − a3/2 2−m
5.3
Coverings
Let us go back to the definition of the covering measure to get some insight on ϕ(r).
We will see that ϕ(r) = 2−αr
−1/2 +g(r)
with g(r) = o(r−1/2 ), α < 2.4. The discussion is
in some given Ik .
But at first, we will evaluate a constant which will be useful later.
X
β!
Q
,
Let cβm =
P
P
j ϑ(j)!
ϑ(j)=β,
Claim.
cβm
jϑ(j)=m
m−1
=
.
β−1
Proof. Here cβm is the coefficient of xm in (x + x2 + · · · + xm )β .
We will prove the statement by double induction on β ≤ m.
First, when m = 2, there are only 2 cases.
If β = 1, then cβm = c12 = 1 = 10 . If β = 2, then cβm = c22 = 1 =
41
1
1
.
So, the statement holds.
Second, assume the statement is correct for all β ≤ m ≤ m0 , where m0 ≥ 2. Then
for m = m0 + 1, c1m0 +1 = 1 = m00 and c2m0 +1 = m0 = m10 . For any 3 ≤ β ≤ m0 + 1
cβm0 +1
=
m0
X
cβ−1
i
m0 −β+1 =1+
X
i=β−1
i=1
m0
m0
i+β−2
∗
=
=
m0 − β + 1
β−1
i
where the ∗ holds by using the equality
n
m
+
n
m+1
=
n+1
m+1
m0 − β + 1 times.
Next, we will deal with a measure 1 subset of H(S) and H(S) separately.
5.3.1
The Measure 1 set G
First, we need this claim to restrict ourself to a much small subinterval.
Claim. [9] P (limn Ln /log2 n ≤ 1) = 1.
Proof. Given any ε > 0 and small, P (`n ≥ (1 + ε) log2 n) ≤ 2−[(1+ε) log2 n]+1 ≤ n−(1+ε) .
P
P
Thus, n P (`n ≥ (1 + ε) log2 n) ≤ n n−(1+ε) < ∞, Borel-Cantelli Lemma implies
P (`n ≥ (1 + ε) log2 n infinitely often) = 0. Thus P (lim supn
Ln
log2 n
≤ 1 + ε) = 1.
On the other hand, still given any ε > 0 small, set k = [(1 − ε) log2 n], then there
are [n/k] blocks of length k on the line of integers from 1 to n, say, Ai . Note that
42
Ai = {`ki+1 ≤ k} and they are independent identically distributed.
P (Ln ≤ (1 − ε) log2 n) ≤ P (A1 ∩ · · · ∩ A[n/k] )
= P (A1 )[n/k]
= (1 − 2k+1 )[n/k]
n
ε−1
≤ exp −n (
− 1)
(1 − ε) log2 n
≤ exp(−nε / log2 n) < exp(−nε/2 )
where the last 2 inequalities hold for n big enough.
Thus
X
P (Ln ≤ (1 − ε) log2 n) =
X
exp(−nε/2 ) <
n
n
X
k!n−kε/2 < ∞
n
for well-chosen k satisfying kε > 2.
Borel-Cantelli Lemma again tells us P (Ln ≤ (1 − ε) log2 n infinitely often) = 0.
In other words, P (lim inf Ln / log2 n ≥ 1) = 1.
Applying the above Claim, we have
log2 n
→P 1
number of consecutive 0’s or 1’s starting from n
As a result, for r small enough, except for some zero-measure set, the longest con
secutive 1 string or 0 string on the interval n ∈ r−1/2 , r−1 has a length at most
p
2 log2 r−1 . But, for all n ≥ (r−1/2 + 1) · log2 r−1 , ρ(n, n − log2 r−1 ) ≥ r. So, we are
43
p
down to the interval Ie = r−1/2 , (r−1/2 + 1) · log2 r−1 and some measure 1 subset
of H(S).
Then we need an upper bound for `. Given ε > 0 small, according to the Strong
Law of Large Numbers, except for some zero-measure set, for any r small enough and
any A = {an }
#{an : an = 0, n ∈ I} 1 − < ε,
N
2
and
#{an : an = 1, n ∈ I} 1 − < ε
N
2
#{an : an = 0}
1 + 2ε
<
. We will use this ratio
#{an : an = 1}
1 − 2ε
here and denote G as the subset of H(S) without the above two measure 0 sets.
