The Elements of Cantor Sets: With Applications

THE ELEMENTS OF CANTOR SETS— WITH APPLICATIONS

THE ELEMENTS OF CANTOR SETS— WITH APPLICATIONS ROBERT W. VALLIN Slippery Rock University

WILEY

Copyright © 2013 by John Wiley & Sons, Inc. Published by John Wiley & Sons, Inc., Hoboken, New Jersey. All rights reserved. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permission. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representation or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services please contact our Customer Care Department within the United States at (800) 762-2974, outside the United States at (317) 572-3993 or fax (317) 572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print, however, may not be available in electronic formats. For more information about Wiley products, visit our web site at www.wiley.com. Library of Congress Cataloging-in-Publication Data: Vallin, Robert W. The elements of Cantor sets : with applications / Robert W. Vallin, Department of Mathematics, Slippery Rock University, Slippery Rock, PA. — First edition, pages cm Includes bibliographical references and index. ISBN 978-1-118-40571-0 (hardback) 1. Cantor sets. 2. Measure theory. 3. Mathematical analysis. 4. Cantor, George, 1941- I. Title. QA612.V35 2013 515'.8—dc23 2013009452 Printed in the United States of America. 10 9 8 7 6 5 4 3 2 1

To Micah, Jessica, Rachel, Sophie, and Sean. And to Jackie, who always believes in me.

CONTENTS IN BRIEF

1 A Quick Biography of Cantor

1

2 Basics

5

3 Introducing the Cantor Set

17

4 Cantor Sets and Continued Fractions

51

5 p-adic Numbers and Valuations

67

6 Self-Similar Objects

91

7 Various Notions of Dimension

119

8 Porosity and Thickness-Looking at the Gaps

143

9 Creating Pathological Functions via C

157

10 Generalizations and Applications

181

11 Epilogue

217

VII

CONTENTS

Foreword

xiii

Preface

xv

Acknowledgments

xvii

Introduction

xix

1

A Quick Biography of Cantor

1

2

Basics

5

2.1

3

Review Exercises

5 14

Introducing the Cantor Set

17

3.1 3.2

17 21 22 29 36

Some Definitions and Basics Size of a Cantor Set 3.2.1 Cardinality 3.2.2 Category 3.2.3 Measure

ix

X

CONTENTS

3.3

4

5

51

4.1 4.2 4.3 4.4

52 59 60 63 65

8

Introducing Continued Fractions Constructing a Cantor Set Diophantine Equations Miscellaneous Exercises

p-adic Numbers and Valuations

67

5.1 5.2

67 72 72 75 80 82 88

Some Abstract Algebra p-adic Numbers 5.2.1 An Analysis Point of View 5.2.2 An Algebra Point of View p-adic Integers and Cantor Sets p-adic Rational Numbers Exercises

Self-Similar Objects 6.1 6.2 6.3 6.4 6.5

7

47 48

Cantor Sets and Continued Fractions

5.3 5.4

6

Large and Small Exercises

The Meaning of Self-Similar Metric Spaces Sequences in (S, d) Affine Transformations An Application for an IFS Exercises

91 91 92 97 106 114 116

Various Notions of Dimension

119

7.1 7.2 7.3 7.4 7.5 7.6

119 123 127 128 131 136 140

Limit Supremum and Limit Infimum Topological Dimension Similarity Dimension Box-Counting Dimension Hausdorff Measure and Dimension Miscellaneous Notions of Dimension Exercises

Porosity and Thickness-Looking at the Gaps

143

8.1

143

The Porosity of a Set

CONTENTS

8.2 8.3 8.4 8.5 8.6 8.7

9

10

11

146 149 150 151 153 154 156

Creating Pathological Functions via C

157

9.1 9.2 9.3 9.4 9.5 9.6

157 161 168 171 173 177 180

Sequences of Functions The Cantor Function Space-Filling Curves Baire Class One Functions Darboux Functions Linearly Continuous Functions Exercises

Generalizations and Applications

181

10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10 10.11 10.12 10.13

181 185 186 193 195 197 198 201 203 205 207 209 211

Generalizing Cantor Sets Fat Cantor Sets Sums of Cantor Sets Differences of Cantor Sets Products of Cantor Sets Cantor Target Ana Sets Average Distance Non-Averaging Sets Cantor Series and Cantor Sets Liouville Numbers and Irrationality Exponents Sets of Sums of Convergent Alternating Series The Monty Hall Problem

Epilogue

References Index

Symmetric Sets and Symmetric Porosity A New and Different Definition of Cantor Set Thickness of a Cantor Set Applying Thickness A Bit More on Thickness Porosity in a Metric Space Exercises

XI

217 219

FOREWORD

Among the delights of mathematics are those moments when it surprises, when something unexpected emerges from a seemingly mundane construction that is pushed beyond the bounds of the finite: staircases of finite length for which one can never climb from one step to the next, continuous and bounded functions that can be in­ tegrated but whose integral does not have a derivative, space-filling curves and frac­ tional dimensions, and the Banach-Tarski paradox: the ability - in theory - to cut a pea into finitely many pieces and reassemble those using only rigid motions (ro­ tations and translations) into a solid object the size of the sun. Such mathematical surprises are the stuff of which great mathematical advances are made. They chal­ lenge us to broaden our understanding of the world in which we live. The Cantor set provides one of these surprising moments. If we take an interval of the real number line, say from 0 to 1, and remove finitely many little intervals from it, we are left with finitely many little intervals. No surprise here. But what if we remove infinitely many little intervals? Are we left with infinitely many little intervals? No less a mathematician than Axel Harnack (1851-88) implicitly assumed this obvious conclusion. He was very wrong. To see how wrong he was, imagine that from our interval [0,1] we remove ev­ ery real number with 7 in its decimal expansion. This takes out everything from 0.7 to just below 0.8, one interval. It also takes out the nine intervals [0.07, 0.08), [0.17,0.18), ..., [0.97,0.98). We are not done. We also have to take out the all of XIII

XIV

FOREWORD

the intervals of length 0.001 that start at a number with two digits that are not 7, followed by a 7, such as 0.237. That adds 92 intervals of length 1/103. This goes on forever. The total amount that we remove is 1

9

92

93

_ 1_ (

_9_

Io + io2 + To3 + To4 + ''' _ To V + i o

+

92

\ _ 1 /

1

\

T o 2 + ' " y ™ i o l^i-9/ioJ ~

It appears that we have removed the entire interval. But, in fact, it is quite easy to come up with numbers that are left behind. We still have 1/2 = 0.5. No 7s in that decimal expansion. We still have 1/3 = 0.333 In some sense, we still have as many real numbers as we started with. Who needs the symbol 7? If we represent our real numbers in base 9, the remaining nine symbols are sufficient to express every real number in [0,1]. Lots of numbers are left, but no intervals. You cannot get from 1/2 to 1/3 without crossing a number with a 7 in its decimal expansion. More than this, between any two distinct real numbers there will always be a real number with a 7 in its decimal expansion. What is left is a very unintuitive set that is so full of holes that nothing is left, and yet it still describes everything. It is what we call a Cantor Set. The name Cantor Set is a bit of a misattribution. Henry J.S. Smith (1826-83) used these sets in a paper published in 1875. Unfortunately, Smith was an Englishman at a time when few expected any significant mathematics to come out of England. Any really important mathematics would have been published in German. The next paper to use these sets was by Vito Volterra in 1881 (1860-1940). Volterra would come to be known as one of greatest of all mathematicians, but in 1881 he was a mere student publishing his results in an obscure Italian journal. No one noticed. When Georg Cantor rediscovered and described these sets in 1883, it was one piece of a major and highly visible assault on our understanding of the structure of the real number line. Now mathematicians paid attention. In this book, Robert Vallin leads us through the many surprises of Cantor Sets. This is only the beginning. The structure of these sets is echoed in much of 20th century mathematics from p-adic numbers to fractals. This tour will explore the connections to continued fractions, to self-similarity and non-integer dimensions, to derivatives that are not continuous and space-filling curves that are, and to one of my favorite mathematical curiosities, the Cantor function, more aptly known as The Devils Staircase. David M. Bressoud

PREFACE

Like probably all mathematicians, I cannot recall the first time I heard about the Cantor Set. I am sure I was an undergraduate, but I just don't remember. What I do know is that I have been fascinated by it since first coming across it. Start with the unit interval, remove what seems like everything, and the Cantor Set is what is left over. There are numerous points in it, despite the fact that so much has been removed. With disbelief I read that it has as many points as the interval [0,1]. Someone told me the number 1/4 is a point in the set. I probably just nodded up and down. I was sure I was being told the truth, but had no clue why it was true. In graduate school, I found out even more. It seemed like everywhere I turned there was the Cantor Set. In Analysis, it was an example of a measure zero set. In Topology, it was nowhere dense. In a class on fractals, it had non-integral dimension. It lived in the line, but had dimension that was neither 0 (like a point) nor 1 (like a line segment). It seemed to be everywhere I turned: analysis, topology, abstract algebra, probability, and the list goes on. It was cool, but I was busy getting my PhD. For awhile the Cantor Set went by the wayside. Once out of school, I gained the freedom to study whatever was my pleasure. Most of my years I've been an analyst/topologist. However, when articles involving the Cantor Set came along, I was sure to read them and enjoy myself. As a great believer in students graduating with a large collection of examples/counter-examples in their quivers, the Cantor Set was one of my go-to topics. xv

XVI

PREFACE

In 2011, I was approached by a student in my analysis course about an indepen­ dent study class the next semester. He was interested in finding out more about the Cantor Set. I was very excited about the idea. Once we got approval for the plan, I set forth to make an outline. The course was written pseudo-IBL (Inquiry-Based Learning). Every week we would meet and I would give him some definitions and some open questions. Sometimes these were theorems which required proofs; some­ times they were false statements that needed counterexamples. We covered a little bit of everything I could think of and the course went well. In putting my outline together I was surprised at how scattered my resources were. There was no central place that had all that I needed. Existing books were too nar­ rowly focused and/or too technical as were the websites around. It occurred to me that the outline was the beginning of something good. So I decided that I would be the person to put this book together. In writing this book, I walked a fine line: The text is not written very formally; but being a serious text, does have the form theorem/proof. However, in writing this manuscript for advanced undergraduates I left out many proof because they were too long and technical for this intended audience. It is my hope that anyone interested in the details will be able to use the bibliography to go to the source and find out the nitty-gritty. ROBERT W. VALLIN Slippery Rock, PA January 2013

ACKNOWLEDGMENTS

Any endeavor like this is not a work done alone, even though it is only my name on the cover. As noted in the Preface, without my student Andy Brown and his interest in an independent study on Cantor Sets, I would not have thought of putting together my notes which then became this manuscript. Those who lent an ear and encouraged me include Lyn Miller, Carol Schumacher, Rich Marchand, and Ron Taylor. Gary Grabner was a huge help with the figures. Rita McClelland and Kelsi Wren were able to find me whatever I needed on interlibrary loan. At Wiley, thanks go out to Susanne Steitz-Filler, my editor, Christine Punzo, for making this readable, and Amy Hendrickson, who patiently answered my ETijXquestions. My biggest cheerleader and fan was my wife, Jacqueline Jensen-Vallin. That said, any and all mistakes are mine and mine alone. If you find any errors, or if I have missed your favorite topic, please don't hesitate to contact me. R.W.V.

XVII

INTRODUCTION

This is not just about the Cantor Set. If it were, there would be much less to say. As an object itself, the set can be fascinating, but what is most appealing is not the set per se, but where the set can lead one. In analysis one uses the notions of measure, category, and cardinality to describe many objects. When teaching those topics the Cantor Set always pops up as an example to fortify understanding of the concept. The text makes for a good ancillary in several different courses: analysis (using Chapters 2, 3, 9 as a start), topology (Chapters 2, 8, 9), fractals (Chapters 6, 7, 10), and algebra (Chapters 2, 5, 4). Many schools are introducing the Capstone Experience Course to their curricu­ lum. This is a Senior-Level class that, while having a unifying theme, draws in material from the previous coursework for their undergraduates. These courses can include analysis, algebra, topology, and number theory. Several Capstone Courses also require an out-of-class project. This book has a unifying theme, of course, and shows how the Cantor Set appears in so many different mathematical topics, thus bringing together the collection of courses that make up the lion's share of the upperlevel math curriculum. Given the wide range of topics covered, there are plenty of places for students to find topics and questions to study outside the classroom for the project portion of the course. This manuscript, which arose from an independent study course, is an example of the type of book needed for an independent study. Cantor Sets are studied from many XIX

XX

INTRODUCTION

different perspectives, with multiple definitions and applications. The book is meant to be something a student can read on his/her own, meeting with a professor from time to time to discuss the topics. There are so many ideas covered that whether a student has a preference for analysis or algebra or something else, that subject can be found. If students enjoy applying what he/she learns, then Chapters 9 and 10 cover so many ideas that one can find a direction to take their projects. Lastly, this book is a resource for professors. As we all know, the "sage on the stage" reading from well-worn notes or overhead slides is an outdated idea. The pro­ fessor, being an expert in their field, should be able to supplement any text, be it with anecdotes, apocrypha, extra examples, generalizations, or applications. Parsed into its separate chapters and sections, this book can be used to add to several different courses in the undergraduate mathematics curriculum. It does have a purpose out­ side the classroom. With the topics covered and detailed bibliography, one can find a myriad of directions to take one's research.

CHAPTER 1

A QUICK BIOGRAPHY OF CANTOR

Georg Cantor Library of Congress, courtesy AIP Emilio Segr Visual Archives The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

1

2

A QUICK BIOGRAPHY OF CANTOR

Georg Ferdinand Ludwig Philipp Cantor was one of the most important mathe­ maticians of the last half of the 19th century. Born 3 March 1845 in St. Petersburg, Russia, he was the oldest of six children. Due to his father's ill health, the family moved to Frankfurt, Germany, hoping to find milder winters. In 1862 he started his higher schooling at the Federal Polytechnic Institute in Zurich. After the death of his father, Cantor moved on to study at the University of Berlin. Among his professors there were Kronecker, Weierstrass, and Kummer. In 1867, he earned his PhD for his thesis entitled "De aequationibus secundi gradus indeterminatis (On Indeterminate Equations of the Second Degree)." Cantor began teaching at the University of Halle, where he would spend his entire career, in 1869. He was promoted to full professor in 1879, which was an achieve­ ment at such a young age. However, Cantor wanted to move to the more presti­ gious University of Berlin, hoping to become chair of the department. This could not happen as Kronecker, who led the department at Berlin, was not supportive of Cantor and his work. Kronecker was one of the founders of the constructive view­ point; that means only dealing with mathematical objects that could be explicitly made. As an example of Kronecker's feelings on this, when von Lindemann proved the existence of transcendental numbers (a type of number defined later on in this book), Kronecker praised the proof, but also said the argument [19] "proved noth­ ing because there were no transcendental numbers." Much of Cantor's work dealt with sets whose members could be described and were well-defined, but not explicit. Kronecker went so far as to try and pressure Heine to not publish Cantor's paper "Uber trigonometrische Reihen" in the journal Mathematische Annalen. In a letter to Hermite, Cantor complained that Kronecker, among, other things, called his work "humbug." The pressure from the non-acceptance of his work would drive Cantor into periods of depression, which he suffered from until the end of his life. He carried on correspondence with some of the greater mathematicians of the day (e.g., Mittag-Leffler, Dedekind), but would at times cease communication. He was very sensitive to even well-intentioned criticisms of his work. In 1894, Cantor suffered his first attack of depression. He even turned his attention away from math­ ematics for a time, lecturing on philosophy and working on proving that Shakespeare did not write the plays attributed to him. Instead, Cantor believed they were written by Francis Bacon. Cantor was hospitalized again for depression in 1899 and then several more times starting in 1904. These sapped him of much of his zeal for mathematics. However, his work continued to be appreciated. In 1912 he was awarded an honorary doctorate from the University of St. Andrews in Scotland. He officially retired in 1913 and died on 6 January 1918 in a sanatorium where he had been for the last year of his life. Cantor's contributions to mathematics are vast, with results in number theory (his thesis topic) and set theory (which he founded). Some of his most well-known ideas are the Cantor Set, the notion of different types of infinities, and the Continuum Hy­ pothesis. Working on the infinite drew the ire of some mathematicians, philosophers, and religious scholars. Philosophers felt there were too many contradictions in deal­ ing with the infinite. For example, if a and b are two positive numbers, then a < a + b and b < a + b. However, oo < a + oo is not true. No less a mathematician than

3

Poincare said that in the future people would look on Cantor's work as "a disease from which one has recovered." To quote from [19]: Christian theologians were also opposed to the actual infinite; for the most part they regarded the idea as a direct challenge to the unique and absolutely infinite nature of God.

However, Cantor was not without his supporters. At the Second International Congress held at the Paris World Exposition of 1900, David Hilbert presented a list of the major unsolved problems of the time, hoping to spur interest in what he believed to be the most important problems of the day. The item at the top of the list was Cantor's Continuum Hypothesis. Hilbert is famously quoted as saying, "No one will drive us from the paradise which Cantor created for us." Hilbert's influence was profound, as can be seen in the vast list of papers written just in the last few years that mention Cantor, Cantor Sets, Cantor Measure, Cantor Spaces, and more.

CHAPTER 2

BASICS

This is the "review" chapter. That means much of what we look at here the reader is familiar with. Thus there will not be a lot of proofs given. Most of these concepts are familiar, even if a bit of dust must be shaken off. Hopefully those ideas which are unfamiliar will be few and easily digested by the reader.

2.1

Review

This chapter begins with sets. A set is a well-defined collection of objects. We will use capital letters to denote a set. There are two typical ways to write a set: List Notation and Set-Builder Notation. As the name suggests, List Notation means the writer lists (some of) the objects in the set. Note that sets are written using braces for enclosure. For example, A = {a, b, c, d} or B = { 1 , 2 , 3 , . . . , 9,10}, where the ellipses (...). means "continue in this manner." For Set-Builder Notation, rather than listing the items in the set, the members are described by some proposition, P(x). This is a useful way of doing things as some sets defy listing. An example of this notation is C — {x : x is a letter in the English Alphabet} The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

5

6

BASICS

which is of the form {x : P(x)}. The colon is usually translated as "such that" or "so that," and then followed by a description. This is read as, "C is the set of x such that x is a letter in the English Alphabet." The empty set, 0, is the set with no elements in it. The textbook definition, which looks a bit strange, is 0 = {a : a ^ a}. An object which is a member of a set is called an element of the set and usually written with a lowercase letter. The symbol for is an "element" of is s. Rather than write, "a is a member of the set X," we just write "a G X." If an object is not an element of the set, then we use the symbol ^ as in t <£ X. We say that the set A is a subset of the set B if and only if every element of A is also an element of B. Notationally, this is A C B if and only if for all x, x € A => x £ B, where => is read as "implies." Using an empty argument, it is possible to prove that for every set B,$ C B. Subsets are used to prove equality of two sets. The sets A and B are equal if and only if A C B and B C A. If we know that A C B and A y^ B, then we say A is a proper subset of B which we write as A C B. Given two sets A and B, there are several ways to create new sets. The union of A and B is the set defined by AU B = {x :x € Aorx

€ B}.

The intersection of ^4 and 5 is the set defined by AnB

= {x : x £ A and x e B}.

If two sets have no element in common (A P\ B = 0), then we say A and B are disjoint. A collection of sets1 {An} is called pairwise disjoint if for i 7^ j we have A, n A,- = 0. The se/ difference of .A and Z? is the set defined by A \ B = {x : x € A and x g B}. A specific set difference which we will encounter many times is the complement of A. This is

Ac = U\A,

where U represents the Universal Set, the collection of all object under consideration. For much of this text U = K. Relating all these operations are De Morgan's Laws: (AUB)C

=ACDBC,

1 We could refer to this as a set of sets; however, there can be issues with this if we are not precise. To learn more on set theory, we suggest [21 ] or [36].

REVIEW

and

(AC\B)C

7

=Acl)Bc.

Union, intersection, and DeMorgan's Laws can be generalized to countably many sets A\,Az, A3, In this situation, UjAi = {x : there exists an i such that x € Ai} and HiAi = {x : for all i, x € At}, while DeMorgan's Laws become ( U ^ ) C = ni(Af)

and (HiAif

= Ut(A?)■

Finally, the product of non-empty ^4 and B is the set of all ordered pairs of elements from A and B, respectively; that is, Ax B = {(a,b)

:a€A,beB}.

If A or B is empty, then A x B = 0. fl

EXAMPLE 2.1 Let A = {a, 6, c} and I? = {a; : x is an English vowel} = {a, e, i, o, u} where the Universal Set is the English Alphabet, U = {a, b, c, d,..., y, z}. Then ■ A U B = {a, b, c, e, i, o, u}. • AC\B = {a}. ■ A\B

= {b,c}.

• Ac = {d,e: / , . . . , y,z}. • Bc = {x : x is an English consonant}. ■ AxB = {(a, a), (a, e), (a, i), (a, o), (a, it), (b, a), (b, e), (6, i), (b, o), (b, it), (c, a). (c,e),(c,i),(c,o),(c,u)}.

■

Specific sets to know include ■ The Natural Numbers: N = { 1 , 2 , 3 , . . . , }. ■ The Integers: Z = {..., - 2 , - 1 , 0 , 1 , 2 , . . . } . ■ The Rational Numbers: Q — {x : x — p/q, where p,q &Z,q j^O}. • The Real Numbers: R = (—00,00).

8

BASICS

Let a . i e K . We shall look at some specific sets in the real line. Intuitively, a set A in R is bounded if there are numbers m and M where no element of A is below m or above M. Now we define intervals as ■ Closed and bounded: [a, b] = {x G R : a < x < b}. ■ Open and bounded: (a, b) = {x G R : a < x < b}. • Open and unbounded: (a, oo) = {x G R : a < x} or (—00, b) = {x G R : x < b}. • Closed and unbounded: [a, 00) = {x G R : a < x} or (—00, 6] = {x G R : x < b}. Generalizing open and closed intervals leads us to open and closed sets in the line. A set A is open if for every x G A, there is an open interval /, such that x G / C A. A set B is closed its complement R \ B is open. Now that we have sufficiently clear ideas of open and closed sets, we can further our understanding of both the interior and boundary of a set. Formally, a: is a point in the boundary of A (written as dA) if every open interval containing x also contains points in both A and Ac. The interior of the set (written A°) can be expressed as A \ dA. Another way to define this is x G A° if there exists an open interval I such that x € / C A. Lastly, for any set A, the closure of A, A is the set along with its boundary, A = A U dA. S

EXAMPLE 2.2 Let A = (0,1) U (2, 00), B = [0,1] and C = Q n (0,1). Then A is open (but not an open interval), B is closed, C is neither open nor closed. dA = {0,1, 2}, d_B = {0,1}, dC = JO, 1]. Meanwhile, A0 = A, B° = (0,1), and C° = 0 and A=[0,l]U[2,oo),B = B = C. u

Starting with the open sets, we will create what is called a a-algebra. A non­ empty collection of sets X is called a <7-algebra of sets if (1) for every A G X, its complement AG G X; (2) if A, G X for i = 1,2,3,..., then U ^ G X. Thus cr-algebra sets are closed under complements and countable unions. The aalgebra created by the open sets is the collection called the Borel Sets (named for Emil Borel). Another extension of the open/closed sets are the Ta and Qs sets. An Ta set is a set which can be written as the countable intersection of closed sets while the Qg sets are described as countable intersection of open sets. Obviously any closed set is J> and any open set is Qs- The interval (0,1) is an example of an J v set that is not a closed set since (0,l) = U neN [0 + l / n , l - l / n ] . Our first theorem, relates Ta and Qs sets and their complements. Theorem 2.1 If A is a Qs set in R, then Ac is an T„ set in R.

REVIEW

9

Proof: This is just an application of the generalized DeMorgan's Law. If A is a Gs set, we can write A = n(Oj) where each Oi is an open set. Then

AC = (HOif = U;(Of). Now each Of is a closed set, thus Ac is Fa. 4k Let A and 5 be sets. A function f : A —> B is a rule which assigns to each element in A exactly one element in B. This means / = {( a , b) : a € A, b € B, and (a, 6) € /, (a, 6') G / implies b = b'}. Of course, usually we do not write (a, b) s / , but instead / ( a ) = b. The set .A is called the domain of / and B is the co-domain. The range of / is the set of elements in B that are the image of something in A; that is {b e B : there exists ana £ A with / ( a ) = 6}. When f : A —¥ B has the additional property that f(x) — f(y) implies x = y, we say / is a one-to-one (1 — 1) function. Sometimes, rather than this definition we use its contrapositive which yields / is a 1 — 1 function if x 7^ y implies f(x) 7^ f(y)If / is 1 — 1, then there exists an inverse function f~l : B —> A defined by y = f~1{x)

if and only if x = f{y).

If, for any set C, we define / ( C ) as {f(x) : x 6 C} we say / is onto if /(A) = B. A function which is both 1 — 1 and onto is called a bijection. Bijections are mathematically important because if / is a bijection, then so is / _ 1 . We will use bijections when we discuss cardinality. The inverse function / _ 1 should not be confused with the inverse image of a set. If / : A —^ B and U C B, then the inverse image of U is the set of all values in A which map to something in U. Notationally, that is f-\U)

=

{xeA:f(x)eU}.

A function does not have to be a bijection or have an inverse in order to talk about inverse images. For example, if f(x) = x2 (with domain E; always assume the biggest domain possible), then / - 1 ([0,1]) = [—1,1]. Given / : A —>■ B and D C A, we say the restriction of f to D is the function / with the domain narrowed down to just the set D. This is written as f\p. These restrictions can have important consequences. The function F : R -> R given by F(x) = sin(x) is not a 1 — 1 function, but if D = [—7r/2, TT/2], then F\D is a 1 - 1 function. One special type of function we will use often is called a characteristic function. Given a set D C R, the value of the characteristic function of D (also called the

10

BASICS

indicator function) has values 1 for points in the set and 0 otherwise. Usually charac­ teristic functions are denoted using a lowercase Greek letter chi, x, for the function name with the set as a subscript. So this means for D XD{X) =

U

X{D.

Some functions have a special property involving the distance between two inputs versus the distance between their respective outputs. A function / : D —> R is called a Lip'schitz function if there exists a constant k > 0 so that for every x,y £ D \f(x)-f(y)\ 0 there exists a J V e N such that n > N implies \xn — x$\ < e. We can write this as lim xn = X(j. n—>oo

If a sequence does not converge, it is said to diverge. A sequence is called increasing (decreasing) if for any pair of natural numbers n and m, if n >TO,then xn > xm (xn < xm). If there are strict inequalities instead, then the sequence is strictly increasing (strictly decreasing). A sequence is called em monotone if it is either increasing or decreasing. Limits of a sequence brings us to limits of a function. The real number L is the limit of / as x approaches xo if for every e > 0 there is a S > 0 such that 0 < \x — xo\ < S implies |/(a;) — L\ < e. This means as the IT—values get closer to, but not equal to, x0 the /(a;)—values are approaching (near or equal to) L. The fact that the x's are not supposed to equal XQ is reflected in the requirement 0 < \x — XQ\. Also, notice that we are not requiring the limit to be the value of / at xo (or even that f(x0) exists). Another important property for a function is continuity. Intuitively we say / is continuous if we can draw its graph without lifting the pencil/pen/chalk/marker. We will need to be much more precise here. So we start with continuity at a point and then expand. This is the typical definition of continuity that one finds in a calculus text. It will be followed by the more precise 6-e definition. We will go beyond these when we study both topological and metric spaces.

REVIEW

11

Let / : A -*■ R be a function and x 0 an arbitrary value. We say / is continuous at XQ if the following three conditions are satisfied: 1. /(xo) exists, 2. lim^^^,, / ( x ) exists, 3. lim x ^ X o f(x) = f{x0). Continuity of a function / at the point x0 is equivalent to the statement that for any value e > 0 there exists a 8 > 0 so that for any x in the domain of / with |x — xo| < S we have |/(x) — /(xo)| < e. A function is called continuous if it is continuous at each point in its domain. If / is not continuous at some point, we refer to that point as a point of discontinuity for / . As an example of how 5 — e proofs are written, let us look at the following two examples, complete with scratch work. While scratch work is a necessary part of coming up with a proof, it should not be included in the proof. That said, many proofs are a matter of writing scratch work backwards, with some verbiage thrown in. fl

EXAMPLE 2.3 Show that lima;_>2(3x — 1) = 5. Scratchwork: If we look at \f(x) — L\ we have |(3x — 1) — 5| or |3x — 6|. This needs to be related t o | x - a | = | x - 2 | . Factoring gives us

|3x-6| = 3|x-2|. So if |3x — 6| = 3|x — 2| < e, then \x — 2| < e/3; and since we need to have |x — 2| < 5, this will be accomplished if 5 = e/3. Proof: Let e > 0 be given. For this e, let 5 = e/3. Then if 0 < \x - 2| < 5, we have |/(x) -L\ = |(3x - 1) - 5| = |3x - 6| = 3|x - 2| < 3<5 = 3(e/3) = s. Since e was arbitrary, this holds for all values of e and thus the proof is complete.

■ fl

EXAMPLE 2.4 Show that f(x) — x2 is continuous at x0 = 3. Scratchwork: This is a little more difficult. Since the function is not linear, we cannot use straight factoring. However, factoring still plays a role here. Starting at the end, we have e > 1/0*0 - f(x0)\

= |x 2 - 9| = |x - 3| • \x + 3|.

12

BASICS

This is good because \x — 3| is our \x — XQ\, but what to do with |x + 3|? Can we just divide both side by \x + 3|? No, although it seems every math major has at one point in his/her education. Instead we must replace \x + 3| with a constant. This is accomplished by making an assumption about \x — 3|. Assuming \x — 3| < 1 (the right side is arbitrary, but usually it is set to 1), this unfolds to — 1 < x — 3 < 1 or 2 < x + 3 < 4. So if \x — 3| < 1 we know \x + 3| < 4. Then |/(a;)-/(a:o)| = | s - 3 H z + 3 | < 4 | a : - 3 | . This takes us to the end of our scratch. If \x — 3| < d, then the right-hand side above is smaller than 46 and we just need to set 5 = e/4. It is vitally important to realize that in this proof we have two restrictions on \x — 3|. It must be less than 6 and less than 1 at the same time. Proof: Let e > 0 be given. For this e, let 5 = min{l, e/4}. Then if \x — 3| < 5, we have I/O) - Z(3)| = \x2 - 9| = \x - 3| • \x + 3| < \x - 3| • 4 < 45 = 4(e/4) = e. Since e was arbitrary, this holds for all values of e and thus the proof is complete.

4

On the real line, the distance between two points is given by d(x, y) = \y — x\. Different spaces may have different idea of distance. These distance functions are called metrics. The requirements for a function to be a metric are as follows: A metric on a set S is a function d : S x S —► [0, oo) which satisfies the following three properties: ■ For all x, y 6 S, d(x, y) > 0 (called nonnegativity) and d(x, y) = 0 if and only if x = y. • For all x,y £ S1, d(x, y) = d(y, x) (symmetry). ■ For all x,y,z S

G S, d(x, y) < d(x, z) + d(z, y) (The Triangle Inequality).

EXAMPLE 2.5 Suppose 5 is a set with three points in it, S = {pi,f2)P3}- We define for non-equal pi the distance to be d{pi,P2)

=d(p2,pi)

= 1,

d{p2,P3) = d(p3,p2)

= 2,

d(p3,Pi)

= 3.

=d(pi,p3)

Of course, d(pi, pi) — 0 for i = 1,2,3.

■

REVIEW

13

A space with a distance on it is referred to as a metric space usually written as the ordered pair (S,d). Every set can have a distance function defined on it as there is always the so-called trivial metric. 0, x = y, d(x,y) 1,

x^y.

Although these definitions and concepts in this chapter are written for the real line, we will explore various generalizations. These include going from R to R", looking at differently structured spaces like metric and topological spaces, and gen­ eralizing our ideas to appropriate types of limits or continuity.

14

BASICS

EXERCISES 2.1

Let A = {1, 3, 5, 7,9} and B = {1, 2, 3,4} in the universal set U = {1,2,3,4,5,6,7,8,9,10}.

Lwf f/ze members of a) AuB b) AC\B c) Ac d) A\B e) Ax B

f) By. A 8) (B\Af 2.2

Explain why for any set A we have ^ 1 x 0 = 0 x ^ 1 = 0.

2.3

Let An = {1, 2 , 3 , . . . , n} for n = 1,2,3,.... Determine UnAn and C\n An.

2.4 G/ve an example of a non-empty, closed set in M. whose boundary consists of a single point. 2.5 set.

Give an example of a non-empty, open set in R whose boundary is the empty

2.6 Determine whether or not the given sequence converges. A proof is not neces­ sary here. a) •&TI = (3n2 - 5n)/(2r7,2 + 3) b) Vn = l + ( - l ) n c) an = sin(n7r/2) d) r ■■= (1 + 1/n)" e) $n '= (y/n + 1 - v/n) n

2.7

Show that [0,1] is a Qs set.

2.8

Prove that f(x) = x3 is continuous at XQ — 1.

2.9

Prove, using an e — N argument, that xn = 1/n converges to 0.

2.10

Prove, using an e — N argument, that yn = sin(n)/n converges.

2.11

Show that f(x) = -^rh is a 1-1 function, but not onto R.

2.12 Explain why f(x) = sin(x) is a Lipschitzfunction. (Hint: This uses the Mean Value Theorem from Calculus.) 2.13 Show that any linear function, f(x) = rax + b, m ^ 0 must be a Lipschitz function.

EXERCISES

2.14

15

Consider the function

jxsin(l/x),

x ^ 0,

a) Show that ifa = 0, then f is continuous at x = 0. b) Prove that for any other value of a f is discontinuous at x = 0. 2.15 Let I be an interval in K.. A function f : I —> R is called uniformly contin­ uous iffor every e > 0 there exists a S > 0 such that ifl x,y £ I with \x — y\ < S, then \f{x)-f{y)\<e. (compare this with the s — 5 definition for ordinary continuity). Show that any linear function f(x) = mx + i, m / 0 must be uniformly continuous. 2.16 Prove that f : [1, oo) —> K given by f(x) = y/x is a uniformly continuous function. 2.17

Prove that any Lipschitz function must be a uniformly continuous function.

2.18

Show that Example 2.5 satisfies the requirement of being a metric.

CHAPTER 3

INTRODUCING THE CANTOR SET

Why is the Cantor Set interesting? Well, we hope to find answers to that question as we continue to read this book. One thing we can say is that the set seems to pop up everywhere. When looking for an example to illustrate a definition some kind of Cantor Set usually does the trick. This can happen in real analysis, topology, abstract algebra, probability theory, fractal geometry, the list go on. In searching for extreme examples, many times one begins, "Start with a Cantor Set and ...." It is this utility of the Cantor Set which makes thorough knowledge of it an important item in a mathematical toolbelt.

3.1

Some Definitions and Basics

There are several ways to create a Cantor Ternary1 Set and we shall first give the standard or canonical way of making one. Start with the closed unit interval (K0 = 'Ternary refers to the fact that we will divide things into thirds. For the most part, we will drop this word and leave it to the reader to understand when we mean this set and when we mean some general Cantor Set. The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

17

18

INTRODUCING THE CANTOR SET

[0,1]). Divide it into three equal sections and remove the open middle third. Thus we now have K\ = [0,1/3] U [2/3,1]. We then continue inductively. At step n, Kn consists of T1 closed subintervals. For the (n + l)st step, divide each of the closed subintervals in the previous step into thirds and remove the open middle third. Continuing indefinitely gives us the collection of sets {Kn}^L0. Finally, the Cantor Set is given by oo

ci/3 - n K n A representation of the first four steps is shown below: K0= Kx= K 2= K3 = Notice that since we are only removing from the middle of an interval, any point that is an endpoint of a interval in Kn will be in C1/3, however, there are many more points than just these. Is there something magical about dividing things into thirds? No, of course not. One way to generalize this is to change the ratio of the interval removed. At each step in this process we have an interval I and from that interval we remove a middle subinterval J so that £(J)/£(I) — 1/3, where £{■) refers to the length of an interval. We can create a new Cantor Set by insisting the middle interval removed have ratio £{,])/£{I) = r where 0 < r < 1. We will denote this Cantor Set by Cr. Other generalizations of the Cantor Set will be considered in Chapter 10. Here is an easy first question. Since we start with the closed unit interval, we are beginning with something of length one. Then we take out the open interval ( 1 / 3 , 2/3) leaving us with two closed subintervals each with length 1/3. From those, we remove one interval each and each of those subintervals has length 1/9. Now we have four closed subintervals of length 1/9 each. So how much is taken away in total? Theorem 3.1 The total length of the subintervals removed in the derivation 0/C1/3 is one. Proof: At each stage n > 0 there are 2n~1 open subintervals removed and each of these subintervals has length (1/3)™. The total length removed is then represented by the infinite geometric series Yl^Li 2 " ~ 1 ^r. From Calculus we know if \r\ < 1, then 5Z^Lo a ■ rn = a / ( l — r ) . Thus 0 0

]T 2"" n—\

1

1

- = 1/3 + 2(1/9) + 4(1/27) + • • • =

1/0 T

^ —

= 1.*

'

So in a way, we start with an interval that has length 1 and remove from it intervals whose aggregate length is also 1, yet there are plenty of fascinating points leftover.

SOME DEFINITIONS AND BASICS

19

Let us refine our characterization ofpoints in (or out of) the Cantor Set. Normally we write numbers in base 10. With base 10 we use the digits 0,1, 2, 3 , . . . , 9 and the position of a digit refers to a power of 10. For example, in 123 the digit 2 means 2 ■ 102 or 200, where for 2103 the 2 means 2 ■ 103 = 2000. While the symbol remains the same, the meaning changes with its position and is interpreted as a multiple of a power of 10. Similarly, a number to the right of a decimal point refers to negative powers of ten; for the number 0.58 the 5 refers to 10 _ 1 while the 8 is for 10~2. If we let b be an integer greater than 1, we can write numbers in base b, using powers ofb rather than 10 and b digits 0 through b. In base 5 the digits are 0, 1, 2, 3, 4; In base 2 (binary) the digits are 0, 1; in base 12 (duodecimal) we use, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, T, and E. We will use the subscript b to say a number is written in that base. So 324s refers to base 5. As for going back and forth between bases b and 10, converting base b to base 10 is done by multiplying out the powers of the base; so 3245 = 3 x 5 2 + 2 x 5 1 + 4 x 5 ° = 75 + 10 + 4 or 89 in base 10. Converting from base 10 into another base can be done by finding the most copies ofbn that "fit" into the numberfor each positive integer power ofn. The number 278 can be written as 2x125+1x25+3x1 = 2 x 5 3 + l x 5 2 + 0 x 5 1 + 3 x 5 ° = (2103)5. A second way to do this is with a division trick, looking at the remainders from repeated division (by 5 in this case since we 're converting into base 5). 278 -T- 5 = 55 with a remainder of 3, 55 4- 5 = 11 with a remainder ofO, 11 -f- 5 = 2 with a remainder of 1, 2 + 5 = 0 with a remainder of 2. We stop here since the dividend is 0. Now if we read the remainders bottom to top we get 2 - 1 - 0 - 3 which tells us 278 = (2103)5. To convert a base 10 decimal into another base, b, we multiply repeatedly by b and collect whole number parts. For example, turning 1/5 = 0.2 into a base 4 number we have 0.2 x 4 which is 0.8 whose whole number part is zero. This means the first spot to the right of the base 4 quaternary point is a zero (of course we cannot say the "tenths place" as we are not in base 10). Then 0.8 x 4 has whole number part is 3 which places a 3 two spots to the right of the decimal. After removing the 3 there is 0.2 leftover and the cycle starts all over again. Thus - = (0.030303...) 4 . o

For a second example of a rational number expansion we look at the number 35/9 and a second technique for computing the base 7 number. This method uses factoring.

20

fl

INTRODUCING THE CANTOR SET

EXAMPLE 3.1 In this method we are factoring out powers of 7 along with converting the nu­ merator into a sum where the first time is a multiple of the denominator 9. 35 9

_

27+8 _ Q _ L 8 _ O , 1 5 6 _ O , 9 9 79

=

3 + y + jz + Yi ( g ) = 3 +

1 54+2 7 9

y + if2 + 73 (-9-) ,

at which point our sequence repeats itself. Thus we arrive at — = (3.613613613... ) 7 .

