Intersections of Deleted Digits Cantor Sets with Gaussian Integer Bases

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Vincent T. Shaw Wright State University

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Athesissubmittedinpartialfulﬁllment

of the requirements for the degree of

Master of Science

VINCENT T. SHAW

B.S., Wright State University, 2017

2020

Wright State University WRIGHT STATE UNIVERSITY

SCHOOL OF GRADUATE STUDIES

May 1, 2020

I HEREBY RECOMMEND THAT THE THESIS PREPARED UNDER MY SUPER- VISION BY Vincent T. Shaw ENTITLED Intersections of Deleted Digits Cantor Sets with Gaussian Integer Bases BE ACCEPTED IN PARTIAL FULFILLMENT OF THE RE- QUIREMENTS FOR THE DEGREE OF Master of Science.

______

Steen Pedersen, Ph.D.

Thesis Director

______

Ayse Sahin, Ph.D.

Department Chair

Committee on Final Examination ______Steen Pedersen, Ph.D. ______Qingbo Huang, Ph.D. ______Anthony Evans, Ph.D. ______Barry Milligan, Ph.D. Interim Dean, School of Graduate Studies Abstract

Shaw, Vincent T. M.S., Department of Mathematics and Statistics, Wright State University, 2020. Intersections of Deleted Digits Cantor Sets with Gaussian Integer Bases.

In this paper, the intersections of deleted digits Cantor sets and their fractal dimensions were analyzed. Previously, it had been shown that for any dimension between 0 and the dimension of the given deleted digits Cantor set of the real number line, a translate of the set could be constructed such that the intersection of the set with the translate would have this dimension. Here, we consider deleted digits Cantor sets of the complex plane with

Gaussian integer bases and show that the result still holds.

iii Contents

1 Introduction 1 1.1 WhystudyintersectionsofCantorsets? ...... 1 1.2 Prior work on this problem ...... 1

2 Negative Base Representations 5 2.1 IntegerRepresentations ...... 5 2.2 Fractals Obtained by Deleting Digits ...... 7

3 Gaussian Integers and Fractals 9 3.1 GaussianIntegerRepresentations ...... 9 3.2 Gaussian Fractals Obtained by Deleting Digits ...... 13

4 Preliminaries 15 4.1 Entropy Dimension of the Intersections ...... 15 4.2 Construction of Deleted Digits Cantor Sets ...... 17 4.3 Investigating T (T + x) ...... 20 \ 5 Proof of Theorem 21

6 Figures and Tables 23

7 Conclusion 32

iv List of Figures

6.1 T , b = 2+i ...... 23 0 6.2 T , b = 2+i, D⇤ = 0, 3 ...... 24 1 { } 6.3 T , b = 2+i, D⇤ = 0, 3 ...... 25 2 { } 6.4 T (T + 3) = , b = 2+i, D⇤ = 0, 3 ...... 26 0 \ 0 ; { } 6.5 T (T +0.3) ...... 27 1 \ 1 6.6 T (T +0.30) ...... 28 2 \ 2 6.7 T (T +0.33) ...... 29 2 \ 2

v List of Tables

6.1 b = 2+i, D⇤ = 0, 3 ...... 30 { } 6.2 b = 3+i, D⇤ = 0, 6 ...... 30 { } 6.3 b = 4+i, D⇤ = 0, 8 ...... 31 { }

vi Acknowledgements

First and foremost, I would like to thank my research advisor Dr. Steen Pedersen for providing me the opportunity to study mathematics under his guidance and leadership. The past two years working with him have given me a new found appreciation for those that contribute to ﬁeld of mathematics. It was only through his encouragement, and profound patience that this thesis was possible. Also, special thanks is in order for Jason D. Phillips for his coauthorship to the predecessor of this paper that laid the foundation for our research.

Thank you to both Dr. Qingbo Huang and Dr. Anthony Evans for reviewing my thesis and providing feedback.

I would also like to take a moment to thank the following individuals that had a positive impact on my personal development as a scholar: Dr. Ayse Sahin, Dr. Ann Farrell and

Peggy Kelly in helping me succeed as a Graduate Teaching Assistant. Thank you to both

Dr. Christopher Barton and Dr. Mateen Rizki for taking me on as a student in their

MatLab course that proved to be an invaluable resource in conducting my research. Finally,

I would like to thank my friend and colleague Joe Sjoberg for rekindling my interest in pursuing a degree in mathematics, as well as my fellow GTA’s for making the last two years an enjoyable and worthwhile expreience.

vii viii Chapter 1

Introduction

1.1 Why study intersections of Cantor sets?

Cantor sets may appear to be rather special. However, they occur in mathematical models of many naturally occuring objects, play a role in number theory, in signal processing, in er- godic theory, and in limit-theorems from probability. We study the “size” of the intersection of two Cantor sets. The signiﬁcance of this problem was noted by Furstenberg [Fur70] and

Palis [Pal87]. Papers dealing with versions of our problem and related applications include:

[Wil91], [Kra92], [PS98], [Kra00], [LN04], [DT07], [DHW08], [DT08a], [ZLL08], [DLT09],

[KLD10], [LYZ11a], [Mor11], [ZG11], [LTZ18], [SRM+20]. The literature in the subject, neighboring areas, and applications is vast. In the list above, we limit ourselves to a small sample of the literature closely related to our problem.

1.2 Prior work on this problem

Most of the prior work related to our problem has been done in the one dimensional case, so we will begin with a summary of that case.