This implies in the entire interval I,
Let ` and β be as defined before. For those particular intervals which result in
missing certain numbers, we have
#{an : an = 0}
=
#{an : an = 1}
P
j
ϑ(j)(2k + j)
2kβ + `
=
> 2k + 1 ≥ 3
β−1
β−1
Therefore, to assume the validity of the Strong Law of Large Numbers, we see that
the number of slots with freedom in I should be O(N ). Actually, we will see later
such slots dominate I.
Let us say ` = cN , then again, by the Strong Law of Large Numbers, we have
2kβ + `
+
N
1
−ε
2
(a + 1)` − 1
1
1−
< −ε
N
2
This leaves us with two extreme cases.
Case 1, those missing numbers are as spread as possible, that is β = `. Plug in to
4ε
4ε
the above inequality, this results in c =
≤
.
a − 1 + 2(a + 1)ε
a−1
44
Case 2, those missing numbers are as crowded as possible. According to section
5.1, the length of consecutive 0’s in I can not surpass 2 log2 r−1 , therefore, we see
`
. Plug in again, we would still get the same estimate as in
that β ≤
2 log2 r−1 − 2k
the previous case.
4εN
Consequently, we conclude that, A only misses up to
numbers for almost
a−1
everywhere. Denote this upper bound by M .
Now we will show that, for G, the covering gauge is ϕ(r) = 2−(1+o(1))r
−1/2
.
Let us consider the 2r-fine cover β of G consisting of ∆(A, r)’s, where each A has
a different set of missing numbers. We have discussed at Remark 5.1 that this is
indeed a covering for G. But, how many such balls are there in β? We first focus on
p
the interval Ie = r−1/2 , (1 + r−1/2 ) log2 r−1/2 .
Fix a, note that N + k ∼ N , if we just consider the number of ∆(A, r)’s where A
misses ` numbers in Ik , it will not change the result. We have
(1 + o(1))
M X
`
X
X
`=1 β=1 {ϑ(j)}
≤(1 + o(1))
M X
`
X
Nβ
X
Q
{ϑ(j)}
`=1 β=1
=(1 + o(1))
N − aβ − ` + 2
β!
Q
β
j ϑ(j)!
1
j ϑ(j)!
M X
` X
`=1 β=1
` − 1 Nβ
β − 1 β!
M X
`
X
N β `β−1
≤ (1 + o(1))
β!(β − 1)!
`=1 β=1
Later, We will use this sum to show that this does not change our result.
N β `β−1 . N β−1 `β−2
N`
N
N
a−1
Note that
=
≥
≥
=
β!(β − 1)! (β − 1)!(β − 2)!
β(β1)
`
M
4ε
` `−1
N `
So, for given `, the sum is (1 + o(1))
.
(` − 1)!`!
45
M
X
N ` ``−1
We take sum with `, and it is (1 + o(1))
.
(` − 1)!`!
`=1
Taking the ratio again, we have
N ` ``−1 N `−1 (` − 1)`−2
N
/
=
(` − 1)!`! (` − 2)!(` − 1)!
`
1+
1
`−1
`−1
So, in extended Ik the number of variation is (1 + o(1))
≥
N
a−1
=
M
4ε
N M M M −1
.
(M − 1)!M !
Next, we take sum with k
log2 r−1
Y
(1 + o(1))
k=1

=(1 + o(1)) exp 
N M M M −1
(M − 1)!M !
log2 r−1
X εr−1/2 k −3/2 ln
k=1
2
e ·k
2ε
− ln εr−1/2 k
−3/2



3
=(1 + o(1)) exp 
εr−1/2 k −3/2 2 + ln k − ln(2ε) − ln ε − ln r−1/2 + ln k 
2
k=1


log2 r−1 h
i
X
−1/2
=(1 + o(1)) exp εr
2k −3/2 − k −3/2 ln(2ε) + k −3/2 ln k 

log2 r−1
X k=1
1+log2 r−1 ≤(1 + o(1)) exp εr−1/2 −k −1/2 (8 − 2 ln(2ε) + 2 ln k) 1
≤(1 + o(1)) exp − r−1/2 (2ε ln(2ε) − 8ε)
Note that we have no control over the interval [1, r−1/2 ], therefore the total number
of constituents is (1 + o(1)) exp r−1/2 ln 2 + 2ε| ln(2ε)| − 8ε .