The construction of a Cantor Set fits right into base 3. When we divide the unit interval into thirds, we can look at this as splitting the numbers up according to their first digits. The left section consists of numbers whose first decimal digit (in base 3) is a zero, the middle section has all numbers whose first decimal is 1, and the right section has numbers whose first digit is 2. Removing the open middle rids us of all numbers whose first digit is a 1. When the left interval is subdivided into thirds and the middle removed, what we are keeping is all the numbers between 0 and 1 which begin in base 3 as .00 and .02. Technically, we have to be a little careful as some numbers are not unique in their representation. For example, in base 10 1.999999... = 2.0000.... Now if we write our numbers between 0 and 1 in base 3, we can say the Cantor Set consists of all numbers which have a base 3 representation consisting only of 0 's and 2 's to the right of the ternary point and a 0 to the left of the ternary point. The endpoints of the intervals in the creation of the Kn, such as 1/3 or 8/9, are points in the Cantor Set and correspond to those numbers in base 3 can be written that way. Quick arithmetic shows

jj = (0.22000... ) 3 and i = (0.1000...) 3 = (0.0222... ) 3 . Both of those are endpoints of intervals in some Kn. There are points in C\/% that are not endpoints of intervals (we can tell a point is at the end of some interval because it will have a finite expansion in base 3). For an example of a non-endpoint in base 3 we have (0.02020202...) 3 , which is 1/4. Let us say a bit more about what goes on with points in the Cantor Set. This comes from [53] where the rational values in C are partitioned.

SIZE OF A CANTOR SET

21

The numbers that are rational in base 3 are the numbers that can be written with finitely many 2 's after the ternary point (the name for the decimal point in base 3). These numbers look like x\

x2

x3

3

2

3

3

3

t

xn

xi3n_1+x23n~2 +

3"

V xn

3"

where each Xi is either a 0 or a 2. These correspond to the left endpoints of the removed intervals in the Cantor Set construction. Purely periodic points have the form Q,(x\X2Xz ... xn)s again with Xi G {0, 2}. Expanding these lead us to x^-1

+ x23n~2

+ ■ ■ ■ + xn (

On

\

, 1 , Qn

1

g2ra

and cleaning up the geometric series gives us xx?,n-1

+ x 2 3"~ 2 + • ■ • + xn 3™-l

Numbers in this category include 1/13, and 1/10. The rest of the rational points are mixed rationals, those whose repeating patterns do not start after the ternary point, but instead become periodic after some terminat­ ing number. For example, (0.220)3. We can find this number's value in base 10 by relating it back to a purely periodic number. Here we see (0.220)3 = (0.2)3 + (0.020)3 = (o.2 + ^ 0 . 2 o ) 3.2

= | + \ ( ^

= ^.

Size of a Cantor Set

The size of a Cantor Set is an important question. Is it a big set or small? Before we can answer this question, we must define what we mean by saying a set is small. This notion comes from Van Rooij and Schikhoff [75]. We say a collection of sets S from K is a collection of small sets if 1. R ^ <S (the whole line is not small). 2. If'A € S and B C A, then B G S (a subset of a small set is small). 3. If A is small, then for every fixed value b so is b + A = {b + a : a G ^4} (the translation of a small set is small). 4. IfAn, n > 1 are each small, then so is UAn (the countable union of small sets is small). Some of the collections we shall look at that qualify as a small set include countable sets, measure zero sets, and first category sets. It is worth noting that finite sets do not meet this idea of a small set. As we shall see, in some ways the Cantor Set is small and in other ways large.

22

INTRODUCING THE CANTOR SET

3.2.1

Cardinality

Cardinality deals with counting the number of elements in a set. Cantor's work in this area opened mathematics up to the ideas of what it means for a set to be infinite and the notion of different types of infinity. First we need to recall two properties a function from Chapter 2, followed by examples. Definition 3.1 Let A and B be sets and suppose f is a function with domain A and range B. We say f is one-to-one iffor a ^ b in A we have f(a) ^ f(b). The equivalent definition one sometimes sees is f(a) = f(b) implies a = b. We say f is onto B if for every b G B there exists an a E A such that f(a) = b. If f : A —> B is one-to-one and onto, then we say f is a bijection.

fl

EXAMPLE 3.2 Let f : R -» K be defined by 1. f(x) = ex. Then f is one-to-one. but not onto. 2. f(x) = x3 — x. Then f is onto, but not one-to-one. 3. f(x) = 2x + 4. Then f is a bijection. u

Now we are ready to look at equivalence between sets and cardinality. As we will see, cardinality, even if not known under that name, is something with which the reader is quite familiar. Definition 3.2 Let A and B be two sets. We say A is equivalent to B if there exists a bijection, f, between A and B. Our notation for equivalent sets is A = B. For example, A = {a, b, c} and B = {x, y, z} are equivalent while there is no equivalence between A and C = {a, /3,7,5}. Our equivalence = is a special instance of an equivalence relation on a collection S. Let us take a step back and define equivalence relations and partitions before we move on. Definition 3.3 Let S be a set. A relation , ~, on S is a subset of S x S. We write

to say, "x is related to y."

SIZE OF A CANTOR SET

fl

23

EXAMPLE 3.3 1. On the set Z say x ~ i y if and only ify — x is an even number. 2. On the set S = {x : x is a person with the first name Sam} we let x ~2 y if and only if x and y are siblings. 3. On the set K we say i ~ 3 j / if and only ifx < y. u

The reflexive, symmetric, and transitive properties are important, especially when a relation ~ satisfies all three at the same time. Definition 3.4 Let ~ be a relation on the set S. We say the relation is 1. reflexive if for every x G S it is true that x ~ x. 2. symmetric if for every x,yESifx~y, 3. transitive if for all x,y,z

then y ~ x.

G Sifwe know x ~ y and y ~ z, then x ~ z.

If a relation is reflexive, symmetric, and transitive, then it is referred to as an equiv­ alence relation. fl

EXAMPLE 3.4 For the relations above, ~ i is an equivalence relation, ~2 '•$ only symmetric, ■ and ~3 reflexive and transitive, but not symmetric.

When the set S has an equivalence relation on it, we can collect equivalent ele­ ments together. These collections are called equivalence classes. Definition 3.5 Let S be a set and ~ an equivalence relation on S. For any x G S we define the equivalence class of x as

\A = {y e s: y ~ x}. [ 8 EXAMPLE 3.5 Let S be the set of cards in a standard deck of 52. For two cards x and y, we say x is related to y if they are both diamonds (or both spaces, both hearts, both clubs). This is an equivalence relation and the equivalence classes are the suits clubs, spades, hearts, and diamonds. If instead x is related to y by them having the same value, then there would be 13 equivalence classes: aces, deuces, treys, ..., queens, and kings. ■

24

S

INTRODUCING THE CANTOR SET

EXAMPLE 3.6 Let P be the set of all polynomials with integer coefficients. We say p(x) is related to q(x) if they have the same y-intercept. This is an equivalence relation and the equivalence classes are the functions which all agree at 0. Using our notation, for a function p(x) we have [p(x)] =

{q(x)eP:q(0)=p(0)}.

Furthermore, if p{x) = 3x2 + 2x + 5 we could see that q(x) = —7x + 5 is a member of \p(x)] while r(x) = x3 — x1 + 4 is not. I The theorem below gives us three important properties for equivalence classes. Casually, these say equivalence classes are like the pieces of a jigsaw puzzle. Each piece contains part of the picture, no two pieces contain the same part, and together they make up the whole scene. Theorem 3.2 Let S be a set and ~ an equivalence relation on S. Then the following are true for the equivalence classes [x] induced by ~. 1. For each x £ S, [x] ^ 0. 2. For each x,y £ S either [x] = [y] or [x] n [y] = 0. 3. UxeS[x] = S. Proof: We will prove part 1 only. Parts 2 and 3 will be left for homework. Pick x £ S and look at [x\. Since ~ is reflexive, x ~ x which means x G {y € S : y ~ x} = [x\. Therefore [x] ^ 0. 4 Now let us get back to cardinality. We say sets A and B are equivalent if and only if there exists a bijection f : A —> B. It is fairly obvious that any set is equivalent to itself so we have that = is reflexive. Any bijection f has an inverse / _ 1 which is also a bijection defined by f~l{b) = a if and only if f(a) = b. Thus the symmetric property is satisfied. Finally, if f is a bijection between A and B and g is a bijection between B and C, then the composition go f is a bijection between A and C. Hence = is transitive and we have an equivalence relation. This means we have equivalence classes and that is where we willfindcardinality. Definition 3.6 Let A be a set. If there exists a natural number k such that A = { 1 , 2 , 3 , . . . , k}, then A £ [{1,2,3,..., k}] and we say A is a finite set with cardi­ nality n(A) = k. Otherwise A is an infinite set. So we have a strict definition to show that A — {a, b, c] is a finite set with cardi­ nality 3. Definition 3.7 A set A is countable if it is either finite or equivalent to N. When the latter is true the set is called countably infinite. If a set is not countable, then it is called uncountable.

SIZE OF A CANTOR SET

25

When a set in M. is infinite, its cardinality is denoted using the Hebrew letter aleph (KJ. For countably infinite we use Ho. The collection of countable sets are our first type of small set using the criteria of [75]. Let us look now at examples of countable and uncountable sets and the proofs of their various cardinalities. But before our first result, let us state a lemma. Its proof is left as a homework exercise, but we shall use it right away in Theorem 3.3. Lemma 3.1 An infinite subset of a countable set is countable. Theorem 3.3 The set Q+ = {x\x is a positive rational number} Z + , q / 0 } i s a countably infinite set.

= {p/q\p,q

Proof: We begin by writing out the positive rationals in the following 1

2

3

4

5

■••

1/2

2/2

3/2

4/2

5/2

•■•

1/3

2/3

3/3

4/3

5/3

•••

1/4

2/4

3/4

4/4

5/4

•■•

G

way:

Starting the upper leftmost corner and moving diagonally left side up to top, we get Y

2^

3*

jP"

^

P?2

2/2

3/2

4/2

5/2

•••

#3

2/3

3/3

4/3

5/3

••■

yf4

$fA

3/4

4/4

5/4

•••

This gives us the one-to-one 1

1/2

1 2

■■■

matchup 2

1/3

2/2

3

1/4

2/3

■••

3

4

5

6

7

8

••■

Thus the first array is equivalent to N. Technically there are repeats (after all 1/2 = 2/4 and 3/1 = 6/2), but regardless, this shows the positive rationals are equivalent to the natural numbers and hence have the same cardinality. 4 Theorem 3.4 If each set An, n > 1 is a countable set, then the union U ^ L ^ n is countable. Proof: Similar to the proof for the rational numbers, each Ai can be written out as M = {ai,i,a1]2,ai,3,---},

26

INTRODUCING THE CANTOR SET

A2 = {a2,l,02,2,02,3,•■■}, A3 = { a 3 , i , a 3 , 2 , a 3 i 3 , . . . } , Ai = {0.4,1,04,2, 04,3, • • •},

where a,ij refers to the j element in the i set. Then a diagonal argument puts these element m l — 1 correspondence with N. 4(t From dealing with functions, we are used to ordered pairs (a, b). In general, given two sets A and B we define Ax

B = {(a,b) : a e Aandb

e

B}.

There is nothing special about having only two sets, so this can be fully from an ordered pair to an n—tuple by the following.

generalized

Definition 3.8 Let Si, i = 1, 2 , . . . , n be a collection of sets. We define the Cartesian product of the Si by Si x ^2 x ■•• x Sn = {{x1,x2,...,xn) Each element {x\, x2 ■ ■ ■ ■, xn) is called an (ordered)

: xt G Si}. n-tuple.

Theorem 3.5 IfSk is a countable set, k = 1, 2 , 3 , . . . , n, then Si x 5 2 x S3 x • ■ ■ x S n is also a countable set. Proof: This is yet another example of a diagonal argument and will be left as an exercise. 4k This extends to countably infinite many countable sets. Theorem 3.6 If Sk is a countable set, k = 1, 2, 3 , . . . , then Si x S2 x S3 x ■ • •

is also a countable set. In 1874 Cantor published a proof that an interval of the form [a, b] is uncountable. Later, in 1891, he published another paper containing his famous "diagonalization" proof of this. We give our own version of the proof now. Theorem 3.7 The unit interval [0,1] is uncountable. Proof: Every number in the unit interval can be written as 0.did2ds ..., where each di is a digit from 0 to 9 and (for sake of having the representation be well-defined)

SIZE OF A CANTOR SET

27

no number ends in an infinite string ofO's, but instead as an infinite string of 9's; for example, 0.135000 . . . will be written as 0.1349999 . . . . This is the only possible redundancy in representations. Proceeding with a proof by contradiction, suppose the numbers in [0,1] are countable. Then we can put the set [0,1] into 1 — 1 correspondence with N as follows: a\ = Q.d\d\d\ . . . ,

a2 = O.dldldl..., a3 = O.dldldl...,

where d\ refers to the ith digit in the decimal expansion of the number a,j. Let us create a number x G [0,1] by x = O.&16263 • • • where bi = 3 if ai 7^ 3 and bi = Aifai — 3. Then x ^ a,j for any j since the two numbers must disagree in the jth spot. This contradicts our list being a complete roster of the numbers in [0,1]. Thus the unit interval is uncountable. 4 The cardinality of the real numbers is 2N°. We note, without proof, that the cardi­ nality of the unit interval is also 2^° Corollary 3.7.1 The set of irrational numbers in [0,1] is uncountable. Proof: We do this by contradiction. Suppose the irrationals, (K \ Q) n [0,1], were a countable set. Then since the union of countable sets is again countable, we have

[o,i] = (Qn[o,i])u((R\Q)n[o,i]) is a countable set, which we know is false, 4k The next result, which we include for completeness, will not be followed with proof. It comes from the fact that if A is uncountable and A C B, then B is uncount­ able. Similarly a subset of a countable set must be countable. Corollary 3.7.2 The set R is uncountable. For an example of something different, we look at another way (other than ra­ tional/irrational) to partition the real numbers. We say x is a algebraic number if it is the root of a polynomial with integer coefficients. So \f2 is algebraic as it is a solution to x2 - 2 = 0. If a number is not algebraic, then it is transcendental. Well-known transcendental numbers include ir and e. There are many, many others. The proofs for n and e being transcendental are non-trivial and will not be presented here.

28

INTRODUCING THE CANTOR SET

Theorem 3.8 The set of algebraic numbers is a countable set. Proof: Each algebraic number a is associated with a polynomial p(x) with integer coefficients. So we will show this collection of polynomials is countable. Let us fix n 6 N . Then every polynomial of degree n with integer coefficients can be matched up with an (n + l)-tuple in the following manner: p{x) = c„x n +c„_ 1 x n _ 1 +c„_2^"~ 2 H

\-CiX+c0 ^

(c„,c„_i,cn_2,...,ci,c0).

These (n + \)-tuples are countable since they consist of integers. Next, the union over all values of n of these (n + \)-tuples is countable and taking this backwards to the polynomials with integer coefficients they are countable. Finally this brings us that the set of algebraic numbers is countable. 4k Corollary 3.8.1 The set of transcendental numbers is uncountable. Proof: This is the same as the proof for the irrational numbers. 4t The Cantor Set is an uncountable set as we shall prove. This really shows the paradoxical nature of the set. We start with the unit interval, [0,1], remove open intervals (themselves uncountable) whose total length is one, yet what is leftover has the exact same cardinality as the unit interval. Theorem 3.9 The cardinality of the Cantor Set is 2N°. Proof: As we have already seen, the Cantor Set can be represented by the set of all numbers between 0 and 1 which have a base 3 representation that consists of only 0's and 2's. We construct a function from C1/3 to [0,1] by using the base 2 (binary) representation of number in the unit interval. Given a number in the Cantor Set we can think of it as x = 0.ai<22a3 • • • with <2j G {0,1} and then we construct

f(x) = O.hhh where b =

f0

ifai = 0

It is left as a homework exercise to show the function is 1 — 1 and onto. So we have C1/3 ^ [0,1] andn(Cl/3) = 2H°. 4 We note here that it is this binary representation of the unit interval that shows us whyn([0,l]) = 2*°. An important application of countability follows in Theorem 3.10. It is used in real analysis to help describe what we refer to as "exceptional sets." This does not mean the set is especially wonderful, it means that a certain property holds except on this set and this is how we can characterize the set. So we give as an example, but without proof the following:

SIZE OF A CANTOR SET

29

Theorem 3.10 If f : M. —¥ M. is an increasing (or decreasing) function , then f is continuous except on a countable set; that is, if A = {x : f has a discontinuity at x}, then A is countable. There are more infinite cardinal numbers than the ones we have looked at. In fact, there is an infinite hierarchy. After Ho there is Hi and H2, in fact, there are H n /or every natural number n, Hw (to is the first infinite ordinal number) and beyond. Let us end this section with one more unintuitive result from set theory and cardi­ nality. The unit interval has cardinality 2K° which seems obviously to be bigger than Ho. The hierarchy of alephs has Hi succeeding Ho- So where does 2K° fit in? The claim that 2*° = Hi is called the Continuum Hypothesis. This was the first of 23 problems that David Hilbertpresented at the 1900 World's Exposition as the most important mathematics problems for the upcoming 20th century. Kurt Gbdel proved in 1940 that using the Zermelo-Frankel axioms of set theory with the Axiom of Choice (together called "in ZFC") one cannot disprove the Con­ tinuum Hypothesis. In 1963, Paul Cohen showed that in ZFC the Continuum Hy­ pothesis cannot be proven true, either. Thus we have a statement that it is impossible to prove or disprove. 3.2.2

Category

Thus far we have one way of looking at the size of a set. A countable set is in some way small, while an uncountable set is large. Another way of small versus large is the idea of category. Category is part of the larger study of topology. In topology we generalize the idea of open sets. The first appearance of a topological space was in Felix Hausdorff's "Grundziige der Mengenlehre" (Foundations of Set Theory) in 1914. Here is the modern definition. Definition 3.9 A topological space is a set S together with a collection T of subsets of S which satisfy the following: 1. %,Se

T,

2. IfUa(a&

A, an index set) is in T, then

\JuaeT,

aeA

3. IfU, V e T, then U C\V

eT.

The sets in T are called the open sets in S and T is called a topology on S. When we refer to a space as a topological space, it is a pair (S,T). So we get to define what it means for a set to be open. Some examples of topolog­ ical spaces include the following.

30

S

INTRODUCING THE CANTOR SET

EXAMPLE 3.7 1. Let S be any set and T = {0, S}. This is called the trivial (also the indiscrete) topology on S. 2. Let S be any set and T = {X : X C S}. This is called the discrete on S. 3. L e f 5 =

topology

{l,2,3,4}fl«dr={0,{l},{l)2},{l)3},{l,2)3},{l)2,3,4}}.

4. Let S = R and T = { i W i : Ul = {a{. fy) : en, bt e R, a% < bt} U 0 U R This is the Euclidean topology on R. 5. Let S = R and T = {U.Ui : Ui = \az, bt) : au bt e R, a, < h} U 0 U R. This is the Sorgenfrey line.

u Of course once we have an idea of what is an open set, we then get closed sets, too. Definition 3.10 A set V is closed in the topological space (S, T) if its complement Vc = S\V is a set in T; that is, ifVc is open. Let us look at how this generalizes what we know about K. A sequence of real numbers {xn} converges to the real number x$ xn and XQ can be made as close as desired by making n sufficiently large. Formally we say the following: Definition 3.11 Let {xn} C M. We say xn converges to .To if for every e > 0 there exists an N G N such that for n > N \xn ~ %o\ < e. However, in a topological space there is not necessarily an idea for "distance" to say closer. Our convergence in R is equivalent to saying that for any positive e there is a natural number N so that ifn is greater than N, xn is in (XQ — e, Xo + s). Since some K the tail of a sequence, this becomes for every e > Ofor we call {xn}n>jifor K large enough all the tails are in the open interval (XQ — e, xo + e). Open intervals are what we generalized to get to open sets in a topological space. This leads us to this definition of convergence. Definition 3.12 Let (S, T) be a topological space and {xn} a sequence of elements from S. Then we say {xn} converges to XQ 6 S if for every set U G T such that xo £ U there is an N G N such that ifn > N, then xn e U. As the examples below show us, strange things can happen depending on the topology on the space. Before the first example we need a definition. We say a sequence is eventually constant if, for some c s S, xn = cfor n> N.

SIZE OF A CANTOR SET

fl

31

EXAMPLE 3.8 Let S have the discrete topology. If{xn} is not eventually constant, then {XQ} is an open set which does not contain any tail of the sequence. Thus the only convergent sequences are the eventually constant ones. ■

S

EXAMPLE 3.9 For S with the trivial topology, every sequence converges to every point. Let {xn} be a sequence in S at XQ an arbitrary point in S. The only open set containing XQ is S and xn € S for all n > 1. Thus since XQ is arbitrary, {xn} converges to any point. ■

For another example of how things extend from R to topological spaces, let us look at the idea of continuity. There is the 5 — e definition of continuity: A function / : R -> 1 is continuous at the point XQ if for every e > 0 there exists a S > 0 such that for all x with \x — XQ\ < 5 we have \f(x) — /(xo)| < £• As with sequences, this depends on absolute value (distance) in R which we do not necessarily have in a topological space. Looking closely, though, we can see that unfolding the absolute values shows that what we are dealing with are open intervals in M. which we can then generalize to open sets. To do so, we need to know about the inverse image of a set. So let f : S —>• V and U C V. The inverse image ofU is the set of all x € S which map into U. The notation for this is

r1(U) =

{xeS-.f(x)eU}.

Although the notation is the same, this is not the inverse function. We are not claim­ ing f is one-to-one so the inverse function may not exist. Definition 3.13 Let (S,Ti) and (V, Ti) be topological spaces, f : S —>■ V, and XQ 6 S. We say that f is continuous at x0 if for every open set U 6 7i with f(xo) € U the inverse image ofU

f-1(U) =

{xeS:f(x)eU}

is open in T\. In addition, we say f is a continuous function if it is continuous at each point in its domain. The example below shows how this works and some of the strangeness involved. fl

EXAMPLE 3.10 Let S be the set R given the Euclidean topology and V beM. given the discrete topology. Let f be the function f(x) = x (the identity function). If the domain is S and range V, then f is not a continuous function: {xo} is open in V, but not open in S. If things reverse themselves and the domain is V and range S, then not only is f continuous, but every function is continuous. m

32

INTRODUCING THE CANTOR SET

Denseness and category are defined for any topological space (S, T ) now. To get the full power of our big result (The Baire Category Theorem) we need the additional property of a space being complete. We are saving that particular property for the chapter on Self-Similar Sets. So in a bit we will switch to only working in M with the usual, Euclidean, topology. Definition 3.14 Let (S, T) be a topological space and let A be a subset of S. We say that A is dense in S if for every U € T there is an a G A such that a £ U. That is, every open set contains at least one point in A. Now we define an idea for a sparse type of set. We are trying to make things so that no part of the set can be thought of as dense. Definition 3.15 Let (S, T ) be a topological space and B a subset of S. Then B is nowhere dense in S iffor every U C S there exists an open setV
= $.

A set can be neither dense nor nowhere dense. It is an exercise in the homework to come up with an example of such a set. This idea is not small in the Van Rooij and Schikhoff way. It is not true that the countable union of nowhere dense sets will be nowhere dense. This is a stepping stone to getting such a smallness. Definition 3.16 A set A is called first category if it can be written as the countable union of nowhere dense sets. So there is a collection Ai where each Ai is nowhere dense and

A= U ^ i . If a set is not first category, then it is called second category. Now let us concentrate on just R with the Euclidean

topology.

A set S in the real line is nowhere dense if, given any open set O there exists an open interval J C O such that J n S = 0. Any finite set will be nowhere dense. The integers are an example of a set which is infinite and also nowhere dense. For a set S to be dense in the line it must be true that for any non-empty interval I, IC\S 7^ 0. The irrational numbers are dense in the line as are the rationals. The notions of dense and nowhere dense are not "opposite " in the strictest sense. Finally, sometimes sets that are not nowhere dense can be broken down (partitioned) into pieces which are nowhere dense. The canonical example of a first category set which is dense in the line is the rational numbers. Each singleton rational number {r} is a nowhere dense set and since the rationals is a countable set Q is first category. However, every non-empty, open interval in the line contains rational points. For the most famous application of category, we need to delve a little more deeply. To start, we say that a set S is residual in K ;/ its complement M \ S is of the first category. So the set~R\Z is a residual set. Then we get to the following theorem:

SIZE OF A CANTOR SET

33

Theorem 3.11 The Baire Category Theorem Every residual subset of R with the Euclidean topology is dense in K. There are other ways to state this. For example, if we have a countable collection of dense, open sets Ak it must be true that C\Ak is dense in R. Alternatively, the complement of UAk must be a nowhere dense set. We will use this idea to prove a result at the end of this chapter. Another way to look at the Baire Category Theorem is to play the so-called Banach-Mazur Game. This game involves two player we will call Player I and Player II. Player I is given a set A in R while Player II then gets R \ A, the comple­ ment of A. In alternating moves the players choose closed and nested intervals. This means Player I chooses a closed interval I\, then Player II chooses a closed interval Ii C I\ after which Player I chooses I3 C I2, and so on. We then look at the point x e nifc. If x € A, then Player I wins, otherwise Player II is the winner. The question is, "Under what circumstances is Player I guaranteed to have a winning strategy?" What Player I needs is the set to be large in some respect which turns out to be In order to be guaranteed a win for Player I, the set A must be second category. Thus some proofs that a set is second category are proofs that Player I can win the Banach-Mazur Game. The origin of this result is worth looking into as a fascinating glimpse into the culture of mathematics. The idea of the game was first posed in the Scottish Book. From 1935 to 1941, a group of mathematicians from the University of Lwow (in Poland at the time, now part of Ukraine) would gather at The Scottish Cafe House. Aside from Banach and Mazur, this group included Ulam, Steinhaus, Borsuk, and others. They would discuss their various works and, if an interesting problem arose, would signal the waiter to bring out a notebook kept behind the bar. Problems would be written in the notebook. Some of the problems had prizes attached to them. For more on this fascinating time see [49]. Problem 43 contained the beginning of the game and proposed by Mazur. It was solved by Banach (with the prize being a bottle of wine), but he never published his proof. The book survived World War II and was published by Ulam in 1957. Let us turn our attention back to the Cantor Set. S

EXAMPLE3.il The Cantor Set is a nowhere dense set in the real line.

■

Proof: Let I be any open interval in the real line. Then there exists a positive integer k such that 3~fc < £(I), recall £(I) denotes the length of I. Recall that in the construction of the Kn+\from Kn any existing interval is divided into thirds and the

34

INTRODUCING THE CANTOR SET

middle subinterval is removed. Hence at stage n + 1 there never remains three sideby-side intervals of length 3 - r a _ 1 . Thus I must contain an open interval J which is removed in the [k + l)st step in the Cantor construction. 4k As for one application of the Baire Category Theorem, it can be used to show that there exists a function / : [0,1] —» R which is continuous everywhere, but has no point at which it is differentiable. Such a proof using this theorem can be done implicitly, that is, without actually having to construct such a function. This is not exactly part of the theme of this book, so we will leave it alone, but it is a very interesting result which should be a part of any course in real analysis. A second application is an interesting property of continuity. For this we will use the more intuitive Calculus I definition that f is continuous at XQ (/" limx_>Xo f(x) exists and equals f(xo). We begin with Thomae's Function (also called the Popcorn Function as well as other names). It is defined by 1/q, F(x)

if

xeQ\{0},

1,

ifx = 0,

0,

ifx

where for rational x, x = 2 with gcd(p, q) = 1. This function is continuous exactly on the irrational numbers. It's easy to realize that for rational x there are irrational numbers arbitrarily close so that l\mx-yXo f(x) does not exist. For irrational x, any rational x nearby must have increasingly larger denominators showing f is continuous on the irrationals. Interestingly enough, we cannot reverse the roles of sets on which a function is continuous. The proof involves the Baire Category Theorem. Theorem 3.12 There cannot exist a function G which is continuous on the rational numbers, but discontinuous on the irrational numbers. Before we prove the theorem, we need a lemma concerning the nature of the ra­ tional numbers. This lemma states that the rational numbers as a subset ofM. do not form a Qs set. Recall a set is Qs if it can be written as the intersection of a countable collection of open sets. Lemma 3.2 The rational numbers Q are not a Qs set of the real numbers with the Euclidean topology. Proof of Lemma We will do this as a proof by contradiction. IfQ was a Qs set, then there would be {Un}, a countable collection of open sets so that

Q = f]Un.

SIZE OF A CANTOR SET

35

As a one-point set, each rational number forms a closed set so that Vq = R \ {q} is an open set. Then we can create the collection of open sets V = {Un : n G N} U {Vq : q e Q}. By the Baire Category Theorem, these dense, open sets in R must have an inter­ section that is dense. However, for V the intersection is empty (one grouping has exactly the rational numbers, the other misses the rationals, one at a time). So our assumption that the rationals numbers form a Q$ set must be false. 4k Proof of Theorem Let f : K —> R be an arbitrary function. We shall show that the set of points at which f is continuous must be a Q§ set and thus, with the Lemma above, the set cannot be the rationals only. For a set A we define its diameter to be the largest of the distances between points in the set; that is, The diameter of A = max{|6 — a\ : a, b € A}. Define the set Wn by Wn = {U : U is open in R and the diameter of f(U) < 1/n}. Each set Wn is open since it's the union of open sets and the points at which f is continuous are precisely

nwn. It is obvious that C\Wn is a Gs set. Thus we have proven our result. tf> One of the key topics studied in topology deals with homeomorphisms between spaces. This is a special type offunction between two sets A and B. A property is referred to as topological ;/ whenever A has the property, and B is homeomorphic to A, then B has the property. Definition 3.17 A function f : A —>• B is a homeomorphism if it is bijective and both f and f~l are continuous functions. We say A and B are homeomorphic if there exists a homeomorphism f : A —» B. An interesting result involving homeomorphism is a relationship between Cantor Sets and compact metric spaces. Metric spaces were (lightly) introduced in Chapter 2. Compact sets are not yet defined, but in the real line any closed and bounded set is compact. We state this theorem without proof, but will revisit it, with proof, in Chapter 6 after more is said about metric spaces. As for the theorem: Theorem 3.13 Any compact metric space is the continuous image of the Cantor Set. This is proved later in Chapter 6.

36

INTRODUCING THE CANTOR SET

3.2.3

Measure

Our last way of talking about small sets uses measure. There are various ways to come up with measures for a set (e.g., Jordan measure, Haar measure, Peano mea­ sure). For now, we will concentrate on Lebesgue measure, which Lebesgue developed in 1902 for his PhD thesis "Integrate, longueur, aire (Integral, Length, Area)". We begin with the basics. It is easy to find the length of an interval, if I — [a,b], then 1(1) = b — a. This also works for open or half-open intervals. We would like to generalize this to come up with an idea of "length "for any set. Instead of length, we call it measure and denote measure of a set E by m(E). The nice properties for measure to ideally have are: 1. Measure is defined for each set E of real numbers. 2. For an interval I, m(I) = (.(I). 3. If En is a sequence of disjoint sets, then m(ljEn)

= ^2 m(En).

4. For any set E, m(b + E) = m(E) where b + E is the translation of E by b units. Unfortunately, it turns out to be impossible to construct a function on sets having all four of these properties. Something must be sacrificed. Before we can talk about measure we need to define an intermediate property called the outer measure. Let E be a set in R. An open cover of E is a countable collection of open inter­ vals, On = (an, bn) so that

EC\J

On.

n>l

We say the outer measure of a set is the finite real number c if c^infi^^Oi)!

[n>i

J

=infi^(6n-an)l,

yn>i

J

where the infimum is taken over all possible open covers of E. We write this as m*(E) = c. Immediately we see that m*(0) = 0. It is also reasonably quick to tell that if E C F, then m*(E) < m*(F). For an example of working with this, let us prove that for any non-empty finite set has outer measure zero. S

EXAMPLE 3.12 If E = {xi, x-2, X's,..., xn}, then m*(E) = 0.

■

Proof: We will show that for any positive e > 0, m*(E) < e and then using "lime_>.o+ "■ S° let e > 0 be given. Since n > 1 is fixed, for each Xj G E, de­ fine Oi = (x,i — e/2n,Xi + e/2n). Each Oi has diameter e/n and E C UOj.

SIZE OF A CANTOR SET

Now

37

n

y ^ £{Oi) = n ■ s/n = e. i=l

This means m* (E) < e and this gives m* (E) = 0. 4 Of course, our focus is on the Cantor Set, and since the total lengths removed is 1 it should be no surprise that our next example is to show m*(Ci/3) exists and, in fact, is zero. fl

EXAMPLE 3.13 The Cantor Set has outer measure zero.

u

Proof: Let e > 0 be given. Recall that at stage n in the construction Kn consists of2n disjoint closed intervals, I\, I 2 , . . . , hn each of whose length is 3~™. Each Ij then can be contained in an open interval Oj where £{Oj) < 3 _ n + 1 . The {Oj} are then a collection of open interval which contain Kn (hence also C1/3) and '2

^ £ ( 0 J ) = 2"-3-™+1 = 3 ( 1 So by picking n large enough we guarantee

m*(C71/3)<3Q) <e. Since m*(Ci/^) is less than an arbitrary epsilon, we conclude m*(Ci/z) = 0. ♦ Now outer measure fails to have all the nice properties we want. It is possible to have m*{yjEn)
m*{A n

Ec),

is the complement of E.

It is trivial to show that both 0 and K are both measurable sets. Note here that the idea behind a proof of measurability of a set usually consists of two pieces. First show that m*(A) <m*{Ar\E)+

m*(A n Ec)

38

INTRODUCING THE CANTOR SET

and then show that m*(A)

> m*(A nE)+m*(An

Ec).

However, part of this we get for free. For any set A and any set E, it must be true that AC (ADE)U(AnEc), which then automatically gives the result m*(A)

< m* (An E) + m* (AC\

Ec).

Thus we only need to show the other direction. To see this in action, we shall prove that any set with outer measure zero is a measurable set. Theorem 3.14 Let E be a set with m* (E) = 0. Then E is, in fact, a measurable set. Proof: Let A be an arbitrary set. Now {A n E) C E hence m* (AnE) = 0. Then A contains (A n Ec) as a subset we have m*{A) > rn*{A n Ec). Those two results together give m*(A) > rn*{A n E) + m*(A C\EC). Since A was arbitrary, this works for all A and we have proven E is measurable. 4ft Corollary 3.14.1 The Cantor Set is a measurable set. That the usual Cantor Set C is measurable is not a shock. It is possible to modify the Cantor Set construction so as to get a Cantor Set with positive measure. One construction of such a set, called a "Fat" Cantor Set, is in the exercises and Chapter 10. What we have sacrificed in creating measure from outer measure is the first cri­ terion, now not every set has a measure. We will derive an example to show that Property 1 does not hold; that is, exhibit a nonmeasurable set. This is a constructive development, but not explicitly so. Before we begin, though, recall the definitions for equivalence relations and partitions. fl

EXAMPLE 3.14 There exists a set E that is not measurable.

■

Proof: We start with the unit interval [0,1]. For any two numbers x and y in the unit — x is a rational number. This is an equivalence interval we say x is related toyify relation (reflexive, symmetric, and transitive (see the exercises)) which means we can partition the unit interval into classes, collections of equivalent numbers. Let S c l be the set which consists of one element from each of the equivalence classes (this can be done through The Axiom of Choice, a heavy-duty concept we are not delving

SIZE OF A CANTOR SET

39

into here). Now for each rational number r we define Sr — (r + 5')(modl). This means for each element s of S find r + s and replace the integer part with 0. For rational numbers r ^ s, SrC\Ss = 0 and the Sr form a partition of the unit interval. Thus LVgQSV = [0,1].

Our claim is that these Sr are nonmeasurable. If they were, then 1 = m([0,1]) = m(UreqSr) Ifm(Sr) = 0, this means "1 = 0" while ifm(Sr) we have a set that does not have a measure. A

=^

m(Sr).

> 0, this means "1 = oo." Thus

Once we have some measurable sets, we can create more by combining them in the usual way. We show this in the theorem below. Theorem 3.15 Let E and F be measurable sets. The the sets Ec, EUF, and ECiF are all measurable sets. Proof: We will prove the result for E U F and leave the others as exercises. Let A C M. As noted above, we really only have to prove m(A) > m(An(EU

F)) + rn{An(EU

F)c).

The rules for DeMorgan 's Laws give us

A = (AnE)u(AnEc) = (AnE)u(AnEc nF)u(AnEc n Fc). Then the subadditivity of measure leads us to m(A)

=

m((A n E) U {A n EG n F) U (A n Ec n Fc))

>

m((A n E)) + m((A n Ec n F)) + m((A n Ec n Fc))

and we have E U F is a measurable set. 4 We also get the following corollaries which will end (for now) our look at mea­ surable sets. Note that these results tell us that measurable sets form a a-algebra. Corollary 3.15.1 Let {En} be a sequence of measurable sets. Then both UnEn and nnEn are measurable sets. Corollary 3.15.2 Every open set and closed set in M. is measurable. Let us end this part of the discussion with an application of measurable sets. As we know from calculus, if a function f is nice, we can find its Riemann integral. In the beginning we have continuous and positive functions for "nice" and introduce this integral to solve The Area Problem. Then our ideas are generalized so that functions neither need to be continuous nor positive-valued. So the question becomes, "What

40

INTRODUCING THE CANTOR SET

types offunctions are Riemann integrable? " A partial answer is the theorem below. This is presented without proof. Theorem 3.16 A bounded function f on an interval [a, b] is Riemann integrable if and only if the set of points at which f is discontinuous is a set of measure zero. Our eventual goal for the end of this section is to introduce Legesbue's integra­ tion. In order to get there, we shall now move on to the idea of measurable functions. Looking at these requires us to use some special functions so we will start with defin­ ing the characteristic function of a set. Characteristic functions are a big part of Lebesgue integration. Definition 3.19 Let Abe a set in R (in general, A is a subset of some abstract space S, but in this section we are just working in the real line). The characteristic function of A, written \A using the Greek letter chi, is given by XAix)

=

(l,

ieA,

\o,

X{A.

Definition 3.20 Let f be a real-valued function with domain E, a measurable set. We say f is a measurable function if and only if for every r G R, the set {xeE:f(x)
=

f-\(-oo,r))

is a measurable set. Right away we see the characteristic function of any measurable set A is a mea­ surable function as f~l(( — oo, r)) is either empty (ifr < 0), A (ifO < r < 1), or E (if r > 1). An unproved by us corollary of this is that any continuous function is measurable. This is a consequence of the following theorem. Theorem 3.17 Let E be a measurable set and let f : E —> K. Then the following are equivalent: 1. For every r G R the set {x G E : f(x) < r} = / _ 1 ((—oo, r)) is measurable. 2. For every r G R the set {x G E : f(x) < r} = f~1((—oo,

r]) is measurable.

3. For every r G R f/ie set {x G E : f(x) > r} = / _ 1 ( ( r , oo)) is measurable. 4. For every r £ R the set {x G E : f(x) > r} = / _ 1 ( [ r , oo)) is measurable. Proof: We will show the first implication. The rest comes from complements. Assume {x G E : f(x) < r} is a measurable set for any r. Then {x G E : f(x) < r + 1/n} is measurable for each n G N. Since measurable is closed under intersection {x G E : f(x) < r} = Dn{x G E : f{x)
1/n}

SIZE OF A CANTOR SET

41

show {x G E : f(x) < r} is measurable, 6 Some quick corollaries of this theorem include: ■ Let f be a real function with measurable domain. Then for each real number r, {x : f{x) = r} is a measurable set. • If f and g are measurable functions on E, then so are f + g, f — g, cf (where c G M.), and fg. • If f is measurable and {x : f(x) ^ g(x)} has measure zero, then g is measur­ able. Now let us move on to the intended application of measure, integration. Lebesgue was one of several mathematicians at the beginning of the 20th century working to find a way to expand the types of functions that could be integrated. The typical home-spun example of how a Riemann integral differs from the Lebesgue is this; a woman looks on her desk, sees a bunch of change lying around (three pennies, two nickels, a dime, and two quarters), and wonder how much is there. Riemann would count the money in the order in which he came across the coins. 0.01 + 0.05 + 0.10 + 0.01 + 0.01 + 0.25 + 0.05 + 0.25 = $0.73. Lebesgue, on the other hand, would group the coins together by denomination as (3 x 0.01) + (2 x 0.05) + (1 x 0.01) + (2 x 0.25) = $0.73. We are all familiar with integrals from the Calculus sequence and (possibly) Ad­ vanced Calculus. Let us start with the some formal definitions. Definition 3.21 A partition P of the interval I = [a, b] is a finite set of elements of [a, b] containing at least one element of (a, b) with P — {to — o,, ti, <2; • • ■ > tn-i,tn

= b},

where tk-i < t^for k = 1, 2 , . . . , n. We will denote the kth subinterval [tk-i, tk] by Ik and let Alk = tk— tk-i- Lastly, the mesh ofP, written mesh(P) is the minimum of the AlkDefinition 3.22 Let f : I —► R be a bounded function. A Riemann sum associated with f on I with respect to P is a sum of the form n

T(/,n,P) = ^/(a:fc)A(/fe) fc=i

where Xk G Ik for k = l,2,...,n. We say that f is Riemann integrable on I if and only if there exists a real number, which we write as R J f, such that for all e > 0 there exists a 5 > 0 such that <e,

42

INTRODUCING THE CANTOR SET

for every Riemann sum associated with f on I with respect to P where mesh(P) S. If f is Riemann integrable on I, we denote R j

<

f as the Riemann integral of f.