1 Let n 3 be an integer. Any real number t [0, 1] has at least one n-ary representation 2

1 t t =0. t t = k n 1 2 ··· nk kX=1 where each t is one of the digits 0, 1,...,n 1. Deleting some element from the full digit k set 0, 1,...n 1 we get a set of digits D := d k =1, 2,...,m with m

1 xk C = Cn,D := k xk D for all k N . (1.1) ( n | 2 2 ) kX=1

We are interested in the dimension of the sets C (C + t) , where C + t := x + t x C . \ { | 2 } Since the problem is invariant under translation we will assume d1 =0. We say that D is uniform, if d d ,k=1, 2,...,m 1 is constant and 2. We k+1 k say D is regular, if D is a subset of a uniform digit set. Finally, we say that D is sparse, if

0 2 for all = 0 in | | 6

:= D D = d d d ,d D . { j k | j k 2 }

Clearly, a uniform set is regular and a regular set is sparse. The set D = 0, 5, 7 is sparse { } and not regular. We abuse the terminology and say Cn,D is uniform, regular, or sparse provided D has the corresponding property.

Previous studies of the sets C (C + t) include: \

When C = C3, 0,2 is the middle thirds Cantor set, a formula for the Hausdorﬀ • { } dimension dim (C (C + t)) of C (C + t) can be found in [DH95] and in [NL02]. \ \ Such a formula can also be found in [DT08b] if C is uniform and d = n 1,andin m [KP91] if C is regular. For sparse C, aformulacanbefoundin[PP13].

Let F + be the set of all t 0 such that C (C + t) is non-empty. For 0 1, let • \   F := t F + dim (C (C + t)) = log (m) . If C is the middle thirds Cantor set { 2 | \ n } + + then F =[0, 1] and it is shown in [Haw75, DH95, NL02] that F is dense in F for

2 all 0 1. This is extended to regular sets and to sets C such that D satisﬁes   n,D d d 2 and d

It is shown in [Haw75, Igu03] that, if C is the middle thirds Cantor set, then the • Hausdorﬀ dimension of C (C + t) is 1 log (2) for Lebesgue almost all t in the closed \ 3 3 interval [0, 1] . This is extended to all deleted digits sets in [KP91].

If C is the middle thirds Cantor set, then C (C + t) is self-similar if and only if the • \ 2yk sequence 1 y is “strong periodic”, where t = 1 k and y 1, 0, 1 for { | k|} k=1 3 k 2 { } all k by [LYZ11b]. This is extended to sparse C in [PP14].P In particular, C (C + t) \ is not, in general, a self-similar set.

Some of the cited papers only consider rational t and some consider Minkowski dimension in place of Hausdorff dimension. It is known, see e.g., [PP12] for an elementary proof, that the (lower) Minkowski dimensions of C (C + t) equals its Hausdorff dimension. \ Palis [Pal87] conjectured that for dynamically defined Cantor sets typically the corresponding set F + either has Lebesgue measure zero or contains an interval. The papers

[DS08], [MSS09] investigate this problem for random deleted digits sets and solve it in the aﬃrmative in the deterministic case.

In two dimensions, where C are certain Sierpinski carpets, the dimension of C (C + t) \ is investigated in [LTZ18] and the dimension is calculated for certain translation vectors t. In [DT07] the authors consider the intersection of a Sierpinski carpet with its rational translates. They use the methods from and obtain results similar to the ones in [NL02].

Both in [LTZ18] and [NL02] the authors study the case where the base is a real number and the “digits” are vectors parallel to the coordinate axes.

In this work, we study the case where the base is a complex number of the form n i ± for an integer n>1 and the set of digits is a proper subset of the set 0, 1,...,n2 .

In Chapter 2 we discuss the one dimensional case, in particular, we show that the case where we use a negative integer as a base does not lead to new results.

In Chapter 3 we show that if we want to use a Gaussian integer as the base and have a

3 digit set of the form 0, 1, 2,...,N ,thenwemusthaveb = n i for some positive integer { } ± and N = n2.

In Chapter 4 we begin to investigate the set C (C + t) and its dimension, in particular \ we identify appropriate conditions on the subset of 0, 1, 2,...,N that can serve as a good { } set of digits.

In Chapter 5 we show that for any 0 ↵ 1, there is a complex number t, such that   C (C + t) has dimension ↵ dim (C) . \

4 Chapter 2

Negative Base Representations

2.1 Integer Representations

Suppose b is an integer such that b 2. Let D = 0, 1, 2,..., b 1 . For d D, we say | | { | | } k 2

M k n = dkb kX=0 is a representation of n in base b with digit set D. When b = 10, 3, 2 this gives the usual decimal, ternary, and binary representations. Here we are also interested in the case where b 2. The case of negative bases is well known, see e.g., [Knu98] for the n =2case, the  proof below is essentially taken from [GG79].

Proposition 1. If b 2, then every integer has a unique representation in base b with  digit set D.

Proof. Existence follows from the usual division algorithm. For a positive base, it is clear that the sequence of quotients is decreasing, hence the division algorithm terminated. Here termination of the division algorithm requires an argument. Suppose n is an integer, positive or negative. Let q = n and inductively q = q b + d , where 0 d < b . 0 k k+1 k  k | | q d q (i) If q > 0 and q d then q = k k k < q . k k k, | k+1| b  b | k| 5 q d d (ii) If q > 0 and q

n = d0 + q1b

= d0 +(d1 + q2b) b

2 = d0 + d1b +(d2 + q3b) b

= ··· = d + d b + d b2 + +(d + q b) bM 0 1 2 ··· M M+1 M k = dkb . kX=0 Thus every interger has a representation in base b.