If φ(r) = 2−f (r) with f (r)r1/2 = c+0(1) where c ∈ (1, ∞), then note that x ln(x) →
0 as x → 0, we can pick ε small enough such that 4ε| ln(2ε)| − 16ε ≤ (c − 1) ln 2. The
46
whole discussion stands when r is small enough and it results in
Cδφ (G) ≤ (1 + o(1)) exp r−1/2 ln 2 + 2ε| ln(2ε)| − 8ε ϕ(r)
c+1
≤ (1 + o(1)) exp
− f (r)r1/2 r−1/2 ln 2
2
c−1
−1/2
≤ (1 + o(1)) exp −
r
ln 2
4
Taking δ → 0, we would have C0φ (G) = 0.
For any G 0 ⊆ G, we can find a similar covering: given δ, for any r < δ/2, if A ∈ G 0
then ∆(A, r) includes all the elements of G 0 missing the same set of numbers. Let β 0
be the collection of all such ∆(A, r)’s, then β 0 is the desired δ-fine cover G 0 . Since β 0
contains no more elements than β, a similar argument shows that C0φ (G 0 ) = 0 as well.
This implies C φ (G) = 0, and φ is too large to be the gauge we are looking for.
If here f (r)r1/2 → ∞, then we can easily see that C0φ (G) = 0. In other words, if
the covering gauge is ϕ(r) = 2−(c+o(1))r
−1/2
, then we have c ≤ 1.
To complete the argument, let us show that c = 1:
Claim. Given c < 1, for any φc (r) = 2−cr
−1/2
, C0φ (G) = ∞. Therefore, φ is too small
to be the desired gauge.
Proof. Let β be any δ-fine cover of G.
Let {∆(Ai , ri )} ⊂ β be a minimal cover of G0 = {E ∈ G : [1, δ −1/2 ] ∩ E = ∅} with
δ > ri ≥ rj if i ≤ j. Then each Ai ∈ G0 as well. Therefore, for any E ∈
/ G0 , for any
−1/2
−1
B ∈ E, σ(A, B) ≥ δ and E ∩ ∆(Ai , ri ) = ∅. Note that 2−ri < µ ∆(Ai , ri ) < 2−ri ,
we have
X
i
φ(ri ) =
X
i
−1/2
−cri
2
>
X
−1/2
−ri
2
!c
!c
>
i
X
i
47
µ ∆(Ai , ri )
> µc (G0 ) = 2−cδ
−1/2
where the first inequality holds since c < 1.
Then, note that G has 2δ
Cδφ (G) ≥ 2δ
−1/2
−1/2
copies of G0 , we conclude that
· 2−cδ
−1/2
= 2(1−c)δ
−1/2
, and C0φ (G) = ∞
Combining the above two arguments, we see that the desired covering gauge for
this measure 1 set is ϕ(r) = 2−(1+o(1))r
5.3.2
−1/2
.
The whole space
Now, the focus moves to the whole K. Then, besides losing the upper bound for ` in
e
each Ik , we also have to go back to the original interval [r−1/2 , r−1 ] rather that just I.
In the following, we will find a specific covering, thus giving an upper bound for
α. To reach this goal, two different construction will be used.
First, we start with Ik ’s where k ≤ 3. Here, we will use the idea of missing numbers
as in the previous setting. Given Ik , the number of constituent in this covering is
N X
` X
`−1
N − aβ − ` + 2
1+
β−1
β
`=0 β=1
=1 +
N N −(a+1)β+2
X
X
β=0
`≥β
N − aβ − ` + 2 !
(` − 1)!
(` − β)!(β − 1)!β! N − (a + 1)β − ` + 2 !
N − aβ − ` + 2 !
(` − 1)!
, to estimate the sum,
(` − β)!(β − 1)!β! N − (a + 1)β − ` + 2 !
P
let us fix β and then compare ` B(`, β) first.
Let B(`, β) =
48
When β = 1, we have B(`, 1) = (N − a − ` + 2) for any ` ≥ 1. Since it decrease
P −a+1
while ` increasing, we see that N
B(`, 1) ≤ (N − a + 1)(N − a − ` + 2).
`=1
1 − β/(N − aβ − ` + 1)
B(` + 1, β)
=
. Note that this
When β > 1, we have
B(`, β)
1 − (β − 1)/`
ratio decreases as ` increases, so where this ratio reaches 1, we get max`≥β B(`, β).
PN −(a+1)β+2
(β − 1)(N − aβ + 2)
This happens at `β =
, and consequently `=1
B(`, β) ≤
2β − 1
(N − (a + 1)β + 2)B(`β , β).