There is an alternate definition which used the supremum and infimum of f each Ik = [tk-i, tk] to create Upper and Lower sums. This integral is called Darboux integral. Since the functions integrable using Darboux's definition and functions integrable using Riemann's definition are the same, we will again use R f notation.

on the the the

Definition 3.23 Let f be a bounded function defined on I, and let P be a partition of I. Define mk{f) = inf {/(a:)}, xeik

M f c (/) = s u p { / ( x ) } . xeik Then we define the Lower and Upper Darboux sums of f on I relative to P by n

L(/,P) = 5>fc(/)A(Ifc), fc=l n

[/(/,P) = ^Affc(/)A(/fc). fe=i

These upper and lower sums use the area under step functions. A step function with domain [a, b] is made by partitioning the interval into a = to < t\ < ti < ■ ■ ■ < < tn = b and having f(x) = Ci, a constant on each subinterval, (i,;_i,t,). tn-i In these cases the supremum and infimum of f on the subinterval form the constant values. Definition 3.24 Let f be defined on I. Then we define the Upper and Lower Dar­ boux integrals of f by [ / ( / ) = inf { [ / ( / , P) : P is a partition of I}, L(f) tfU(f)

= s u p { L ( / , P) : P is a partition of I}.

— -^(/)> then we say f is Darboux integrable on I and denote this common

value as R J

f.

It is a fairly straighforward Advanced Calculus exercise to prove that the Riemann and the Darboux integrals are the same. Although the Riemann definition is typically what one sees in a Calculus course, the example below will show us the usefulness of the Darboux definition. Notice that there are some restrictions in these definitions. The interval of inte­ gration must be closed. The function f has to be bounded. These two conditions, while necessary, are not sufficient to make a function integrable.

SIZE OF A CANTOR SET

fl

43

EXAMPLE 3.15 Let A = Q n [0,1]. Then the characteristic function

XA[X)

fi, \o,

ieQn[fl,i], ^Qn[o,i]

« nof a Riemann integrable function. On any subinterval Ik we have Mk{f)

= landmk(f)

= 0.

Thus for every partition P n

n

L(f, P) = 52 mfc(/)A(Jfc) = ] T 0 ' A ( ^ ) = ° fc=i fc=i

and n

n

fc=i

*;=i

u(f, p) = J2 Mk(m(h) = E 1 ■ A ( 7 *) = ^ giving us 0 = L ( / ) < £/(/) = 1, hence f is not integrable.

■

Lebesgue integration uses a special type offunction to approximate the measur­ able function f called a simple function. These are basically sums of characteristic functions. Definition 3.25 The real-valued function g is called simple if it assumes only finitely many values a\, 02, ■ • •, an. If a function is simple, then it must have the form 9 = ^2ai

-XAi,

where Ai = {x : g{x) = a{\. For our use, we will insist that each Ai be a measur­ able set. This is called the canonical representation of g. There are other ways to represent g. For example, if g is 1 on A = [0, 2), we could also say it is ] on AiD A2 where A\ = [0,1) and Ai = [1, 2). Although this seems unusual, we will have need of this in proving theorems. S

EXAMPLE 3.16 A step function, like 2,

m = { 1, 10,

ifx > 3, if l < x < 3, ifx < 1

44

INTRODUCING THE CANTOR SET

is an example of a simple function.

■

However, not every simple function is a step function. We see this in the next example. fl

EXAMPLE 3.17 Let A be any set. Then \A {x) is a simple function. Consider A to be the integers. Then this simple function is not a step function. ■

// g is a simple function that is nonzero on a set of finite measure, then we can define the integral of g. Before we begin, recall that a collection of sets {Ai} are pairwise disjoint if for all i ^ j Ai n Aj = 0. Definition 3.26 Suppose g is a simple function which is 0 outside of a set E with finite measure. Furthermore say g(x) = Yl't-i ai' XAt {%) where the Ai are pairwise disjoint, measurable, and UAi = E. Then we define the (Lebesgue) integral of g to be P

n

I 9 = V] alm(Al

JE

The usual results regarding sums and multiplying by constants holdfor integrating these simple functions. Before we prove it, though, we need this lemma. Lemma 3.3 Let g = YH=i aiXE% where we have Ei are pairwise disjoint. If each Ei is measurable and m(Ei) < oo, then n

/

9 = ^2alm(El)

Proof: This is a consequence of two facts: (I) Measure is subadditive, and (2) Ai = Ua=aiEi. Then a ■ m(Ai) — ^

a, • m(Ei).+

ai=a

Theorem 3.18 If g and h are both simple functions which are nonzero on sets with finite measure, then

(ag + Ph) =a

g+ 0

h,

where a and /3 are real numbers. Proof: Suppose that {Gi} and {Hi} are the sets from the canonical representations of g and h, respectively. Define the finite collection of sets Ei by taking all the

SIZE OF A CANTOR SET

45

possible intersections Gi fl Hj. These Ek are measurable and pairwise disjoint with the additional property that

9 = YLakXE*' and

n h =

Yl hkXEk ■ fe=l

This gives us

{ag + ph) = Y^iaak + 0bk)XEkk=i

Then it is an application of the Lemma that j {ag + j3h) = ^T(aak

+ 0bk)mEk

=a

g+0

h.

Corollary 3.18.1 Iffor simple functions g and h, g > h almost everywhere on E, a set with finite measure, then J g > J h. Proof: This proof is left as a homework exercise. We are now ready to define a Lebesgue integral for a bounded function f. We do this by approximating with simple functions similar to the way step functions are used to approximate the Riemann integral. Definition 3.27 Let f be a bounded, measurable function defined on a measurable set E, where the measure of E is finite. Then the Lebesgue integral of f over E is defined by

f f = mf / 5,

IE JE

J^9JE f<9JE

where g is a simple function. This is an extension of the Riemann integral. Every step function is a simple function which give us the result L(f) < sup /

h < inf /

h
f^9J[a,b]

g

< u(f),

where h, g are simple functions. Theorem 3.19 Let f be a bounded function on the interval [a,b]. If f is Riemann integrable on [a, b], then it is measurable and

!

f'-J

Ja

J\a,b

46

INTRODUCING THE CANTOR SET

As we can see, there are extra requirements placed on our situation aside from the expected requirement of E being a measurable set and f a measurable function. So have not yet arrived at the finished Lebesgue integral. The extras that we must dispose of are the requirements that m(E) is finite and that f is bounded. We can rid ourselves of both, but at the cost of a new condition. Definition 3.28 Let f be a nonnegative measurable function defined on a measur­ able set E. Then we define the integral of f over E by / / = sup / h, JE

h
where h is a bounded, measurable function with m({x : h(x) ^ 0}) < oo. Adding new functions and sets to our quiver of integration does not change the usual results. We state officially (but without proof) that if f and g are nonnegative measurable functions on the measurable set E, then cf = c f , JE

c > 0,

JE

and

f(f + 9)= I f+ I 9JE

JE

JE

Finally, we are ready to define this integral for any measurable f. Definition 3.29 Let f be a measurable function on the measurable set E. By f+(x) we mean max{/(x), 0} and f~(x) is max{—f(x), 0}. Both of these are nonnega­ tive functions and f(x)=f+(x)-f-(x) for all x g E. We define the Lebesgue integral of f by

r

L'-ls-L As an example of how the Lebesgue integral is more powerful than the Riemann, let us look at the "easy" function f = XQ> the characteristic function of the rational numbers. This function is has upper sum U(f, P) — lfor any partition P and lower sum L(f, P) = Ofor any P. Thus U(f) ^ L(f) and so it is not Riemann integrable. On the other hand, f is constant at 1 except on a set with measure zero. So / / = 1 x m([0, l ] n ( R \ Q ) ) + 0 x m([0,1] n Q) = 1. Jo

LARGE AND SMALL

3.3

47

Large and Small

In this chapter we have looked at various ways to call a set large (and, of course, small). These notions are mostly unrelated to one another: A set could be small in one way (category) yet large in another (cardinality). We wish to end this chapter with a striking result that can be found in Oxtoby [58], taking the entire space we are working in and decomposing it into two small pieces. Theorem 3.20 The real line K can be partitioned into two disjoint sets A and B such that A is first category and B is measure zero. Proof: Start with r\, T2, r$,..., a listing of the rational numbers (it is left as a home­ work exercise to show Q is a countable set). For each Tj, let Uj,n be the interval

(

I T' —

i 3

1 2J+n '

V " ~~f~~ J

M

I

2-?+n) '

For each j let Oj = yj^L-JJ^n and then define B — f l ^ O j . For any e > 0 we can find an n so that 1/2" < e so B C UjUjtU and 2_j \Uj,n\ = Z^ 2^+"

=

2™

<

£

'

j

So B is a measure zero set. Each Oj is an open set (since it is the union of countably many open intervals) and Q C Oj. Thus its complement Of = R \ Oj is nowhere dense (by the Baire Category Theorem). So let A = Bc = LijO'j, a first category set. 4

48

INTRODUCING THE CANTOR SET

EXERCISES 3.1 Draw the first few iterations for constructing C1/5. Find the sum of intervals removed for C1/5. 3.2

Find a formula for the amount removed in creating Cr, where r is a fixed ratio.

3.3 Prove that 1/4 is a point in the Cantor Set, but not the endpoint on a removed interval. 3.4 Give two examples of non-empty, open sets which are not open intervals. Sim­ ilarly, find two examples of non-empty, closed set which are not closed intervals. 3.5

Is M. an open set or a closed set? What about 0 ?

3.6 Determine (with proof) whether the Cantor Set C1/3 is a closed or an open set1. 3.7

Prove Theorem 3.5.

3.8

Prove that Z = N.

3.9

Explain why finite sets do not meet our definition of small.

3.10

Show that the set Q is a countable set.

3.11

Show that m(Qn[0,1])

3.12

Prove Lemma 3.1.

= 0.

3.13 Prove that the Cantor Set C and the unit interval have the same cardinality. (Hint: This also has to do with a base other than 10. This time it is base 2.) 3.14

Show the function from the proof of Theorem 3.9 is I — I and onto.

3.15

Give an example of a set which is neither dense in K nor nowhere dense.

3.16 Prove that the Cantor set C1/3 is nowhere dense. Thus a set can be small in one way (category) while large in another (countability). 3.17 Find an example of a set which is not nowhere dense, but is first category (and prove it!). 3.18

Supply the missing proofs for Theorem 3.2.

3.19

Give a proper proof that if E C F, then m*(E) < m*(F).

3.20

Show that the interval [a, b] has outer measure b — a.

3.21

Show that the rational numbers have outer measure zero.

2

A set does not have to be open or closed, it can be neither. There are also sets which are both open AND closed. They are called "clopen."

EXERCISES

3.22

Supply the missing parts of the proof for Theorem 3.15.

3.23

Prove that the Cantor set is measure zero.

49

3.24 Let Crn denote a Cantor Set where the ratio removed at each step is not constant. Find the total length removed ifrn = l / 2 2 n . This is the Smith-VolterraCantor Set, an example of a "fat" Cantor Set. 3.25

The questions below go back to our construction of the nonmeasurable set S. a) Give three examples of values in [0,1] that are equivalent to 7r/4. b) A relation is called reflexive iffor every x we have x is related to itself. Show that this is true for =. c) A relation is called symmetric if for every x and y, if we assume x is related to y, then we can prove y is related to x (the relation "greater than " on the real numbers shows this is not always true). Show that this is true for =. d) A relation is called transitive iffor every x, y, and z, if we assume x is related to y and y is related to z, then we can prove x is related to z. Show this is true for =.

CHAPTER 4

CANTOR SETS AND CONTINUED FRACTIONS

Five survivors of a shipwreck end up on an island deserted except for one coconut tree and a mischievous monkey. The survivors spend the day gathering a pile of coconuts. They will divide the coconuts up the following day. During the night one man woke up and started worrying about the coconuts. He decided to take his share, and divided the coconuts equally into five piles. There was one left over which he gave to the monkey, then he took his share and hid them away, before going back to sleep. A little later a woman woke up and had the same fear. She too divided the coconuts into five piles, and had one left over that she gave to the monkey, before hiding her share and going back to sleep. And so the night went on, with each person waking up, dividing the remaining coconuts into five piles, hiding one share, and giving the one extra to the monkey. In the morning, they awoke and divided the remaining coconuts, which came out tofiveequal shares. No one said anything about the missing coconuts as each thought he/she had been the first to hide a share and was thus the best off. How many coconuts were there to start with?

The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

51

52

4.1

CANTOR SETS AND CONTINUED FRACTIONS

Introducing Continued Fractions

We are all familiar with the concept of a fraction, -, where a and b are integers. In Precalculus or Intermediate Algebra we even see what are sometimes called com­ pound fractions, a fraction that contains a fraction in its numerator or denominator. A continued fraction is similar to that idea of fractions within fractions. Definition 4,1 Let a-i, i = 0 , 1 , 2 , 3 , . . . and bi, i = 0,1, 2 , 3 , . . . denote collections (possibly finite) of real numbers} A continued fraction is an expression of the form ^

bo ^

bj_ 02

a3 To keep our investigations easier, we will stick with the simple continued fractions, where all 6, = 1. Definition 4.2 Let ai, i = 0,1, 2, 3 , . . . denote collection (possibly finite) of integers with cii is positive for i > 1. A simple continued fraction is an expression of the form 1

ooH

1

ai

a2 + a3 + '• To simplify our notation we will write this expression as [a0 :

ai,a2,a3,...}.

The individual values ao, a\, a2,. ■. are referred to as the partial quotients of the simple continued fraction expansion. S

EXAMPLE 4.1 The finite simple continued fraction [2:3,6] represents 2+

1

1 r = 2 + - r 7 r = 2H

3+i

f

6

19

44 = —.

19

This does, of course, work in the opposite direction, too. If we start with any rational number - where a, b £ Z, a ^ 0, then we can express it as a finite simple 1 Some sources allow complex numbers while other restrict the values to integers. Read these books and papers carefully.

INTRODUCING CONTINUED FRACTIONS

53

continued fraction. This requires just an application the Division Algorithm.2 For example, start with the number 32/7. We know 32 -f- 7 — 4 + 4/7 which is the same as Performing long division on 7/4 we have 1 + 3/4 so now our original number is 1 !+473

Next up, 4/3 = 1 + 1/3 and thus 32 7

1+

iil

[4:1,1,3].

77iere are two important features to point out. First, for a rational number in lowest terms this process is guaranteed to terminate. Notice that in our dividing process the remainders are strictly decreasing. This means they must at some point become the number 1. So although we are not writing this as a formal theorem, every rational number can be written as a finite simple continued fraction. Secondly, this is not the only way to represent 32/7. It is also correct that 32 - y

= 4 +

1 i

i

=[4:1,1,2,1].

However, the Division Algorithm, which guarantees that division gives unique re­ mainders, tells us that as long as we do not write the partial quotient an as an — 1 + \ we cannot have two representations. We shall then require in our representation [a0 : ai, a 2 , . . . , an] that an > 1. If our number ^ is negative, we change the first step so that we are multiplying the divisor by a negative number which will give us the smallest positive remainder. So, for instance, the continued fraction expansion of — 15/7 starts with —15 = —3 x 7 + 6. This returns to us -15 _ 1 ~ ~ ~ 3 + 7 / 6 ' and we continue to arrive at —15/7 = [—3 : 1,6] As we would expect, there are infinite simple continued fractions, where the pro­ cess continues indefinitely. It turns out that these very intriguingly partition into different types. An irrational number is one which cannot be written as p/q where p and q are integers, with q =/= 0. A special type is the quadratic irrational, a number of the form

p±Vd 2

Given two integers a and b with b j= 0, there exists unique integers q and r with 0 < r < \b\ such that a — bq + r.

54

CANTOR SETS AND CONTINUED FRACTIONS

where p, q, d are integers and d > 0 and not a perfect square. The name comes from the fact that it comprises the solutions to q2x2 - 2pqx + (p2 - d) = 0. The value TT, the all-purpose number, is an example of an irrational number that is not a quadratic irrational. Let x be a positive irrational number. Due to the Archimedean Property, there exists a largest integer ao such that x = a$ + — where 0 < (xi)^1 < 1. Then xx =

>1 x — ao and is irrational (after all, x is irrational and a$ is an integer). Repeat this process: Starting with x\, find a,\, the greatest integer such that 1

x\ = ai H

,

where 0 < ( a ^ ) - 1 < I. As we continue we generate the continued fraction for x [a0 : a i , a 2 , a 3 , . . . ] , where this time the sequence of'a, does not terminate as x is irrational. S

EXAMPLE 4.2 The easiest example is \/2. Given \/2 = 1.4142 . . . we see that ao = 1. Then \jx\ = \/2 — 1 or, taking reciprocals and rationalizing the denominator Xl='=-J—.y^±l

\/2-

1

V2-

1 A/2

+ 1

= V2 + l = 2.Ul2....

Thus ai = 2. The process then starts repeating, as X2 =

1 an-2

1

= —F=

A/2-1

Xi,

leading us to ai = 2 for i > 1. Hence V/2 = l +

^ 2

+

= [1:2,2,2,2,...].

1" 2+

2 + ■-

To shorten up the notation, since it is repeating, we follow the convention of using an overline, \/2 = [ 1 : 2 ] . ■

INTRODUCING CONTINUED FRACTIONS

55

This example shows why the quadratic irrationals are easy to deal with, as we can rationalize the denominator to simplify the process. Lagrange proved in 1779 that the continued fraction expansion of any quadratic irrational will eventually become periodic. It is also worth noting here a way in which continued fractions and rational numbers differ. If the decimal expansion of a number is periodic, then the number is rational. Having the continued fraction expansion repeat (as it does for \/2) does not guarantee a rational number. How about going the other direction ? When we have an ordinary real number but with a repeating decimal representation, we are able to find its rational number representation, can we do the same with a repeating continued fraction ? The answer is yes. Working backwards, we start with 1

[1:2] = 1

x7

where x is an unknown value. Using a trick similar to what we do with irrational numbers, then

x-i =

2+

.

K 2+ 2+ ••

We cannot just substitute into our equation willy-nilly. Using x — 1 to replace the entire second term in our definition of x leads to x = 1 + (x — 1) a tautology. This does not help us solve for x. Instead, we will substitute x — lfor the parenthetical part of 1

1

V 2+ 2+

• ■ . /

Thus we have the equations 1 2 + (a: - 1)

1 +

This equation simplifies to x2 = 2; and since we know we are dealing with the positive solution, it must be that x = \/2.

56

CANTOR SETS AND CONTINUED FRACTIONS

As another example, let us start with [1 : 1,1,1,...]. We set

i+

=i+ i

'

1+

x

I ^

1+ - 1 1 + '--

2

That yields x — x — 1 = 0 and the positive solution to this equation is

1 + V5

x =

, '

2

which is the golden ratio, usually denoted by . Thus
1+

52 2 + ••

and

l2

n = 3+

32

6+

52

6+ 6+

72

6 + '-. To discuss continued fractions a little more, we need to add to our terminology: We need to bring convergents to the mix. Definition 4.3 Let x have the simple continued fraction expansion (finite or infinite) [an : ai,a,2,a,3,...}. The convergents of'x are the sequence offinite simple continued fractions c0 = a0,

ci = a0 H

1 ai

,

c2 = a0 -\

1 ai

r

+ -

,

and, in general, cn = [ao '■ CL\, 0,2,0,3, • • •, an\. Typically, we represent these convergents as the rational numbers that they are. For example, Cn = an = —, Ci = an + — = —, Co = an H TT - = —. and so on,

INTRODUCING CONTINUED FRACTIONS

57

where each Pi,qt G Z and qt j ^ 0. The relationship between the values of the p's, q 's, and a's is an inductive one, summarized in the theorem below. Theorem 4.1 Start with the simple continued fraction [ao : a,\, a2, ^3, • • •]• Let C{ be the ith convergent which in rational number form is written ^r, pi is an integer and qi a positive integer. Then the values of these numerators and denominators satisfy the equations Pi

=

CliPi-l+Pi-2,

qi

=

cnqi-i + qt-2

for i = 2, 3,4, 5 , . . . with initial values Po = ao,

Pi = ai«o + 1,

go = 1,

qi~ai.

Proof: It is obvious for i = 0 that Co = a 0 / l = Po/qo and that C\ = (a\ao + l ) / a i = Pi/qi- At i = 2 we have c2=a0-\

1 aoa1a2 + ao+a2 ——.— = — aia2 + 1 ai + l / a 2

=

a2p\ + po P2 ■ = —. axqx + q0 q2

Assume the relationship has been shown for i — 0 , 1 , 2 , . . . , k. Then Cfc+i —

Pk+l Ofc+l

r

— [ao : o,i,a,2, ■ ■ ■

i

,ak,ak+\\.

This can be rewritten as Cfc+i = [a 0 :

ai,a2,...,(ak,+l/ak+i].

Now we have a continued fraction with only k terms. Thus via the induction esis

Cfc+i

=

r

a 0 : a2,a2,...,ak-i,

/

,

i \i

{ak + ^ + j )

hypoth­

(ofc + ^)pfe-i+Pfc-2

= -, r (afc + ^ T r J 9fe-i + 9fc-2

_

(afcafc+i + l)Pfc-i + afc+iPfc+i _ ak+i(akpk-i (akak+ipk-i + afc+iafc_2 ak+i(akqk^j

_

a-k+iPk +Pk-i a/c+igfc + qk-i

+ Pk-2) + Pk-i + qk~2) + qk-i

the last step making use of the induction hypothesis. 4b An application of this theorem is the following study of irrational numbers.

result, which is important in our

58

CANTOR SETS AND CONTINUED FRACTIONS

Theorem 4.2 If the convergents C{ are written as rational numbers Pi/qi, then PiQi-i

-Pi-iqt

= (-i)J

holds for i = 0,1, 2 , 3 , . . . . Proof: This is left to the reader. It is another induction proof. A consequence of Theorem 4.2 shows how these convergents approach a limit. If we have two consecutive convergents Ci and Cj+1, we see that _ Pi+i Cj+1 ~ Ci —

Pi _ pi+iqi -Piqi+i —

—

Qi+i ii qi+iqi and, following similar ideas for cn and c n _2, we obtain ^n

_ (-i)l+1 qi+iqi

Cn — 2

<7n9n-2 Recalling that the definition of convergent insists that qi is always positive, we have Theorem 4.3 For any simple continued fraction [ao : 0,1,0,2,0,3,...} the convergents Ciform a sequence of real numbers where for all i ' cii > c2i~i and " C2i_l > C2?;-2-

Thus the sequence of convergents has the property Ci < C3 < C5 < • • • < C2i-1 < • • • < C2i < • ■ • < C4 < C2-

This brings us, finally, to a theorem that says that any infinite simple continued fraction has meaning as a point on the real number line. Theorem 4.4 Let [ao : a,\, a2, 03,...] represent an infinite, simple continued frac­ tion. Then there is a point x on the real line such that x = [ao : aj, 112,03,...] Proof: Most of this proof has already been shown. We know that the two subse­ quences of convergents, {c2i} and {c2i-n}, are bounded and monotone. A wellknown theorem from real analysis (e.g., see [33]) is that monotone sequences con­ verge if and only if they are bounded. Thus {c2i} and {c2i+i} converge to limits Li and Li, respectively. Our job is to show that Li and L^ are the same. If not, then there exists an e > 0 such that Li — L2 > e, which means that for all i we have C2i ~ C2i+1 = l/(<72i £•

However, Theorem 4.1 shows us that for all i we have qi = aiqi-i + qi-2, where o-i, qi-i, and qi-2 aw all positive integers. This implies that {qi} is itself an un­ bounded increasing sequence. Thus 1
>0,

CONSTRUCTING A CANTOR SET

59

a contradiction. Thus L\ = L 2 , and the convergents approach the same value. 6 We will skip the technical proof of the next theorem, which wraps up our look at continued fraction representations. We know that for rational x the simple continued fraction expansion we get from x is equal to x when simplified. With an irrational x, since the continued fraction does not terminate, we look at the sequence of conver­ gents. The theorem below explains that the sequence converges to x. Theorem 4.5 If x is an irrational number and we expand it into the simple, infinite continued fraction [ao : ai, a 2 , a 3 , . . . ] , then the convergents { Q } of this continued fraction are a convergent sequence in R with limit x.

4.2

Constructing a Cantor Set

In this section we will introduce another way to construct a type of Cantor Set, this time by using continued fractions in the definition. In general, we can think of a Cantor construction by starting with a closed interval [a, b] and removing an open middle portion of the interval, thus creating two closed subintervals. We then re­ peat without end, at each stage removing an open middle interval from each closed interval. Define our initial closed interval as A = [x,x + a] where x and a are some real numbers and a > 0. We then remove a middle (not necessarily centered) open subinterval from A that has the form (x + a\, x + a\ + 012). This leaves us with two closed subintervals: Ai = [x, x + ai] and A2 = [x + a,\ + ai2, x + a}. Continuing this process leads us to a Cantor Set. The point of this section is to show that we can make such a construction utilizing continued fractions. In fact, for integers k > 2, let the set S(k) be the set of all x in [0,1/fc] where the continued fraction expansion of x contains no partial quotient less than k, is a Cantor Set in [0,1/fc]. Since the left endpoint is zero, ao = 0 in the continued fraction expansion. There are two types of intervals with rational endpoints that we need to consider when looking at what is in S(k). The first are of the form (call it Form I) [[0 : a i , a 2 , a 3 , . . . , a n ] , [ 0 : ax, a2, a 3 , . . . , a„ + 1 ]], where n is an even number and a, > kfor all i while the second type (call it Form II) has the form [[0 : a i , a 2 , a 3 , . . . , a n ] , [ 0 : ai,a2,a3,...

,an-i]],

where n is a positive even natural number and a^ > k. Once we have a closed interval, when we remove the middle subinterval, of the type Form I we remove the interval ([0 : a i , a 2 , a 3 , . . . , a n + i ] , [ 0 : a i , a 2 , a 3 , . . . ,an+1, k})

60

CANTOR SETS AND CONTINUED FRACTIONS

whereas removing intervals of type Form II we take open subintervals of the type ([0 : ai,a2,a3,...,an,k],[0

: oi, a 2 , a 3 , . . . , a „ _ i , a „ + 1]).

In both instances, the removal of the open interval leaves an interval of the first form on the left side and an interval of the secondform on the right. Thus we can continue the process, and the resulting set consists of the numbers described by S(k). Marshall Hall, in [35], published an interesting application of these construc­ tions. In his paper he referred to F(N) (N a positive integer) as the set of continued fractions [ao : ai, a2, ■ ■.], where ai < N for i > 1. Notice here ao can have any value. If ao is fixed at n, then the resulting set is referred to as F(n, N). He then proves that F(0,4) +F(0,4) is the interval ((\/2 - l)/2, 2\/2 - 2) and continued to explain how this means that F(A) + F(4) = K. In 1973, Divis and, independently, Cusick ([16] and [17]) proved that the value of N for which F(N) + F(N) — R cannot be lowered as F(3) + F(3) ^ K. In addition, Hall shows that if x > 1, then x £ F(A) ■ F(A) where A ■ B = {xy : x £ A and y 6 B}. 4.3

Diophantine Equations

No discussion of continued fractions is complete without a foray into diophantine equations. So let us take a quick glimpse into these, starting with an example and a definition. There are many solutions to an equation like 8a; + 15y = 31. There are the ordered pairs (2,1), or (17, —7), or (1,23/15). As a matter offact, for any x, the matching y is the value 31 -8x y 15 However, things get sticky if we insist that the solution consist only of integer pairs (a similar interesting effect occurs when making the transition from Linear Program­ ming Problems to Integer Programming Problems in Operations Research). This integral solution problem is the heart of the study of diophantine equations. Definition 4.4 A linear3 diophantine equation is an equation of the form ax + by = c, where a, b, and c are fixed integers, a and b are not both zero and have no factors in common, and the solution is a pair o/integers. The name honors Diophantus, a Greek mathematician who wrote about these equations in a series of books called the Arithmetica. 3

Typically these can be polynomial equations, but we shall restrict ourselves.

DIOPHANTINE EQUATIONS

61

One way to solve a diophantine equation is through repeated applications of the Euclidean Algorithm, the way we do long division. So let us solve 8x + 15y = 31 for x (Why? Because x has the smaller coefficient.). Then we get X

_ 31 - 15M _ (3 • 8 + 7) - (y ■ 8 + ~ 8 " 8

,o

7M)

~[6

, , V)+

7-7y 8 •

Since we want our solutions to be integer pairs, it must be that (7 — 7y)/8 is equal to some integer t. This means we now have another diophantine equation 8t + 7y = 7. Solving this for y (again, due to the smaller absolute value of the coefficient) we have 7-8* , t , y = —^— = i - t - - = i - t - u , where u = t/7, and we need u to be an integer since y and t are integers. Since there is no constant term in u = t/7 we are done and can begin going back to find our answers. Back-substituting with this gives us t = 7u and thus y = \ — 7u — u = \ — 8u and x = 3 - (1 -

8M)

+

7M

= 2+

15M.

So letting u run over all possible integer values, we create a family of solutions to this equation. This includes (2,1) when u = 0 ( — 13, 9), u — —1, and (17, 7) when M = 1. The Euclidean Algorithm is also what we use when we find the integers we need for the continued fraction representation of a number. Therefore, we cannot be too surprised that we can talk about the solution to these equations using continued fractions. We build up in stages, starting with one specific type of equation and leading up to the general ±ax ± by = c. Thefirsttype of equation we solve is ax — by = 1 where gcd(a, b) = 1. This equation is guaranteed to have infinitely many solutions by the Euclidean Algorithm. Since a/b, b ^ 0 is a rational number in lowest terms, we can convert it to a finite, simple continued fraction. - = [a0 : a i , a 2 , . . . ,an]. From this we get the convergents CQ, C\ , . . . , cn. The last pair of convergents Pn-l

c n _i =


,

Pn

and cn = — In

are what is important here. From Theorem 4.2 we know that PnQn-l

~ qnPn-l

=

(-1)"-

62

CANTOR SETS AND CONTINUED FRACTIONS

For the moment, assume that n is even. Using that assumption and plus the fact that pn = a and qn = b and we arrive at aqn-i

- bpn-i

= 1,

as a particular solution. This depended on n being even. That giving (qn-i,Pn-i) is not an issue as if n is odd we can rewrite our continued fraction expansion as [a0 : ai,a2, ■ ■ ■ ,an - 1,1] ifan > 1 and [a0 : a\,a2,. ■ ■, a „ _ i + 1] if an = 1.

Now we have to turn our particular solution into a general solution. Thus far we have ax — by = 1 and axo — byo = 1, where XQ = qn-i, yo = Pn~\- Subtracting the equations takes us to a{x-x0) = b(y-y0). This is where gcd(a, b) — 1 comes into play. The right side is a multiple ofb, but no factor ofb can be a factor of a. Thus x — Xo = bt where t is an integer. This forces y — yo = at. Thus the general solution is x =

XQ

+ bt and y = yo + at

for all integer values oft. The equation ax — by = — 1 is solved similarly, but this time we insist that the continued fraction have an odd number of' convergents. Then aqn^\ —bpn^\ = —1, but we can again find the general solution as we did before. If ax — by = c, so the right side is no longer ± 1 , we can go back a step and solve ax — by — Ifor XQ and yo, then multiply those by c. The solution we are looking for is then x = CXQ + bt and y = cyo + at for all integers t. Changing the operation on the left side from subtraction to addition we arrive at the problem ax + by = c where gcd(a, b) = 1. First, writing a/b as a continued fraction with an even number of convergents, we can find pn-\ and qn-i so that aqn-\

~ bpn-i

= 1.

This can then be substituted in by ax + by = c = c-l

= c- {aqn-i

- 6p„_i).

Now we shuffle things around to get a ( c g n „ i -x)

= b(y +

cpn-i)-

This takes us to the solution x = cqn-\ — bt and y = at — cpn-\. Now we are done as any equation of the form ±ax ±by = c can be changed into one of the two forms ax + by = ± c or ax — by = ± c . Let us now apply this idea to our beginning example 8a; + 15y = 31.

MISCELLANEOUS

fl

63

EXAMPLE 4.3 Starting with 8x + 15y = 31, this gives us a/b = 8/15. Expanding this into a simple continued fraction yields

41 0 = ° + 11 + , \

T+T78

=[0 = 1,1,8],

an expansion with an even number of terms. The convergents are 1 8 Co = 0, ci = 1, c2 = - , arcdc3 = —. Then n — 1 = 2 awrf we ftave 92 = 2 and p 2 = 1- 77iM5 o«r solutions are x = 31(2) - 15i = 62 - 15* and

y = 8y-31(l) = 8 t - 3 1 , where t G Z. TTiis is f/ie same solution as before.

■

This is where the monkey problem at the beginning of the chapter comes in. If we originally started with x coconuts and if the five shipwreck survivor A, B, C, D, and E each hid a, 6, c, d, e coconuts, respectively, then = 4a == 46 == 4c == Ad --= 4e == X

-

5a + 1, 56 + 1, 5c+1, 5d + l, 5e + l, 5y,

where y is how many coconuts each received the following day. Backsubstituting leads us to the Diophantine Equation 1024x - 15625y = 8404.

4.4

Miscellaneous

Two last results. The first one joins together continued fractions, and self-similar sets from Chapter 6 and comes from [30]. The proof of this result can be found in that book.

64

fl

CANTOR SETS AND CONTINUED FRACTIONS

EXAMPLE 4.4 Let F be the set ofpositive real numbers x whose continued fraction expressions have partial quotients 1 or 2. Then F is a fractal with 0.44 < dim// a u s (F) < 0.66. ■

The second result is about ordinary (not simple) continued fractions and a for­ mula for them. Theorem 4.6 Let 0,1,0,2,0,3,... be nonzero real numbers with a,k-i ^ afc for all k. Then for any n £ N

£ 1 . 1

(-i) f c

1

a k

k=1

ai +

a2 - ai + a 3 - a2

al_1/(an

-an_i)

Letting n —> 00 we get an infinite continued fraction expansion, assuming both sides converge. Let us apply this theorem in an example. The theorem will enable us to find a beautiful representation for a nontrivial number. fl

EXAMPLE 4.5 From Calculus II we know that the infinite Maclaurin series for ln(\ + x) is

Therefore 1 , ,„N Y^ ( - I ) * " 1 ln(2 - = y ) = > > ^ k 1 2 k= l

1 1 1 H 1----. 3 4

This finally leads us to 1

ln(2)

1

1+

1+

~

! + -£■

EXERCISES

65

EXERCISES 4.1

Convert each of the following into a simple continued fraction. a) 15/11 b) .19 c) -37/44 d) 2.145

e) V6 4.2

Find p/q if the continued fraction is a) [1:2,1,3]. b) [2:3,1,4,2] c) [2:2,4,2,4,...]

4.3 Find the simple continued fraction expansion of the conjugate of the golden ratio

4.4

Solve (ifpossible) the linear Diophantine equation A7x + 15y = 4.

If there is no solution, explain why. 4.5

Solve (if possible) the linear Diophantine equation 42x + 27y = 47.

If there is no solution, explain why. 4.6

Solve (ifpossible) the linear Diophantine equation llx + 42y = 35.

If there is no solution, explain why. 4.7

Prove Theorem 4.2.

4.8

Verify that y/7 = [2 : 1,1,1,4]

4.9

Find the first three convergents in the simple continued fraction expansion of

VI. 4.10 e.

Find the first three convergents in the simple continued fraction expansion of

4.11 Determine the first five convergents for the value r = do you notice with the numbers involved? 4.12

1+

2

. What pattern

Use Theorem 4.6 to find the continued fraction expansion for 7r/4.

66

CANTOR SETS AND CONTINUED FRACTIONS

4.13

Show that ifa>b

and --[a0:

ai,a2,a3,...],

then

4.14 4.15

b , - = 0 : a0,ai,a2,... . a Find a continued fraction expansion for 4/TT. Show that F in Example 4.4 is both closed and bounded.

CHAPTER 5

P-ADIC NUMBERS AND VALUATIONS

As noted earlier, one of the amazing aspects of the Cantor Set is how it insinuates itself into so many branches of mathematics. Our study of p-adic numbers will merge from two different directions. On one hand, it can be looked at as part of abstract algebra, and for that we will review some basics in the topic; on the other hand, p-adic analysis is a generalization of "normal" Euclidean analysis on the real line. And, of course, in the end either direction brings us around to the Cantor Set.

5.1

Some Abstract Algebra

As with any review, the key is to start small and work our way up to the larger results. Abstract algebra (sometimes called modern algebra) arose from many sources: the study of linear equations, the rise of non-Euclidean geometry, and work on solving Diophantine equations, to name a few. Our main concern here is to introduce groups and then move on to rings and fields. A group consists of a set G and a binary operation, *, on the set. The operation will be a way to take two elements ofG, say a and b, and produce a result we will call a* b. Many books refer to the operation as multiplication, although this does not necessarily mean multiplication of numbers The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

67

68

P-ADIC NUMBERS AND VALUATIONS

as we are used to. That said, we will eventually drop the * and write a*b as ab, as with usual multiplication. Definition 5.1 Let G be a non-empty set and * an operation on G. We say the G is a group under * if the following four criteria are met: • G is closed under *; that is, ifaEG

and b G G, then a*b G G.

• G contains an identity, e; that is, there exists an e G G such that for every a G G we have a*e = e*a = a.

G contains an inverse for every element; that is, for each a G G there exists a - 1 G G such that a* a 1 = a 1 * a = e. ■ * is associative; that is, for any a,b,c G G we have a * (b * c) — (a * b) * c. We will denote the set and its operation as an ordered pair, (G, *). The abstract notion of groups first appeared in papers by Arthur Cayley in 1854. S

EXAMPLE 5.1 The following are the somewhat canonical examples. 1. (Z, +) the set of integers with the usual addition is a group. 2- (Q + , ■)> the positive rational numbers with multiplication is a group. 3. (Q, •) is not a group, not every element has an inverse. 4. Let C be the set of all bijections (1 — 1 and onto functions) with domain and range M. Then C is a group with the operation of composition, f * g = f o g. 5. Let S = {a, b, c] be a set with multiplication * defined as follows: *

a

b

c

a

a

b

c

b

b

c

a

c

c

a

b

This is a group. In fact, this is an example of a finite group since it has a finite number of elements. The number of elements in G is called the order of G and is denoted by \G\.

SOME ABSTRACT ALGEBRA

69

One important property a group may or may not have is commutativity. This is a property of the operation *, not the set G. We say * is commutative on G if for every a,b £ G we have a * b = b * a. In the examples above, all groups but (C, o) have the commutative property. If a group has this property, it is called an abelian group (after Neils Henrik Abel). If a group is not abelian, it is called non-abelian. A simple introductory theorem is as follows: Theorem 5.1 IfG is a group, then 1. There is only one identity element. 2. For each a G G, the inverse of a is unique. 3. For a e G, ( a - 1 ) " 1 = a. 4. For a,b£G,(a*

b)'1 = b'1 * a" 1 .

Proof: We will prove the first and fourth parts, leaving the remaining two as an exercise. Suppose G has two identity elements, e and e. Thus for every a € G we have a* e = e* a = a and a*e = e*a = a. So we have e = e * e = e,

the left side from e being the identity and the right halffrom e being the identity. The fourth part has two parts to the proof. First, we must show that b~l *a~l G G. We get that from the definitions of what makes a group. Since a,b £ G it must be that a~l and 6 _ 1 are both members ofG. Because G is closed under *, then b~l * a - 1 is in G. b~l * a - 1 . Second is a direct proof that b^1 * a - 1 does what (a * b)^1 is supposed to do. This uses the associativity for *. So (a*b) * (b^1 * aT1) = a* (b* fr_1) * a - 1 = a * e * a - 1 = a * a~l = e and {b~x * a - 1 ) * (a * b) = 6" 1 * ( a - 1 * a) * b = b~x * e * b = 6 _ 1 * b = e.4^ We now move on from groups to rings. Rings seem a little more usual (read: like the integers) in that they involve a set and two operations, + and ■. As there is more structure to a ring, there are also more requirements.