Suppose d D, e D, and M d bk = M e bk. Then k 2 k 2 k=0 k k=0 k P P M k 1 b (d e ) b = e d k k 0 0 kX=1 so the left hand side is an integer divisible by b and the right hand side is an integer < b . | | | | Hence, the right hand side e d =0. Therefore d = e . 0 0 0 0 M k 1 M k 1 Using d0 = e0 it follows that k=1 dkb = k=1 ekb . Repeating the previous M k 2 M k 2 argument shows d1 = e1 and k=2 Pdkb = k=2 ePkb .

Continuing in this mannerP we see that dk P= ek for all k =0, 1,...,M.

6 2.2 Fractals Obtained by Deleting Digits

Fix an integer b > 1 and let D = 0, 1,..., b 1 as above. Let f (x)= 1 (x + j) for | | { | | } j b j D. 2

Suppose b>1. Let ↵ =0and =1, then j D fj [↵, ]=[↵, ] . In fact, fj [0, 1] = 2 j j+1 b 1 j j+1 0 b 1+1 b , b , so j D fj [↵, ]= j=0 b , b = Sb , b . If D⇤ is a subset of D, then 2 ⇥ ⇤ ⇥ ⇤ ⇥ ⇤ the compactS non-empty set Cb,DS ⇤ = CD⇤ satisfying j D fj (CD⇤ )=CD⇤ is the deleted 2 ⇤ k digits Cantor set C = 1 d b : d D⇤ . InS [PP12] the authors studied the fractal D⇤ k=1 k k 2 dimensions of the intersectionsP of CD⇤ with its translates. b 1 Suppose b< 1 and let D = 0, 1,..., b 1 as above. Let ↵ = 1 b and = 1 b , { } then ↵ < 0, > 0, and j D fj [↵, ]=[↵, ] . In fact, ↵ =1, f0 [↵, ] is an interval of 2 1 1 length b , and fj [↵, ] isS obtained from fj 1 [↵, ] by a translation by b . Hence the intervals fj [↵, ] and fj 1 [↵, ] intersect at endpoints, and we just need to check that the right hand endpoint f0 (↵) of f0 [↵, ] is and the left hand endpoint f b 1 () of f b 1 [↵, ] is ↵. Now b 1 b 1 f (↵)=f = = = 0 0 1 b b · 1 b 1 b ✓ ◆ and

1 1 1 b 1 b f b 1 ()=f b 1 = + = = ↵ 1 b b · 1 b b 1 b ✓ ◆

Again, if D⇤ is a subset of D, then the compact non-empty set Cb,D⇤ = CD⇤ satisfying

j D fj (CD⇤ )=CD⇤ is the Cantor set 2 ⇤ S

1 k CD⇤ = dkb : dk D⇤ . ( 2 ) kX=1 In this case, the reﬂection caused by b being negative means we cannot simply copy iterated function systems arguments from [Gau32] when studying fractal dimensions of the intersection of CD with its translates. However, it is easy to see that as a set C b ,D can be ⇤ | | ⇤ b | | obtained by translating the set C b ,D⇤ by 1+ b . For any integer b, such that b > 1, the | | | | | |

7 union of translated k (CD + k) is a tiling of the real line. 2Z S

8 Chapter 3

Gaussian Integers and Fractals

3.1 Gaussian Integer Representations

The set of Gaussian integers is the set Z [i]= x + iy : x, y Z . To extend the study of the { 2 } dimension of the intersection of a deleted digits Cantor set with its translates we will study the case where the base b is a Gaussian integer and our Cantor set is a subset of the complex plane determined by a set of digits as above. We consider a digit set D = 0, 1, 2,...,N { } and a Gaussian integer b = m + in. We say a Gaussian integer z has a representation in base b, with digit set D, if z = M d bk, with d D. In analogy with Proposition 1, k=0 k k 2 the ﬁrst question is: Can we chooseP m, n, N such that every Gaussian integer has a unique representation in base b = m + in with digit set D = 0, 1, 2,...,N ? { } We begin by showing that a necessary condition for every Gaussian integer to have a representation in base b = m + in, is that the imaginary part of b has modulus one.

Lemma 2. If every Gaussian integer has a representation in base b = m + in, with digits set D = 0, 1, 2,...,N , then n = 1. { } ±

Proof. Since (u + iv)(x + iy)=ux vy + i (uv + vx) , it follows by induction that for any integer k, n is a factor of y, when bk = x + iy where n divides y in Z. Consequently, if M k M k z = k=0 dkb , then n divides the imaginary part of z in Z. By assumption i = k=0 dkb , P P

9 for some dk D, so n divides 1 in Z. Thus n = 1. 2 ±

Our next result establishes a connection between representations in base b and congruence classes modulo b.

Lemma 3. If every Gaussian integer has a unique representation in base b, then every

Gaussian integer is congruent to exactly one element of D, that is D is a complete set of representatives for Z [i] / (b) .

M k 1 Proof. Let z be a Gaussian integer. By assumption z = d + b d b ,hencez d 0 k=1 k ⌘ 0 (mod b). So, if every Gaussian integer has a representation in baseP b, then every Gaussian integer is congruent to some element of D.