There are two things worth noticing. First, N − aβ − `β + 2 =
β
`
β−1 β
and `β ∼
N + 2 − 2aβ 2
1
(N
−
aβ)
=
O(N
).
This
will
be
used
later.
Second,
`
−
`
=
.
β+1
β
2
2−1
4β
p
When β < (N + 2)/2a, `β increases as β increases. But, for bigger β, `β decreases
as β increases. Now we can compare the sums for different β.
p
Pick any β < (N + 2)/2a − 1, let b = `β+1 − `β , then we see that `β ∼
aβ = O(N ) and
1
2
N−
B(`β+1 , β + 1)
B(` + b, β + 1)
=
B(`, β)
B(`, β)
b−2
a+b−1
Y β−1
`β + b − 1 Y
β+1
1+
=
(N − aβ − β − `β + 1)
1−
β(β + 1) m=0
`β − β + 1 + m
N − aβ − `β − m + 2
m=0
=O(1)
`β (N − 3β − β − `β + 2)
β(β + 1)
≥O(N ) 1
p
(N + 2)/2a − 1 and to find the maximum
p
among all the β, we need to turn to the case β ≥ (N + 2)/2a − 1.
N + 2 − 2aβ 2
a
For bigger β, now `β decreases, so we let b = `β − `β+1 = −
≤ .
2
4β − 1
2
In below, we have two estimates, one for the upper bound, while another one for the
Therefore, B(`β , β) increases for β <
lower bound. According to these two estimates, when this ratio is close to 1, we need
49
β = O(`), which implies β = O(N ).
B(` − b, β + 1) (` − β) · · · (` − β − b)
1
=
B(`, β)
(` − 1) · · · (` − b) β(β + 1)
(N − aβ − β − ` + 2) · · · (N − aβ − β − ` − a + b + 2)
·
(N − aβ − ` + 2) · · · (N − aβ − ` + 3 − a + b)
b β
(` − β)(N − aβ − β − ` + 2) Y
1−
=
β(β + 1)
`−k
k=1
Y
β+1
·
0 1−
N − aβ − ` + k + 2
k=−a+b+1
(` − β)2
β2
(` − β)2
3β ` − 1 3(β + 1)
N − aβ − ` + 2
≥2 (1 + o(1))
exp − ln
−
ln
β2
2
`−b
2
N − aβ − ` − a + b + 3
2
3β(a − 2)
(` − β)
exp
−
≥ (1 + o(1))
β2
2`
≤1 (1 + o(1))
Here inequality 2 holds because ln(1 + x) ≥ x/2 for x ∈ [0, 1] and
p
` ≥ β = O( N/2a).
We will work on these two inequalities separately. Write N = cβ. We have
(β − 1)(N − aβ + 2)
c−a
`=
= (1 + o(1))
β.
2β − 1
2
(c − a − 2)2
According to the inequality 1, the ratio is
, which decreases as c de4
(x − a − 2)2
3(a − 2)
creases. Now we turn to the inequality 2, let f (x) =
exp −
,
4
x−a
we see that f (x) is also decreasing while x decreases. This implies, at some specific
c, this ratio reaches 1 and that is the place where B(`, β) reaches the maximum.
50
Let (`, β) be the pair corresponding to this c, then ` =
B(`, β) = (1 + o(1)) h
c−a
β and
2
`!(` − 1)!
i
(` − β)! 2β!(β − 1)!
`2`−2
β 2β−2 (` − β)2`−2β−1
2−2` 2β−1
2
c−a
= (1 + o(1))2πβ 1 −
−1
c−a
2
2 c−a−2
= (1 + o(1)) exp −(2` − 2) ln 1 −
+ (2β − 1) ln
c−a
2
−1/2 a−c
r
2
2 c−a−2
√
ln 1 −
+ ln
= (1 + o(1)) exp
c
c−a
c
2
2 k
= (1 + o(1))2π
To estimate the last formula, we let
ga (x) =
a−x 2 2 x−a−2
ln 1 −
+ ln
x
x−a
x
2
Study g 0 (x), we found that g 0 (x) is a decreasing function. Use the fact that ln(1+x) <
x, we see that
ga0 (a
+ 3) > 0 while
ga0 (5a
+ 2) < (3a + 2)
−2
1
+ 2 ln 2 − ln 4a < 0
2
Therefore, ga (x) reaches its maximum in the interval (a+3, 5a+2). Direct calculation
gives the corresponding maximums for each a and the upper bound 2(1+o(1)).61r
−1/2
for
the number of constituents of the covering in I1 ∪ I2 ∪ I3 .