70

P-ADIC NUMBERS AND VALUATIONS

Definition 5.2 A non-empty set R is a ring if R has with it two operations +, called addition, and *, called multiplication, such that 1. If a, b e R, then a + b £ R. 2. For all a,b E R, a + b = b + a. 3. For all a, b, c E R, a + (b + c) = (a + b) + c. 4. There exists an element 0 E R such that a + 0 = 0 + a = afor every a E R. 5. For every a E R there exists an element —a E R such that a + (—a) = (—a) + a = 0. 6. For all a,b E R, a * b E R. 7. For all a,b, c E R, a * (b * c) = (a * b) * c. 8. Finally, between + and * there is a Distributive

Law:

a* (b + c) = a*b +

a*c

(b + c)*a

c*a.

and = b*a +

This says that R is an abelian group under +. Also, we note that we do not claim R has an identity element for multiplication, an element 1 so that a ■ 1 = 1 ■ a = a. If the ring does have such an element, then we say R is a ring with unity. Additionally, the multiplication does not have to commute. If we have a-b = b-a, then we say it is a commutative ring. Here are two more properties, important enough to get their own definitions. The first brings up an important property emphasized in Intermediate Algebra as the Zero Product Property or Zero Factor Rule. Definition 5.3 A commutative ring R is called an integral domain ifa-b = 0 implies either a~0orb~0. The integers are an integral domain. Some simple results about rings are in the theorem below. The proofs are left as an exercise. Theorem 5.2 Let R be any ring and let a.b E R. Then 1. a*0

= 0*a

= 0.

2. a* (~b) = (—a) *b = —(a * b). 3. (—a) * (—6) = a * b. 4. If R contains a unity, 1, then ( — 1) * a = —a.

SOME ABSTRACT ALGEBRA

71

Definition 5.4 A ring R with unity is called a division ring iffor every a ^ 0 in R there is an element b € R such that a ■ b — b ■ a = 1. Usually we write b using the notation a~l. The integers are not a division ring. An example of a ring that is not an integral domain are the set of 2 x 2 matrices with matrix addition and multiplication. The rational numbers, Q, with the usual operations of addition and multiplication are a division ring. Finally we come to what we really need, namely, fields. Definition 5.5 A ring R is said to be afield ifR is a commutative division ring. The rational numbers, real numbers, and complex numbers are all examples of fields. Let us look at an extremely important example used in abstract algebra and num­ ber theory, modular arithmetic. We start with the integers Z. Given two integers a and b, we say a divides b if there is an integer k such that b — ak. We usually write this as a\b. Thus 4|12 and 6|42 while 5 j 24. Notice that this definition uses multiplication and not division. Furthermore, 2|8 says, "Two divides eight," while 2/8 is a fraction equivalent to 1/4. Anyway, we can then go on to define an equivalence relation on Z by first choosing a fixed positive integer n and then we say a is equivalent to b modulo n if and only ifn\(b — a). We then write this as a = 6(modn). If we work mod 4, we see 8 = 0(mod4), 5 = l(mod 4), and 9 ^ 2(mod 4). Since this is an equivalence relation, we collect the integers together in equivalence classes, [k] = {m 6 Z : m = fc(mod n)}. The collection of equivalence classes is a finite set as each integer ends up being congruent to its remainder on division by 4. So the only equivalence classes are [0], [1], [2], and [3]. We can then define addition and multiplication on the classes by [a] + [b] = [a + b] and [a] ■ [b] = [a ■ b}. So we have a set of equivalence classes and two operations + and ■. Do we then have afield? It depends on n. The set of integers mod 4 is not afield: [2] ~ l does not exist as [2].[0], [2]-[l], [2]-[2], and [2]-[3] are all not equal to e — [1]. However, if n is a prime number (usually primes are denoted by p) we do get afield. The set of equivalence classes in this case is denoted by Z/pZ (other sources use Z p , but we have another use for that notation). Theorem 5.3 Let p be a prime number. The set of integers modulo p, Z/pZ, is a (finite) field.

72

S

P-ADIC NUMBERS AND VALUATIONS

EXAMPLE 5.2 Let p = 7. Then Z/7Z = {[0], [1], [2], [3], [4], [5], [6]}. With regard to arith­ metic, we see [3] + [4] = [0], [3] • [4] = [5], and [3]-[5] = [l], the last example showing [2>]~l = [5].

■

5.2 p-adic Numbers 5.2.1 An Analysis Point of View Let us recall from Chapter 6 the idea of a metric or distance function on a set S. Definition 5.6 A metric on a set S is a function d : S x S —> [0, oo) which satisfies the following three properties: • For all x,y G S, d(x, y) > 0 (called nonnegativity) and d(x, y) = 0 if and only ifx = y. • For all x,y £ S, we have d(x, y) = d(y, x) (symmetry). • For all x,y, z 6 S, we have d(x, y) < d(x, z) + d(z, y) (The Triangle Inequal­ ity). Given a metric space (S, d), we can then develop the ideas of Cauchy sequences, convergent sequences, and complete metric spaces. The major example of a complete metric space is the set M of real numbers as the completion of the set of rational num­ bers Q. Usually this is shown by starting with Cauchy sequences of rational num­ bers. We then say that two sequences {xn} and {yn} are equivalent if\xn—yn\ —> 0. This equivalence relation creates equivalence classes and M. is the collection of all possible limits for these classes. What is hidden in all of this is the dependence on using the notion of Euclidean distance \ ■ \. This section develops another idea of distance, creating a new way of looking at rational numbers and their completion. In the next section we will then relate the p-adic integers to Cantor Sets. We start by fixing a prime number p. Then, we can write any number x in Q as X = P

b

P-ADIC NUMBERS

73

where n G Z, p] a, p\ b, and a, b have 1 as their greatest common divisor1. Then a size on Q can be defined by

10,

x = 0.

Intuitively, \x\p when x is a fraction in lowest terms "measures" the factor x must be multiplied by to "cancel" all p's from its numerator and denominator. [ji EXAMPLE 5.3 Let us fix p = 3. Then |12| 3 = 1/3, | f § | 3 = 9, and |8| 3 = 1.

■

A few things worth pointing out about this idea of size: ip

takes on a discrete set of values, {0} U {p n} where n G Z.

■ One name for | • \p is a valuation . In algebra, a valuation is a function that measures the size of elements in afield. ■ | • |p is, in fact, a norm. We define norms right now. For any field F, a norm, || • || on F is a map from F into [0, oo) such that (1) \\x\\ = 0 if and only if x = 0, (2) \\xy\\ = \\x\\ \\y\\for all x,y G F, and (3) for all x,y G F we have \\x + y\\ < IMI + \\y\\ (the triangle inequality). • For different values pi and p2, the norm \ ■ \Pl is not equivalent to | • \P2. This means they do not generate the same idea of size and we can see that with the following example. Suppose p\ — 2 and pi = 3. Let xn — (2/3)". Then \xn\2 ->• 0, while \xn\3 ->• 00. ■ The collection of norms is exhaustive. A theorem in [9] states that the Euclidean absolute value, | • |, and the p-adic valuations, \ ■ \p, p = 2, 3,5, 7,11,..., are all the possible nonequivalent norms on the field Q. We now take this norm and turn it into a metric on the rational numbers, the same way we would with the ordinary absolute value. Definition 5.7 Let x and y be rational numbers and let p be a fixed prime number. We define the p-adic distance between x and y by dP(x,y) =

\x-y\p.

Thus we have the metric space as (Q, dp). 1 For two integers a and b, not both zero, the greatest common divisor of a and b is the largest integer k that is a factor of both a and b. This is usually denoted by k = gcd(a, b). For example, gcd(12, 27) = 3. Two numbers are called relatively prime if their greatest common divisor is one.

74

P-ADIC NUMBERS AND VALUATIONS

It is obvious that \x — y\p satisfies the first two requirements to be a metric. Let us look at the third. We claim the following is true (proof will follow): for any a, b 6 Q, \a + b\p < m&x{\a\p,\b\p}. Given this, then for any x, y, z £ Q \x - y\p = \x - z + z - y\p < max{|x - z\p, \z - y\p} <\x - z\v + \z - y\p. Theorem 5.4 For rational numbers x, y we have \x + y\p < max{ja:|p.|y|p}. Proof: We will actually prove a different statement: that \x\p < 1 implies | l + x | p < 1. So let us show these two ideas are equivalent. If\x + y\p < max{|a:|p.|y|p}, then given \x\p < 1 we have \l + x\p < max{l, \x\p} = 1. Conversely, start by assuming that \x\p < 1 implies jl + x\p < 1 is true. Without loss of generality, assume y ^ 0 (if' y — 0 it is all trivially true) and \x\p < \y\p. Then \x/y\p < 1 and then I1 + x/y\p < 1 and thus \x + y\p < \y\p = max{|a:|p.|y|p}. So, those two statements are equivalent. Now we just have to show \x\p < 1 implies \l + x\p < 1 works. This is trivial if x = 0, so assume without loss of generality that \x\p < 1 and x ^ 0. We know x can be written as x = c/d where p \ d and gcd(c, d) = 1. Then c c+d 1 + x = 1 + - = —— d d and, given p does not divide d, it must be jl + x\p < 1. 4 We now have shown that a p-adic metric is actually stronger in one sense than the Euclidean metric. This is actually what is called an ultra-metric. Definition 5.8 Let Sbea set. A function d : Sx S —> [0, oo) is called an ultra-metric ;/('/ satisfies the following three properties: • For all x,y £ S, d(x, y) > 0 (called nonnegativity) and d(x, y) — 0 if and only ifx = y. • For all x,y e S, d(x, y) = d(y, x) (symmetry). • For all x,y,z equality).

e S, d(x, y) < max{d(x, z),d(z, y)} (The Strong Triangle In­

So an ultra-metric space is a metric space satisfying the Strong Triangle Inequal­ ity. Ultra-metric spaces have many interesting properties, but we will leave you with this one to ponder: In an ultra-metric space, all triangles are isosceles. The position we are now in is a familiar one from a course in real analysis. In that course, we see rational numbers with the valuation given by the usual absolute

P-ADIC NUMBERS

75

value and d(x, y) = \x — y\ has issues. There are numbers missing from the set of rational numbers -for example, the limit of the sequence that begins #1 = 1 and xn+i

= -^ H

for n > 2.

One way to fix this is to relate Cauchy sequences. This leads us to the completion of the rational numbers, which we call the real numbers, R. We follow this idea with Q andp-adic distance, \x — y\p and find ourselves with a new completion of the rational numbers. As alway, start with a fixed prime number, p. We say that two Cauchy sequences (xn) and (yn) are equivalent if lim \xn -yn\p

= 0.

n—>oo

This is an equivalence relation (reflexive, symmetric, and transitive), which leads us to equivalence classes of sequences, which we shall denote by [xn]. Each equiva­ lence class [xn] can then be associated with a number x defined to be the limit of the sequence. The set of all the limits of Cauchy sequences is the completion o/Q. We will denote this set by Q p . More about rational and irrational numbers in Q p will be in Section 5.4. 5.2.2

An Algebra Point of View

This section focuses on a specific set ofp-adic numbers, the p-adic integers. We will see how they differ from our usual concept of integer and, in the end, are a non-trivial example of an abelian group. In our usual base 10 numeration system, we can represent any number x using a series of decreasing powers of 10 via oo

x = y^afc10~,c k—n

where n G Z and for each k, a^ e {0,1, 2 , . . . , 9}. Recalling infinite series, what we are dealing with here is the limit of a sequence of partial sums. Let xm = Y^k=n a fcl0 -fc . This is a rational number and x — lim

xm.

For example, x = 1/3 is the limit ofxm = Yl™=i 3 ' 10~fc and \x — xm\ < 3/(10 m ). In the world ofp-adic numbers, the idea of convergent becomes different from the one used above. Recall that for p a fixed prime \x\p = -~ where n is the largest power ofp that divides x and this leads to the distance d(z, y) = \x — y\p. For our examples, letusfixp = 3. Then |3|3 = |6|3 = |15|3 = 3 _ 1 = 1/3. In a similar vein,

| 5 - 1 4 | 3 = l/9.

76

P-ADIC NUMBERS AND VALUATIONS

Let us expand this into the limit of a sequence of 3-adic numbers. Now

26

1+3, 1+32, - 1 + 32,

which implies which implies which implies

| 2 - ( - l ) | 3 == 1/3, | 8 - ( - l ) | 3 == 1/3 2 , 3 | 2 6 - ( - l ) | 3 = l/3

Before we continue, a few things to note: • In the 3-adic numbers, we see 8 is closer to —1 than 2 is, and 26 is even closer than 8. ■ Just like with the integers mod p, it seems unreasonable to use numerals like 8 or 6 in our 3-centric numeric expansion. So what we will use from now on is the idea that in p-adic expansions the numerals available are {0,1,2,... ,p — 1} and any positive integer x will be written as n

x = ^2ahpk fc=0

where the a^ come from the available set of numerals. We will use a subscript to denote the numbers in p-adic form. • Some sources allow digits beyond p — 1, but we prefer this idea of restricted digits (called the canonical form) so that our way of representing p-adic is unique. Note that in our usual system we do not have uniqueness as 1.00000... = 0 . 9 9 9 9 9 . . . . ■ As an example, 8 = 2 - 3 + 2 becomes 22 3 and 26 = 2 • 3 2 + 2 • 3 + 2 = 2223. Just to throw in something different, 47 = 1 • 3 3 + 2 • 3 2 + 0 ■ 3 1 + 2 • 3° = 12023. Now our chart becomes 2 3 = —1 + 3, 223 = - 1 + 3 2 ,

which implies which implies

|2 3 — (—1)3|3 = 1/3, |223 - ( - 1 ) 3 | 3 = 1/3 2 ,

2223 = - 1 + 3 3 , 22223 = - 1 + 3 4 ,

which implies which implies

|2223 - ( - 1 ) 3 | 3 = 1/3 3 , |22223 - ( - 1 ) 3 | 3 = 1/3 4 .

So obviously some interesting things are going on here. Let us now make this official with some definitions. Definition 5.9 Let p be a fixed prime number. Then for any positive integer x, the p-adic expansion of x is of the form n fc=0

P-ADIC NUMBERS

77

where a* £ {0,1, 2 , . . . ,p — 1}. This definition restricts us to the positive integers. So our first task is to expand things to as big a collection of numbers as possible. We begin with what we have already been looking at, the 3-adic numbers and — 1. Our restricted numerals means we cannot use the symbol — 1, and the pattern we have established says the more 2 's we use, the lesser the difference we get in \ • \$ between a string of2's and the idea of — l p . So it is reasonable to guess that - 1 3 = ...22222 3 . There are many things to point out here. Firstly, to talk about certain p-adic numbers, we must use infinitely many digits. However, unlike decimals, things have turned around and the infinite string goes to the left, not the right. This brings us to our next definition. Definition 5.10 Let p be a fixed prime number. Then for any integer x, the p-adic expansion ofx is of the form fc=0

where a* £ { 0 , 1 , 2 , . . . ,p — 1}. In some literature, the p-adic expansion ofx is written as x = p^x) (a 0 + axp + a2p2 H

),

with a,i £ {0,1, 2 , . . . ,p — 1}, ao > 0, andy(x) £ Z. Before we present more, let us look at some properties of p-adic numbers. The first, which we present as a theorem, again shows how surprising things can be. Theorem 5.5 The set of p-adic integers is not a countable set. Proof: Since the p-adic integers have infinite decimal expansion to the left, we can see that there is a one-to-one correspondence between these integers and the product l[{0,l,2,...,p-l}

=

{0,1,2,...,p-if.

n>0

This is equivalent to 2^°, which is uncountable. 4 Another way to show uncountability is to mirror the Cantor Diagonal argument. Assume the p-adic integers are countable, list them, and develop a "new" integer x by making sure the ith digit of x does not match the ith digit of the ith number on the list. Now we wish to look at addition of two p-adic numbers. Similar to the algorithm for our usual number system, we will line things up and carry. This time, however the addition is modulo p.

78

S

P-ADIC NUMBERS AND VALUATIONS

EXAMPLE 5.4 Letp = 3. The sum 210123 + 112013 is 210123 +

112013 1022203

■ This works for finite or infinite expansions. The precise definition is as follows: Definition 5.11 Let a, = X^fc>o akPk and b = J2k>o bkPk be two p-adic integers. Then their sum a + b is defined to be X^fe>o CkVk wnere (1) ci € {0,1. 2 , . . . ,p— 1}, and (2) for each m = 0,1, 2, 3 , . . . m

/ m

\

k

b

k

mod

J2°kp = Z > * + ^p fc=0

\fe=0

(p m + 1 )-

/

A straightforward consequence of this definition is that the set of p-adic integers is closed under addition. Also, addition is commutative and associative. There is also a zero element,

Op = ]Toy\ k>0

and a 1 l p = 1 + Op + Op2 + Op3 + ■ ■ ■ . Now we need to show that every element has an additive inverse. We have already looked at —1 for 3-adic integers. The same argument works for any p:

~iP = J2(p~l)pkfc>0

We can use the definition of addition to see that l p + ( — l p ) = Op, so lp has an additive inverse. Finally, if we let n he a positive integer, then we know

fc>0

By hand calculation, we can see that if

fe>0

then np + n* = —lp; that is, np + (n* + l p ) = 0 P and we have found the additive inverse for np.

P-ADIC NUMBERS

Definition 5.12 Let p be a fixed prime number. For any positive

79

integer

np = ^2 nkPk fe>o

it must be that

-np = ^2{p - 1 - nk)pk + lp. k>o

So we have done the work to give us the following

theorem.

Theorem 5.6 Let p be a prime number and Z p represent the p-adic integers. ( Z p . + ) is an abelian group.

Then

Let us move on now to multiplication of p-adic integers. As one would expect, this is a lot like ordinary multiplication with the exception of multiplication (and addition) being carried out modulo p. fl

EXAMPLE 5.5 The product 21012 3 x 11201 3 is 21012 3 x

2013 21012 3 OOOOOO3

+

II2IOIOO3 12001112 3

■ Computing exact values becomes much more complicated when the integers have infinitely many digits, but the idea is the same. As multiplication m o d p commutes, we have that ( Z p , + , •) is a commutative ring with identity (lp = ... 00001). Here is another interesting fact for p-adic integers. Let a = ...0403020100 be a p-adic integer with ao 7^ 0. Since p is prime, we can find a value bo s { 0 , 1 , 2 , . . . ,p — 1} such that 0060 = l(modp). And, assuming ao&o — c o l (so Co is carried to the next column), there is a value b\ € { 0 , 1 , 2 , . . . ,p — 1} so that (ai&i + Co) = 0(modp). We can continue this indefinitely so that a^b^ + Ck-i = 0(modp) where Ck-i is the carryover from the previous place. This gives us our next theorem. Theorem 5.7 Let a = ... 03020100 be a p-adic integer, where p is a prime number. If ao 7^ 0, then there is a p-adic integer b so that ab — lp. Finally, we note that with Z p we still have the Zero Product Property. Recall that this means that if a,b 7^ 0, then ab 7^ 0. Thus our final result from our algebraic view of the p-adic integers is the theorem below. Theorem 5.8 (Z p , + , •) is an integral domain.

80

5.3

P-ADIC NUMBERS AND VALUATIONS

p-adic Integers and Cantor Sets

There are a lot of more interesting properties of p-adic numbers, and we will look at p-adic rational numbers in the next section, but let us not forget the focus should be on how this relates to Cantor Sets. We will start by reviewing some topological ideas. Let (X, d) be a metric space and choose x G X. The open ball about x of radius r is the set of points in Xwhose distance from x is less than r; that is, B(x,r)

= {y G X : d(x,y)

< r}.

We refer to x as the center of the open ball. From this, a set U is open in X if for every x G U there is an open ball containing x where the ball is a subset of U. So there exists an r > 0 and an a G U where x G B(a, r) C U. A set is then closed if its complement is open. This brings us to some surprises with the p-adic completion

Theorem 5.9 The sphere S{x, r) = {y G Q p : \x - y\p = r) is an open set in Qp. Proof: Let y G S(x,r). We will show B(y,r/2) C S(x,r). Choose a point z G B(y,r/2). Then \z — y\p < \y — x\p = r. However, | • \p is not just a norm, but a non-Archimedean norm. That means \\a + 6|| < max{||a||, ||6||}. A consequence of this is that if\\a — b\\ < \\a\\, then \\a\\ = \\b\\. So \z — x\p = \y — x\p = r, which means \z — x\p = r or z € S(x, r). Thus B(y, r/2) C S(x, r) and we are done. 4* Corollary 5.9.1 Any open ball in Qp is both open and closed. Proof: The fact that B(a, r) is an open set is trivial as for any x s B(a, r) the ball itself is an open ball containing x and a subset of the ball. In order to show that it is also a closed set, we must look at the complement of B(a, r). The complement can be written at Qp \ B(a, r) = {y€Qp:\x-y\p>r}

= EU F,

where E = {y G Q p : \x - y\p > r} and F = {y E Qp : \x - y\p = r} = S{a, r). Theorem 5.9 takes care ofS(a, r). To show that E is open, pick an arbitrary y G E. Then y — a\p = r* > r. Then let us look at B(y,r* —r). We claim B(y,r* —r) C E Otherwise there is a z G B(y, r* — r) with \z — a\p < r. But then r* = \y-a\p

< \y-z\p

+ \z-a\p

< m&x{\y-z\p,\z-a\p}

= m a x { r * - r , r } < r*.

Thus we have a contradiction. So B(y, r* — r) C E, which means that E is an open set and the union of two open sets is open, proving our corollary, ft

P-ADIC INTEGERS AND CANTOR SETS

81

Previously we have seen continuous functions. In topology, what is important is not just continuity, but that the function f and the inverse function / _ 1 be continuous. Then f is a homeomorphism. We repeat the definition from Chapter 7: A function f : A —► B is a homeomorphism if it is bijective and both f and f~l are continuous functions. We say A and B are homeomorphic if there exists a homeomorphism f : A —* B. One big result for this section is relating p-adic integers and the Cantor Set by means of a homeomorphism. The argument uses our ability to completely describe points in the Cantor Set via base 3 numbers that do not contain the numeral 1. Theorem 5.10 Let p = 2. The set of2—adic integers (also called dyadic), Z2, is homeomorphic to the Cantor Set, C. Proof: In this proof, we will give an explicit function and show that it is a homeo­ morphism between the two sets. Every integer for p = 2 can be written as

Y,ak2k k=0

where ak G {0,1}. Every point in the Cantor Set can be written as

E

Ok Qfe

fc=l

where bk £ {0, 2}. Let F be the function defined by

^E^ f c -£jS fc=0 fc=0

(note the change in index in the image). Because p-adic integers and elements in the Cantor Set have unique representa­ tions, we immediately get that this function is both 1 — 1 and onto. Hence we have a bijection. Now we will show continuity for both F and F_1 by showing that if we start with points that are close together, their images must also be close. First let us be more specific about when we say close. Of course in the dyadic integers if the first N digits of X\ and £2 agree, then \x\ — X2\v < 1/2^ For points in C, at stage N in the construction we have K^ which consists of2N disjoint subintervals of [0,1] and so |j/i — 2/2I < 1/3N means y\ and 2/2 agree for the first N digits. In the other direction, if their first N digits agree, then y\ and yi must be in the same component. Since both 1/2W and 1/3N converge to zero as N —> 00, we can say that both F and F~l are continuous functions. 4k Now let us turn our attention to p-adic integers in general and not just the dyadic ones. We can find a similar homeomorphism between %p and a general form for

82

P-ADIC NUMBERS AND VALUATIONS

a Cantor Set. We create the generalized set as follows: Fix p. Stage 0 is the unit interval [0,1]. In Stage 1, the unit interval is divided into 2p — 1 equal subintervals and every second open subinterval removed. What is left is called K\. At stage n there are pn disjoint closed intervals making up KM- Each of these is divided into 2p — 1 equal subintervals and every other open subinterval removed, leaving behind KM+I- Our new Cantor Set is then Cp = ClK^. We can characterize Cp as the set of all numbers 0.a\a2a?, ■ ■ ■ such that each a^ is an even number. Then we have a homeomorphism, Fp, between Zp and Cp given by

Expanding on these ideas, we can prove that for any prime numbers p and q the sets Z p and Z ? are homeomorphic. 5.4

p-adic Rational Numbers

So far we have looked at p-adic integers and have seen them to be quite different from our ordinary integers. The rational numbers will also be a different world. The example we will use is the 1—adic rationals. In order to get used to this, let us quickly review arithmetic in TLjlTL. Here our available digits are 0, 1, 2, 3, 4, 5, and 6. The corresponding addition and subtraction tables are

+ 0

1

0 0 1 1 2 2

2

3 3 4 4 5 5

1 3 4 5 6

6 6 10 fl

2

3

4

2

3 4

4

5

6

5 6 5 6 10 5 6 10 11 5 6 10 11 12 6 10 11 12 13 10 11 12 13 14 11 12 13 14 15 3 4

X

0 1

0 0 0 1 0 1 2 0 2 3 0 3 4 0 4 5 0 5 6 0 6

2

4

5

6

0 0 0 3 4 5 6 11 13 6 12 15 21 11 15 22 26 13 21 26 34 15 24 33 42

0 6 15 24

0 2 4

3

33 42 51

EXAMPLE 5.6 The following are reminder examples of the four basic operations in base 7. The reader is welcome to practice by verifying the answers. 1. 23547 + 631 7 = 33157 2. 23547 - 631 4 = 14237 3. 3527 x 23 7 = 120567

P-ADIC RATIONAL NUMBERS

4. 120637 + 34 7 = 2567

83

■

We begin here with the rational number —1/4. The reason will hopefully become clear. To convert this to a 7—adic number we will use a sequence of approximations and see —1/4 is the limit of the sequence. The terms consist of ratios using even powers of 7, converting the numerator to base 7 (the denominator is already a base 7 digit) and performing the division. fl

EXAMPLE 5.7 In this example, we find the 7—adic expansion of the number ~ | . We will use powers of 7. How do the powers of 7 help us? Keep in mind, for 7—adic num­ bers, as n increases |7"| 7 gets smaller. So 72-l 4 1 4

48 4 2400 4

66 7 4 66667 4

ir

15 7 , 15157

76 - 1 117648 6666667 1515157 4 4 4 We are using only even powers because the odd powers of 7, minus 1, is not evenly divisible by 4 and we would like to avoid the decimals. The interested reader may compute on his/her own that (7 3 — l)/4 = 151.333 .. .7. General­ izing the pattern, we have 7n 7n - 1 4 = 4 + ~

4^°+~

1 _1 4 ~~ 4'

Our established pattern shows that in the 7-adic number system _1 ..151515 7 4 ~ ' We can verify that this is the correct value. ~l7 = ... 666667 = (• - ■ 1515157) x 4 7 . This gives us our first surprise: —1/4 is an integer. Two more rational numbers that we can easily find - - = 3 x - - = ...0003 7 x ...151515 7 = ...535353 7 and

1 3 - = 1 + " - = . . . 5353547. 4 4

84

P-ADIC NUMBERS AND VALUATIONS

Similar to what we experience in M with the rational number, we have this theorem in Q p . Theorem 5.11 Let x G Q p where the canonical p-adic expansion of x has the form

J2akpk k—n

with cik G { 0 . 1 , 2 , . . . . p — 1} and n £ Z . Then x is a rational number if and only if the digits of x are eventually periodic: There is an N > 0 and an m, > 0 such that a-n+m — «« for all n > N. Thus we have x

=

anpn +an+1pn+l +pn+k+l +pn+k++j

\-an+kpn+k

H

b3p>-1)

(bx+b2p+...+ +l (6i

+ b 2 p +

...

+

bjpi-1)

The proofforthis can be found in [2]. It is fairly obvious that if x G Q, then x G Qp. In fact, given that there are only finitely many elements in { 0 , 1 , 2 , . .. ,p — 1} it is true that if x is a rational number in the Euclidean sense, then x must be rational in Q p . The question then becomes, what else is in Qp and how is this completion different from M.? The next two results help show the difference. Let us look at solving an equation in Q5. Really we are looking at \/6, but we frame this as solving the equation x2 = 6 where x = ao + aj ■ 5 + a2 ■ 5 2 + • • • and at G { 0 , 1 , 2 , 3 , 4 } . So we need (a0 + o-i • 5 + a2 ■ 5 2 H

) 2 = 6 = 1 + 1 • 5.

Expanding the left side and comparing coefficients we see the constants give us the equation a 2 = 1 (mod 5). This has two solutions, ao = 1 and CLQ = 4. We will stick with the former. Thus we now have (1 + aj • 5 + a 2 • 5 2 + • ■ ■ ) 2 = 1 + 2ai • 5 + • • • = 1 + 1 • 5

(mod 25)

or 2ai = 1 (mod 5 2 ) Then a,\ = 3. This brings us to (1 + 3 • 5 + a 2 ■ 5 2 + • • • ) 2 = 1 + 2ai • 5 + 2a 1 • 5 2 + • • • = 1 + 1 • 5

(mod 5 3 ).

So 2a 2 = 0 (mod 5 3 ), thus a2 = 0. Continuing, 03 = 4 and 04 = 2. This continues with each a^ uniquely determined, thus

V& = . . . 2 4 0 3 1 5 .

P-ADIC RATIONAL NUMBERS

85

Although the ai are unique after we chose ao, we did have a choice there. This makes sense because in a nonzero field a square root, if a solution exists, will have two solutions. The second y o is ...20414 5 . Yet another integer! In the next example, we again look at square roots in Q5, but this time see that things are definitely not what we are used to. S

EXAMPLE 5.8 In Q5, \fl does not exist. If there were such a number, we would have x = ao + a\ ■ 5 + a-i ■ 5 2 + • • • and (a 0 + ai • 5 + a 2 • 5 2 + • • • ) 2 = 2 + 1 • 5 would have a solution in mod 5. The first comparison, though, yields a 2 = 2. Given the possible digits {0,1,2,3,4} this is impossible. ■

Finally, with a method similar to what we did with p-adic integers, we present a theorem relating Qp and a Cantor-like subset o / R + = [0, 00). As always, p is a fixed prime number and every x € Qp is written as 00

where a^ e {0,1, 2 , . . . ,p — 1} andn £ Z. Let cf) : Qp —> E + be the function defined by oo

4>{x) =

\x\p^akp~2k.

k=o

In order to look at properties of (f> it is necessary to put an order on the set Q p . For two numbers x, y € Qp we can write them as oo

oo

x = pnx Y^ akPk and V = Pny^2 k=0

hpk,

fc=0

where ao ^ 0, bo ^ 0 and nx, ny 6 Z. Then we say x < y if either \x\p < \y\p or \x\p = \y\p and there exists nonnegative integer j such that a0 = b0, ax = bi,...,

aj_i — bj-i,a,j < bj.

This makes Qp a totally ordered set. Let us also point out that for any x £ Q p \ {0}, 0 <x.

86

P-ADIC NUMBERS AND VALUATIONS

Now we show that (j> is an increasing function. Lemma 5.1 cf)(x) > <j>{y) if and only ifx > y. Proof: If' <j){x) > y. Assume \x\p > \y\p > 0. This means \x\p > p\y\p. We then see oo

oo

h 2k


^

k=0

(p-i)\y\PY,P~2hfc=0

Then we have (*) - <j>{y) > \x\p - ^ \ y \

p

> p\y\p (l - -

^

=*

> 0.

On the other hand, if\x\p = \y\p, but a® = bq.ai = b\,... , a^-i = bj-x,a,j < bj, then j

(f>{x) >

\x\pJ2akP~2k fc=0

while 0(y)

< \y\pEi=0ykP"2k + \y\P^p-2U+1) = \x\P Ei=o akP~2k + \X\PVJP~2J + \AvP~2j Tpp-

This means {x) - 0{y) > \x\pp-2j

(XJ - y3 - —-jj

> 0.

So regardless, must also be a one-to-one function. In addition \4>{x) -4>{y)\


for all x,y £ Qp. This means we are dealing with what is called a Lipschitz function. This also means (f> is continuous. Now we wish to look at the image of the whole domain, Qp. It turns out that K = (p(Qp) is related to a Cantor Set. We state the following without proof: In Qp, we can define the sphere with center a and radius pr to be Sr(a) = {x £ Qp : | a — x\p = pr}. Because the only distances in Q are {0} U pr where r £ Z there are countable many radii with which to be concerned. Then we decompose Qp into a collection of spheres all with center a = 0. Since the centers are the same, we will just call these spheres Sr. Now let Kr = 4>(Sr), and K = \Jr^j,Kr. lfr\ ^ r(S0) = Hi* £ Q P : k - 0| p = p0}).

P-ADIC RATIONAL NUMBERS

87

For any x € So, 1 <
= [0,p)

and K0\I leaves a Cantor Set. The details are in [76].

88

P-ADIC NUMBERS AND VALUATIONS

EXERCISES 5.1

Supply the missing proofs for Theorem 5.1

5.2 Prove that ifG is an abelian group, then for any positive integer n, (a*b)n an * bn. Here xn = (x * x * ■ ■ ■ * x).

=

n times Let G be a set with operation # such that

5.3

• G is closed under # . ■ # is associative. • There exists an element e G G such that e # a — a for all a G G. • For every a G G, there is ab e G such that b#a = e. Prove that G is a group. 5.4

Determine whether or not 7L with a*b = a — b is a group.

5.5 A non-empty subset, H, of a group G is called a subgroup of G if, using the product for G, H itself is a group. Let G be the integers 1 under +. Let H be the set of even integers. Prove that H is a subgroup. 5.6 Let G be a group and a G G. If n is a positive integer an is defined as in Problem 5.2; a0 = e, an ■ am = an+m, a~n = ( a " ) ' 1 . Prove that A = {an : n G Z} is a subgroup ofG. This is called the cyclic subgroup ofG generated by a and is denoted by < a >. 5.7

Prove that a cyclic group is abelian.

5.8

Show that any field is an integral domain.

5.9

Prove Theorem 5.2.

5.10 Show that the Euclidean metric on M. is also a norm. Prove that for all x, y G R, \xy\ = \x\ ■ \y\. 5.11

Convert each integer below into thep-adic version for the value ofp specified. a) 23, given p = 2 b) 23, given p — 3 c) 117, given p — 5 d) —6, given p — 2

5.12

Find the p-adic expansion of'1/3 ifp = 2.

5.13

Find the p-adic representation for 1/p.

5.14

Provide a (short) proof that (Z p , +) is, in fact, abelian.

EXERCISES

89

5.15 Our example of a sequence in Q that is Cauchy, but does not converge in the rationals, is X\ = 1 and xn+i = -^ -\ for n>2. L

Xn

What is the real number that is the limit of this sequence? (Note: This is not asking for a proof of convergence, which is assumed.) 5.16 Determine the p-adic distance between a = 252 and b = 140 for p = 2,3,5,7 5.17

Find the representation of a) - 1 / 5 in
5.18

Find the values ofp — 2,3, 5, 7 for which —1 has a square root in Q p .

5.19

Find the value of |90| p ifp — 2,3,5,7.

5.20

Prove that in Q 2 , if\x\2 = \y\2, then \x + y\2 = (l/2)|x| 2 .

5.21

Fill in the details for the diagonal argument to prove Theorem 5.5.

5.22 In many sources, the absolute value of a real number is written using the notation \X\OQ. Using this notation, explain why it is true that n2
Show that for any prime p, lp + Ylk>o(P ~ ^)Pk

=

*V

5.24 Show that in contrast to Theorem 5.7 if'ao = 0, then a = ... 0302^100 cannot have a multiplicative inverse. 5.25 Fill in the missing piece in Theorem 5.9. Suppose we have a non-Archimedian norm. So\\a + b\\ < max{||a||, ||6||}. Show that if \\a - b\\ < \\a\\, then \\a\\ = ||6||.

CHAPTER 6

SELF-SIMILAR OBJECTS

If we look at the construction of the Cantor Set, we see that for any k, a scaled copy of the picture kth level appears in subsequent levels. For instance, when k = 1, the set K\, which is the union of the intervals [0,1/3] and [2/3,1], is represented by the picture

For k = 3, the image of K3 looks like

which can be interpreted as four shrunken copies of K\. This leads us to the ideas of sets being self-similar which, in turn, takes us into the world of fractals and later, in Chapter 7, to various notions of dimension. 6.1

The Meaning of Self-Similar

We will be more formal later, but before we start we should say a little more about what makes a set self-similar. A set that is self-similar has the property that at any The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

91

92

SELF-SIMILAR OBJECTS

magnification there are pieces of the set that have the same shape as the whole. A line segment is a trivial example of such a set, but for a nontrivial example let us look at a fern branch

These are typically generated by repeatedly plugging an image into a system of func­ tions called similarities and letting the sequence approach a shape called an invari­ ant. The study of these sets really took off in the last half of the 20th century as re­ searchers were able to use computers to approximate pictures of these sets and the study of fractals rose in prominence.

6.2

Metric Spaces

The Cantor Set, of course, is a set in the real line, so we will begin this section with K. The usual idea for the distance between two points on the line is the absolute value of their difference. Symbolically we say that ifx,y e K the distance between x and y is d{x,y) = \y-x\. This is referred to as the Euclidean distance on M. and we will sometimes write this as | • |. A metric is the official name for a distance function, and we will begin by looking at the properties that make something a distance function. This was quickly remarked on in Chapter 2. We go into more detail here. Definition 6.1 A metric on a set S is a function d : S x S —»■ [0, oo) which satisfies the following three properties: ■ For all x,y £ S, d(x, y) > 0 (called nonnegativity) and d{x, y) = 0 if and only ifx = y. ■ For all x,y 6 S, d(x, y) = d(y, x) (symmetry) • For all x, y, z 6 S, d(x, y) < d(x, z) + d(z, y) (The Triangle Inequality) Some examples of spaces and accompanying metrics are as follows:

METRIC SPACES

93

1. LetS — R", where x = (xi,x2, ■ ■ ■ ,xn) and y = (yi, y2, ■ ■ ■ ,yn), and d(x, y) = V22/!=i(^fc — Uk)2y the Euclidean metric on R™. 2. Let S be any set and let

[0 ifx = y The function d is referred to as the trivial metric. 3. Let S be the plane R 2 with points written as ordered pairs; so x = (x\, x2) and y = (yi, y2). We may define distance as ~d(x,y) = |x/i - x i | + \y2 - x2\. This is referred to as the taxicab metric. 4. Let S = (—7r/2,7r/2) and define distance as D{x, y) = \ tan(y) — tan(x) |. We should give an example of a proof showing a function on a space is a metric, so we will show one way that we can define distance on the space of continuous functions. fl

EXAMPLE 6.1 Let C[0,1] = {/ : [0,1] —>■ K : / is a continuous function} and define p{f,g)=8w{\f{x)-g{x)\:x£[0,l]}. Then p is a metric (called the sup metric) on the space C[0,1].

■

Proof: First, since absolute value is nonnegative, we know \f(x) — g(x)\ > 0 for all x and hence so is its supremum. Furthermore, the supremum is equal to zero if and only if \f(x) — g(x)\ — Ofor every x £ [0,1] which, thanks to the properties of absolute value, means f(x) — g(x) = 0 or f(x) = g(x) for all x. The second criterion is straightforward. Since \f (x) — g(x)\ = \g(x) — f(x)\ these two quantities have the same supremum and p(f, g) = p(g, / ) . Finally, for any fixed value x and continuous functions / , g, and h, \f(x) - g(x)\ < |/(x) - h(x)\ + \h(x) - g(x)\. Thus sup{|/(x) - g(x)\} < sup{|/(:r) - h(x)\ + \h(x) - g{x)\} < sup{|/(x) h(x)\} + sup{\h(x)-g(x)\}.Sop(f,g) = sup{|/(x) -g{x)\ where x S [0,l}}fits all the requirements to be a metric. 4* We can now take some of the familiar properties ofM. with the Euclidean metric | • | and rewrite them in terms of a metric space, (S,d).

94

SELF-SIMILAR OBJECTS

Definition 6.2 For a point x £ S and a fixed r > 0, the open ball about x with radius r is given by B(x,r) = {s £ S : d(s,x) < r}. Note: What is contained in an open ball depends on the metric (see homework for some interesting examples), but our notation B(x, r) does not state what metric is being used. This should be clear from the context. S

EXAMPLE 6.2 Let S be any set and the metric the trivial metric, d. Then for any point s £ S there are two possible open balls around that point. If 0 < r < 1, then B(s, r) = {x}, while ifr > 1, then B{s, r) = S. m

Just as open intervals in the line generalize to open sets, we can generalize simi­ larly in a metric space. Definition 6.3 A set A C S in (S, d) is called open iffor every a in A there exists a value r so that B(a, r) C A. Definition 6.4 A set B is called closed if its complement Bc — S \ B is open. S

EXAMPLE 6.3 1. An non-empty open interval in (M, | • |) is an open set. (To generalize this, in any metric space (S, d) pick a point x and positive e. Then B(x, e) is an open set.) 2. The empty set 0 is both an open and closed set regardless of the metric space. 3. The set (—1, 0) U (2,3) in (R, | • j) is an open set. 4. Let S be any set and the metric the trivial metric, d. Then any set A C S is an open set since (for example) for all a £ A, B(a, 1/2) = {a} C A. 5. The set (1, 2] is neither open nor closed in (K, | • J). 6. The set of continuous functions f with /(0) = 0 is a closed set in (C[0,1], sup).