M k M 0 k Uniqueness of representation means, if k=0 dkb = k=0 dk0 b , then dk = dk0 for all k.

Where if MM0, we set k  d0 =0for all M 0

D. So suppose z is a Gaussian integer and z is congruent to d D and to e D. Then 2 2 z = d + bu and z = e + bv for some Gaussian integers u, v. Now v u is a Gaussian integer, so v u = M d bk for some d D. Hence, k=0 k k 2 P M d = e + b (v u)=e + d bk+1.. k kX=0

By uniqueness, d = e and dk =0for all k.

Due to Lemma 3 we ask: For a Gaussian integer b = m + in, does there exist an N, such that D = 0, 1, 2,...,N is a complete set of representatives for the congruence classes { } Z [i] / (m + in) . Gauss showed that if m, n are relatively prime the answer is aﬃrmative. We reproduce Gauss’ argument below.

Lemma 4 ([Gau32, Theorem 40]). If b = m + in is a Gaussian integer and m, n are relatively prime, then D = 0, 1, 2,...,m2 + n2 1 is a complete set of representatives for the congruence classes Z [i]/ (b) .

10 Proof. Since m, n are relatively prime, there are integers ↵, , such that ↵m + n =1. Note

↵n m + b ( + i↵)=↵n m +(m + in)( + i↵) = ↵n m + m +(m↵ + n) i n↵ = i.

Hence, if x + iy is any Gaussian integer, then

x + iy = x +(↵n m) y + b (y + i↵y) . (3.1)

Let h, k be integers such that x +(↵n m) y = h + k m2 + n2 and 0 h

x +(↵n m) y = h + bk (m in) , (3.2) since b (m in)=m2 + n2. Combining Eqns. (3.1) and (3.2) we see

x + iy = h + b (y + km + i (↵y kn)) .

Thus x + iy h (mod b). ⌘ To complete the proof we need to show that no Gaussian integer is congruent to more

2 2 than one element of D. Suppose x+iy is a Gaussian integer, h, h0 D = 0, 1, 2,...,m + n 1 , 2 x + iy h (mod b), and x + iy h0 (mod b), then h h0 = b (u + iv) for some Gaussian ⌘ ⌘ integer u + iv. Hence,

2 2 (h h0)(m in)= m + n (u + iv) . So

2 2 2 2 (h h0) m↵ = u↵ m + n and (h h0) n = v m + n .

11 Adding these equations and using ↵m + n =1it follows that

2 2 h h0 =(u↵ v) m + n . 2 2 Since h, h0 0, 1, 2,...,m + n 1 and u↵ v is an integer, h = h0. 2 Recall, we are interested in when every Gaussian integer has a unique representation in base b = m+in using digit set D = 0, 1, 2,...,N . By Lemma 2, we may assume b = m i. { } ± and by Lemma 3 and Lemma 4, we may assume D = 0, 1, 2,...,m2 . To complete this line of reasoning we need to go back to the original question of representing every Gaussian integer uniquely in base b = m + i with digits in D = 0, 1, 2,...,m2 . Lemma 5. Suppose b = m + i, D = 0, 1, 2,...,m2 , and every Gaussian integer has a unique representation in base b with digit set D, then m<0.

Proof due to [KS75]. The proof is by contradiction. If m =0, then D = 0 and we can { } only represent 0 in base b with digit set D. 2 If m>0. Suppose 1 m + i = m d bk. Multiplying by 1 b =1 m i we get k=0 k P m2 2 k m2+1 (1 m) +1=d0 + (dk dk 1) b dm2 b . (3.3) kX=1

The left hand side m2 2m +2is in D and the right hand side is d (mod b), so using ⌘ 0 Lemma 3 and Lemma 4 we conclude m2 2m = d . So, by Eqn. (3.3), 0

m2 k 1 m2 (dk dk 1) b dm2 b =0. kX=1 Either d d or d d is in D, so we can repeat the argument to get d = d . Continuing 1 0 0 1 1 0 in this manner we get

2 dk = dk 1, for k =1, 2, 3,...,m , and dm2 =0.

Consequently, d0 = d1 = d2 = = dm2 1 = dm2 =0. Thus, we get the contradiction ···

12 (1 m)2 +1=0, from Eqn. (3.3).

We have shown, that if m, n, N are integers, b = m + in,andeveryGaussianinteger has a unique representation in base b with digit set D = 0, 1, 2,...,N , then m<0, { } n = 1,andN = m2. Converserly, it is known [KS75], that if n>0 is an integer, then ± every Gaussian integer has has unique representation in base b = n i with digits form ± D = 0, 1, 2,...,n2 . 3.2 Gaussian Fractals Obtained by Deleting Digits