Next, we move on to the rest of [r−1/2 , r−1 ].
√
Before starting the construction, we first look at the following interval (r−2/3 / 2−
r−1/3 /2, r−1 ], which is disjointed from all the Ik ’s. Since in this interval, k > r−1/2 /2,
51
r−1/2
√
2 k
< 2k. To better study this interval, we cut it into subintervals again.
1
Let ιk = 1 −
with k ∈ Z+ , and
3k
we have
I00
=
−ιk −ιk+1 −1
r−2/3 r−1/3 r−2/3
r
r
r
0
0
−1
√ −
,
, Ik =
,
, I∞ =
,r
2
2
2
2
6
2
ln r
.
−3 ln 3
Now, this following discussion is in both Ik ’s and Ik0 ’s except for I1 , I2 , I3 . The
We just need r−ιk+1 /2 < r−1 /6, which gives k <
idea for construction here is to find a specific A = {an }, where ∆(A, r) will play an
important role. And then, we will finish this part of covering by adding some other
constituents.
To construct this A, we need a sequence of integers, denoted as {n0 }.
r−2/3
r−1/3
First of all, by letting √ −
∈ A, we guarantee A does not miss any
2
2
r−2/3 r−1/3
is the only integer in {n0 } ∩ I00 .
number in I00 . Then √ −
2
2
Secondly, inside any Ik , we construct A in the following way:
We find {n0 } by requiring between any such two consecutive n0 s, the distance is
less than r. A direct calculation shows that, for each n ∈ I4 , ρ(n, n + 3) < 3ς −2 r,
1
1
1
where ς = 1 + + √ + √ Let this bound be r1 . Then we divide Ik and Ik0 into
2 2 2 2 3
subintervals with length r1 and resizing each subintervals accordingly to make sure
the endpoints are integers and the length is less than r. This is possible since the
choice of r1 guarantees that, for any n ∈ Ik , (n, r1 ) is properly contained in (n, r)
with at least 1 integer short on both side. These endpoints are in {n0 }.
Then according to remark 5.1.3, this will not only make A miss no number in Ik0 ,
but also, more importantly, for any B does not any such n0 , σ(A ∩ Ik0 , B ∩ Ik0 ) < r.
52
0
, we use the same idea by letting
Last, for I∞
1
1
1
1
1
0
= { r−1 , r−1 , r−1 , r−1 , r−1 }
{n0 } ∩ I00 = A ∩ I∞
2
3
4
5
6
0
This will allow A miss no number of I∞
.
Sum up, we have
√
∆(A, r) ⊃ {B : B does not miss r−2/3 / 2 − r−1/3 2, r−1 /2 and any n0 }
To complete the covering, we need a subcover for the other B’s.
Denote N0 as the set of {n0 } that B missed, and let A(N0 ) = A \ N0 , then
1. If n0 is inside some Ik0 or Ik and, in the corresponding subintervals where n0
served as endpoint, B does not miss any other numbers, including the other two
endpoints. Then because the length of either such interval is less than r, and both
the endpoints are in A, we see that A(N0 ) still does not miss n0 . But, restricted to
these two consecutive subintervals, we have σ(B, A(N0 )) < r.
2. If n0 is inside some Ik0 or Ik and B misses some m within the corresponding
intervals where n0 served as endpoint. Then B will miss all the numbers in [n0 , m]
since ρ(n0 , m) < r. Let n01 < n0 < n02 be the closet numbers to n0 in N0 such that B
does not miss any of them, then restricted to (n01 , n02 ),
{Numbers A(N0 ) misses} ⊆ {Numbers B misses}
This implies, restricted to (n01 , n02 ), σ(A(N0 ), B) < r.
53
Therefore, we found a covering here, consisting of A ∪ {A(N0 )}N0 . Here, the length
of this entire interval is from r− 1 to the left starting integer of I4 , that is
(1 + o(1))ρ r−1 , ςr−1/2 = (1 + o(1))ς −1 r1/2 − r
0
, we see that the size of {n0 } is
Sum up by k, and include the six n0 from I00 and I∞
less than
6+
ς −1 r1/2 − r
ς
= (1 + o(1)) r−1/2
r1
3
ς −1/2
So, the new cover has 2(1+o(1)) 3 r
elements.
By similar arguments as in the previous subsection, we see that α ≤ 1+0.61+
2.4.
54
ς
<
3
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55