■

A couple of things are worth noting in these examples: First, let us take a closer look at the last example. If A = {/ £ C[0,1] : /(0) = 0}, then we will show A is closed by proving A = C[0,1] \ A is open. Pick a function g £ Ac. So we know
METRIC SPACES

95

Theorem 6.1 For any metric space (S, d), the countable union of non-empty open sets is an open set. Proof: Note that a finite collection of objects is countable. If there are only finitely many A\, A2,.. ■, Am we can extend this to countably infinite by saying An = 0 for n > m, and so our proof has a union of infinitely many sets. Let A = W^L-^An where each An is an open set To show A is open, let x £ A. Then x G Aj for some j . Now Aj being open means there exists an £ > 0 such that x G B(x,e) C Aj C uAn. This shows A is an open set. 4 Corollary 6.1.1 For any metric space (S, d), the countable intersection of closed sets is a closed set. Proof: This is really the proof above and an application ofDeMorgan 's Law

(uAnf = n(A%). A Definition 6.5 A set A C S in (S, d) is called bounded if there is a point s € S and a number r > 0 such that d(s, a) < r for all a e A. Related to this is the idea of a totally bounded set. This will be useful in our study of fractals. Definition 6.6 Let A C S in (S, d). We say A is totally bounded iffor each e > 0 there is a finite set ofpoints {eti, «25 • • ■, a m } in A so that for any a & A there is an aj with d(a,a,j) < e. fl

EXAMPLE 6.4 In the real line, with the Euclidean metric, the closed interval [0,1] is totally bounded. Given any distance e > 0, let N € N such that l/N < e/2. We can use the set of points given by Xi = ei, where i = 0,1,2,..., 2N. For any x € [0,1] there is an IQ such that ei0 < x < e(i0 + 1) so \x -xio\ < \xio+1\ = e/2 < e. The metric is important. In the trivial metric [0,1] is not totally bounded as any two points must be a distance 1 from each other. ■

96

SELF-SIMILAR OBJECTS

Finally, we extend the notion of continuity to functions whose domain and range are metric spaces. Definition 6.7 Let f be a function with domain (S, d) and co-domain (T, p) (usually written f : (5, d) —> (T, p)). Then f is continuous at XQ £ S iffor every e > 0, there exists a 5 > 0 such that ifx £ S with d(xo, x) < 6, then p{f{x0),f(x))

<£.

Furthermore, f is a continuous function on S if it is continuous at each XQ £ S. Recall the definition of continuity we used when we dealt with only topological spaces, without a definition of distance. These spaces used the idea of open sets and their inverse images. The theorem below shows that idea still holds with metric spaces. Theorem 6.2 Let f : (5, d) —>• (T, p). Then f is a continuous function if and only if for any open setV £ T its inverse image f~l{V) is an open set in S. Proof: Since this is an "if and only if" theorem, we have two directions to prove. Note that we will deal with open balls in both the metric d and metric p. We are not differentiating between them in our notation. First, assume that f : (S, d) —> (T, p) is a continuous function and that V is an open set in T. If f~l{V) = 0, then we are done since 0 is an open set. Let us assume there is an XQ € S such that /(.To) = v £ V. Since V is open, there is an e > 0 such that v £ B(v,e) C V. That f is continuous means that for this e there is a 5 > 0 such that for all x £ B(XQ, S) we have p(v, f{x)) < e. This makes B(XQ, 5) a subset of f~l(V) hence f~l(V) is an open set. In the other direction, we start with the fact that the inverse image of every open set is an open set. Pick XQ £ S and let e > 0. Then V = B(f(xo),e) (1 T is an open set in T. Our hypothesis given us that f~y{V) is an open set in S. The definition of open set means there is a positive 5 such that B(XQ,6) C f~l(V). Thenify £ B(x0,5) we have p(f(xo),f{y)) < e. This proves f is continuous at xo and since this point was arbitrary, f is a continuous function. 4k Let us now provide proof of Theorem 3.13 from the last chapter. Recall the theo­ rem tells us, "Any compact metric space is the continuous image of the Cantor Set." Before we can proceed we need the following definition and fact. A space Y is called a Hausdorff space iffor any two points x,y £ Y there are open sets U and V in Y with x £ U, y £V and

t / n y = «. Any metric space is a Hausdorff space. However, this is a topological property, so a metric is not required to be on the space. There are topological spaces that are not Hausdorff. Fact: Let / : X —> Y where X is a compact metric space and Y is a Hausdorff space. If / is 1 — 1 and continuous, then / is a homeomorphism between X and the co-domain, f(X). The proof can be found in most topology books.

SEQUENCES IN (S, D)

97

Proof: (for Theorem 3.13) This proof is due to Rosenholtz (see [65]) and consists of a collection of small results put together. First, let I" be the product of countably infinitely many unit intervals; Iu — [0,1] x [0,1] x [0,1] x ■ • •. Then any compact metric space (X, d) is homeomorphic to a subset of IM. Without loss of generality we can assume the metric is bounded by 1. After all, d is equivalent to the metric d/(l + d). Since X is compact, it contains a countable dense set, call this xi, X2, ■ ■ ■■ Define H : X —>■ Iu by H(x) = (d(x,Xi),d(x, X2),d(x,X3),...). This function is continuous because each d(-, Xj) is a continuous function. H is one-to-one because if x 7^ y then there is some point Xj that is closer to one of x or y than it is to the other. Hence H(x) ^ H(y). Since X is compact and Iu is Hausdorff (being the product of Hausdorff spaces) our fact above shows that H is a homeomorphism onto H(X). We have already seen that the unit interval is the continuous image of the Cantor Set. This is the function g{^2a>i/3l) — J2ai/2l+l where a* — 0 or 2. The Cantor Set itself is homeomorphic to the countable product of sets of the form {0, 2}. The function h : C -* {0, 2} x {0, 2} x {0, 2} x • • • is the function where h(^2 di/31) = (0,1,0,2,0,3,...). This is a rather straightforward choice of function. Because the countable product of countable sets is itself countable (Theorem 3.6) we can extend our result to this: The Cantor Set is homeomorphic to the countable product of Cantor Sets. The set Iu is the continuous image of the Cantor Set. From the result above, any point in C can be thought of as (xi,X2,x-s,...), where each X{ € C. Thenwe create the function F : C —> I" where F(x1,x2,x3,...)

=

{g(xi),g(x2),g(x3),...).

Now let K C C be closed. Then K is the continuous image of C. The proof of this uses the "obvious" fact that C is homeomorphic to C, where C" = {x = ^2 bz/6l where bi = 0 or 5}. This new C has the additional property of being nonaveraging; if x,y E C, then (x + y)/2 £ C. Assume K' is a closed subset of C Then if x € C there is a unique point k £ K' where d(x,k) = d(x,K) = inf {d(x, y) : y £ K}. The function T : C —>• K' given by T{x) = k is continuous and onto, and is the function which we were working to find. There are other ways to prove this result, using heavier topology concepts. See [43] and [69]. 4(t 6.3

Sequences in (S, d)

A sequence in a metric space (S, d) can be defined as a function f with domain N and co-domain S. However, rather than writing f : N —» S we usually rely on the typical notation {i„}™=1 (or {xn} if the index is understood). Colloquially, we say {xn} converges to a point x, if as n grows, the distance between xn and x becomes arbitrarily small. To talk about distance precisely we need to pay attention to the metric used. So officially: Definition 6.8 Let (S,d) be a metric space. We say the sequence {xn} C S con­ verges to the point x € S iffor every e > 0 there exists a natural number N such

98

SELF-SIMILAR OBJECTS

that ifn > N, we have d{xn,x) S

< e.

EXAMPLE 6.5 In (R, j • |) the sequence given by xn = ^ - j - converges to 1. To prove this, let e > 0. For this e, set N £ N such that N < l/e. Then ifn > N, we have

ra+1

1

-1 ra + 1

1 n+1

1 1 < n- < — N <e

and we /jave our proof. S

■

EXAMPLE 6.6 /n 5 vw'tfi ?/ze trivial metric, d, every sequence is a convergent sequence. The proof of this will be left as a homework exercise. ■ In keeping with our example of the sup metric we have the following example:

S

EXAMPLE 6.7 The sequence fn(x) = cos(ir/n) converges in (C[0,1], p) to the function f(x) = I. M

Before we go too far, let us take a step back to work on sequences of functions and seeing from where their limits come. Let {fn} be a sequence of real-valued functions all with the same domain D. Being real-valued means for any x £ D, the image f(x) £ K. Thus if we fix a value x G D, the sequence {fn{x)} is a sequence of real numbers and, if lira f(xn) exists, we can call that the image of x under the the limit function f. Rephrasing that we have the next definition. fl

EXAMPLE 6.8 Let {fn} be a sequence of real-valued functions defined on the set D. We say {fn} converges pointwise to f iffor each x G D, the sequence {fn(%)} con­ verges in K to f{x). Notationally, we write this as fn —» / . ■

S

EXAMPLE 6.9 The sequence gn[x) = xn with domain [0,1] converges pointwise to

1,

X = 1.

SEQUENCES IN ( 5 , D)

99

Let us go as far as to look at the proof of this pointwise convergence. We wish to show that if we fix XQ, then for every e > 0 there is a N £ N such that if n > N, then \gn(x0) - g(x0)\ < e. Proof of Example 6.9: Fix XQ e [0,1] and let e > 0 be given. If x0 = 1, then gn(xo) = 1 = g(xo) and so \gn(xo) ~ 9{xo)\ < e for all n and the conclusion is trivially true. Assume XQ ^ 1. For this XQ and e there exists an N e N such that ln(xo) Then for n > N, \§n{xo)

~ g(,X0)\

= XQ < XQ < 6.

Thus gn converges pointwise to g. 4k Two things worth pointing out here. First, the definition ofN in the proof involves both e and XQ. This dependence on XQ is a clue that the convergence is pointwise. Secondly is how important the space is for Example 6.9. If the space is the set of all functions with bounded range, we will write this space BD[0,1], and we are fine, g fits that bill. However, g is not an element o/C[0,1], since g is not a continuous function despite all the gn being continuous. If we have put the sup distance on the set BD[0,1], gn still does not converge to g G BD[0,1] as that supremum of \g{%) - 9n{x)\ = 1 for all n. In order to create the fractal-type objects that are our goal, we will need to take sequences and transform them via a set offunctions. The following theorem shows that if we know the function is continuous, then the property of being a convergent sequence is preserved. Theorem 6.3 The continuous image of a convergent sequence is convergent. Proof: Suppose that f : (S, d) —► (T, p) is a continuous function and that {xn} is a sequence in S which converges to x 6 S. We claim {f(xn)} converges to the point f{x). Pick an e > 0 and consider V = B(f(x),e). Since V is an open set in T, f~l{V) is an open set in S that contains x. Because {xn} converges to x, there is an N e N such that xn £ / _ 1 ( ^ ) ifn > N. This, in turn, means f(xn) € V for n > N, which proves {f(xn)} is a convergent sequence. 4k The difficulty with convergence (or lack thereof) is that to prove convergence one needs to know the limit ahead of time. Under the right circumstances we can actually get around having to know the limit. First, we have to have a particular type of sequence and a certain type of space. Definition 6.9 Let (S, d) be a metric space. A sequence {xn} in S is called a Cauchy sequence iffor every e > 0, there exists an N £ N such that n,m> N implies &\Xnj

Xm)


100

fl

SELF-SIMILAR OBJECTS

EXAMPLE 6.10 Let xn = 1/n. Then {xn} is a Cauchy sequence in R with the Euclidean met­ ric, m

Proof: Let e > 0 be given. For this e, let N £ N such that N > 1/e. Then for n, m > N n—m 1 1 1 1 < —r-r T < — < £• n m nm mm{n, m\ N Since e was arbitrary, this holds for all e and thus Cauchy sequence. 4t fl

EXAMPLE6.il A sequence x = {xn} of real numbers is bounded if there exists a finite real number M such that \xn\ < M for all n. Let B denote the collection of all bounded sequences. On this set we put the distance d between sequences as

[2 *

ifx^y

where K represents the index of the first entry where x and y disagree. So if x = (1,1,1,2,1,1,...), y= (1,1,1,1,2,1,1,...) a « ^ = (1,1/2,1/3,...), then d*(x7 y) = 2~ 4 , d*(x, z) = 2^ 2 . If nn is the sequence which is zero in all entries except the nth where that entry is 1, then {K„} is a Cauchy sequence in (B,d*). * As we can see by these examples, the most obvious examples of Cauchy sequences all look like they should converge. The following theorem shows that this is trueconvergent sequences will be Cauchy sequences. This is not an "if and only if" statement though. Theorem 6.4 If {xn} is a convergent sequence in (S, d), then it is a Cauchy se­ quence. Proof: Suppose {xn} is a convergent sequence. Thus there exists a point x € S such that for any fixed e > 0 we can find a N g N such that if n > N, we have d(xn, x) < e/2. Then ifn, m > N, we have, via the Triangle Inequality, d{xn,xm)

< d(xn,x)

+ d(x,xm)

< e/2 + e/2 = e.

Since e is arbitrary this proves that the sequence is Cauchy. 4b As we said, the converse is not true. Below are examples ofnonconvergent Cauchy sequences. Thus having the property that all Cauchy sequences converge is a very special property for a metric space. So it gets a particular name. Definition 6.10 A metric space (S, d) is called complete if every Cauchy sequence {xn} converges to a point in S.

SEQUENCES IN (S, D)

S

101

EXAMPLE 6.12 Look at the sequence {1/n}. In the space S\ = K with the absolute value metric, this sequence converges to 0. In the set 5*2 = (0,1) with the same metric, the sequence does not converge. ■

S

EXAMPLE 6.13 In the metric space (Q, \-\) the sequence x\ = 2, xn+\ = ^f- + ■£- /or n > 2 is a nonconvergent Cauchy sequence. Thus Q is «of a complete metric space. ■

77je rea/ /ine WJY/J Euclidean metric is a complete metric space. In fact it is "the completion of the rational numbers." This phrase comes from one way of developing M. from Q. The reals can be shown to be the set of all possible limits of Cauchy sequences made up of points in Q. The space C[0,1] with the sup metric is also complete. Think about how this changes things: If we have a sequence of functions and we can prove they are Cauchy in this space, then we can conclude the sequence converges to a function f even if we don not have an explicit formula or graph for the limit function on hand. In order to start our discussion of self-similar objects and fractals, we need our sets to be compact. Without compactness we would not be able to define a metric that gives us the necessary convergence of our sequence of sets. In order to discuss compact sets we need the following: We are taking aspects of the real line (and R n ) and re-writing them in metric space terms. For these next definitions, let (5, d) be a metric space and A a subset ofS. However, the examples are in M. with the Euclidean metric. Definition 6.11 A point a is called a limit point of A if there is a sequence {xn} of points in A \ {a} such that xn —> a. fl

EXAMPLE 6.14 In (M, | ■ |) the point a = 0 is a limit point of the sets A\ = [0,1], A^ = (0,1), and A3 = {1/n}. The point 0 is the only limit point for A3. There are lots of other limit points for A\ and A2. Any point a with 0 < a < 1 has a sequence in A\ (or A2), not using a, that converges to a. ■

Limit points do not have to be members of the set. If we adjoin the limit points to the set, we get what we refer to as the closure. Definition 6.12 The closure of the set A, written as A, is the collection of the points of A and the limit points of A; that is, A = A U {x : x is a limit point of A}.

102

5

SELF-SIMILAR OBJECTS

EXAMPLE 6.15 In (R, | • |) the three sets above have A\ = [0,1], A2 = [0,1], and A3 =

{l/n}U{0}.

■

We have already defined closed sets, saying that closed is the complement of open. We can also define that in terms of limit points. A set A is closed if A contains all of its limit points. Definition 6.13 The set A is closed if A = A and is perfect if A is equal to the set of limit points (that is, every point in the set is a limit point of the set). fl

EXAMPLE 6.16 In (R, | ■ |) for the three sets above, A\ is perfect (hence closed) and neither A2 nor A3 is closed or perfect. The set A3 is closed, but still not perfect, no nonzero point is a limit point. There is no sequence in, for example, A3 \ {1/2} which converges to {1/2}. ■

6

EXAMPLE 6.17 The Cantor Set is a perfect set.

■

This is actually the historical origin of the Cantor Set. The set was a solution to the problem of finding a non-empty, perfect, nowhere dense set. Cantor was actually not the first person to solve this problem. Henry J. S. Smith, an English mathemati­ cian, first constructed such a set in 1875. However, his work was not well-known. Vito Volterra published his version of this work in 1881, but was still a graduate stu­ dent at the time, publishing in a somewhat obscure Italian journal. Finally Cantor's set was published in 1883. So sometimes this set is referred to as the Smith--VolterraCantor Set. Now we move on to sequences and compact sets. A beforehand, a sequence in a set S is a function with domain N and co-domain S. We will write a sequence as an ordered, countably infinite list {xn}'^L1 or just {x„}. Definition 6.14 Given a sequence {xn} in a space (S, d), a subsequence is an in­ finite sequence of elements taken from {xn}, in order. For the sequence {xn} the notation for a subsequence is {xnk} and the understanding that the elements are taken in order translates to the fact that n^ > k. S

EXAMPLE 6.18 The sequence {1/2,1/4,1/6,...} is a subsequence of{l/n}.

■

It is worth pointing out that in R where there is a notion of order, there is the notion of monotone subsequences (see Chapter 2 for the definition of monotone). The

SEQUENCES IN (S, D)

103

proof of the famous Bolzano-Weierstrass Theorem, which states that every bounded sequence in M. has a convergent subsequence, is based on the fact that every bounded sequence must have a monotone subsequence; this is where every sequence has a convergent subsequence. 8

EXAMPLE 6.19 The sequence {1, - 1 / 2 , 1 / 3 , - 1 / 4 , . . . , ( - l ) n + 1 / n , . . . } has {1,1/3,1/5,...} as a decreasing, hence monotone, subsequence. Yes, there is an "obvious" in­ creasing one, too. m

Definition 6.15 The set A is compact if for every sequence {xn} in A, there is a subsequence {xnk} which converges to a point in A. The notion of compactness is a topological concept. In a general topological space, the topology defines what makes an open set. We say that the collection {Ua} is an open cover of a set X if each Ua is an open set and X C \JUa. A finite subcover ofX is a finite collection {C/,}™=1 from the set ofUa where n

Xc{JUi. Now the topological definition of a compact set is that X is compact if and only if every open cover of X has a finite subcover. Since we are only dealing with metric spaces, we are using the equivalent idea of sequential compactness. Theorem 6.5 In a complete metric space (S, d) a set A is compact if and only if it is both closed and totally bounded. Proof: Suppose that A is a closed and totally bounded set in S. Let {a^} be a sequence of points in A. Since A is totally bounded, we can find a finite collection of closed balls of radius 1 such that A is contained in the union of those balls. The Pigeon-Hole Principle l implies that since there are more points in the sequence than balls in the collection, one of the closed balls must contain infinitely many points out of the sequence. Say that is the ball £?i and select a point XNX in B\. Since B\ n A is a subset of A, it too is totally bounded; thus we can cover it with a finite collection of closed balls of radius 1/2. Repeating our Pigeon-Hole argument, there exists one of these balls (call it Bi) containing infinitely many {xi} where i > N\. Choose a out £JV 2 of B2- Continuing in this fashion we construct a nested sequence B1 D B2 D B 3 D ■ ■ • , 'The Pigeon-Hole Principle states, in essence, that if you have N holes in which the pigeons may roost and more than N pigeons, there is at least one pigeon-hole that has two or more pigeons in it.

104

SELF-SIMILAR OBJECTS

where each Bn has radius 1/n. It is easy to see that the subsequence {XN .} is a Cauchy sequence in (S, d). Since A is closed, the sequence converges to a point in A. In fact, it converges to the point

X= H

B

n-

So we have that A is compact. In the other direction, assume A is compact. Let e > 0 be given. Suppose we cannot cover A using finitely many open balls of radius e. Then there is a sequence of points {xn} such that d(xn,xm) > e for i ^ j . Since A is compact, there must be a convergent subsequence {xn,}. But every convergent sequence is Cauchy, hence there is a pair of integers N\ and N2 such that d(xN1,XN2) < e, a contradiction.

Thus A is totally bounded.

4

One nice property of compactness is that under the right type of transformation, this property is preserved. Here, the right type of transformation is a continuous one. Theorem 6.6 The continuous image of a compact set is compact. Proof: Let f : (S, d) —> (T, p) be a continuous function and suppose A is a compact set in S. To show f(A) is compact, let {£/„} be an open cover of f(A). Then is an open cover of A. Since we know A is compact, there is a finite {fl{Un)} subcover. Without loss of generality, call it

{f-1(Ul)J-1(U2),f-1(Us),...J-1(Un)}. Our claim is {U\. U2, ■ ■ ■, Un} is a finite subcover for f(A). This is because y G f (A) means there exists x G f^1(A) with f(x) = y and x 6 f~1(Uj) for some j . Hence y £ U3 so f(A) C U]=1Uj. 4 Finally we come to what Barnsley in [5] calls "The Space Where Fractals Live." We will start with the space S where S is a non-empty closed subset o/R™. By H(S) we mean the collection of non-empty, compact subsets of S. Now that we have the space, we need to define the distance. We will adopt the ideas from [30]. Let Abe a set in our collection H(S). We define a (5-body of A as the set of points at most 8 far away from A; that is, As = {x e K" : \x - a\ < 6 for some a € A}. Given two sets A and B in rl{S), it is fairly obvious that for a large enough 5, B C A§. Our distance between two non-empty compact subsets will be the least distance so that we do get containment in both S-bodies.

SEQUENCES IN ( 5 , D)

105

Definition 6.16 Let H(S) denote the collection of non-empty, compact subsets of S C K n , where S is a non-empty closed set. We define the Hausdorff metric on H(S) by d(A, B) = inf{<5 : A C Bs and B C As}. That d(A, B) > 0 is obvious and, since there are two containments to take into consideration, d(A, B) = 0 if and only if A = B. The symmetry is obvious. We will leave proof of the Triangle Inequality as an exercise. Another way to frame this idea of distance is the following: Given a point b and a compact set A in the metric space (S, d) define D(b, A) as the minimum of all the dis­ tances between b and a where a £ A. Notationally, that is D(b, A) = min{D(&, a) : a G A}. Then if we have two compact sets A and B we say the distance from B to A is the largest of all the D(b, A); i.e. D(B,A)

=

max{iam{D(b,a)}}.

This idea, however, does not define a metric. It is another homework problem to show that it may not be true that D(B, A) = D(A, B). That is why we emphasize the words from and to in the language. Tofinallyget to the Hausdorff metric we have d{A, B) = max{D(^, B), D(B, A)}. As a consequence of this definition and A and B being compact sets, we have that there are points a £ A and b G B such that d{A,B)

=d(a,b),

where on the left the metric is the Hausdorff metric and on the right the metric is the original metric on S. As with the canonical construction of the Cantor Set, when we create these selfsimilar objects we have a process which leads to the final result, but the object in question is the result of infinitely many applications of the routine. This means that we have a sequence of iterations and must know that this sequence converges in the correct sense. In other words, we need ('H(S), d) to be a complete metric space. Theorem 6.7 The metric space (T-l(S), d) is a complete metric space. So if{An} a Cauchy sequence in (7i(S),d) there exists a set A £ "H(5) such that

is

A = lim An, n—»oo

where the limit is taken under the Hausdorff metric. Proof: To prove this, we must start with a metric space (S, p) yet work in the space (H(S), d). Suppose we have a sequence An which is Cauchy in (7i(S),d). Let A = {x : there is a sequence {XJ } with Xj G A, and Xj —> x}.

106

SELF-SIMILAR OBJECTS

(Note: Xj —> x in the metric p, not d.) We will show that the limit of the An is in fact A. Let e > 0 be arbitrary and fixed. Since Am is Cauchy in (%(S), d) there exists an N £ N such that for n,m > N we have d(Am, An) < e/2. Let n be a value greater than N. If x £ A, then there is a sequence {XJ} such that Xj £ Aj and for large k p(xk,x) < e/2. Now because d(Am, An) < e/2 when k > N there is a y £ A.m such that p(y, Xk) < e/2. By the Triangle Inequality, this means p(x, y) < p(x, xk) + p(xk, y) < e/2 + e/2 = e. Thus A C (Am)e. Now take y £ Am. Since {Am} is a Cauchy sequence we can choose integers j \ < J2 < J3 < ■ ■ ■ with ji = m and d(Ajk,Am) < £ for all m > j k . Now we define a sequence yj with each y.j £ Aj. For j < n, pick any y} £ Aj and let yn = y. Forjk < j < j k + : , choose y0 £ Aj where p(yjk,yJ) < ~. Then {yk} is Cauchy in (S, p) and converges to some x £ A. Then p(y, x) = limp(y, ijj) < e. So y £ Af and since y was arbitrary, Am C A£. Thus d(A7 Am) < e and {Am} converges to A. 4 6.4

Affine Transformations

While many of our notions will work in any metric space (S, d), in order to have illustrations of what is going on, we will mostly restrict ourselves to S — R or M2 with their respective Euclidean metrics along with TL(S) with the Hausdorff metric where S is a compact subset of X; however the proofs will tend to be for general metric spaces. We wish to apply a certain type of function to the compact subsets of S. This function is capable of doing three things (separately or in tandem) to a closed and bounded set. It can (I) shrink the object, (2) rotate the object, (3) translate the object. Such a function is called an affine transformation and is defined in full generality below. Definition 6.17 A function T : X —>■ Y is called an affine transformation iffor any s,t £ X and any c & [0,1] it is true that T(cx + (1 - c)y) = cT(x) + (1 -

c)T(y).

Theorem 6.8 An affine transformation of an interval [a, b] into M. must be of the form T(x) = mx + b, where m, b £ R. Proof: Suppose a < b (it is possible that a = b, but then the interval is trivial as is the result). For any x £ [a, b] we can rewrite it as x =

b— x x —a a+ b. b— a b— a

AFFINE TRANSFORMATIONS

107

Since x is between b and a, this is of the form ca + (1 — c)b where c G [0,1]. Using the fact that T is affine, we get h X x a T( T{x)A= ~ Tt T(a)\^+~ T(h\ T(b) = T(b)-T(a) x -\ b— a b—a b- a This shows that T is indeed linear. 4t

fl

bT(a)~aT(b) . b- a

EXAMPLE 6.20 The function T(x) = \x + 2 is an affine transformation on the set A = [0,1]. If we let T(A) — {T(x) : x e A}, we see T(A) = [2, 2.5], which is a shrunken version of the original interval that has been then shifted to the right 2 units. ■

Theorem 6.9 Every affine transformation from M.2 into M.2 has the form T(x1,x2)

= (ax i + bx2 +e,cxi,dx2

+ /),

where a, b, c, d, e, / are real numbers. For those who know matrix multiplication, this comes from the formula

Although transformations move points around in the space, there are points that do not move under T. These points are very important in the study of dynamical systems and will help us in creating fractals. Definition 6.18 Let T : S —>■ S be a transformation on a metric space. A point XQ E X is called a fixed point (or attractor or invariant) ifT(xo) = xo. fl

EXAMPLE 6.21 /. IfT\(x) = 3x + 1, then the fixed point is the solution to T(x) = 3x + 1 = x which is x — —1/2. 2. IfT2(xi,X2) = ( | x i —X2 + I, |^2)» then the fixed point is (3/2,0) (check this to see that it works). u

In order to create the self-similar objects we are heading for, we need a particular type of transformation, called a contraction, which ensures us that as we apply the transformation points are getting closer together. Definition 6.19 Let (S, d) be a metric space. A transformation T : S —)• S is called a contraction mapping if there is a constant r G [0,1) such that d(f(x)J(y))
108

fl

SELF-SIMILAR OBJECTS

EXAMPLE 6.22 The transformation T on (M, J • j) given by T{x) = x / 3 + 1 is a contraction. We see that \T(x) - T(y)\ = \\x - y\. ■

S

EXAMPLE 6.23 0 « M2 r/ie transformation T(x, y) = ( x / 4 + 2 / 3 , y/2) shrinks things by 1/4 in the x-direction and 1/2 in the y-direction, and then it translates the shrunken im­ age by 2/3 of a unit to the right. This is a contraction with d(T(x\, yi),T(x2,2/2)) < ■ ±d{{x-l,y1),(x2,y2))-

Theorem 6.10 IfT is a contraction on (S, d), then T is a continuous

function.

Proof: Given e > 0, we must show the corresponding S > 0 such that d(x, y) < 6 implies d(f(x), f{y)) < e. So let us start with e > 0 and suppose T has contraction factor r > 0. If we let 6 = ^rh-, then for d(x. y) < S d(T(x),T(y))

< rd(x.y)

< rS = r - ^ - < e r + 1 and we are done. (This is really a proof that T is uniformly continuous, but we shall not go into that distinction here.) 4 Contractions give a special quality to their invariants. While we cannot guarantee the existence of a fixed point (it depends on the space), if there is an invariant, we know there cannot be more than one. Theorem 6.11 Let T be a contraction on (S, d) with contraction factor r. Then the fixed point for T (if one exists) must be unique. Proof: Suppose not. Say T has two distinct fixed points x and y. Then we have the following: d(x,y) = d(T{x),T(y)) 0 and we arrive at the contradiction 1 < r. Thus there can be at most one fixed point. A When can we guarantee a invariant? Exactly when we can guarantee that se­ quences in (S, d) that are Cauchy are going to be convergent. That means the space must be complete and gives us the Contraction Mapping Theorem. A bit of notation for this: Given a function T and a positive integer n we denote the composition ofT n times as

rW(s) = (ToTo..-or)(s). v

v

n copies ofT

'

Theorem 6.12 The Contraction Mapping Theorem Let (S, d) be a complete met­ ric space and let T : S —> S be a contraction with factor r. Then for any point s £ S, the sequence {T^ (,s)} converges to a point s and s is the invariant for T.

AFFINE TRANSFORMATIONS

109

Proof: Let s be a point in S. Then the triangle inequality gives us d(s, T^is))

< d{s, T(s)) + d(T{s), r ( J ' ( s ) ) < d(s, T(s)) + rd(s, T(s)).

In fact, for any positive integer n d{s,T^{s))

<

d(s,T(s))+d(T(s),r( 2 )(s)) + --- + d(T("- 1 )(s),T(")(s))

<

d(s, T(s)) + rd{s, T(s)) + ■■■ + rn~ld{s, 2

=

(1 + r + r + --- +

r ~ )d(s,T(s))

=

l=rld(s,T(s))

jh-A^ns))-

<

T(s))

n 1

In a similar vein, for positive integers m and n d{T^\s),T^m\s))

<

r^{m,n}d(s,T{s))

This shows that {T<~n\s)} is a Cauchy sequence. Since (S, d) is complete, it has to converge to some point which we will call s. Due to the fact that T is a continuous function, we can now see T(s) = T( lim T^n\s)) n—►oo

= lim T^n+1\s)

= s.

n—>oo

Thus s is the invariant. A Now we are ready to move on to the space (H(S), p) and apply our ideas. Theorem 6.13 Let T : S —> S be a contraction on the metric space (S, d) with contraction factor s. Then T is also a contraction on % with metric p (where T(A) — {T{a) : a £ A}) with contraction factor s. Proof: From Theorem 6.10 we have that T is continuous on S and Theorem 6.6 tells us the image of a compact set is compact, hence T maps % onto itself. The metric p looks for the largest distance between any two points in the compact sets U and V. Since the sets are compact, these are actual members of the set; that is, there exists u G U and v E V such that if 8 = d(u, v) we have U C VsandV C Us. Take A, B € H(S). Then there exist a e A and b € B so that d(T(A),T(B)) = d(T(a),T(b)) < sd(a,b) = sd(A,B). This means via the definition of the Hausdorff metric T(A) C T(B)S* where 5* = s ■ d{A, B). Similarly, T(B) C T(A)S. where S* = s ■ d(B, A). Thus p(T(A),T(B))<s-p(A,B), which is what we needed to prove. ♦

110

SELF-SIMILAR OBJECTS

By an Iterated Function System (IFS), we mean a finite collection {Ti}™=1 of contractions, each with corresponding contraction ratio Ti. With this collection we create the transformation

r(A) = T1(A)uT2(A)u-urn(4 where A is a non-empty, compact set. This is itself a contraction with contraction ra­ tio r = maxjri, r<2, ■ ■ ■, rn}. The invariant for T is guaranteed to exist by Theorem 6.13 and will be self-similar.

Let us give some examples of these systems and their invariants. 9

EXAMPLE 6.24 On the interval [0,1], ifJ\(x) — x/2> and T2{x) = x/3 + 2/3, then the IFS {Tj, T2} has the Cantor Set as its invariant. Note: We use [0,1] here and not the whole line because in order to get uniqueness of the invariant, we really need to work in the Hausdorff Space of non-empty, compact sets with the Hausdorff metric. If we allowed noncompact sets, then R = 7\(R) U T2(M) and we lose uniqueness of the fixed point. ■

9

EXAMPLE 6.25 In the unit square, let our three contractions be

Tx(x,y) = (x/2,y/2), T2{x,y) = {x/2 + l/2,y/2), T3(x,y) = (x/2+l/2,y/2 + l/2).

So a compact set (such as a triangle inside the unit square) is shrunk by a factor of 1/2 in both directions, one time shrunk and then translated a 1/2-unit to the right, and another time shrunk and then translated a 1/2-unit to the right and a 1/2-unit up. The resulting invariant is called the Sierpinksi Triangle.

AFFINE TRANSFORMATIONS

H

*

*

■

»

*

•

■

*

•

»

■

■

*

'

,m-

*

"

N »

»

* " " "* -

C\. tx Cx ; .

tx N

\

'

X

■

*

H

»

l

*

i

l

"

111

*

IF

i

Several Iterations of T = T\ U T2 U T3 /f is wort/i noting here that the end result being a triangle has nothing to do with the initial image being a triangle. The idea of invariant is that the end result will be the Sierpinski Triangle regardless of what shape we begin with. The only difference would be the speed with which the iterates converge.

S

EXAMPLE 6.26 Again we start in the unit square. Our starting image will be the unit inter­ val which we will turn into a curve. The Koch Curve (named for the Swedish mathematician Helge von Koch) first appeared in a 1904 paper entitled "On a Continuous Curve without Tangents, Constructed from Elementary Geometry." This example, unlike the others, uses rotation of the object. There are four transformations here. The first is shrinking by 1/3 in both directions. The second also shrinks by 1/3 in the x- and y-directions, but then translates over by 2/3 horizontally. The next shrinks by 1/3, then rotates by 60°, then translates 1/3 to the right. The final transformation is shrink by 1/3, rotate —60°, and then translate 2/3 to the right and up by ^ . For those who know matrix multiplication, a rotation by the angle 9 is given by

112

SELF-SIMILAR OBJECTS

Our four transformations are then

1/3

Ti{x,y)

'l/3

T'2(x,y)

0\

x\

/2/3N

0 \ fx\

, 0 1/3 i I J =

J

/(l/3)cos(60°) ^(l/3)sin(60°)

I 0, (-l/3)sin(60°)\M (l/3)cos(60°)J ^

/l/3\ ^ 0 ) '

(l/3)sin(60°)\ AA (l/3)cos(60°W \y

/ 1/2 N 1^3/6,

'Ll

/ (l/3)cos(60°) l-(l/3)sin(60°)

T^y)

=

T2(x,y)

=

(a:/3 + 2/3,y/3),

T 3 (x,y) T4(xV£/)

= =

( i / 6 - N / 3 ! / / 2 + l/3 ) a;v/3/6 + j//6), (x/6 + >/3!//2 + 1 / 2 , x V 3 / 6 - y / Q + >/3/6).

r

=

or

(x/3,y/3),

One interesting fact about this curve is that for at each step, k, the length of the curve is (4/3) k . Thus the final product has infinite length; but, since it fits inside the unit square, it covers finite area.

AFFINE TRANSFORMATIONS

113

„.r^.f\.

ft*

i]£

D.S

9.7

19

ftt

The first two iterations and a representation of the final curve. u When we defined the Cantor Ternary Set, we discussed one way to tweak it. This brought us to a generalization we wrote as Cr, 0 < r < 1. These sets are also the attractorfor an IFS, using the contractions T\{x) = rx andT2(x) — rx + (1 — r). This brings up the question, "Is every Cantor Set the attractor for an IFS?" The answer is, "No," but requires a different idea of metric. This comes from [15] where a Borel Probability Measure is used and the following theorem applied. Theorem 6.14 There exists a Cantor Set X% and a Borel probability measure p supported on X\ such that for every contractive map f : X\ —> X\, we have M/(A-i)) = 0. The set X\ has the property that for any iterated function system {fi, f2, ■ ■ ■, fn} the set Ufi (X\) has measure zero while X\ has full measure. Then X\ cannot be the attractor of an IFS. More ways to make Cantor Sets are covered in Chapter 10.

114

6.5

SELF-SIMILAR OBJECTS

An Application for an IFS

As an application of iterated function systems, let us consider the following. A theo­ rem from Calculus I states that if a function is differentiable at XQ, then it is contin­ uous at XQ, too. After the proof (an application of the limit definition of derivative), the next thing to show is an example to show the converse is not true. Continuity does not imply differentiability and the function f(x) = |a;| at XQ = 0 illustrates this. The question we wish to answer here is, "How much can we disconnect continuity and differentiability? " Since we can do this for one point, can it be done for two? Five? Countably infinitely many points? The answer is that this can be done for the entire line. There are functions which are continuous on all of R, yet differentiable at no point. The first example of a function with this property is credited to Weierstrass. His result, published in 1872, is defined by oo

/(X) = J > " C 0 S ( 6 " 7 T X ) , n=0

where 0 < a < 1, b is a positive integer, and ab > 1 + =y. Pictures of the graph are available nowadays thanks to computers. It is difficult, though, to see how the graph develops the properties we need. We will produce an example due to Katsuura [42] in which we can see the function developing because of the self-similar nature of the graph and the iterative process from which it is derived. We will work in the space of the closed unit square ([0, 1] x [0,1]). Our three contraction mappings are T^y)

=

T2(x,y)

=

T3{x,y)

=

(x/3,2y/3), (2-x/3,l+y/3), (2 + x/3,l

+

2y/3).

The first transformation takes an object and shrinks it by 1/3 in the x-direction and 2/3 in the y-direction. The second shrinks the image by 1/2 in each direction, trans­ lates it 2/3 to the right and up 1/2, then rotates it about the line y = 1/2. The last shrinks the image with the same ratios as the first, but then translates the image 2/3 to the right and 2/3 up. If we use the graph of y = x as the initial image (fo) then each successive application of T(A)=T1(A)UT2(A)UT3(A) will also yield the graph of a function. first three images are below.

We name them fn(x)

= Tn(fo(x)),

and the

AN APPLICATION FOR AN IFS

115

The initial graph and first two iterations Due to the least contraction being by 2/3, we see that for any k and any value x e [0,1], \fk{x) - /fe+i(x)| < (2/3)fe. This means for n < m sup{|/ m (x) - fn(x)\

:xe[0,1]}

< (2/3)11,

which shows that {fn} is a Cauchy sequence in (C[0, l],p). This is a complete metric space, hence the sequence must converge to some continuous function f. We still need to show that this function f does not have a derivative at any point. Rather than repeat all the technical details, we will say that the interested reader can find them in Katsuura's paper and we present the heuristic proof. The interval [0,1] can be split into two types of points: The first are points like x = 1/3, or x — 4/9; rational numbers in base 3. These points are where the fn have cusps (sharp corners). Notice that this never changes. Once a cusp appears at x, fn(x) will continue to have a cusp there. Thus f is not differentiable at that point. The other type of point, where there is not a cusp, are points like x = 1/2. At these locations the graph alternates direction. For / Q ( 1 / 2 ) the slope is positive 1. For / i ( l / 2 ) the slope is — 1. This alternating keeps going forever, meaning that there cannot be a limit to the sequence of slope values, which is the same as saying /'(1/2) does not exist. Thus f is a continuous function that is nowhere differentiable.