Fix an integer n>0 and let D = 0, 1,...,n2 . Then every Gaussian integer has a unique representation in base b = n + iwith digits set D. Every complex number is within 1 p2 [i] 1 of some Gaussian integer Z .Henceeverycomplexnumberiswithinp2 of some complex number of the form m d bk,d D. Now, let z be a complex number and pick a k=0 k k 2 m k 1 1 m k m 1 j Gaussian integer k=0Pdkb ,dk D within p of bz. Then b k=0 dkb = j= 1 dj 1b is 2 2 1 z.P z, P m Pd bk,d D, within b p2 of Similarly, given any complex number we can ﬁnd k= l k k | | 2 1 z. P within b lp2 of It follows that every complex number has at least one representation of | | the form m m k k 1 k z = dkb = dkb + d kb ,dk D. 2 k= k=0 k=1 X1 X X m We say that z has radix expansion (or equivalently radix representation) z = k= = 1 dm ...d1d0.d 1d 2 ....Whenbaseb = n + i,theradixexpansionofz does notP share the same properties when b Z.Forexample,ifb = 10, it is well known that the geometric 2 1 k series 0.999 = k= 9(10) =1.But,ifb = 2+i with digits set D = 0, 1, 2, 3, 4 , ··· 1 { } 1 k 2 then 0.444 = Pk= 4( 2+i) = 5 ( 3 i) =1. Nevertheless, we will use radix ··· 1 6 representation withP Gaussian integer base b where convenient. We are particularly interested in the set of complex numbers whose radix expansions have integer part 0.So,thedeﬁne the set

1 k T0 = d kb : dk D . ( 2 ) kX=1 This takes the place of the unit interval. See Figure 6.1 of T0 when n =2.

13 We deﬁne some terms that will be useful in proving upcoming lemmas that will be necessary in proving the main theorem. First, let D⇤ = d ,d ,...,d D denote a set of { 1 2 t} ⇢ 2 2 at least two integers such that 0=d1

Deﬁnition 6. We say that D⇤ satisﬁes the separation condition if d d 2, for all i = j | i j | 6 if b = 2+i,or d d 2n, for all i = j if b = n + i, for all n 3. | i j| 6

We then deﬁne our analagous Cantor set in the complex plane.

Deﬁnition 7. The set T = T = z T d D⇤ is a deleted digits Cantor set b,D⇤ { 2 0 | i 2 } with Gaussian integer base b.Here,T is obtained from T0 by restricting attention to those representations that only contain digits from the set D⇤.

We wish to reﬁne the set T0 to construct the deleted digits Cantor set T in the same way [PP12] constructed their deleted digits Cantor set. For that we have the following deﬁnition.

Definition 8. For all k 0,thesetT = 0.d d ... d D⇤, i k is called a refinement k { 1 2 | i 2 8  } of T0 at the kth stage.Here,Tk is obtained from T0 by restricting attention to those representations whose first k digits past the radix point are from the set D⇤.

See ﬁgures 6.2 and 6.3 for reﬁnements T1 and T2 when n =2. The fractal set T0 is self- similar, meaning that T0 has the same shape as one or more of its parts. [Fal85] refers to these parts as similitudes. In [PP12], similitudes of the unit interval were called subintervals.

Hence, it is natural to introduce the following term.

k k k k Definition 9. Let b T0 = b x x T0 ,andb T0 + x = z + x z b T0 , x C. | 2 | 2 2 k k i Then a subtile of Tk refers to a subset of Tk of the form b T0 + d ib for fixed i=1 d i D⇤. P 2 In order to prove the theorem, we must first turn our attention to the entropy dimension of a subset of Rn.Tocalculatetheentropydimensionofintersections,weneedtoextend the definition of possible covering sets.

14 Chapter 4

Preliminaries

4.1 Entropy Dimension of the Intersections

In this chapter, we set up some necessary foundation that will aid in the proof of the theorem. Due to the structure of T ,westudyitsentropy dimension (also called box-counting dimension, Kolmogorov dimension,orMinkowski dimension.)

Deﬁnition 10. Let E Rn such that E is nonempty. Then the entropy dimension of E is ⇢ deﬁned as log N (E) dim E =lim log !1 where N(E) denotes the smallest number of sets each of diameter at most needed to cover E.

The lower and upper entropy dimensions are deﬁned respectively as follows:

log N(E) dimE =lim !1 log

log N(E) dimE = lim . !1 log If the limit does not exist, then we can talk about the upper and lower entropy dimensions

15 obtained by replacing the limit by the limit superior and limit inferior, respectively. [Fal85] considers various deﬁnitions of N(E),namely

the smallest amount of closed balls of radius that cover E; •

the least amount of cubes of side length that cover E; •

the number of mesh cubes that intersect E; •

the smallest number of sets of diameter at most that cover E; •

the largest number of disjoint balls of radius with centers in E. •

j We extend the list to include the smallest number of subtiles of T0;setsoftheformb T0 +

0.d1d2...dj for some integer j,thatcoverthesetE.

Lemma 11. If E Rn such that E is nonempty, then the entropy dimension of E is ⇢ log N (E) lim where N (E) is the smallest number of subtiles of diameter at most !1 log needed to cover E.

Proof. Denote the diameter of T0 as diam(T0).LetNj (E) denote the smallest number of subtiles of T each of diameter from a decreasing sequence 1 , where = 0 j { j}j=0 j j 2 pn +1 diam(T0). Let Lj (E) be the smallest number of closed balls of radius j needed to cover E.Obviously,L (E) N (E). Let M (E) be the smallest number of mesh j  j p2 log Mp2(E) cubes that intersect E. Now, Mp (E) Nj (E). If p2j < 1, then 2  log(p2)  log Nj (E) .Asj 0 when j , we get the following, log(p2) log() ! !1

log Nj (E) dimE =lim 0 j ! log j and

log Nj (E) dimE lim 0 .  j ! log j Any set of diameters at most is contained in 32 =9 mesh cubes. Therefore, we get j j N (E) 9M (E) and taking logarithms leads to the opposite inequalities of the equations j  j