116

SELF-SIMILAR OBJECTS

EXERCISES 6.1 For the sequences xn = ( — 1)™ and yn — \jn give an example of monotone subsequences. 6.2 Prove that the function d*(x,y) = \x\ — x%\ + \y\ — 3/21 is a metric on the plane ]R2. 6.3

Show that in the trivial metric, d, every sequence is a convergent sequence.

6.4 A unit circle centered at the origin is the set of all points whose distance from (0,0) is equal to I. What this circle looks like depend on the metric. Draw two unit circles with center (0, 0) in the plane: one using the usual Euclidean metric and the other using d* from the exercise above. 6.5

Let S = C[0,1]. Given two functions f and g, define d(f,g)=

I Jo

\f(x)-g(x)\dx.

Show that d is a metric on S. Find an example to show this is not a metric if S is the set of Riemann integrable functions on [0,1]. 6.6 Show f(x) = rax + b is a similarity on the real line. If \m\ < 1, find the location of the fixed point. 6.7 Show that if'T is not a contraction, then the set of fixed points can have more than one element. 6.8 Although we could work with any non-empty, compact interval, we will stick with [0,1]. a) Let fi(x) = x/2 and f2{x) = 3a;/4 + 1/4. Show that both of these are similarities and

[o,i] = / 1 ([o,i])u/ 2 ([o,i]). b) Now come up with two different similarities,
Prove the Triangle Inequality for the Hausdorff metric. Show that for a finite set A, HS(A) = 0.

6.11 The Hausdorff metric needs the sets in the space to be compact and non­ empty. a) Find an example of two sets A and B so that if we allowed non-closed sets, we could have D(A, B) = 0 with A / B. (Hint: Keep things simple, stay in R.) b) Explain why d({0}, 0) = 00.

EXERCISES

117

6.12 Let p be a fixed prime number. Consider the following on the set of integers, Z. For i £ Z the value oford(x) is the largest natural number k such thatpk divides x. Define a function p : Z x Z —s- [0, oo) by

S P(X,V)

s _ J0' ~ \p-or^-y),

x = y, x^y.

Show that p is a metric on Z. This is an example of a metric that takes a countably infinite number of values. 6.13 Let A, B s M. A function f : A —> B is called uniformly continuous if for every e > 0 ?/?ere existe a (5 > 0 .SMC/Z that for all x,y £ A // |x — y| < 5, f/zen | / ( x ) — / ( y ) | < £■ 77HS, of course, does generalize to metric spaces A and B. a) Prove that f(x) — y/x is uniformly continuous when the domain is [1, oo). b) Prove that g{x) = x2 is not uniformly continuous on (—oo, oo). c) Prove that any linear function is uniformly continuous.

CHAPTER 7

VARIOUS NOTIONS OF DIMENSION

In Chapter 3 we looked at indicators of size (measure, category, cardinality). How­ ever, these did not show the differences in Cantor Sets of various ratios r. That is, for any fixed r, Cr is measure zero, nowhere dense, and has cardinality 2N°. So how is C1/3 different from C1/4 or C2/5? For that we turn to the notion of dimension.

7.1

Limit Supremum and Limit Infimum

Before we go into dimension, we must extend what we know about limits. Sometimes, such as for the sequence {(—1)"}, there is not a limit; however, we can look for a limit of a part of the sequence. What we mean by that is explained below. Definition 7.1 Given a set A of real numbers, the number u is an upper bound of A ifa< ufor all a £ A. Similarly, the number v is a lower bound of A if v < a for all a e A. S

EXAMPLE 7.1

The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

119

120

VARIOUS NOTIONS OF DIMENSION

LetA

= { 1 , 2 , 3 , 4 , 5 } , B = [0,1), and C = {x : x eQandx2

< 2}. Then

1. The upper bounds for A include 5, 7, and \/30, while some lower bounds are 0, —3, and —n. 2. For the set B, 0 is a lower bound and a member of the set and 1 is an upper bound, but not in B. 3. C has upper and lower bounds, but no such bound is a member of C.

u As we can see, upper/lower bounds are not unique nor do they have to exist. If a set has a lower bound, the set is called bounded below; and if it has an upper bound, the set is bounded above. Sets that are both bounded above and bounded below are just referred to as bounded. Sets without an upper bound or lower bound (or both) are unbounded. Definition 7.2 Let A c R. The number U is the supremum (also known as least upper bound) of A ifU is an upper bound of A, and for any other upper bound x we have U <x. If no such number exists, then we say sup(yl) = oo. Similarly, V is an infimum (or greatest lower bound) of A ifV is a lower bound of A, and for any other lower bound x we have

x
fl

EXAMPLE 7.2 For the sets A, B, and C in Example 7.I, sup( A) = 5, inf (A) = 1, sup(i?) = 1, inf (B) = 0, sup(C) = \ / 2 , and C has no infimum. ■

The reader may be familiar with maximums and minimums. Suprema and infima are different in that the max/min of a set must be a member of the set while sup/inf need not. The set B above does not have a maximum, but does have a supremum. Now we wish to combine this idea of sup/inf with the well-known concept of limit. Here we will stick with limits of real numbers. In spite of having already generalized limits to metric spaces (S, d), sup and inf need the space to have a total ordering to it. What we mean by that is for any two elements x and y, there needs to be a way to determine if' x < y, x > y, or x = y. A space having a metric attached to it does not say anything about order existing. Even in such a usual space as IR2 we usually do not refer to order, there is no typical rule to determine how the points (1, -\/3) and (0, 2) compare.

LIMIT SUPREMUM AND LIMIT INFIMUM

121

Suppose we have a sequence of real numbers, {xn}^=1. Associated with this se­ quence is the set of values of the sequence. We can then look at the sup and infofthe set. To avoid extra notation we will not distinguish between the sequence and the set of values of the sequence and so we will write this as sup n > 1 {x n } and'm£n>i{xn}. For example, the sequence tn = 1/n has sup n > 1 {£ n } = 1 and mfn>i{tn} = 0. = 1 and mln>i{yn~} = - 1 . Note With the sequence yn = (-1)™, supn>1{yn} that in these examples tn has an inf which is not a member of the set of sequence values and yn is an infinite sequence with a finite set of values. Lastly, we point out that for tn having a different starting index affects the value of the supremum; that is sup n > 1 {£„} = 1, sup„> 2 {£„} = 1/2, sup n > 3 {t n } = 1/3, etc. This brings us to the notion of limit supremum and limit infimum. Definition 7.3 Let {xn} be a sequence of real numbers. We definite the limit supre­ mum and limit infimum of the sequence as follows: lim sup x n = lim (supx^) and lim inf xn = lim (inf £&).

fl

EXAMPLE 7.3 1. Let xn = (—1)™. Then l i m s u p , ^ ^ xn = 1, while lim.infn_j.oo xn = —1.

{

n,

n is even ,

sin(l/n),

n is odd .

Then lim sup y = oo while lim inf y = 0. n

n

3. Let rn be an ordering of the rational numbers in [0, 2]. Then lim s u p ^ ^ ^ rn — 2, while lim inf^^oo rn = 0.

■

As the next two theorems show, our intuition is correct. The first shows that a lim sup is at least as big as a lim inf, and the second that if a sequence converges, then its lim sup and lim inf will converge to the same value. Theorem 7.1 Let {xn} be a sequence of real numbers. Then limsupa; n > lim inf xn. n->oo

n—>oo

Proof: If either lim sup xn = oo or lim inf xn = — oo, there is nothing to prove, so let us assume that there are real numbers a and b such that lim inf xn = a and

122

VARIOUS NOTIONS OF DIMENSION

lim sup xn = b. Consider a < a and j3 > b. By the definition of supremum and infimum, there exists an N G N so that for n > N we have a < xn < j5. But a < xn for infinitely many n means a < lim sup xn = b. Since a is arbitrary, letting a approach a gives us a < b which is our desired result, jt. Theorem 7.2 Let {xn}

be a sequence of real numbers and let c £ t . Then

lim xn = c if and only if lim sup xn = lim inf xn = c. Proof: Let e > 0. If lim sup xn = c, then there exists an N\ G N such that n > N\ implies xn < c+e. Similarly, there is an N2 G N s o thatn > N2 implies c — e < xn. Putting these together gives us, for n > max{7Vj. A^2 }, —e < xn — c < e, which is the same as \xn - c\ < e. Thus linin-^oo xn = c. In the other direction, assume lim„_>.00 xn = c. Then given e > 0 there exists N G N such that n > N implies \xn

-c\<e.

Thus for n > N we obtain c — e < lim inf xn < lim sup xn < c + e. But e is arbitrary. Letting it approach zero gives lim inf xn = lim sup xn = c. 6 This notion of'lim sup and lim inf extends to functions, the same way the ordinary idea of limit does. We will assume that our function has domain M. although we could be more general and have domain E C U . For more information, see [11]. Definition 7.4 Let f : M. —> R. and suppose XQ is fixed. Then lim sup f{x)

= inf { s u p { / ( x ) : x G {XQ — 5, x0 + S),x ^

xo}}

lim inf / ( x ) = supjinf{/(re) : x G (XQ — S, XQ + 5), x ^

xQ}}.

and

Continuing in a parallel fashion to what we had with sequences, we start with an example and two theorems (without proof this time) showing the same intuitive results.

TOPOLOGICAL DIMENSION

S

123

EXAMPLE 7.4 Let f(x) -1.

= sin(l/x). Then limsupa._>o f(x)

=

1> while liminf^^o

f{x)

Theorem 7.3 Let f : R —>• K and /ef £0 € M be fixed. Then lim sup/(x) > liminf f{x). Theorem 7.4 Let f : R —> R, /ef XQ € K be fixed, and let c 6 R. 77ie« lim /(#) = c if and only if lim sup /(x) = liminf/(x) = c.

Wfe wi// 500« utilize these to come up with various, non-integer notions of dimen­ sion. Let us begin with an easier idea, topological dimension, whose possible values are —1, 0,1, 2,....

7.2 Topological Dimension As we have seen earlier, topology concerns itself with the study of open and closed sets. We repeat the definition here. Given a space S, a topology is T is a collection of sets which satisfy certain rules:

1. %Se T. 2. IfUa(a&

A, an index set) is in T for all a £ A, then

\JuaeT.

aeA

3. IfU,V eT.thenUHV

e T.

The sets in T are referred to as the open sets. If A is an open set, then its complement S \ A is closed. A topology is generated by something called a basis. These are the sets in T which are apart of every open set. More specifically, we have the definition: Definition 7.5 Let T be a collection of open sets and let B be a family of subsets from T. We say that B is a basis for the topology iffor every open set A G T and every x G A, there is a set U G B such that x G U C A.

124

fl

VARIOUS NOTIONS OF DIMENSION

EXAMPLE 7.5 For M. with the Euclidean metric, the family B = {(a,b) : a,b G K, a < b} forms a basis. This is not the only one. The collection B* = {(a, b) : a, b G , a < b} is also a basis. ■

It is this idea (unstated) that we used when describing the real line and Sorgenfrey line in Example 3.7. Topological dimension is so named because it is based on the topology of the space and does not need the notion of distance (metric). There are three types of topological dimension: small inductive dimension, large inductive dimension, and Lebesgue covering dimension. In a normal space with a countable base (and we are staying in K™ with the Euclidean metric, which meets these criteria) the three notions have the same value. We first define small inductive dimension . As the name implies, it is defined inductively. For these sets we use the idea of boundary. Recall for a set A, the boundary of A is the set of points in the closure of A, but not in the interior of A. We write that as DA = A\A°. Definition 7.6 Begin with 0. Its inductive dimension is set to be — 1 (the only set with negative dimension). For a set A, the inductive dimension of A is the smallest value n such that for every x G A and every open set U with x G U we can find an open set V where x G V and V CU and the inductive dimension ofdV is at most n — 1. We will write this as

dimtop(A).

In our K", this reduces to V being an n-dimensional ball centered at x and the boundary ofV will have dimension n — 1. Thus a countable set inductive dimension 0, a line has inductive dimension 1, a plane 2, and so on. For a Cantor Set, given any point x and open U containing x there is an open interval I inside ofU so that x G / and I C U. The boundary of the interval has dimension 0 (just being singletons), thus dim top (C) = 0. Another way to define this is with separating components. To begin with, we say a topological space S is disconnected if there exists two non-empty, disjoint, open sets U and V so that S = UUV.Ifa space is not disconnected, then it is connected. A disconnected space consists of pieces called components. By the definition each component is a closed set. This seems a little mysterious, so before we go on to some examples, let us talk about inheriting a topology. Let S be a topological space with topology T and ACS. Then A inherits its own topology by TA = {A n T : T G T } . So as a subset ofM. with the Euclidean

TOPOLOGICAL DIMENSION

125

topology, the interval [0,1] is a closed set. However, in the space Y = [0,1] U [2, 3] with the inherited topology, [0,1] being equal to [0,1] (~l (—1, 2) is an open set in Y(!!). This tells us how each component can be open. The complement of the compo­ nent must (by being contained in one of U or V) must be an open set. The idea of connected and disconnected can also be inherited. For the set A to be disconnected, look atl/nA and V n A and see how A = (UnA)U(Un V).. fl

EXAMPLE 7.6 1. The interval [0,1] is a connected set in K with the Euclidean metric. 2. The interval [0,1] is a disconnected set in M with the discrete topology. Since each singleton is an open set we can just partition the interval into the open sets {0} and (0,1]. We can do even more and divvy this into uncountably many components. 3. The set A = [0,1] U [2, 3] is a disconnected set in K with the Euclidean metric that consists of two components. 4. The Cantor set in K and Euclidean is a disconnected set also with uncountably many components. In fact, it is what we call totally disconnected because the components are just single points. u

So the topological dimension of a set A can also be described this way: The dimension of the empty set is — 1. The dimension of any other set is one more that the dimension of the object with the smallest inductive dimension that can be used to separate the set into components. The condition of minimality is important. A finite set can also be separated by a single point that is not in the set. Let us redo some examples with the new point of view: If we have a finite set all the points are separate components and it takes nothing to separate the set. Hence the inductive dimension is — 1 + 1 = 0. Similarly, this argument shows a countable set has topological dimension zero. If we have an interval [a, b] we need a single point to separate it. Thus the interval has dimension 0 4- 1 = 1. A line is divided using a single point, which has dimension 0, so the line has topological dimension 1. A rectangle in the plane needs a line to separate it, so the topological dimension of a rectangle is 2. The Cantor Set is totally disconnected. Hence it need no points to separate the set into components. This means dim t o p (C) = 0. Topological dimension does have some nice properties. As the proofs bring us a little too far into topology, we will omit them. They can be found in [26]. Theorem 7.5 Let AC. B. Then dim top (A) < dimtop(_B). Theorem 7.6 If An are closed and d\mtop(An) d\mtop(UnAn)

= Ofor all n € N, then = 0.

126

VARIOUS NOTIONS OF DIMENSION

Recall that a homeomorphism between two sets is a bijection f where both f and f~l are continuous functions. The topic of topology concerns itself with properties that are conserved when a homeomorphism is applied. Theorem 7.7 Topological dimension is preserved by homeomorphism. are homeomorphic, then dim t o p (yl) = dim t o p (i?).

If A and B

Thus all types of Cantor Sets Cr have the same topological dimension. This is not helpful as we are looking for ways to distinguish Crfrom Cs. Now that we have the concept of homeomorphism and see how objects can be topologically equivalent, let us close this section with one of the more fascinating properties of Cantor Sets, the equivalence of compact, perfect, totally disconnected sets. The next theorem shows that all of these are the same as the Cantor Set. Theorem 7.8 Let Abe a compact, perfect, totally disconnected set. Then A is home­ omorphic to the Cantor Set C. Proof: In this proof, we do not create an explicit homeomorphism, F, but instead describe the construction of such a function. The set A is compact, hence bounded above and below. Let m be the greatest lower bound of A while M is the least upper bound of the set. Our construction, F, will be a function with domain [ra, M] and range [0,1] with the additional property that F(A) = C. Since A is totally disconnected, it cannot contain an interval, hence it is a nowhere dense set. This means the set [0,1] \ A is the union of countably infinitely many disjoint open intervals with the additional property that for any two intervals their closures are disjoint. Why? The intervals must be countable since each one contains a rational number. There cannot be finitely many since A is nowhere dense. Because A is a perfect set it cannot have any isolated points, hence the closure property. This collection of intervals will by denoted by I. The collection X is a set of bounded intervals. Take an interval I\ of maximal length (do you see why at least one must exist?). We match this up with the Cantor interval [1/3, 2/3] as an increasing line segment. Now let In and I\i represent the longest intervals to the left and right of 1\, respectively. Map these to the Cantor intervals [1/9, 2/9] and [7/9, 8/9] in an increasing linear manner. This process can continue with { / m , /112, I121 > ^122} being paired with {[1/27, 2/27], [7/27,8/27], [19/27, 20/27], [25/27, 26/27]} via increasing lines, et cetera. So now we have a strictly monotone function F : [m, M] \ A —> [0,1] \C which is clearly one-to-one and onto. To finish this, we must define F at each a € A. Define the set La as {x £ [m, M] \ A : x < a and x is an endpoint of an interval in I}. These points are to the left of a at which F has already been defined. We can then determine the value for F(a) by F(a) = sup{F(x)

:x e

LA}.

SIMILARITY DIMENSION

127

Thus we have a monotone bijection between [TO, M] and [0,1] and F(A) = C. Now we must show the continuity of F and F~l. First, note that since F is monotone, so is F~l. Second, F_1 is a bijection since F is one. So we will show that F must be continuous, the proof for F^1 being the same. Suppose (a, b) is an open interval in [0,1]. Then F" 1 ((a, b))

= = =

{x e [m, M] : F(x) £ (a, b)} {x e [TO, M] : a < F{x) < b} {xe[m.M}:F-l(a)<x
=

(F-\a),F-\b)),

which is open. This means the inverse image of open sets are open, hence F is continuous. 6 The last part of the proof above is that our function F and its inverse F^1 are continuous. This is a specific example of a much more powerful result: Iff : A —> B is a monotone bijection, then both f and f~l are continuous functions.

7.3

Similarity Dimension

We mention similarity dimension as it is easy to calculate and the equation involved will reappear later, but we shall see that similarity dimension does have a problem with it. As we noted in Chapter 6, the Cantor Set is an example of a self-similar object. It is the limit ofthe iteratedfunction systemTi(x) = ^xandT2(x) = j ^ + f Given an iterated function system with similarity ratios {r\, T-I, ..., rn} the simi­ larity dimension for the invariant K is the number s such that r{ + rs2 + ■ ■ ■ + rsn = I. So for our Cantor Set we have

,+

G) G)'which leads us to dim S j m (C) — |^-|. The sets to which we can apply this idea are somewhat limited. For a set A to be the invariant for an iterated function system, there must be a set of similarities {/$} such that A = Utf(A). This is not possible for some reasonably simple geometric shapes. This cannot be done for a unit circle. A second issue has to do with overlap. The unit interval I — [0,1] is self-similar and the invariant for transformations U\(x) = \x and U2 = \x + \ leads us to

128

VARIOUS NOTIONS OF DIMENSION

d i m s i m ( J ) = 1, which seems correct. However, it is also true that I = where V\(x) = ^x and V ^ x ) = | x + | . Since

Vi(I)UV2(I)

does not have s = lfor a solution, which counts as the "real" similarity dimension ? {UUU2} With the IFS {VUV2} we can see that VX{I) n V2(I) = [|, §] while has Ui(I) (~l {72 (-0 — {5}- So perhaps what we need is a condition that says the following for the invariant F: Ti(F) Pi Tj{F) must be small in some way. This will be formalized later with the Open Set Condition (Section 7.5).

7.4

Box-Counting Dimension

This type of dimension is popular since it is relatively easy to calculate. It has gone by different names over the years including capacity dimension, logarithmic dimen­ sion, and Kolmogorov dimension. At each stage n in the construction of the Cantor Set there are 2™ closed subintervals making up Kn. We can cover Kn using that many closed subintervals of length (1/3)™. Or we could use 4™ closed subintervals of length (1/6)™, or many oth­ ers. Box counting dimension looks at the relationship between the number of boxes used to cover a set and the diameter of the boxes involved (notice, unlike topological dimension, this requires us to have a metric in place). Let us look at this idea in generality. Definition 7.7 Let A be a non-empty, compact subset ofM.n. Let S be a positive number and let N$(A) be the smallest number of sets of diameter 5 needed to cover A. We then define the upper and lower box-counting dimension of A by dimbox(A)

=limsup S^o+

-ln(d)

and dimbox(A)=limmf <5-+o+

- \n(S)

When these two values agree, we have the box-counting dimension of A and write it as dimbox(A)

fl

= hm d~>o+

HNS(A)) -ln(<J)

.

EXAMPLE 7.7 We will not prove this; but as we would expect, diraf,ox({a}) = 0. In the line, dira(,ox([a,b]) = 1 where a < b; and in W', if B is the open ball of radius r > 0, d\m.box{B) = n (in fact, if A is an open set in W1, then dimb oa; (A) = n ■ since A must contain an open ball).

BOX-COUNTING DIMENSION

fl

129

EXAMPLE 7.8 For our Cantor Set, using the fact that at each stage there are 2n boxes (closed intervals since we are in the line) of diameter (1/3)™, it would seem we are dealing with ln(2") ,. nln2 In 2 lim ——r^-^ = -— n = lim n->oo — ln((l/3) ) n^oo n In 3 m3 (and this is the case); however, there is more work to be done before we can switch from the limits with functions to limits of sequences. ■

It is easy for us to see that it makes no difference whether the boxes we use on covering the Cantor Set are open or closed. This actually works in the definition, too. Open or closed does not change the dimensions. Are there other concessions to make ? Yes, many. To keep our arguments simple, let us stay in M.2 although all of this holds in M.n. Given 6 > 0 we can make a grid of side length 6 by considering all the squares of the form [niS, (ni +1)6} x [ri2<5, (n^ + 1)6} where n\ andn-i are integers. These squares have diameter 6\/2 and sides parallel to the axes. Any square in the grid with diameter 6 has side length 6/\/2. This means that an arbitrary set with diameter 6 needs at most 9 elements the grid to contain it (consider three rows of three boxes with diameter 6/y/2). So if N 1(A) is the number of grid boxes needed to cover A, we have N$(A) < 9Ng(A). On the other hand, a square of diameter 6 gives 5NS(A) < N%(A). Together H5NS(A)) -ln((J)

"

ln(JV|(A)) -ln(<5) "

\n(9N6(A)) -ln(6) '

and using ln(xy) = \n(x) + ln(y) and \n(6) —> —oo as 6 —> 0 + we have dinibox(A) A\mhox(A) are the same using open sets or grids. There are other equivalences (which we shall not prove) summarized below [30]. Definition 7.8 The upper and lower box-counting dimension for a set A C M.n are given by -PHNs(A)) dimbox(A) = limsup and dJmtoai(>l) = h m i n f — j ^ - , where Ng(A) is any of the following: • the smallest number of sets of diameter 6 that cover A; ■ the smallest number of closed balls of diameter 5 that cover A; ■ the smallest number of grid cubes of side length 6 that cover A; ■ the largest number of disjoint balls of radius 6 with centers in A.

130

VARIOUS NOTIONS OF DIMENSION

Lastly, we note here that our 5 approaching zero can be done through a specific sequence rather than all possible ways to go to zero. Specifically, for some c £ (0,1) we can let 5n = cn and use "linin^oo ." This brings us back to our Cantor Set example and the fact that dinifcox(C) = |^|. Let us show that there is a need to have both upper and lower Box-Counting dimensions. Sometimes the limsup and liminf do not agree. To find this example, we go back to the Cantor Set construction. The same as before, at each stage we have 2k closed intervals. However, this time we do not always have the same ratio at each stage. fl

EXAMPLE 7.9 There exists a set E in the line such that dim;,ox(.E) ^ dm\box(E).

■ fc

Proof: We begin with the unit interval [0,1] and the sequence n^ = 10 , k = 0,1, 2, 3 , . . . . For stages 0 , 1 , . . . , n\ we do our Cantor construction where the mid­ dle interval removed from each A is 1/3 of the length of A. Then from stage ri\ + 1 , . . . , ri2 the middle interval removed has length 9/10 the length of A. We continue to alternate, taking away 1/3 on stage 712 + 1 to rij, and 9/10 from 77.3 + 1 to 714, etc. Let E be the resulting set. This brings us to dui\box

( £ ) > 1^5 >&mbox(E) from using the fact that 1/3 > 1/5 > 1/10. 6 Now let us look at some of the properties (good and bad) for the Box-Counting Dimension. If we use closed sets to cover up A, it does not make a difference in the value ofNg(A). Nor will it matter if A is open or closed. The count will be the same. This brings us to the following theorem: Theorem 7.9 For a set A, d\vabox{A) = d.\m.box{A) and dimkcx(A)

=

dimbox(A).

Applying Thereom 7.9 leads us to the following example. ffl EXAMPLE 7.10 Let A be the set of rational numbers in [0,1]. This set is dense in the interval so its closure is [0,1]. This means dimfcocc(A) = dim{,o:r([0,1]) = 1. ■ This example serves to show that Box-Counting dimension does not have the prop­ erty of countable stability; that is, for a countable collection of sets An it need not be true that dimi,ox(\jAn) = sup n {dim(, oa; (j4 n )}. The case of a finite collection of sets has a little more subtlety. There are examples to show that we can have d\mbox{A U B) > max{d\mbox(A), dimbox(B)}. On the other hand, for upper dimension it is true that d\mbox{A UB) =

m&x{dimbox(A),dimbox(B)}.

HAUSDORFF MEASURE AND DIMENSION

131

The following theorem is stated without proof, but the idea is very straightforward. It states that Box-Counting dimension is monotonic. Theorem 7.10 If A C B, then d\mbox(A)

< dm\box(B)

anddimbcx(A)

<

dimbcx(B).

We end this section with one more example. This example is a little counterin­ tuitive. It shows that a countably infinite set (so small in measure, category, and cardinality) can have nonzero dimension. W

EXAMPLE7.il Let A = {l/n}£° =1 U {0}. This set has dimbox{A) = 1/2.

■

Proof: Take an open set U of diameter 5 < 1/2. Then there exists an integer k such that 1 1 g (k-l)k> ~ k{k + l)' Thus U can cover at most one of the points in { 1 , 1 / 2 , 1 / 3 , . . . , 1/fc}. So at least k sets of diameter S are required to cover A, giving us MN5{A)) -In 5

-]n(k{k

lnfc + l))'

This leads us to dimbox(A) > 1/2. In the other direction, if 1/2 > S > 0, again take k so that l/(k(k — 1)) > S > \/k(k + 1). Then it takes k + 1 intervals of length S to cover [0,1/fc], leaving k — 1 points to be covered by another k — 1 intervals. Thus \n(Ns(A)) -In 5 ordh^box(A) 7.5

ln(2fc) -ln(fc(fc-l))'

< 1/2.*

Hausdorff Measure and Dimension

We have already seen measure in terms of Lebesgue measure. We now go on to explore a related idea (Hausdorff measure) and show how this measure gives us another idea of dimension. Let Abe a set in R n ; and by a 5-cover of A we mean a collection of sets £/» (finite or countably infinite), each of whose diameter is less than S for which A C UC/,. For Lebesgue measure on the real line, we looked at

inf (fScE/O J ,

132

VARIOUS NOTIONS OF DIMENSION

where £(I) means the length of the interval I and the infimum was over all possible covers. It is reasonable to see that if we restricted ourselves, we would have gotten the same value with 6—covers. Moving into M.n we will change notation a bit. Since we no longer talk about intervals and length, given a set U we define its diameter as \U\ =swp{d(a, b)

:a,b&U}

with distance using the Euclidean metric. Definition 7.9 Let 4 c l " and s > 0. We define the Hausdorff s-outer measure of A with respect to 5 by

Hss(A) = mf{j2\Ut\s} where the infimum is taken over all 5-covers of A. This value depends on both s and 6 and is not yet what we want. As we shrink the value of5 the collection of acceptable covers gets smaller so the value of the infimum gets larger. Taking the limit we get Definition 7.10 Let A C R" and s > 0. Then we define the Hausdorff s-measure of A by US{A) = lim HI(A). We see right away that W (0) = 0; and if A C B, then any cover ofB also covers A so HS(A) < HS(B). When s = 1, this is the definition of Lebesgue measure. For R™, this is a constant multiple of Lebesgue n-dimensional measure , the constant being the volume of the unit ball in W1. Let us fix the set A and let S be small (usually, S < 1 is required). Then for s < t,

n\{A)

<

inflEI^I'l^mfiEl^r^8} IS^-^nf

{j:\U.n

<nss(A).

Taking the limit as 5 approaches zero gives us the following theorem. Theorem 7.11 Let A C M.n be fixed. For t > s, U\A) decreasing function of s.

< HS(A); that is, Hs is a

There is, in fact, more here. Suppose A has HS(A) < oo. Then for t > s, ft (A) < lim,5_>.o 5t~s'Hs(A). The right side here is zero, though. In an "opposite" way, for t < s the value ofTLt(A) = oo. At some point, the value of the Haus­ dorff measure jumps from infinity to zero and that is where we find the Hausdorff dimension. 1

Definition 7.11 Let A C R™. We define dim# atis (^4), the Hausdorff dimension of A, by Aim.Haus{A) = sup{s : HS{A) = 0} = sup{s : lis{A) = oo}.

HAUSDORFF MEASURE AND DIMENSION

Ifs = dim.Haus(A), it may be true that lis(A) 0 < %s (A) < oo, then we say A is an s-set.

133

is 0, infinite, or positive and finite. If

Let us move on to the following example, which shows that this dimension, much like measure, behaves in an expected way with finite sets having dimension zero. fl

EXAMPLE 7.12 Let A be a finite set. Then dim/f ous (A) = 0.

■

s

Proof: This is an immediate consequence ofH being a measure as for any value of s > 0, HS(A) = 0. Let us note, though, H°{A) ^ 0 if A ^ 0. * Two other quick properties that fall out ofHs being a measure are: ■ For the countable collection {Ai}, dimHausiViAi) -

swp{d\mHaus{Ai)},

which is called countable stability. ■ If A C B, then dimi^ aus (A) < dim# ous (.B), called monotonicity. This shows a big difference between Hausdorff and Box-Counting dimensions since we have dim# a u s(Q fl [0,1]) = 0 and dimBo:r(Q H [0,1]) = 1. One aid in computing the dimension is what is referred to as the Open Set Con­ dition (OSC). It is a characteristic that shows the set, in a way, consists of separate pieces that do not overlap too much. We use this with an Iterated Function System {/i, / 2 , . . . , / „ } . The IFS satisfies the OSC if its invariant set contains a set O that is open relative to the metric space on the attractor A, such that

■ ■ Uf=1f(0)

fi(0)nfj{0)=9ifi?j,and C O.

If the IFS does not satisfy both criteria, then we call it overlapping. fl

EXAMPLE 7.13 1. For the IFS fi(x) = l/2(x) and f2(x) = 3x/4 + 1/4 on E, the attractor is the unit interval. This is an overlapping IFS. 2. On R, / i (x) = x/Z and f2(x) = x/3 + 2/3, whose attractor is the Cantor Set, and the IFS satisfies the OSC. Just use the open set (0,1). 3. In the plane, the IFS 7i(x, y) = (x/2, y/2), T2{x, y) = (x/2 + 1/2, y/2), and Ta(x, y) = (x/2 + 1/4, y/2 + 1/2) from Example 6.25 satisfies the OSC using the open unit square.

134

VARIOUS NOTIONS OF DIMENSION

If a set satisfies the OSC, then computing dimensions becomes easy. The proof of this is a little lengthy for us, though. Theorem 7.12 Let A be the invariant for the IFS {/i, / 2 , • • •, fn} with ratios {c\, C2, c%,..., cn}. Suppose the system satisfies the Open Set Then = dimBox{A) = s, dimHaus(A)

contraction Condition.

where s is the solution to the equation c\ + c | + c | + • • • + csn = 1. Also, for that value ofs, 0 < rls(A) < oo. Let us return to our continuing example of the Cantor Set and find its Hausdorff dimension. S

EXAMPLE 7.14 Using Theorem 7.12, we find that the Hausdorff dimension of the Cantor Set is also In 2 / In 3, just like its Box-Counting dimension. ■

Because Hausdorff dimension is based on measures, various techniques have been devised for determining the measure and, from that, the dimension. In computing Hausdorff measure for a set, A, there is usually a two-pronged ap­ proach. We look at open covers of the set to find an upper bound for the measure and then find a lower bound. When they agree, we get HS(A). Lower bounds can be difficult to find. One technique used to get this involves the idea of a mass distribu­ tion. We define this for a measure /J, on the space M". The support of a measure /i is the smallest closed set X such that /i(R™ \ X) = 0. This set, supp(p), is closed. It is also true that x G supp{p) if and only if for every positive r, p,(B(x, r)) > 0, where B(x, r) denotes the open ball with center x and radius r. Now we define a mass distribution. Definition 7.12 A mass distribution is a measure on a bounded subset ofW1 that 0 < fi(Rn) < oo.

such

This is often thought of as taking a finite mass and spreading it out on a set X. S

EXAMPLE 7.15 Let a be a point in the set W1. Let Abe a subset ofW1.

^)

=

We define fi by

{ 0 a^ A

This is a mass distribution where all of the mass has been concentrated on a single point. m

HAUSDORFF MEASURE AND DIMENSION

fl

135

EXAMPLE 7.16 Let L be a line segment in R. Then a mass distribution with support L is given by fi(A) = m(L n A), where m is Lebesgue measure. m

A typical method for constructing a mass distribution follows closely with the canonical Cantor Set construction. Recall from Chapter 2 that a Borel set is formed by the a—algebra (collection of countable unions and complements) of open sets. We start with a Borel set A and an initial collection, EQ which consists solely of A. Inductively, Ek is a collection of disjoint Borel subsets of A such that for U € Ek is a subset of one of the sets in Ek-i and contains finitely many of the sets in Ek+i, call them {U\, U2, ■ ■ ■, Uj}. It is also required that for the sets in Ek, the maximum diameter of the sets decreases to zero as k —► 00. Finally, the mass in set U is divided among the {Ui, U2, ■ ■ ■, Uj} so that n(U1) + n(U2) + --- + n(Uj) = fi{U). Define £k to be the union of all the sets in Ek- We see that p(Rn \£k) = 0. We now get the result we are seeking, without proof. Theorem 7.13 Let p be defined on a collection of sets £ as above. Then p can be extended to all subsets in the space R™ by p(A) = p(A n £) and this is a measure on R™. If A is a Borel set, then this number is unique. The support of p is contained in C\£kNow we move on to applying this result to Hausdorff measure and dimension. This is the so-called Mass Distribution Principle found in [30]. Mass Distribution Principle Let \ibe a mass distribution on a set A C R n and suppose that for some s there are positive values c and 5 such that p(U) p(A)/c and s < dimHaus(A)

< dimboa;(A) <

dimbox(A).

Proof: The result is an application of the following. If {Ui} is any cover of A, then

0 < p(A) = piUiU,) < 5>(£^) < cJ2 l^lsi

i

Apply the infimum process to each part and we see ~ris(A) > p(A)/c and for small 8 we have Ha (A) >

p(A)/c.^

136

VARIOUS NOTIONS OF DIMENSION

Let us apply this principle to the Cantor Ternary Set. For our mass distribution on C, let us suppose the unit interval has mass 1. At Step k, we have 2k subintervals of length 3~k. On each of those subintervals, let the mass be 2~k. Essentially, the mass then belongs to subintervals I is split equally among subintervals Ii and Ir at the next step. Let U be a set with \U\ < 1. Let k be the integer such that 3-k+1

< \U\

<3-k.

Then U can intersect at most one interval at Step k so H(U) < 2~k = (3"fc) l n 2 1 "3 <

(3\U\)ln2/ln3

and so %la2l l n 3 ( C ) > 0 by the Mass Distribution Principle. Thus dim Haus{C) < In 2 / In 3. In Chapter 10, we will apply mass distributions to a generalization of the Cantor Set construction to get a fix on the dimension of this new type of set. Some other results on Hausdorff dimension are the following. Theorem 7.14 If S is a complete metric space, dimin(i(S) [26]).

< dimHaus(S)

(see

Theorem 7.15 Let f : X —> Y be a similarity with ratio r > 0. If A C X is a Borel set, then "H s (/(^4)) = rsT-L(A) for any positive s and dirri# a M S (/(A)) = d.\mHaus(A). Proof: For any set U, since f is a similarity, \f(U)\ = r ■ \U\. Thus the sets which make a 5-cover of A form an rd-cover off(A). The first result follows from this and the second from the first. 4* Some of this will be especially useful when looking at products of sets.

7.6

Miscellaneous Notions of Dimension

Packing dimension is similar to Hausdorff dimension, but involves requiring the covering set to be disjoint. Let A be a set and give S > 0, let B = {Bi} be a collection of disjoint balls which cover A where each Bi has radius at most 6 and is centered at a point in A. Define the the S-packing number as

PZ(A) = sup I. Y,\Bi\a\, where the supremum is taken over all possible covers. As 8 approaches zero, the value of VI {A) must decrease and therefore has a limit. Let VZ(A) =

llmVZ(A).

MISCELLANEOUS NOTIONS OF DIMENSION

137

This, unlike for Hausdorff, is not a measure. If we have a countable, dense set each individual element has Vo({xi}) = 0, but as a whole Vo{U{xi}) > 0. We modify this by looking not just at the set A, but at sets containing A. So we come to the s-dimensional packing measure of the set A:

P°(A) = inf J5> 0 S (^) : A C U ^ ^ I , where the infimum is over all covers {Ai}. As with Hausdorff, as s increases, the value of Vs goes from positive infinity to zero, and that is where we find the packing dimension: dimp(A) = sup{s : VS{A) = 00} = inf{s : VS{A) = 0}. As the covers are more restrictive than we use for Hausdorff, yet do count as what we use for box-counting, we find that dimHaus(A)

< dimp(A) <

dimbox(A).

In a similar vein, we can change our definition of box-counting by using countable covers of the set. Our result will give us countable sets have dimension zero (just as with topological dimension), even when they are dense. The upper and lower modified box counting dimension of A are given by dhnMB(A)

= inf I sup{dim Bo:c (A i )} : A C U ^ ^

and

dirnMB(A) = inf Lup{dmBox(Ai)} ■ A Q ^ZiAi I. Now for any set A C M™ we have 0 < dimHaus{A) < d i m M B ^ ) < dim M B(-4) < d\mBox{A)

< n.

Now we turn to Besicovitch-Taylor dimension which first appeared in [7]. This is based on the gaps in the set and is defined for closed sets in the real line. There are some attempts to expand this to M.n ([64]), but we shall forego them here. Let E be a bounded and closed set in R and, without loss of generality, let us assume E C [0,1]. Then [0,1] \ E is a countable union of open intervals, {On}. This countable union can be ordered in such a way that the sequence {an} where an = (-{On) forms a nonincreasing sequence of positive numbers called the gap sequence. For the Cantor Set, this sequence is

138

VARIOUS NOTIONS OF DIMENSION

1/3,1/9,1/9,1/27,1/27,1/27,1/27, If E has Lebesgue measure zero, then

E°" In the other direction, ifty = {an} satisfies ^an = 1, then there exists closed sets B c K such that [0.1] \E consists of intervals with lengths in ^. Then we say E belongs to ty. If we let rn = 5Z7-_„ aj, then rn is a decreasing sequence with rn —> 0. This leads us to the following results about Hausdorff measure and dimension. Theorem 7.16 Suppose ty is a nonincreasing sequence of positive numbers satisfy­ ing Y^ o-n = 1 and E is a closed set in R belonging to *&. Then for 0 < s < 1 HS{E) oo

Proof: Define X(s, <3>) by A(s,*) = liminf n

. r, n

There is nothing to prove if\{s, *) — oo, so we will assume this is not the case. So lim infn^oo [n ( ^ ) ' ] < oo and let E be a fixed set belonging to Vf and [0,l}\E

= UnQn,

where the Qn are disjoint open intervals with £(Qn) = an. Define En recursively as Ei = [0,1] and En = Ei \ U™=1 Qi, n = 2, 3,4,.... Then En consists ofn closed intervals of total length rn and E is a subset of Enfor each n. Given e > 0 and r\ > 0, there exists an integer m such that rm < V and

mC-^-X
Vm ) Then each Em contains m intervals of total length rm, and we can see that the sum of the lengths of the intervals in Em is bounded above by X(s, \&) + e; and since r\ and e are arbitrary, we get our result. 4b To convert this to a result about Hausdorff dimension, let an be defined by (rn\an

i

MISCELLANEOUS NOTIONS OF DIMENSION

and let

139

a($>) = liminf an.