16 above. As a result, we can calculate the entropy dimension using the subtiles of Tk. As

2 vector spaces, R is isomorphic to C. So if T C is deﬁned as above and N(T ) is the ⇢ minimum number of subtiles in T = z T d D⇤, i k needed to cover T ,then, k { 2 0 | i 2 8  } by [Fal85], the entropy dimension of T is equal to log 2 D⇤ . pn +1 | |

Let

F = x C T (T + x) = { 2 | \ 6 ;} be the set of all x C such that the intersection of T and its translate by x, T + x = 2 z + x z T ,isnon-empty.Weconsiderallx in F such that T (T + x) has entropy { | 2 } \ dimension ↵ log 2 D⇤ where 0 ↵ 1. Under appropriate assumptions, we show pn +1 | |   that for any ↵ [0, 1],thereexistsx F such that dim (T (T + x)) = ↵ dim T. 2 2 \

4.2 Construction of Deleted Digits Cantor Sets

In this section, we consider how the self-similarity construction of T can be used to study

T (T + x). \ Let b = n + i Z[i] and n N 1 . Let D = 0, 1,...,n2 be the digit set. Let 2 2 { } D⇤ = d ,d ,...,d D be a set of at least two distinct integers satisfying the separation { 1 2 t} ⇢ condition. Let T0 and Tk be deﬁned as above. The set Tk+1 is obtained by reﬁning the set

Tk;thatis,byremovingcomplexnumbersinTk with digits in the k+1th position to the right

k of the radix point that are not in D⇤. If we take b T0 = 0.0...0d k 1d k 2... di D { | 2 } k and consider its translates b T0 +x = 0.0...0d k 1d k 2... + x di D ,x C,thenwe { | 2 } 2 k can write Tk = b T0 +0.d 1...d k .Whatthismeans,isthatforeachk,thesetTk d D i2 ⇤ k S 1 consists of D⇤ subtiles each the area of T . It should be clear that T T | | (n2 + 1)k 0 k ⇢ k+1 for all k,andthatT = k1=0 Tk. In order to use subtilesT in calculating the entropy dimension of T ,weneedtoshowthat the subtiles of Tk are pairwise disjoint. To prove this, we require the following lemma.

Lemma 12. Let T0 be given. It follows that

17 i) if n =2and D⇤ satisﬁes the condition d d 2 for all i = j, then T (T +d)= | i j| 6 0 \ 0 ; for d =3, 4;

ii) if n 3 and D⇤ satisﬁes the condition d d 2n for all i = j, then T (T +d)= | i j| 6 0 \ 0 for all d D⇤ 0 . ; 2 { }

2 Proof. Let z = n D. Denote c =0.sz where s D is ﬁxed. Then c represents the d 2 e2 s 2 s 1 “center” of the subtile b (T + s) T as 0 ⇢ 0

n2 1 2 1 2 n j n j 2 x cs = (xj )b b = d e = ✏ | | d 2 e  d 2 e | | n2 +1 pn2 +1 j=2 j=2 X X 1 1 for all x b (T0+s). Denote B✏(cs) as the ball centered at cs with radius ✏.Asb (T0+s) 2 n2 n2 ⇢ 1 B✏(cs),wegetT0 B✏(cs).Similarly,b (T0 +t)+d B✏(ct +d) and T0 +d B✏(ct + ⇢s=0 ⇢ ⇢t=0 d) where t D is ﬁxed.S For any t, s =0, 1,...,n2, c (c +d) = 0.sz d.tz = d.S(s t) is 2 | s t | | | | | the distance between the centers of the balls B✏(cs) and B✏(ct +d). Therefore, it is suﬃcient to show that the ball centered at cs with radius ✏ is disjoint from any ball of radius ✏ centered at c + d;thatis,B (c ) B (c + d)= . Figure 6.4 provides a visual representation when t ✏ s \ ✏ t ; b = 2+i,andD⇤ = 0, 3 . { } The proof reduces to show that d.(s t) > 2✏. Algebraically, we get | |

1 d.(s t) = d +0.(s t) = d +(s t)b | | | | | | (s t) (s t)( n i) (s t)n (s t) = d + = d + = d + + i n + i n2 +1 n2 +1 n2 +1 ✓ ◆ ✓ ◆ (s t)( n i) (s t)n (s t) = d + = d + + i n2 +1 n2 +1 n2 +1 ✓ ◆ ✓ ◆ 1/2 (s t)n 2 (s t) 2 2dn 1 = d + + = d2 (s t)+ (s t)2. n2 +1 n2 +1 n2 +1 n2 +1 ✓ ◆ ✓ ◆ ! r

2 2 n n2 +1 Also, 2✏ = d 2 e . We will take a closer look at small n2 +1 pn2 +1  n2 +1 pn2 +1 values of n.

2 2 5 12 If n =2, then D⇤ = 0, 3 ,or 0, 4 and we have (2✏) =4✏ < 9 (s { } { }  5 p5 5

18 t)+ 1 (s t)2 = d.(s t) 2 . Table 6.1 compares 2✏ with 3.(s t) for all possible s and t 5 | | | | values. Note that 2✏ < 3.(s t) 4.(s t) .Takingthesquarerootyieldsthedesired | |  | | result. d D if d 2n If 3 n<5,thenD⇤ = d1,d2 where di = 8 2 . The reason di =  { } > <>0 if i =1

dn 1 > 2 d 2n is because d2 2 (s t)+ (s:> t) pd2 2dn +0+1 1 .So, n2 +1 n2 +1 2 r ⇡ 2n + p4n2 +3 d2 2dn 3 > 0 implies d = which leads to d 2n. Clearly, 4 2

n2 +1 dn 1 (2✏)2 =4✏2

2✏ and 6.(s t) when n =3.Again,notethat2✏ < 6.(s t) d.(s t) for all digits | | | |  | | d>6.Table6.3compares2✏ and 8.(s t) when n =4.Again,notethat2✏ < 8.(s t) | | | |  d.(s t) for all digits d>8. | | If n 5, and D⇤ satisﬁes the separation condition, then we have,

n2 +1 dn3 dn(s t) 1 (2✏)2

Now, for an important corollary.

k k 0 0 Corollary 13. If b T0 +0.d 1...d k and b T0 +0.d 1...d k are subtiles of Tk and d i = 6 0 d i for at least one i, then the subtiles are disjoint.

k k 0 0 0 Proof. Assume b T0 +0.d 1...d k and b T0 +0.d 1...d k are subtiles of Tk and d i = d i 6 i k i k for at least one i. Then b b T0 +0.d 1...d k = b T0 + d 1...d i.d i 1...d k and i k 0 0 i k 0 0 0 0 0 0 b b T0 +0.d 1...d k = b T0 + d 1d 2...d i.d i 1...d k.Notethatd j = d j for all ⇣ ⌘ 0 i k jd.Bytranslating,wegetb T T and i i 0 ⇢ 0

19 k i k 0 j k 0 b T0+ (d j d j)b T0 +(d j d j).Bytheresultoflemma12,theintersections j=i ⇢ of these twoP subtiles are disjoint.

4.3 Investigating T (T + x) \

Assume D⇤ satisﬁes the separation condition. Consider the translation T + x = z + x z { | 2 T,x F .Realizethat 2 } 1 T (T + x)= (T (T + x)). \ k \ k k\=0 For x =0.d d d ... let x denote the truncation of x to the ﬁrst k places; i.e., 1 2 3 b ck

x =0.d d d ...d . b ck 1 2 3 k

We will investigate T (T + x ),sincebothT and T + x consists of subtiles. k \ k b ck k k b ck Remember a subtile in Tk is short for a similitude in Tk;thatis,oneofthetilesobtained from the refinement process. We apply similar terminology to a subtile in Tk + x. Consider how T (T + x ) is beget from T (T + x ) for k 0.T is k+1 \ k+1 b ck+1 k \ k b ck k+1 obtained from T by refining each subtile in T . Consequently, T + x is obtained k k k+1 b ck+1 from T + x by refining each subtile in T + x and then translating the resulting subtiles k b ck k b ck k+1 in the direction of dk+1/b . Figures 6.5, 6.6, and 6.7 provide a visual demonstration when b = 2+i,andD⇤ = 0, 3 for the first two possible refinements. The intersections are { } circled.

We study the subtiles of T and T + x as we transition from T (T + x ) to k k b ck k \ k b ck T (T + x ). The subtiles in T + x either coincide with the subtiles in T , k+1 \ k+1 b ck+1 k b ck k share boundaries in common with subtiles in Tk,oraredisjointfromthesubtilesofTk.

Provided that the digits of x are chosen from the set D⇤,thesubtilesofT and T + x k k b ck either coincide, or are disjoint. Notice that once the subtiles are disjoint, they stay disjoint upon further transition as a result from corollary 13.

20 Chapter 5

Proof of Theorem

And now for the pièce de rèsistance.

Theorem 14. If D⇤ satisﬁes the separation condition d d 2 for all i = j, b = 2+i | i j | 6 or d d 2n for all i = j, b = n + i, n 2, then given any 0 ↵ 1, there exists | i j | 6   x F such that T (T + x) has entropy dimension equal to ↵ log 2 D⇤ . This x may 2 \ pn +1 | | be chosen to admit a terminating radix expansion.

Proof. Let 0 ↵ 1 be given. We use the transition process to construct an x such   that T (T + x) has entropy dimension ↵ log 2 D⇤ .Webeginthereﬁnement \ pn +1 | | process starting with the set T (T + 0). The idea is if x =0, then transitioning 0 \ 0 j+1 from T (T + x ) to T (T + x ) multiplies the number of subtiles by D⇤ , j \ j b cj j+1 \ j+1 b cj+1 | | and if xj+1 = dt the transition multiplies the number of subtiles by one. Using the procedure outlined in [PP12], let h := [j↵] . Then h is a positive integer such that h j↵ < 1+h , j j j  j and consequently, h /j ↵ as j . Since 0 ↵ 1 we have h h 1+h . j ! !1   j  j+1  j Suppose 0 < ↵ < 1. For j 1 set

dt if hj = hj 1 xj = 8 . > <>0 if hj =1+hj 1 > :> h Then the number of subtiles in T (T + x ) is D⇤ j .SinceT (T + x) is a subset of j \ j b cj | | \