Corollary 7.16.1 Suppose E C R is closed and belongs to ^ with a(^) as above. Then 0 < dimHaus(E) < a ( # ) . Besicovitch and Taylor also show the following result about Cantor Sets Ca: If t < dimHaus{Ca), then for any 7 > 0 there is an arrangement of E, the gaps in [0,1] \ Ca, such that ri^E) = 7. This work on gap sequences gives us another type of dimension. This is called the Besicovitch-Taylor dimension. Given £ c l with gap sequence <3/ = {an} we let <jp = XI a n where 0 < /3 < 1 since we are on the real line. Then the BesicovitchTaylor dimension of the set E, written dim.B-T(E), is the smallest (3 for which that sum is finite; that is, &\YO.B-T(E)

= inf{/3 : 073 < 00}.

For the Cantor Set, the Besicovitch-Taylor dimension agrees with Hausdorff and others, diuiB-T(C) = In 2/ In 3. We will revisit this topic in Chapter 10. The following gives us another criterion for Hausdorff s-measure zero sets in the line. Corollary 7.16.2 Let E be closed in M. where E belongs to * and assume ^ a* < 00 for some a s G (0,1], Then Hs{E)=0. The gap sequence also relates to upper Box dimension for a set. If a set has Lebesgue measure 0 and is compact in R, then the gap sequence determines the value ofdimBox(E). The proof of this can be found in [71]. Theorem 7.17 Let E be a compact subset ofM. with Lebesgue measure 0 and let {an} be the gap sequence of E. Then -— Inn d\m.Box{E) = limsup — . n—>oo

For the Cantor Set, - 4 p ~ = — In otn

lnO!ra

' n 2 " „ = In 2/ In 3. — In i

n

'

As with many of the topics we cover, there are different paths down which to travel, yet we cannot look at much. Doing so would bring us too far from the topic of Cantor Sets. However, the interested reader is encouraged to learn more about the different notions of dimension and how to compare and contrast them.

140

VARIOUS NOTIONS OF DIMENSION

EXERCISES 7.1 For each set below, determine if it is bounded above. If so, give three examples of upper bounds. If not, declare the set not bounded above. Then go over them again, this time for lower bounds. a) [-1,1] b) {1/n : n e N } c) j r e Q : r < 2 ] d) {xeQ:x2<2} e) UneN(n,n+ 1/n] f) n~ = 1 [0,l/n] 7.2 For the sequences xn = ( — 1)", yn = 1/n, lim sup and lim inf. 7.3

an

d zn = sin(n7r/2) find the

Let an and bn be the sequences defined below: {a n } = {0,l, 2,0,1,2,0,1, 2 , 0 . . . } {6„} = {-l, 0 , 1 , 1 , - 1 , 0 , 1 , 1 , - 1 , 0 , . . . }

Find a) b) c) d) e) f) 7.4

lim sup an + lim sup bn lim sup{a„ + bn} lim inf an + lim inf bn lim inf {an + bn} lim sup a„ + lim inf bn limsup{anfe„}

The following statement is incorrect:

Let {xn} be a bounded sequence and k £ K. Then limsup{fcx„} = fclimsup.T„. Find a counterexample that shows this is wrong. Determine changes in the hypothe­ sis that will make the conclusion true. 7.5 What do the solutions to the two problems in Exercise 6.8 say about similarity dimension s ? 7.6

Show that if A is a finite set, then &uni,ox(A) = 0.

7.7 Suppose E is a bounded set and f : E —» K has the property that for all x and y, \f(x) — f(y) | < k\x — y\ where k is constant (such a function is called a Lipschitz function). Show that (MLBox(f(E)) 7.8

< dmBox(E)

anddimBox(f(E))

<

dimBox(E).

Let E = {x\,X2, X3,... ,xn}. Show that'H0(E) is the number ofpoints in E.

EXERCISES

7.9

141

Prove that if {E^} is a collection of sets

us{uEk)
Suppose f is a Lipschitz function. Show that dimHaus(f(E))

<

dimHaus(E).

7.11 Let f : R -» R be the function f(x) = x2. Show that for any set E C R, dim H a u s (.E) = dim f f o u s (/(£)). 7.12

Find the Hausdorjf dimension of the set {0,1,1/4,1/9,1/16,...}.

7.13

Show that % 0 (C) — oo and V} (C) — 0, where C is the Cantor Set.

7.14 Find the Box-Counting dimension of a general Cantor Set, Cr. What happens asr -> 0+ ? 7.15 Compute the Box-Counting dimension for the Sierpinski Triangle (see Exam­ ple 6.25).

CHAPTER 8

POROSITY AND THICKNESS-LOOKING AT THE GAPS

The main purpose of this chapter is to introduce readers to some reasonably new ideas and results which involve the Cantor Set. The sources for much of this are some papers which are very technical in their definitions, proofs, and constructions. To keep things at the correct level, there will be many statements here that are not matched with proofs. The list of references in the back should help the interested reader find out more.

8.1

The Porosity of a Set

Porosity was first studied by Denjoy in his work on trigonometric series in 1920 [20]. The subject arose again, this time with the modern nomenclature, in the 1960's with Dolzenko's work in cluster sets [25]. Similar to the definition of continuity, we begin with a pointwise idea and then expand it. Porosity can be defined for any metric space (S, d), but we shall begin with sets in the real line, moving later on into metric spaces. By starting in R, we will have an additional restriction beyond just looking pointwise. Wefirstlook only to one side ofxQ, finding the right porosity. The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

143

144

POROSITY AND THICKNESS-LOOKING AT THE GAPS

Definition 8.1 Let £ c l and let XQ £ £ ' . For a fixed r > 0 let X(E, XQ, r) denote the supremum of the lengths of the open intervals in Ec fl (XQ, XQ + r) (this can be any number in [0, r]). Then the right porosity of E at XQ is given by + /TP

\

v

\{E,x0,r) r->o

r

Similarly, there is the left porosity of E at x$, written p (E,XQ) r,xo) n Ec. Finally, the porosity of E at x0 is given by p(E,x0)

m&x{p+(E,xo),p~{E,x0)}.

=

\(E,x0,r)

x 0_

fl

and using (XQ —

EXAMPLE 8.1 Let E = {0} U { l / 2 n } ~ = 1 . Then p+(E,0)=

lim

^ " 2 ^

1 ~ 2

Notice that although the definition is written as a l i m s u p of a function (ofr), we could find our answer as a limit of a sequence. This is often the case. The following are properties and definitions that emerge from the concept of right porosity. 1. Porosity at a point is a value between zero and one. Ifp(E, XQ) = 0, we say E is nonporous at XQ. Ifp(E, Xo) — 1 we say E is strongly porous at XQ. 2. A set E is a called a porous set ifp(E, x) > Ofor each x 6 E. 3. If a set A = U'^L1En and each En is a porous set, then we say E is a-porous. The following theorem and its corollary are presented without proof, but details on this and other properties can be found in [80]. Theorem 8.1 If E is a porous set, then E is both measure zero and nowhere dense. Corollary 8.1.1 If E is a a-porous set, then E is both measure zero and first cate­ gory. Some definitions do not require XQ to be a point in E, but we are only interested in points in the set.

THE POROSITY OF A SET

145

These results say that porosity has both an analytical (measure) and topological (category) nature to them. The advantage to using porosity arguments is that we get two results via only one proof. As stated before, these ideas generalize to metric spaces. In such a setting a porous set will be nowhere dense, but measure zero depends on whether or not the underlying space has a measure on it. These results do not reverse themselves. There are sets that are measure zero and nowhere dense that are not porous. In fact, they are not even o -porous. Now let us look at porosity and the Cantor Set. As we can see in the picture A(C,0,2/3) = l / 3

0

" -v-

r=2/3

and via the self-similar nature ofC, we have P+(C,0)>\. Of course, p~ (C, 0) = 1 which leads to p(C, 0) = 1. This is actually the typical value. The following theorem is due to Denjoy and shows that most of the points in a set are strongly porous from both sides (bilaterally strongly porous). Theorem 8.2 If E C R is closed and nowhere dense, then the set of points in E at which E is strongly porous from both the left and the right sides is a set which is residual in E. The literature also contains what is now called porosity index . This was what Denjoy was working with in [20]. Definition 8.2 Let £ c l and let XQ S E. For a fixed r > 0 let h, k be positive real numbers such that (XQ + h, XQ + h + k) D E = 0 and h + k < r. Then the right porosity index of E at XQ is given by i+(E,xo)

k = lim sup —. r->0

h

Similarly, there is the left porosity index of E at XQ, written i~(E, XQ). Finally, the porosity index of E at XQ is given by i{E,x0)

=

max{i+(E,x0),i~(E,x0)}.

The index can take any value in [0, oo). In a similar vein as before, E is small ifi(E,xo) > 0 for all xo £ E. As we now see, porosity and porosity index are equivalent notions. Relating the two concepts is the following lemma whose proof is obvious.

146

POROSITY AND THICKNESS-LOOKING AT THE GAPS

Lemma 8.1 Let p = P+(E,XQ)

andi = i+(E,xo). p—

1+i

and i

Then

l-p

We should at some point look at an application of this idea. One of the most com­ mon uses of porosity is in the description of an exceptional set in the real line. The quick definition of an exceptional set would be when something happens everywhere except on some set E. What we then need is a good way to classify E. The usual ways are measure, category, and cardinality. Since there are measure zero, nowhere dense sets that are not porous, describing E as porous is a stronger result that even both of the other two. Recall a quick definition before our example theorem, we all know that for f : R —► K to be differentiable at XQ we must have that H

f{xo + h) - f(x0)

fr->o

h

exists and is finite. In a similar vein, we can use h —» 0 + (h —> 0~) to define the right (left) derivative of f at XQ. If f is differentiable at XQ then f has equal left and right derivatives at XQ, while f(x) = \x\ and XQ — 0 shows that having finite left/right derivatives does not imply differentiability. So our application of porosity is this theorem from [6]. Theorem 8.3 For an arbitrary function f : R —> R, the set of points of which f has a finite right derivative, but not a two-sided derivative, is a a-porous set.

8.2

Symmetric Sets and Symmetric Porosity

One aspect of Cantor Sets that we have not looked at yet is the symmetry of the set. If XQ is a point in C, then so is its reflection about the point 1/2, call this yo To find that reflected point, it is the same distance from 1/2 as XQ, that means either y0 = 1/2 + (1/2 - x0) (ifx0 < 1/2; or y 0 = 1/2 - (x0 - 1/2) (when x0 > 1/2). Either way, that brings us to yo ~ 2(1/2) — XQ. This leads us to a new definition. Definition 8.3 A set S is called symmetric if there exists a point t (not necessarily in S) such that for all x 6 S the point 2x — s is also in S.

&

EXAMPLE 8.2 An open (or closed) and bounded interval is symmetric about the midpoint of the interval. ■

fl

EXAMPLE 8.3 The Cantor Set is a symmetric set.

■

SYMMETRIC SETS AND SYMMETRIC POROSITY

147

In fact, all Cantor Sets Cr are symmetric about t = 1/2 by virtue of the con­ tiguous interval being taken from the center of each subinterval. We move from symmetric sets to a symmetric version of porosity for the real line. A word on nota­ tion here, for a point x and a set A, the symmetric copy of A about the point x is 2x - A = {2x - a:a e A}. Definition 8.4 Let £ c l and let XQ G E. For a fixed r > 0 let j(E, XQ, r) denote the supremum of the lengths of the open intervals A such that A n E = 0 and (2xo — A) fl E = 0 (this can be any number in [0, r]j. Then the symmetric porosity of E at xo is given by sfjp

N

,-

p (is, XQ) = limsup -

l(E,x0,r)

Just as before, this porosity can be any value between 0 and 1, inclusive, if ps(E,xo) > 0 we say E is symmetrically porous at XQ- The set E is symmetri­ cally porous if it is symmetrically porous at each of its points. In addition, there is the notion of a-symmetrically porous. If a set is symmetrically (asymmetrically) porous, then it obviously is porous (aporous). This idea does not flow in the other direction. The following result is from Evans [28]. S

EXAMPLE 8.4 There exist sets which are porous, yet not even asymmetrically porous.

■

We change our focus a little in this section. When we refer to the number r, it will reference the relative since of the interval removed; that means for each r, the Cantor Set in question is C(l-r)/2-

By the way that the contiguous intervals are removed, one would expect the Cantor Set to be a symmetrically porous set.This is true for the ternary set. In fact, for r > 0 fixed, any C'(i- r )/2 will be a symmetrically porous. Suppose though we let the size of the interval removed change at each iteration. Definition 8.5 Let rn be a sequence of real numbers with 0 < rn < 1. We define the Cantor Set C(i_ rn )/2 by the following construction. At Stage I, the interval 1% = [0,1] is divided into three pieces IQ, O, and I\ such that I = J0 U O U h and

£(I0) £{i)

=

1(h) e(i)

=

1 2y

l)

'

At stage k, there are 2k subintervals of size YljZi(~ip~)- ^a^ tnese {la}, where a is a string ofk — 10's and 1 's and we write the length of the string by \a\. Take each

1 48

POROSITY AND THICKNESS-LOOKING AT THE GAPS

Ia and divide it into three pieces Ia,\, Oa, and 7^2 such that I =

IafiUOaUlatl

and

KLfi) 1{I„)

=

*(/„,!) _ i nU l{h) 2

,

Tkh

The Cantor Set C( 1 _,.)/ 2 is then defined to be

c
\\
j

Now we have the following result relating Cantor Sets and asymmetric porosity. The proof is very detailed and will be left out. Theorem 8.4 For the Cantor Set C(i_ r )/2 the following are equivalent: 1. C(i_ r )/2 is porous. 2. C(i_ r )/2 is a-porous. 3. C( j _ r ) /2 is asymmetrically porous. 4. limsup n _ >(X) r n > 0. In the list above, symmetrically porous is left off, This is due to the following example by Evans, Humke, and Saxe [27]. S

EXAMPLE 8.5 There is a asymmetrically porous Cantor Set which is not symmetrically porous. I We will not prove this example, but the proof begins with the statement Let S = {k2/2 + 9k/2} and for each n set

(_ 0

otherwise

This means we are removing intervals, but not at every step and with quickly decreas­ ing frequency. We will still be able to break this apart into symmetrically porous pieces, but have so many O's in the sequence that we destroy the possibility of being symmetrically porous as a whole. As a comparison with another indicator of a set's size, in [74] a strongly symmet­ rically porous set with Hausdorff dimension 1 was constructed. This was done in a fashion similar to that of the above example.

A NEW AND DIFFERENT DEFINITION OF CANTOR SET

8.3

149

A New and Different Definition of Cantor Set

Previously, we used a fixed ratio r to create the set Cr. This point of view looked at the relative size of the open interval removed versus the original interval. Here we shall just collect together the countably many intervals removed (0\, 02,03,...) and focus on the order in which these Oi are listed. Let us define a Cantor Set C by

c = i\\Jot, i>l

where I is a finite closed interval and {Oi\i > 1} is a (finite or infinite) collection of disjoint open intervals contained in I. Notice here that this is a broader defini­ tion than we have had before; that is, there are sets which are Cantor Sets by this definition which are not Cantor Sets by the previous one. Some terminology and notation: We are going to restart our notation, beginning at ground zero in order to match the notation the interested reader would find in the literature on this topic. By a tree T> we mean a collection of strings of O's and 1 's where for our purposes, 0 denotes the left and 1 the right. Each finite string of O's and 1 's, UJ, is called a word and \UJ\ will denote the length of the word. Let I be the root of the tree. We'll refer to {1} as the zeroth level of the tree. Now suppose the tree has been defined to the nth level. We define the (n + l)st level as follows: Let Iu be an nth level vertex (a word consisting ofn O's and 1 's) of our tree. Assume first that

^n IU °A ±0Let Oju be the interval in the set {Ot\i > 1} of least index which is contained in 1^, and let I^0 and I"1 be the closed intervals with

I" = f ° u O , - U f 1 . We then let I"0 and 7""1 be subvertices ofIu in V. If

l n

" ( U °\ = 0'

then we set I"0 = Iu and let I"0 be a subvertex of I" in V. We repeat this pro­ cess for every vertex in 1^ in the nth level ofV. The (n + l) s * level of the tree is the set of vertices in Iu in V with \u\ = n + 1, where \u\ denotes the length of the word v. We continue this process inductively, creating the infinite tree V. The intervals 1,1°,I1,!00,101,... are called the bridges while the open intervals Oj, Ojo, Oji,... are called the gaps. Note that {O/c \I" is a bridge ofV} = {0{\i > 1},

150

POROSITY AND THICKNESS-LOOKING AT THE GAPS

hence oo

/

Any tree with this property is called a derivation of the Cantor Set C from I. While the end result will always be the same, the set one has at the nth step is determined by the order in which the Oi are labeled. In the picture below, which is the construction of the ordinary Cantor Set, on the left side the interval (1/9, 2/9) is the first interval removed, while on the right side the interval (1/3, 2/3) is removed first.

Thus different orderings of {Oi}, called derivations , give one different bridges and gaps. Comparing the relative sizes of bridges and gaps and looking across all order­ ings it the idea behind the thickness of a Cantor Set.

8.4

Thickness of a Cantor Set

This idea of constructing a Cantor Set in more than one fashion by using different ordering of the intervals removed leads us to the idea of thickness. Thickness is yet another way of measuring the size of a set. Using C to represent a Cantor Set, we use T(C) to denote the thickness of the set. If {Oi : i > 1} = 0, then we set T(C) = oo. If{Oi : i > 1} ^ 0, then r 6 ) = sup inf mm < ——F, ——r } , ■D A&V \\0A\ \OA\) where the supremum is over all derivations V ofC, the infimum is over all bridges A ofV, and \A\ means the length of the interval A. This supremum is always attained. Determining the thickness of Cantor Sets is not as intimidating as it sounds, thanks to the following Definition and Lemma. Definition 8.6 We say a derivation is ordered iffor intervals A and B in the deriva­ tion with A = A0 U OA U A1, B = B° U 0B U B\ and B C A, then \OA\

> \OB\.

Lemma 8.2 Let T> be any derivation ofC from I. We shall denote the thickness ofC with respect to a fixed derivation V as r-p. Then there exists an ordered derivation T>o of C from I with MC)
APPLYING THICKNESS

151

Furthermore, ifT>\ and T>2 are two ordered derivations of C from I, then TVl(C)
8.5 Applying Thickness Now let us apply some of this. For two sets A and B, their sum A + B is defined as A + B = {a + b: a<= Aandb e B}. This idea generalizes something already known, translation of a set. If we have a set P and add the number k to each point in P, the result is just P translated over k units. Typically this is written as k + P. So if P = [0,1], then 2 + P = [2,3] and if S = {-1,2}, then S + P= [-1,0] U [2,3]. Another interesting fact about the Cantor Set C has to do with the sum of C and itself. As we saw in the previous chapter, C is nowhere dense and measure zero and thus very small in two different ways. It turns out, however, that in another way this set is big. Theorem 8.5 For the Cantor Ternary Set C, C + C =[0,2]. Proof: Recall that C = { i £ [0,1] : the tertiary expansion of x consists of only 0's and 2 's}. Then 1/2(C) = {\x where x E C) are the numbers whose base 3 expansion con­ sists of just 0's and 1's. It is easy to see that if x € [0,1], we can decompose its expansion into the sum y + z where y and z consist of only 0's and 1 's in base 3. This shows

\c + \c = M and our desired result comes from multiplying both sides by 2. 4 This is the typical proof one finds in the literature. There is also a visual proof, the so-called "proof without words." In this, we see that any line of the form x + y = k

152

POROSITY AND THICKNESS-LOOKING AT THE GAPS

where k £ [0, 2] will, by use of the self-similar nature ofC, pass through a point in C x C, giving us our result. J L

\

J U

3 ^

P

:: d

n c.

I B I r

J L_ "1 C

-a r:

aL

U L

i

■ I'

\

\

n r a Ei

a ■

\ fl ■

\

\ \

J L 9 p

\ "i r \

n r nn

V«wa/ Proof that C + C ={0,2} Now we look at the sum of two Cantor Sets through the lens of thickness and porosity. This leads in the obvious fashion to the sum of a Cantor Set and itself. Theorem 8.6 For j = 1,2 let C0 be a Cantor Set derived from Ij with 0} a gap of maximal size in Cj. Assume that |Oi| < \I2\ and \02\ < \h\. IfT(C\) ■ T(C2) > 1, then C\ + C2 = h + hUsing the relationship between thickness and porosity, [73] has the following, which is a corollary of the corresponding thickness result in Theorem 8.6. For this, the Cantor Sets, Cj, have a constant ratio, Tj. Corollary 8.6.1 For j = 1,2 let Cj be a Cantor Set derived from I3 with Oj a gap of maximal size in Cj. Assume that |Oi| < |/ 2 | and \02\ < \h\. Ifp(Ct) + p(C2) > 1, then C1+C2

= Il+

I 2.

This result is not an "if and only if" type. Falconer, in [29] has an example of Cantor Sets C\ and C2 that are strongly porous (p(Cj) = I) yet C\ + C2 is an interval. In [56] and [61] the following question was asked and answered: For any k £ [0, 2], what is the cardinality of the set ofpoints (x,y) e C xC such that x + y = k. The short answer is, depending on k, the set is either finite or uncountable. The detailed answer, given by Nymann, depends on the base 3 expansion of h = k/2. For this we also need to define a value c(k). Let c(k) be the cardinality of the set

A BIT MORE ON THICKNESS

153

{n € N : an = 1}. Suppose h has a unique ternary expansion. Then the expansion cannot end in a string of zeroes. Define h as

where an = 0,1, or 2, an ^ Ofor infinitely many n, and an ^ 2 for infinitely many n. Now ifx, y £ \C, then x = Y,eJTandy

=

Y,Zn/3n

with £n,£n = 0,1. Given x + y = h, then if an = 0, it must be en = e'n — 0, and if an = 2, it must be en — e'n = 1. However, when an = 1 we can have either en — \ and e'n = 0 or en = 0 and e'n = 1. Thus there are 2c(k' choices for (x, y) (this is uncountable ifc(k) is infinite). Ifh has two different ternary expansions, then they are of the form h = Q,a\a2 ■ ■ ■ ar222 ... = Q.a\CL2 ■ ■ ■ ar~ibr, where a-y, a,2, ■ ■ ■, ar-i are 0, 1, or 2, and ar = 0 or 1, with br = ar + 1. Then we have either h(2c{k)) if ar = 0, [ ] * \3(2 C W- 1 ) if ar = l. For more on sums of Cantor Sets, see Section 10.7. 8.6

A Bit More on Thickness

The sum of two sets (our big application of thickness) can also be looked at as an intersection of sets issue. For a value x, S + x = {s + x : s G S} and S + T = {s + t: s e S andteT}. Then S +T =

{x:Sn(x-T)^®}.

In Theorem 8.6 we assume that \0\\ < \I2\and\O2\ < |/i|. The idea behind this is that neither of the Cantor Sets can lie in an open interval from the construction of the other. When this condition is met, the two Cantor Sets are said to be inter­ leaved. Williams, in [78], shows the counterintuitive result that intersections can be small, even when the thickness of the two Cantor Sets greater than 1 and the sets are interleaved. In [39], the authors answer the question, When is the intersection of two Cantor Sets empty, non-empty but possibly a singleton, and non-empty, but large in terms of thickness? Let C\ and C2 be Cantor Sets with thicknesses T\ and T-I, respectively. We will call this a thickness pair (T\ , T2). First we need the following two functions ofr.

154

POROSITY AND THICKNESS-LOOKING AT THE GAPS

/ T) =

r 2 + 3r + 1

andg(j) =

(2r + l) 2 7

.

Afext, we define three regions in the first quadrant (since r is non-negative) as fol­ lows: one region (called Region A) is in the quadrant, but below T\ ■ T2 — 1; the second is above Region A and below the area defined by n > T2, TI > / ( r 2 ) , andr2

>

g(n).

T"2 > n , T2 > f{n),

>

g{r2).

or andri

This is Region B. Finally, Region C is the area above Region B. This leads us to Theorem 8.7 Let C\ and C2 be Cantor Sets with associated thickness pair (TI, T2). If the pair lies in Region A, then C\ fl C2 can be empty. If the pair lies in Region B, then C\ fl C2 is non-empty, but can be just a singleton. If the pair lies in Region C, then the C\ fl C2 must be non-empty and can be either an interval or a Cantor Set, but in either case the intersection must contain a set of positive thickness. One last quick note using thickness. In [55] there is a lower bound result on Hausdorff dimension. This is important as upper bounds can usually be quickly obtained using covers of the set, but lower bounds are harder to come by. This paper shows that if the thickness of a Cantor Set C is r, then dunHaus(C)

>

^Jf^.

The bound is sharp as this is exact for the Cantor Ternary Set.

8.7

Porosity in a Metric Space

Like measure, category, and cardinality, porosity lends itself to the idea of a small set from Chapter 3. For porous sets, the small sets are the sets which are a-porous and when written as UE^, p(Ek) = 0. The fact that this is small becomes obvious when we recall that p(E) = 0 implies the set is measure zero. Just as with those other indicators of size, porosity can be taken to other spaces. For this we do need the idea of distance, so we cannot have just a topological space, but need a metric space. Definition 8.7 Let (X, d) be a metric space. Furthermore, let At C X and x £ M. For R > 0, we define A(x, R, M) as the supremum of all r > 0 so that there is a z e X such that B(z,r) C B(x,R) and B(z,r) fl M = 0, where B(z,r) is the open ball with center z and radius r. We then define the porosity of M at x by , w s ,. 2A(x, R.M) p(M,x) = Inn sup — .

POROSITY IN A METRIC SPACE

155

As before, we have the following: • p(M, x) is a value between 0 and 1. ■ The set M is porous if it is porous at each of its points. • The set M is a-porous if it can be written as the countable union ofporous sets. The appearance of the "2 " in the numerator of the definition seems a bit out of place, but recall from the definition in K the ratio is a diameter versus a radius. The following lemma from [80] is useful. Lemma 8.3 Let M c K and let N = M x Rn c R n + 1 . Then N is a porous subset (or a-porous subset) ofM.n+1 if and only if M is also as a subset ofM.. We leave this section pointing out one other type of porosity that appears in lit­ erature. In this version, there is a real-valued function associated with the distance A. Definition 8.8 Let g : K —> M. be a function which is increasing on the interval [0, h) for some positive value h with g(0) — 0. Let (X, d) be a metric space and M C X and x £ M. We say x is (g)-porous at x if limsup H->O+

—

> 0, R

where A(x, R, M) is defined as before. There are numerous other definitions that generalize the notion of porous. These include shell porous sets and superporous sets. Superporous sets also lend them­ selves to creating a topology. As with dimension and fractals, there are many roads upon which to travel, and the interested are encouraged to find out more.

156

POROSITY AND THICKNESS-LOOKING AT THE GAPS

EXERCISES 8.1 In the definition of Cantor Sets that we use for thickness, the set {Oj} can be finite. Explain how this opens us up to Cantor Sets that are not allowed in the canonical definition of a Cantor Set. 8.2 Note in the definition it does not say {0{\ has to be non-empty. What happens if it is empty? 8.3 One more difference. For an interval I, the closure of I, written I, is the in­ terval union its limit points2. So [0,1] is the closure 0/(0, 1) (and also the closure of (0,1], [0,1), and [0,1}). Anyway, other people, in their definition of this construction, insist that when i ^ j . Find an example of a "Cantor Set" that meets Astels definition [1] but where for some i.j. 8.4

Determine the value of

*HM}) for the Cantor Ternary Set using the usual ordering of the intervals removed (the typical derivation). 8.5

Show that the usual derivation for the Cantor Ternary Set is ordered.

8.6

Prove

T(C1/3)=1.

8.7 Find r(Cr) where 0 < r < 1. 8.8 Use the thickness, r(Cr), to determine the values of r so that Cr + Cr is an interval. 8.9 What about Cr and Cs ? Ifr is fixed, find the values of s which create a Cs so that Cr + Cs has an interval. 8.10

Give an example of a set in K2 that is porous, yet connected.

8.11

A set E is called very porous from the right at x if p^(E,xo)

= nm mi r—>0

. T

Similarly there is very porous from the left, very porous at a point, and very porous sets. Give an example of a set E and a point x such that E is porous at x, but not very porous there. 2

A limit point for a set S is a point x such that there is a sequence sn E S with limn-joo s„ = x.

CHAPTER 9

CREATING PATHOLOGICAL FUNCTIONS VIA C

Let / be a function. We say / has the Intermediate Value Property if for any two values a and b in the domain of / , and any y between / ( a ) and f(b), there is some c between a and b with /(c) = y. For a time, having the Intermediate Value Property was thought to be equivalent to being continuous. We will present counterexamples to this thought and use the Cantor Set to show a function with this property can have lots of discontinuities.

9.1

Sequences of Functions

Functions that illustrate a property, or more often where a nice property goes wrong, are sometimes referred to as pathological functions. For instance, continuity is a nice property. The pathological function then would be a nowhere continuous function. An example of this is the characteristic function of the rationals; the function is

11, ifx e Q, [0, ifx £ Q. The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

157

158

CREATING PATHOLOGICAL FUNCTIONS VIA C

Differentiability is also a nice property to have. A function which is not differentiable at any point (nowhere differentiable) is easy, just use the f above. This comes from the theorem "If f is differentiable at XQ, then f is continuous at XQ." The contrapositive then says, "If f is not continuous at XQ, then f cannot be differentiable there." The absolute value function shows that functions can be continuous at a point, but not differentiable there. So the real pathological example in this case is a function that is continuous everywhere, but nowhere differentiable. In Chapter 6 we used iterated function systems to create such a function. However, we actually did more. It showed an important technique in creating pathological functions. We could not explicitly draw the graph of such an f so instead we created a sequence offunctions which converged to our goal function. That is very typical when creating pathological functions. We create a sequence and show that the limit gives us what we want. There are, however, many meanings to the idea of "limit" and that is where we begin. We will start with a repeat of some earlier ideas on sequences of functions, the pointwise limit of a sequence offunctions. This is probably the easiest type of con­ vergence since it takes a problem involving functions and changes it to a problem involving real numbers. Definition 9.1 Let D be a set in the real line and fn : D —> Hfor n = 1, 2, 3 , . . . and f : D —> R. We say fn converges pointwise to f iffor every point x G D, we have liirin-Kjo fn(x) = f(x). We write this as fn —» / . fl

EXAMPLE 9.1 If D = [0,1] and fn — xn, then fn converges pointwise to the function

/(*)

For this type of convergence, we choose a point, XQ, then look at fn(xo) and find the limit as n approaches infinity. We do this for each XQ € D and we have the limit function, f. In Example 9.1,if0<x 0, while ln —> 1, giving us our f. fl

EXAMPLE 9.2 If we let D = R and gn{x) = x ^ n + 1 , then gn —> g where (0,

g{x) = I 1/2, ll,

(fare (-1,1),

ifx = ±1, otherwise.

SEQUENCES OF FUNCTIONS

159

What can happen, and does happen in these examples, is that even though the sequence offunctions converge, they do so at different rates for different points. For example, with /io, (1/2) 10 is much closer to zero than (99/100) 10 . If the values do converge at more or less similar rates, we say that fn converges uniformly to f and define that rigorously now. Definition 9.2 Let D be a set in the real line and fn : D —> Hfor n = 1, 2, 3 , . . . and f : D —> R. Then fn converges uniformly to f (fn —>• f) iffor every e > 0 there exists an N G N such that n > N implies that for all x G D we obtain \fn{x)fl

f{x)\

<e.

EXAMPLE 9.3 Let D = [0,1] and fn(x) — x/n. We claim fn —» 0. This is seen as follows: Given any e > 0 choose N G N so that N > \je. So for any XQ G D and any n > N we have

1/nM - 0|

1

1

1

So for all the x in the [0,1], the fn(x) and f{x) values are e-close at the same time (time here means after the same index N). Note how important the domain is in this example. If D = R, the sequence still converges pointwise to the zero function, but the convergence is not uniform. Uniform convergence brings with it some important properties. For instance, it preserves the notion of continuity. If we know each fn is a continuous function, then the uniform limit must also be continuous. Theorem 9.1 Let fn : D —>ffibe a sequence of continuous functions. If {fn} converges uniformly to some function f : D —> K, then f must be a continuous function. Proof: Pick an XQ G D. We will show that f must be continuous there, and since XQ was arbitrary, this must hold for all XQ. We will use the typical e/3 argument. Let e > 0. Since fn —> f, for this e there is an N G N such that if n > N, then \fn(x) — f{x)\ < e/3 for every x £ D. We will then concentrate on f^. We know /JV is continuous at XQ, hence there exists a 5 > 0 so that for all x G D with \x — XQ\ < 5 we get \!N{X)

- fN(x0)\

< e/3.

160

CREATING PATHOLOGICAL FUNCTIONS VIA C

This implies that when \x — XQ\ < S we obtain \f{x)-f(x0)\

=

\f(x)-fN{x) x

+fN(x)-fN(x0)

<

\f( )

<

e/3 + e/3 + e/3 = e.

- /N{X)\

+ \!N{X)

+

- /WOo)

fN(x0)-f(xQ)\ + |/JV(ZO) -

/Oo)j

So we have shown that given e > 0 there exists a 5 > 0 such that for all x e D if \x — XQ\ < 5, then \f(x) — f{xo)\ < e. Hence f is continuous at XQ. 4 This fits right in with the continuous, nowhere differentiable function from Chap­ ter 6. Then we were using the fact that C[0,1] with the sup metric was a complete metric space. This metric is also called the uniform convergence metric/norm. The convergence of a sequence of functions using this metric is equivalent to the conver­ gence being uniform. Are there other properties which are held together by uniform convergence? How about differentiation? Integration? It turns out that the uniform limit of differen­ tiable functions does not have to be differentiable, but the uniform limit of integrable functions must be integrable. fl

EXAMPLE 9.4 Let fn be the function on [0,1] defined by l>|,

\x\>l/n,

We can see that fn converges to the absolute value function and the convergence is uniform since \fn(x) — f(x)\ < l/n. Each fn is differentiable at every point in R while f is not differentiable at zero. ■ Now we turn our attention to integrable functions and a positive result on conver­ gence. Theorem 9.2 If fn : [a, b] —> R is a sequence of Riemann integrable functions and suppose {fn} converges uniformly to f, then f is a Riemann integrable function and

Sa f = lim »^oo C /„. Proof: We shall use the equivalent idea of the Darboux Integral (Definition 3.24) found in Chapter 3 to show that f must be integrable. Again we have an e/3 argument. Let e > 0. Since {fn} converges uniformly to f, there is an N £ N such that \fn{x) — f(x)\ < e/(3(& — a)) for all x <E [a, b] and for all n > N. Thus I

SUP{/W(.T)}

- sup{/(a;)}| < e/(3(6 - a))

and \mf{fN(x)}-m{{f(x)}\<e/(3(b-a))

THE CANTOR FUNCTION

161

where the supremum and infimum are taken over all x 6 [a, b]. Now since JN is an integrable function, let P = {a = XQ,X\,X2, ■ ■ ■ ,Xk-i,Xk b} be a partition of [a, b] with the property that U(/N, P) — L(fN,P) < e/3. Using this partition on f, we obtain \U(f, P) - U(fN, P)\

< | sup{fN(x)} - sup{/(x)}| E K + i - *k)
Similar reasoning gives us \L(f, P) — £(/yv, P)\ < e/3. Thus \U(f,P)-L(f,P)\ <\U(f,P)-U(fN,P)

+

< \U(f,P) - U(fN,P)\ + \U(fN,P) < e/3 + £/3 + e/3 = e.

U(fN,P)-L(fN,P)+L(fN,P)-L(f,P)\ - L(fN,P)\

+

|L(/JV,P)

-

L(f,P)\

Taking the limit as e approaches zero gives us that f is an integrable function. Due to | J g\ < J \g\ and the uniform convergence of fn, we have that {J / „ } is a Cauchy sequence in K. Call the limit of this sequence L. Another application of the Triangle Inequality will show that \ J f — L\ < efor any positive e. This means

la f = l i m n^°° !l fn- * These results answer some questions about interchanging operations (limits, dif­ ferentiation, and integration). They say that for limits and integration, we can switch the order if we know the convergence is uniform; while for limits and differentiation, even uniform convergence is not enough to let us move things around. We will not venture into more here, but we can modify things so that differentiabil­ ity is preserved. We just need more conditions than just uniform convergence. This theorem is stated without proof and comes from [33]. Theorem 9.3 Let {/«} be a sequence of dijferentiable functions on the interval I with the property that f is a continuous function (such functions are called contin­ uously differentiable). Suppose fn converges pointwise to f on I and f'n converges uniformly to g on I. Then f is differentiable on I and f = g. 9.2

The Cantor Function

The first real construction that we will do is the Cantor Function , also known as the Devil's Staircase. It is a function F : [0,1] —» [0,1] which is increasing and onto (so by necessity, F(0) = 0 and F(l) = 1), but essentially flat. There is an explicit definition for F which we will see later. It is more instructive to look at this as an iterative process. So we will define Fn : [0,1] —> [0,1] and show that these do what they need to do. At each step Fn, will be defined on the closure

=

162

CREATING PATHOLOGICAL FUNCTIONS VIA C

of the intervals removed at the nth step of the construction of the Cantor Set along with Fn(0) = 0 and Fn(l) = 1. The rest of the function is made by joining points linearly. Stage 1: Fx{x) = 1 /2 ^ £ [ 1 / 3 , 2 / 3 ] .

'1/4, Stage2:F 2 (x) = < 1/2, ^3/4,

ifxe ifxe ifxe

[1/9,2/9] [1/3,2/3] [7/9,8/9]

7 1 - 1 jf _ J Stage n: Fn (x) = 2 72k l J ^ -Lm,k) where k represents the stage at which the contiguousinterval is removed and m refers to the position (first, second, third, etc.) of the interval removed at stage n ordered left to right on the number line.

The Cantor Function It is (somewhat) obvious that when Fn is defined on the closure of the contiguous intervals from Stage n, then the values on those intervals are determined for Fk where k > n and that each Fn is a continuous function. Less obvious is that the limit function F is a continuous function. This can be accomplished in a few ways. One is to find a way to explicitly describe^ the function F and then show that Fn —>• F. We then use the result that the uniform limit of continuous functions is a continuous function. However, the explicit formula is difficult to work with. Instead we would like to take a page from sequences of real numbers and use the idea of Cauchy sequences and the fact that (C[0,1], sup) is a complete space. Definition 9.3 Let {fn : [0,1] —► R} be a sequence offunctions. This sequence is uniformly Cauchy iffor every e > 0, there is an N £ N such that for all x £ [0,1] ;/ 'For example, let x g [0,1] have base 3 expansion x = 0.12101203 . . . and let A^ be 00 if x = 0 and otherwise N is the smallest number such that ajv = 1. Then define b„ = an/2 for n < N and fejv = 1. Finally, N ,

F(x) = E

THE CANTOR FUNCTION

163

n,m > N, we have \fn(x) - fm(x)\

< e.

Then basically the same theorem as before holds true: If {fn : [0,1] —> R.} is a uniformly Cauchy sequence of continuous functions, then fn converges to a continuous function f in C[0,1] with the supremum distance. So showing that our {Fn} is a uniformly Cauchy sequence is enough to prove that the Cantor Function F is continuous. This is usually done by showing \Fk(x)-Fk+1(x)\<2-k. The graph of F, which we shall denote by Q{F) has some additional interesting properties. Due to the symmetric nature of the contiguous intervals and of the choice of y-values on those intervals the graph is symmetric about the point (1/2,1/2). In addition, there is a self-similar nature to the graph. Despite this look to it, the graph does not have a fractional (Hausdorjf) dimension. We will not prove this, but dim.Haus{G{F)) = 1. The next result we will prove. This is the value of the length

ofG(F).

Theorem 9.4 The graph of the Cantor Function, as a subset of the plane, has length 2. Proof: At the kth step of the construction of C, there are 2h+l endpoints for the contiguous intervals, write them as {0 = X\, x2, x^,..., x2k+i = 1 } - The length of Fk, the approximation at step k, is 2k+i

Lk = Yl [^ - **-i)2 + (^fc) - ^fc-i)) 2 ] 1/2 • i=2

Now on contiguous intervals x2i+\ - x2i = (l/3) fc while Fk(x2i+1) - Fk(x2i) = 0. On the complement of those intervals, F{x2i) — Fk(x2i-i) = (l/2) fc . This means Q{Fk) consists of2k horizontal intervals of total length 1 — (2/3)fc and 2k intervals where Fk increases that have total length [(2/3) 2 + l] 1 / 2 . Then the length ofQ(F) is at least lim 1 - (2/3)fc + [(2/3)2fc + l] fc-4-OO

=2.