21 1 T (T + x ) this provides an upper bound for the number of subtiles of scale j \ j b cj (n2 + 1)j of T needed to cover T (T + x): 0 \

h N (T (T + x)) D⇤ j . j \ | |

To calculate the entropy dimension of T (T + x) it remains to check that any subtile in \ T (T + x ) leads to points in T (T +x) so that the upper bound for N (T (T +x)) is also j \ j b cj \ j \ j 2 alowerbound.Notethatthediameterofthecopiesaregivenbyj = pn +1 diam(T0) for each j.ButbythereﬁnementprocesseachsubtileinT (T + x ) transitions to one j \ j b cj or D⇤ subtiles in T (T + x ). Hence, it follows from Cantor’s Intersection | | j+1 \ j+1 b cj+1 Theorem that each subtile in T (T + x ) has inﬁnitely many points in common with j \ j b cj T (T + x). Using h /j ↵, we conclude \ j !

hj log Nj (T (T + x)) log D⇤ lim \ =lim | j | ↵ dim(T ). j 0 log j 2 ! ! j !1 log pn +1 diam(T ) 0 ⇣ ⌘

We can change some of the digits xj =0to xj = dm or visa versa, as long as the limit remains unchanged. That is, we can make changes of this nature on a sparse set of j’s. For

↵ =0or ↵ =1,wecanchoosedigitsdj such that the subtiles are disjoint at some jth reﬁnement, or let x =0,respectively.Weleavethedetailstothereader.

22 Chapter 6

Figures and Tables

Figure 6.1: T , b = 2+i 0

23 Figure 6.2: T , b = 2+i, D⇤ = 0, 3 1 { }

24 Figure 6.3: T , b = 2+i, D⇤ = 0, 3 2 { }

25 Figure 6.4: T (T + 3) = , b = 2+i, D⇤ = 0, 3 0 \ 0 ; { }

26 Figure 6.5: T (T +0.3) 1 \ 1

27 Figure 6.6: T (T +0.30) 2 \ 2

28 Figure 6.7: T (T +0.33) 2 \ 2

29 Table 6.1: b = 2+i, D⇤ = 0, 3 { } s t 2✏ 3.(s t) | | 4 1.80901699437495 4.6690470119715 3 1.80901699437495 4.24264068711928 2 1.80901699437495 3.82099463490856 1 1.80901699437495 3.40587727318528 0 1.80901699437495 3 1 1.80901699437495 2.60768096208106 2 1.80901699437495 2.23606797749979 3 1.80901699437495 1.89736659610103 4 1.80901699437495 1.61245154965971

Table 6.2: b = 3+i, D⇤ = 0, 6 { } s t 2✏ 6.(s t) | | 9 1.46247529557426 8.74642784226795 8 1.46247529557426 8.43800924389159 7 1.46247529557426 8.13019064967114 6 1.46247529557426 7.82304288624318 5 1.46247529557426 7.51664818918645 4 1.46247529557426 7.21110255092798 3 1.46247529557426 6.9065186599328 2 1.46247529557426 6.60302960768767 1 1.46247529557426 6.30079360080935 0 1.46247529557426 6 1 1.46247529557426 5.70087712549569 2 1.46247529557426 5.40370243444252 3 1.46247529557426 5.10881590977792 4 1.46247529557426 4.81663783151692 5 1.46247529557426 4.52769256906871 6 1.46247529557426 4.24264068711928 7 1.46247529557426 3.96232255123179 8 1.46247529557426 3.68781778291716 9 1.46247529557426 3.42052627529741

30 Table 6.3: b = 4+i, D⇤ = 0, 8 { } s t 2✏ 8.(s t) | | 16 1.3201941016011 11.8022928978677 15 1.3201941016011 11.563125976696 14 1.3201941016011 11.3241023537253 13 1.3201941016011 11.085231298497 12 1.3201941016011 10.8465228909328 11 1.3201941016011 10.6079881110749 10 1.3201941016011 10.3696389409037 9 1.3201941016011 10.1314884801558 8 1.3201941016011 9.89355107841348 7 1.3201941016011 9.65584248616593 6 1.3201941016011 9.41838002805903 5 1.3201941016011 9.18118280218071 4 1.3201941016011 8.94427190999916 3 1.3201941016011 8.70767072251614 2 1.3201941016011 8.47140518936221 1 1.3201941016011 8.2355041990011 0 1.3201941016011 8 1 1.3201941016011 7.76492869555534 2 1.3201941016011 7.5303308262613 3 1.3201941016011 7.29625205962945 4 1.3201941016011 7.06274400931936 5 1.3201941016011 6.8298652126912 6 1.3201941016011 6.5976823024988 7 1.3201941016011 6.36627141776926 8 1.3201941016011 6.13571991077896 9 1.3201941016011 5.90612842234035 10 1.3201941016011 5.67761341741819 11 1.3201941016011 5.45031029877577 12 1.3201941016011 5.22437724968812 13 1.3201941016011 5 14 1.3201941016011 4.77739776570517 15 1.3201941016011 4.55683068396807 16 1.3201941016011 4.33860915637312

31 Chapter 7

Conclusion

M All complex numbers z can be expressed as a radix expansion z = dM ...d0.d 1d 2... = k= 1 d bk such that b = n + i is a Gaussian integer, n 1 and d is a digit in the digitsP set k k D = 0, 1,...,n2 for all k. The set T = 0.d d d ... d D forms a tile in the complex 0 { 1 2 3 | i 2 } plane. The subset T = z T d D⇤, i k is a reﬁnement of T by restricting d , k { 2 0 | i 2 8  } 0 i 2 1 i k,tothedigitsinD⇤ = d ,...,d D such that 0=d

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