L

When a and b are real numbers, [a2 + b2}1/2 < \a\ + \b\. Thus the summation above has the property

E t r [te - Zi-i)2 + (Fk(Xi) - Fk{x^))2}112 2k+1

< Ei=2 Kxi - xi~l) + (Fk(Xi) ~ = {b-a) + {f{b)-f{a)) = 2.

Fk(xi-i))}

Thus the total length of the Cantor Function, which is 2 fc+i

sup ] T [{Xi - x^)2 i=2

+ (Fk(Xi) - F f c ^ . i ) ) 2 ] 1 / 2

164

CREATING PATHOLOGICAL FUNCTIONS VIA C

where the supremum is over all partitions cannot exceed 2 and our proof is done. 4ft We can also look at this function with the ideas of differentiation and integration. Firstly, the derivative of F is nearly always zero. From Calculus I, we know that if F' = 0 everywhere, then the function must be constant. Here the function is not constant, in fact it is onto [0,1], yet the derivative is zero where the graph is horizontal and that happens on all of [0,1] \ C. In measure terms, we say that F'(x) = 0 almost everywhere. At the points of C, F' does not exist. Secondly, since F is continuous on [0,1], it must be integrable there. Given the symmetry of the function, we see it divides the unit square into two equal-sized pieces and that /o F{x) dx = 1/2. This function is also used when the topic of absolutely continuous functions is brought up. This is one of many generalizations of continuity for functions. Definition 9.4 Let f : [a, b] —> R.. Then f is absolutely continuous if for each e > 0 there exists a 5 > 0 such that if {[an,bn]} is any countable collection of non-overlapping closed intervals in [a, b] with Yln°=i(bn — (in) < 8, then oo

£I/(M - / M I

<e

-

On a closed interval, most of our usual "nice " functions are absolutely contin­ uous; e.g. x2, sin(a;), ex. Functions that have a discontinuity are not absolutely continuous. It is not too difficult to see a function from [0,1] into R like

<1A /(*)' = {°' * \l, x> 1/2 V

is not absolutely continuous. If £ = 1/2, then for any 8 > 0 the interval [1/2 — <5/4,1/2 + 5/4} has length less than 5 yet | / ( l / 2 + 6/4) - / ( 1 / 2 - 5/4)\ > 1/2. This is a jump discontinuity. The function f above has a jump discontinuity. The function

fsin(l/x), 5(x) =

1n 10,

i/O, n x = 0

has an essential discontinuity at the origin (an essential discontinuity at XQ means lim x ^ X o f(x) does not exist). This time for any value M > 0 and any 5 > 0 there is a finite collection of closed, non-overlapping {[an, bn}} in [0, 5} with

J2\f(bn)-f(an)\>M. This brings up the topic of bounded variation.

THE CANTOR FUNCTION

165

Definition 9.5 Let f : [a, b] —> K and let P = {a = XQ, X \ , a?2, ■ ■ ■, xn-i,xn be a partition of [a, b]. Then the variation of f with respect to P is n

V(f,P) =

Y,\f(*k)-f(xk-i)\. fc=i

Finally, the variation of f on [a, b] is defined by V(/, [a, b]) = sup{V(/, P) : P is a partition of [a, 6]}. IfV(f, P) < oo, then we say f is o/bounded variation. S

EXAMPLE 9.5 The Cantor Function, F, is an example of a continuous function that is not an absolutely continuous function. ■

Intuitively, we can see this from the fact that we know F(\) — F(0) = 1, but F is constant on the contiguous intervals. Thus all the increasing goes on in the Cantor Set and that has measure zero. So continuous functions do not have to be absolutely continuous. The follow­ ing result tells when a continuous function will also have the absolute continuity property. A function is absolutely continuous if and only if it is continuous and has bounded variation. Now let us look at some further applications of this function. The first is in probability theory. Probability is the measure of the likelihood of the occurrence of an event. For example, if we roll a single die there are six (simple) outcomes, {1, 2, 3,4, 5, 6}. Each of these has probability 1/6, assuming the die is fair. We can summarize this in a chart. Outcome, 0

Probability, P(0)

1 2 3 4

1/6 1/6 1/6 1/6

5 6

1/6 1/6

This is an example of a probability distribution for a discrete random variable. The rules are: (1) For each outcome Oi, P{0{) > 0. (2) ^ P(Oi) = 1, where the sum is over all possible outcomes of the experiment. A random variable can also be continuous. In this instance an event is an interval of values the random variable can take, and the probability of the event is determined

= b}

166

CREATING PATHOLOGICAL FUNCTIONS VIA C

via integration. So f(x) is the continuous density function and for a < b in the domain of f the probability that a randomly chosen X falls in the interval [a, b] is given by

P(a < X < b) = [ f{x) dx. Ja

The Cantor Distribution is a probability density function f(x) with the property that the cumulative distribution function is the Cantor Function F; that is,

F(x)= f

J — oo

f(t)dt.

This is an example of a singular distribution, where all the probability is concen­ trated on a set of measure zero. To see another application of the Cantor Function, we go back for a moment to metrics. Many texts (including this one) have a problem in the homework to show that if d, is a metric on a set S, then so is d 1+ef The purpose of the exercise is to show how any metric space can be turned into a bounded metric space, as the distance between any two points is between 0 and 1. One can look at this in terms of composition of functions. The bounded metric is the composition f o d where f(x) = x/(l + x). This brings us to metric preserving functions. Definition 9.6 Let f be a function from [0, oo) into [0, oo). We say f is metric pre­ serving if for every metric space (S, d), the composition f o d is still a metric on S. fl

EXAMPLE 9.6

x is a metric preserving function. ■ 1+ x There are lots of properties and interesting examples of metric preserving func­ tions, including continuous, nowhere differentiable examples and examples with (ex­ tended) derivative infinity at No many points, but we will just stick with the following: a function is metric preserving iff(0) = 0 and f is subadditive (for all a, b G [O.oo), f(a + b) < / ( a ) + f(b)). In [24], it was shown that the Cantor Function F is subadditive, therefore it is an example of a metric preserving function. The function f(x) —

The concept of a Cantor Function can be generalized using the so-called Fat Cantor Sets. Recall that a Cantor Set can have positive measure if at each stage the proportion removed is tending toward zero. Consider 0 < A < 1. At the kth step, rather than removing an interval of length fc l/3 we now remove an interval of length A/3fc. This is a bit of notational abuse,

THE CANTOR FUNCTION

167

but easier than writing A/3fc. To clarify, when we use the Greek letter X we mean A/3fc is the proportion removed (£(Ij)/£(I), where j = 0,1) at step k, where using the English letter a means at every stage the proportion removed is a. Then the total lengths of the intervals removed in the construction of this Cx is A/3 + A/3 2 + • ■ • + 2 fc - 1 A/3 fc + • • • = A, resulting in a Cantor Set with positive measure. If we fix a A < 1, then we can create a Cantor Function, F\. Again, this function is increasing with F\(0) = 0 and Fx(l) = 1, and F\ is horizontal on the contiguous intervals. Arguments similar to before lead us to the length of F\, A + v / ( l - A ) 2 + l. Note when A = 1 we have the usual Cantor Function and length 2. On intervals of the form [0, x) we conform a measure, px, given by fix([0,x)) = Fx(x). The details of the proof of the length ofF\, which are found in [18], include the fact that Fx{y)-Fx{x)< 1 ~ 1 — A' y—x this means that (ix is an absolutely continuous function. Thus for every measurable (/jx-measurable) set E we have

px(E)=

f F x. JE

Since F^(x) = 0 on the complement of the set Cx, measurable subset of I,

/JLX(I

\CX)

= 0, and, if E is a

where m(E) is Lebesgue measure. Now if E is a measurable subset ofCx, then \i(C — A) = 1 — A and 1

and so m(E)/{\

= = < = =

Lix(I) = v\(CX) + pX(I\Cx) [i\(Cx) = fix(E) + px(Cx\E) JEF'x + m(Cx\E)/(l-X) JEFx + (m(Cx)-m(E))/(l-X) JEFx + l-m(E)/(l-X)

— A) < JE Fx. This proves that Fi(x) =

for almost every point in Cx.

l/(1-X)

168

9.3

CREATING PATHOLOGICAL FUNCTIONS VIA C

Space-Filling Curves

In 1878, Cantor showed that the cardinality of the unit interval, I = [0,1], and the unit square, I x / = [0,1] x [0,1], were the same. In modern notation, we show that these two sets are the same as follows: Obviously, any horizontal line segment in the unit square can be put into 1 — 1 and onto correspondence with [0,1] so n(I x I) > n(I). In the other direction, any point in I can be written as 0.010203 . . . where (since there is a choice) we say any representation that can be written in terminating fashion is instead written with a tail of9's. Then the function P(0.010203 . . . ) = (0.010305 . . . . 0.020406 . . . ) is an injection from I to I x I. Thus n(I) > n(I x / ) and all together n{I) =n(I

x I).

Cantor used continued fractions for actually constructing a bijection (see [34]) and was not his first try at proving this result. Cantor's original attempt was based on the decimal expansion of numbers, but Dedekind showed him that his function was, in fact, not onto. The trouble he ran into was due to some numbers having more than one expansion. The gist of his successful argument was thus: Every real number x in (0,1) can be expressed as a continued fraction 1

x= a\ H

j

, i

where the a,i are all positive integers. The two keys to Cantor's success here is that the continued fraction representation is infinite if and only if x is irrational and, when the representation is infinite, the representation is also unique. He then used a method similar to the function P defined above to show the unit square has the same cardinality as the irrational numbers in (0,1). Finally he showed there is a bijection between [0,1] \ Q and [0,1]. Almost immediately the question of continuity arose. Is it possible to construct a continuous bijection between I and I x I? The answer is no. This was proven by Eugen Netto in 1879. So, as mathematicians do, the question was changed to something similar, but with slightly different hypotheses. The requirement of being one-to-one was dropped. So now the question is, "Is there a continuous function from I onto / x I?" The answer to this is yes, with different constructions coming from mathematical luminaries like Hilbert, Peano, Sierpinski, and Lebesgue. All of these appeared in the early 1900 's. The first construction we will show is due to Hilbert. This is the most well-known of the ways to find a space-filling curve. As with many of these constructions, we create things in steps and the actual object we need is the limit of a sequence of iterations. We start with the unit square. At Stage 1, we bisect each side of the unit square to create four subsquares. Since the square is in four pieces, we divide the

SPACE-FILLING CURVES

169

unit interval into four parts. Each of these subintervals is mapped continuously to one of the subsquares in such a manner that adjacent intervals map to squares that abut (first interval to the lower left corner, second to the upper left, third to the upper right, last to the lower right) still preserving the continuity. This is illustrated via the U-shaped notation in the figure below. This mapping can be done because we know that bijections from an interval to a square exists.

Step 1 We repeat at the next step. Each subsquare is divided into four pieces, the unit interval is divided into 16 subintervals, and we piece things together as illustrated. The next two figures show steps 2 and 3. Our pattern ensures that if subinterval In is mapped to subsquare Qn, then the subsquares that Qn are divided into at the next step are the images of the subsets that In is divided into. This continues inductively.

Step 2

Step 3

So we have a sequence offunctions hn from the unit interval onto the unit square. Let h : I —> I x I be the pointwise limit of hn. Any point (2/1,2/2) G [0,1] x [0,1] is the limit of a nested sequence of subsquares Qn. The inverse images of the Qnform

170

CREATING PATHOLOGICAL FUNCTIONS VIA C

a nested sequence of subintervals in I. These subintervals will converge to a point i £ l and so h(x) = (2/1,2/2)As for continuity ofh, each hn is a continuous function onto I x I and suppose we pick x\ < xi such that \x\ — x^\ < 1/2 2 ". Then the interval [x\, X2} is contained in at most two subintervals from step n. This means \hn(x\) — hn(x2)\ < v 5 / 2 2 n , the diameter of two adjoined subsquares. Thus h is continuous. In [50] and [51], McClure took Hubert's space-filling curve and wrote it as h(x) = (X(x),Y(x)), a pair of coordinate functions. The self-similar nature of the coordinate functions were analyzed, and it was shown that for X and Y the graphs of X(x) and Y(x) d i m / / a u s ( X ) = dimBox(X)

= dimHaus(Y)

= dimBox(Y)

= 3/2.

As mentioned before, Henri Lebesgue (of Lebesgue Measure fame) also produced a space-filling curve. His construction is peripherally related to the Cantor Function (Section 9.2). Lebesgue was interested in the differentiability properties of such a curve. As is known, just because a function is continuous at a point does not ensure that it is differentiable at that point. In this instance, the 2 2 n subintervals of I that are mapped onto I x I to create gn are contiguous intervals from the 2nth step in creating C, the Cantor Set. Also, the four subsquares are not connected by a Ushaped pattern, but by an N-shaped pattern. The end result, g, is a space-filling curve that is continuous and differentiable on [0,1] \ C. Lance and Thomas, in [48], construct a new type of space-filling curve. One feature is that this does not involve any rotations of step k in the construction of step k + 1. Another is that one can see in the construction the making of a Cantor Dust in the plane, thus giving insight to the Lebesgue measure of the arcs in the construction. At each step in the recursion there is a subsquare which will be decomposed into four subsquares. Given that the original square has side length L, the subsquare will have side length s/2 ■ L, where s G (0,1]. The four subsquares are then linked together by line segments as shown:

For the recursion, each subsquare will have a copy of the graph from the previous step where the graph begins in the upper left corner and ends in the lower right, thus

BAIRE CLASS ONE FUNCTIONS

171

giving us a connected arc. We write the recursion operator as Rs (
The function with domain [0,1] that describes this arc is found as follows: On the intervals [0,1/7], [2/7, 3/7], [4/7, 5/7], and [6/7,1] the function maps to the upper left, upper right, lower left, and lower right squares, respectively. The three intervals left in [0,1] are mapped to the connecting line segments to give us a continuous function. The picture above is (of course) from step 1. One continues to iterate, and these functions converge uniformly to a limit function. Because the convergence is uniform, the limit is continuous; and because the construction starts with choosing a value s, we will denote this as fs. Officially the function we have depends on both the step (n) and the width used in R at this step (s). With / s o the diagonal between (1,0) and (0,1), we then have fs,n

= R(l +

s)/2(f2s/(l+s),n-l)

and fs^n converges uniformly to fi- Due to the Cantor Dust nature of this construc­ tion, it can be seen that the Lebesgue measure of the graph of fs is s2. Our ultimate result here is f\, which is a space-filling curve.

9.4

Baire Class One Functions

Since the Cantor Set is in [0,1], in this section we will restrict ourselves to func­ tions with domain the unit interval. These can, through translations, be extended to functions with domain K. The Baire Functions are defined as a hierarchy offunctions starting with Class 0. The Baire Class 0 functions are the continuous functions. A function f is called Baire Class One (or of the first Baire Class) if it is the pointwise limit of Baire Class 0 functions, but not Baire Class 0. This means f is not continuous, but there is a sequence of continuous functions, fn, with fn —t f.

172

S

CREATING PATHOLOGICAL FUNCTIONS VIA C

EXAMPLE 9.7 The function 1,

J/X

= 1

is B\. This is the pointwise limit of fn{x) = xn.

m

We can, of course, generalize this to any finite ordinal number k. We say f is in Baire Class k if it is not in Baire Class j for j < k and there exists /„ in Baire Class k — 1 such that fn —> f. This idea actually goes even further, to the transfinite ordinals, but we will not proceed in that direction. Our notation for Baire Class k functions will be BkS

EXAMPLE 9.8 The function

JO,

ifxt®,

\l,

ifxeQ

is B2- It is the pointwise limit of the function Jl, = <

9n(X)

10,

ifx =

qi,q2,q3,...,qn,

otherwise,

th

where qi refers to the i rational number which is a B\ function for each n. The fact that XQ is not in B\ will be left as a homework exercise. ■ If you recall, in Chapter 2 we looked at the Borel Sets. These are the collection of sets generated from the open sets via union, intersection, and complement. We start with the open sets and then, via complement, the closed sets. Next, a set is called Qg if it is the countable intersection of open sets while a set is Ta if it is the countable union of closed sets. Continuing onward, we call a set which is the countable union of Gs sets will be considered to be in the collection (G$)a, which for convenience sake, we will write at G$a. Following this pattern we have Tas sets being the countable intersection of Fa sets. The classes then make up two sequences yS ? y5a i y5aS i • • • ; J~a 7 J~ad : ^aSa

, ••■ •

For a function f, one way to show that the function is continuous is to look at A\ = {x : f(x) > a} and A-i = {x : f(x) < a}. Then f is continuous iffor all a £ M., both A\ and A2 are open sets. This extends into the Baire Classification of functions on the real line. It can be shown that f £ B\ implies that both A\ and A2 are J>. At the next level, f £ B2 means that A\ and A2 are both Q$a. The rest of the hierarchy (at least for finite n) can be extrapolated by the reader. For ease in determining what functions are or are not in Baire Class One, we shall rely on the theorem below.

DARBOUX FUNCTIONS

173

Theorem 9.5 The following are equivalent. • f is in Baire Class One. • For any closed set, D, the restriction of f to D, f\r), has a point of continuity. • For any closed set P and real numbers a < b, at most one of the sets {xeP:

f{x) >b} {xeP:

f{x) < a]

is dense in P. S

EXAMPLE 9.9 Let C denote the Cantor Set. The function

(l,

ifxec,

Xc(x) = < lu,

otherwise

is in Baire Class one, while the function H(x)Ji,

ifxeC\E, lO,

otherwise

where E = {x : x is a left endpoint of a contiguous interval ofC}, is not. This can be seen by using the second part of Theorem 9.5. Given any closed set D, if D\C 7^ 0, then\c is continuous at a point in that set difference. IfD C C, then the function is constant when restricted to D. However, for H, the restriction H\D has no point of continuity since E is dense in C and C is dense in E (this is saying the sets are co-dense). ■ 9.5

Darboux Functions

In Calculus I, we have the Intermediate Value Theorem which states, "If f(x) is continuous on a closed interval [a, b] and f(a) ^ f(b), then for any C between /(a) and f(b), there exists at least one c E (a, b) such that /(c) = C." A function need not be continuous for this to be true. Let us start then, by turning the Intermediate Value Theorem into the Intermediate Value Property. Intermediate Value Property A function f has the Intermediate Value Property (IVP) iffor any two points a < b with f{a) ^ f(b) and for any value C between f(a) and f(b), there exists a c £ (a, b) such that /(c) = C. A function with the Intermediate Value Property is called a Darboux ("Darboo") function. Typically the set of Darboux functions is denoted by V.

174

CREATING PATHOLOGICAL FUNCTIONS VIA C

For awhile, it was thought that being Darboux was equivalent to being continu­ ous. This is not true as the following simple example shows. S

EXAMPLE 9.10 I sin 1/x, i / O , The function f (x) = < is not a continuous function, yet has I 0, x = 0 the Intermediate Value Property. ■

As we can see, this function has a special type of discontinuity. This is not a jump discontinuity, but an essential one, for every y E (liminf^^o f(x), n m s u P : c ^ o f(x)) there is a sequence xn converging to 0 such that f(xn) = y. While this is a good example, we can to better. It is straightforward to have a function with two points where it is not continuous, yet is Darboux. Just use y = f(x) + f(x — 1). This idea can be extended to any finite set of points. We can even extend this to an uncountable set of points, the Cantor Set. This is a good example because we can write it as the pointwise limit of fn and show pictures of some of the steps. fl

EXAMPLE9.il Let / i be the function which is zero everywhere except the first interval removed in the Cantor construction. On the interval [1/3, 2/3] the function is linear, join­ ing the points (1/3, —1) and (2/3,1). The next function fa looks like /\ except on the two intervals removed at Step 2 (1/9, 2/9) and (7/9, 8/9)) there are line segments connecting the left endpoint and y = — 1 with the right endpoint and y = 1. This process continues indefinitely.

The third iteration, fs We can actually write the limit function explicitly.

DARBOUX FUNCTIONS

175

otherwise, where (a, b) is an interval removed during the Cantor Set construction. Using Theorem 9.5 we can see this is not a Baire Class One function. The Cantor Set is a closed set, but f\c has no point of continuity. We could convert this to a Baire Class One function by making the linear parts be on the open intervals (a, b) and therefore f\c = 0, however, this function is no longer Darboux. ■ Still this function is continuous almost everywhere. Is it possible to keep the Inter­ mediate Value Property yet the function not be continuous? Yes, to do so just requires switching our numbers to base 2. S

EXAMPLE 9.12 Let x € [0,1] and let x be written in base 2 as x = 0.010203 ..., where a^ € {0,1} and if x has two representations we use the one that ends in a string of 0's. Now define f(x) by f(x) = limsup

~y^Qfc fc=l

This function has the property that on every nontrivial interval (a, b), f takes on every value between 0 and 1. We'll give a heuristic argument for this. For any y € [0,1] the tail of the decimal expansion is what influences the f(x) value and gives us the ability to have f(x) = y. The beginning of the decimal expansion then is what can be fixed in order to place x in the interval (a, b). ■ When we define a new collection offunctions, it is natural to ask if there is some old collection offunctions which is either equivalent or contained in this new idea. Here we show that a very familiar set offunctions is a subset ofTJ, the collection of derivatives. Be careful here, we are not saying that the differentiable functions (those f for which an / ' exists) are Darboux, but that the derivates (the / ' themselves) are inV. Theorem 9.6 Iff is a derivative, then f is the pointwise limit of continuous functions and / has the Intermediate Value Property. Proof: If f is a derivative, that means there exists a function F such that for each x in the domain of f, we have

lim n^h)-F(x) /i-s-o

= K

h

'

Since F is differentiable, then it must also be continuous. So let fn(x)=n{F(x

+

l/n)-F(x)).

176

CREATING PATHOLOGICAL FUNCTIONS VIA C

Each fn is continuous and fn —» / . In some books, this is considered proof that derivatives are in Baire Class One, but since we are saying that continuous functions are in Baire Class 0 and Baire Class 1 functions do not contain Baire Class 0, we need to be a bit careful. As for IVP, suppose that we have two values a and b and that without loss of generality we have f(a) < f(b) (if f is constant, there is nothing interesting going on as far as intermediate values go). Take a C between f(a) and f(b). The function G(x) = F(x) — Cx is continuous on [a, b] and differentiable on (a, b). Furthermore, G'(a) < 0 < G'(b). Since G is not linear (why?), it must reach an extreme value in the interior of [a, b}. So there must be a c, with a < c < b, such that G'(c) = 0. This gives us f(c) = C as needed. 4 This gives us a quick way to discount some functions as derivatives. For example,

X{0}{X)

=

fl, \0,

x = 0, x^O

cannot be the derivative of any function as it does not satisfy the IVP. So derivatives are a subset of both Baire Class One and Darboux functions. There is a lot of classical real analysis devoted to the functions in both categories. These are the Baire One, Darboux (B\D) functions. When put together, these two proper­ ties lead to some interesting results. First, some equivalences. Theorem 9.7 Let f : [a,b] —» ~R be a Baire One, Darboux function. This is equiva­ lent to ■ For each x, there exist sequences xn and yn where xn < x < yn and lim f(xn)

n—>oo

= f{x) = lim n—>oo

f(yn).

• The graph of f is connected as a subset of [a, b] x R ■ The sets A\ = {x : f(x) dense in themselves2.

> a} and A2 = {x : f(x)

< a} are (bilaterally)

Now we can proceed with an example to show that there are more than just deriva­ tives in B\ V. fl

EXAMPLE 9.13 Begin with the Cantor Set and {(an.b„)} the set of intervals removed in the construction. Define cn = (an + bn)/2, the midpoint of the intervals. We define

2

A set E C K is bilaterally dense in itself if for any e > 0 and any x € E, both (x,x + e)C\ E ^ % and (x — e, x) n E 7^ 0; that is, for any x in E, there are points in E that are arbitrarily close to x on both the left and right sides of x.

LINEARLY CONTINUOUS FUNCTIONS

177

f(x) to be 0 on C, arbitrary on cn, and linearly on [an, cn] and [c„, bn\. Then f £ B\D (any of the equivalencies in Theorem 9.7 will do). However, f is not continuous at x = 0 unless liniu^oo f(cn) = 0. Thus we can make sure that f is not a derivative. I Another pathological example can be found in [10]. There a Baire One, Darboux function is constructed which is zero on all but a measure zero set. A property which is true except on a measure zero set is said to be true almost everywhere, abbreviated a.e.

9.6

Linearly Continuous Functions

For this section, we move off of the real line and look at functions whose domain is in the plane with the Euclidean metric. For f : R2 —> R to be continuous at the point (xQ,yo), it means that for any e > 0 there exists a 6 > 0 such that for all (x,y) 6 M2 with d((x, y), (xo, yo)) < 6 we have \f(x, y) — f(xo, 2/o)| < £• Thus there is an open disk around (xo, yo) with radius S so that everything in the disk is mapped e—close to /(xo, yo). There is an alternate way to look at this involving sequences. Definition 9.7 Let f : R2 —> R. We say that f is continuous at (xo, yo) iffor every sequence {(xn, yn)} C R 2 that converges to (XQ, yo) we have f{xn,yn)

-> f{X0,V0)-

Since it is difficult to look at all possible sequences, this is not often used to prove continuity. Instead, this definition is mostly used to show a function is not continuous at a point. For example, the function

f(xv) = {^i" [0,

(*'»>* <0'°>' (x,y) = {0,0)

is discontinuous at (0,0) because the sequence (1/n, 1/n) converges to (0,0), but

/ ( l / n , l / n ) = l.

The function f above is the canonical example of a special type offunction, the separately continuous function. Definition 9.8 Let f : R 2 —>■ R. For a fixed value ofx0, the function fXo : R —> R defined by fXo (y) = f(x, y) is called the x-section of f at XQ. Similarly there is the y-section at yo, fyo (x). We say that f is separately continuous if each fXo and fyo is a continuous function for all Xo,yo £ ROur example f is separately continuous. The only possible trouble spot being at the origin. However, fo(x) = 02_? J2 if x ^ 0 and 0 at the origin, so trivially continuous. The same is true for fo{y).

178

CREATING PATHOLOGICAL FUNCTIONS VIA C

The generalization of this idea would be to have f\e, the restriction of f to a fixed line £, be a continuous function. This is what is known as a linearly continuous function. It is here that we will bring in our Cantor Set. That said, as with many constructions, we will start small and build our way up. We begin by defining the function g : M2 —> R by

g{x, y)

0, y/x2, x2/y, 1,

x 0 0 y

= 0 or y = 0, < x2 < y, < y < x2, = x2.

This function is continuous everywhere except the origin. At the origin is it linearly continuous (intuitively, for any line through the origin, if we look close enough the line is not involved with the parabola y = x2 and so the parabola does not interfere with the continuity at the origin). Let (a, b) be any point in the plane. We can then create a function with the dis­ continuity at (a, b) by using translation and reflection. Let go(x, y) be zero on the lines x = a and y = b. If a < x and y < b, go(x, y) = g(x — a,y — b). To the left ofx = a and above y — b, find go by reflecting go about the vertical line. Finally, on the rest of the square, go is the reflection of what we have so far about the horizontal line y = b. Now let {(an, 6„)}^Lj be any countable collection of points in the plane. Then each of the functions gn(x, y) = go(x — an, y — bn) is continuous everywhere except (a„, bn), where it is linearly continuous. We can then define the function G by oo

G{x,y) =

1

J2^9n{x,y). 71=1

Since each gn is bounded by 1, the Weierstrass M-Test 3 gives the convergence is uniform. We know that G(x, y) — gn(x, y) is continuous at (x, y), and this means that G is linearly continuous at (an,bn). Thus we can have a linearly continuous function with countably infinitely (even dense if we want) points of discontinuity. Can we have uncountably many such points? Yes, but for that we will need a Cantor Set. First, for a, b £ K with a < b, let (a, b) be an interval in the real line and let M denote the midpoint of that interval. Let Fo(x, y) be the function which is 0 outside of {a, b) x M., 1 on the parabola y = (x — b)2, and, on the strip bounded by [M, b), FQ(X, y) = while Fo(x, y) —

3

(x - b)2

if x is above the parabola,

y

=— if x is below the parabola and above the line y = 0. (x — b)2

Let {Mfc} be a sequence of nonnegative real numbers such that "^2 Mk < oo. If \gk(x)\ < M^ for all x 6 S, then Y2 9k converges uniformly on S.

LINEARLY CONTINUOUS FUNCTIONS

179

The values for FQ on the other three parts of the strip are found via reflection about the x-axis and x = M. This FQ is continuous everywhere in the place except at (a, 0) and (b, 0)) where it is linearly continuous. Letting (a\, b\), (02, 62), («3, bs),... be the intervals removed during the con­ struction of the Cantor Set, C. Let Fn be defined as above with respect to the interval (an,bn). The function „,

,

\Fn{x,y), I 0,

x £ (an,bn), otherwise

is continuous everywhere except C x 0, an uncountable set. This can be extended to a set dense in the plane, via C x q where q G Q. See [79] for the details.

180

CREATING PATHOLOGICAL FUNCTIONS VIA C

EXERCISES 9.1 Draw the graph of Fi{x), i = 1, 2,3, where F^ refers to the functions converge to the Cantor Function. 9.2

Show why fn = xn does not converge uniformly on the domain [0,1].

9.3 Prove that the sequence fn(x) converges uniformly if D = [0,1].

— x/n convergespointwise

9.4 Explain how if we know \Fk(x) — Fk+i(x)\ {Fk} is a uniformly Cauchy sequence. 9.5

which

if D = [0, oo), yet

< 2~k, then we can claim that

Provide the proof for Example 9.8 showing that \Q is in Baire Class 2.

9.6 Let D he a finite set in [0,1]. Show that XD(X) is Baire One. (Proof by picture is not rigorously legitimate, but it is okay here.) 9.7 Prove that \c is a Baire One function (thus showing that Baire One functions can have uncountably many points of discontinuity). 9.8

Divide the Cantor Set into two pieces: A = {x : x is an endpoint of a contiguous

interval}

and B =

C\A.

Show that for any a £ A and any £ > 0 there is ab £ B such that \a — b\ < e. Then show this is true if the roles of A and B are reversed. This means that A is dense in B and B is dense in A (that is, they are co-dense in each other). 9.9

Divide the Cantor Set into two pieces: A = {x : x is an endpoint of a contiguous

interval]

and B =

C\A.

Prove that XA is not a Baire One function. 9.10 Draw a graph of the Cantor Function F on the contiguous intervals ( 1 / 3 , 2/3), ( 1 / 9 , 2 / 9 ) , and (7/9, 8/9). 9.11

Explain why F has the Intermediate Value Property.

9.12

Prove that F is not B\.

CHAPTER 10

GENERALIZATIONS AND APPLICATIONS

There are many ways to tweak a Cantor Set. The literature is rife with papers on these ways and their consequences. Many of these results are intriguing, but highly technical. Hence in this chapter we present several short, and mostly "proof-less" sections. These are ideas which may serve to whet the appetite for further study by the reader. The author strongly encourages such studies, whether in class or out.

10.1 Generalizing Cantor Sets Our first generalization of the Cantor Ternary Set was seen in Chapter 3, where we dropped the "ternary" requirement. This construction begins with a fixed number r, 0 < r < 1 and at each step in this process we have an interval I and from that interval we remove a middle subinterval J so that £(J)/£(I) = r. This left two subintervals behind. We wrote this Cantor Set by Cr. This next generalization appears in [14]. We still take an interval and divide it into three pieces, keeping the outer two, but this time the ratios of the kept pieces do not have to be the same. Specifically, each closed interval I is partitioned into 1$, O, The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

181

182

GENERALIZATIONS AND APPLICATIONS

and 11, where I and IQ share a left endpoint, I and I\ share a right endpoint, and

with r\ + r2 < 1. Then K0

=

[0,1],

Kx

=

[0,ri]U[l-r2,l],

K2

=

[0,rf]u[r1(l-r2),r1]U[l-r2,l-r2(l-r2)]U[l-r1r2,l].

As before, each Krl is the union of2n disjoint closed intervals. We then set Cri t,.2 = n^L0-K„. This is again a perfect, uncountable, nowhere dense set in the unit interval. The most obvious question, as we had in the beginning, is, "How much is taken away in this method?" Well, informing K\ we remove from KQ an open interval of some length, call it r. Making E2 we remove two open intervals, one of length r\T, the other of length r2r. Continuing, at Step n+l we remove from the set En exactly 2" open intervals: one has length T\T, then n open intervals of length r™~lr2r, next n(n — l)/2 of length r\r2x^2T, etc., ending with one interval of length r2r. The total of these sum is ^J'

j=0

This is at Stage n. Thus the total length removed becomes oo

^

l-(ri+r2)

as a convergent geometric series. This set satisfies the Open Set Condition, and therefore the Hausdorff dimension can be found easily as the solution to r« + r«

= i.

We can use a similar idea to find the Besicovitch—Taylor dimension ofCri
f

\

/ \ ■ i 3=0 \J<

T

r

i

r2

=

rrj

where ri = r\ + r^. Thus sum of all the terms is r' 3 (1 + 77 + rj2 + r/3 + • • •), which simplifies to ?7< 1, ■q > 1.

GENERALIZING CANTOR SETS

183

This idea was further generalized by Baekfor what he referred to as Perturbed Cantor Sets . Let {an} and {bn} be real sequences where 0 < an,bn < 1/2. As always, let I® = [0,1]. At step 1 we take away an open subinterval, leaving behind the closed subintervals Too and loi where

KIo)=a^ndKLj-h and define E\ — loo U IQ\. At stage k we have a set Ek that consists of the union of sets Ia, where a is a string ofk 0's and 1 's (we will also say this as a s {0, l} fc and the length of a, \a\, is k). For stage k + 1, each Ia is divided into Iafl U O U I„tl where O is an open interval and

—

=ak+1and1(I-y=bk+l.

Then Ek+1 = U\a\=k+\I
For our results here, we make the assumption that an,K,dn

= 1 - (an + bn) > a > 0.

In [3] and [4], these sets were investigated in terms of (Hausdorff) measure and dimensions. In order to present these results, we introduce two new functions: n s

fc (C0„,fcn) = l i m i n f T T K + &£) = liminf V fc=l

*\I°),

\a\=n

and n

Qs{Can,bn) = limsup Y[(a% + bsk) = limsup V n

^°°

fc=l

t(Ia).

\a\=n

With these we can then define the upper and lower Cantor dimension for C 0 i i j n by dim

cSS5F( C '«»,6n) = S U P { S > ° : QS(Can,bn) = oo}

and dimcaritor(Can,bn)

= SUp{s > 0 : hs (Can,bJ

= Oo}.

Now we can present our first result. Lemma 10.1 Let {sn} be the sequence defined by solutions to a^1 + 6^" = 1. Then 0 < liminf sn < d\ra.cantor(Can,bn)

< ^mcantor(Ca„,bn)

< lim sup s„ < 1.

184

GENERALIZATIONS AND APPLICATIONS

Proof: Since {an} and {&„} are bounded away from 0 we know that inf sn < sups„ < 1. Now choose s and s' such that 0 < s < s' < lim inf sn. Then there exists an N such that for k > N, Sk > s' > s. We can choose a number j5 where 1 > j3 > a^, bkfor all k and this gives us n

n

n

k=N

k=N

k= N

This gives us liminf'n^oo rifc=/v(afc + ^fc) = °°- Thus hs(Caii_t,„) = oo ifO < s < liininfs n . This means lim inf ,sn < dinic a „ t o r (C a ? i _{,,„). In a similar manner, we can prove qs{Canbn) = 0 for t > lim sup sn and arrive at dim^r^-^:(C an :f,?i) < lim sup sn . 4 Recall mass distributions from Chapter 7. We can apply this concept to our con­ struction for Can ,6„ and get our results in measure and dimension. For the dimension result we see that Cantor dimensions are just a new way of looking at Hausdorff and packing dimensions. These are the following two theorems. Theorem 10.1 dinica„tor(Car,,b,:)

(Ca„,6 n ).

and Theorem 10.2 dim Contor ((?„„, b J = dimp(C a , lj6n ). In terms of measure, [4] shows the equivalence of hs(Canib„) andT-l{C'a„.b„) via two results which we combing here into one theorem. Theorem 10.3 Ifhs{Can^rl) = oo, then'H{Can,bn) ~ oo and if 0 < hs(Ca„,b„) < oo, then 0 < "H(Ca?Mbn) < oo. Now we move on to more than one subinterval, all with the same size. Let m, be an integer greater than 2 and fix r* so that 0 < r* < 1/m. The change to the steps in the construction is that at each stage k, the closed interval I is replaced by m equally spaced subintervals of length r*\I\, the ends of the far left and far right subintervals coinciding with the left and right endpoints of I. It is an exercise in arithmetic and geometric series to show that with fixed m and r* the set will be measure zero. As far as dimension goes, in [30] it is shown that the resulting set F has I n 777,

dim W a u s (F) = dmiBox(F) = — — lnr* and

0
Below is the picture when m = 3.


FAT CANTOR SETS

10.2

185

Fat Cantor Sets

In the homework in Chapter 3 we saw one example of a Cantor Set that had positive measure. We looked at another type offat construction in Chapter 9. This construc­ tion was used in both [8] and [18]. Below is an interesting result from the former paper, but first we repeat a construction we have for this type of set. Pick A G (0,1]. At stage k of the Cantor Set construction, rather than removing 2k subintervals of length 3~k from the middle, we remove middle subintervals of length A/3fe. Then the total length removed is A/3 + 2A/3 2 + • ■ • + 2 fc - 1 A/3 fe + • • • = A. We will denote such a set as C\ (A bit ofnotational abuse, but easier than writing C\/3k. When we use the Greek letter A we mean A/3fc is the proportion removed at step k, where using the English letter a means that at every stage the proportion removed is a.). When A < 1 the set C\ must have positive measure. The question asked and answered in [8] was about the type of numbers that are captured by these Cantor Sets. Define the set [x] as [x] = {A e [0,1] : x e Cx}. It turns out that these sets are small in terms of category. Theorem 10.4 For each x, the set [x] is a closed, nowhere dense subset of[0,1]. Proof: For any point x G C\, we can think ofx as the limit of a nested sequence of closed intervals. The intervals come from keeping either a left or right subinterval at each step. For X fixed there is a one-to-one correspondence between the point x in C\ and the elements S G <S, where S is the set of subsets of the natural numbers. This is done by saying for x G C\ the positive integers n is in Sx if x was in a right subinterval at Step n. When A G (0,1] we use <j)\ to be the mapping taking Sx to x. When A = 0 (then Co = [0,1]) there can be more than one subset ofN which corresponds to an x. For instance, the point x = 1/2 corresponds to {1} and {2, 3 , 4 , . . . } . Still we can define Co. Fix a set S C N. We define two functions on S: T](S) = 2 ^2neS 3 _ n and KS) = Enes 2~n with VW) = A*(0) = 0. Then JJ(5) = d and fi(S) = C0. We note that V ( 2 ~ n - 2 - 3 - n ); = - - - . , *-" V 3 2 Kn
186

GENERALIZATIONS AND APPLICATIONS

SoifliS^%, then r](S) < /i(S) while i f l € S / N , then p(S) < r]{S). After Step 1 of the construction, we have two intervals of length I\ = 2 _ 1 (1 — A/3); after Step 2, we have four intervals of length I<± = 2 _ 1 (I\ — A/3 2 ). In general, Step n has 2n intervals of length In

=

2-1(/n-i-A/3")

= =

2~ n (l - A/3)[l + (2/3) + ■ • • + (2/3)"- 1 ]) 2 - n ( l - A ) + 3-"A.

Now ifnG S € S and ifx&C\ xn+1 -xn=In

is the point that corresponds to S, then + A/3 = 2-™(l - A) + 2 • 3 " n • A,

where xn denotes the left endpoint of the interval that contains x at the nth stage. So x = 4>\(S), where \ = (1 — \)ji + Xr). Thus 4>\ takes S onto C\. Since

sup{|0 Q (S)-<MS)|} = |a-/9|/6 the set [x] is a closed subset of [0,1]. Define a linear ordering < on S as follows. We say E < F if there exists a positive integer n such that En^\ = Fn-\ and En C Fn, where Hk = H D {0,1, 2 , . . . , k}, H G S, and k > 0. We proceed to use this to show that [x] is nowhere dense in [0,1]. Suppose a, (3 € (0,1) and 4>a(E) = x = 7(G), 4>1{Fn)) is a component of'[0,1] \ C 7 . Addition­ ally, 4>p(G) < 4>p{Fn) < x < 4>a{G) < 4>a(Fn). Thus, as as 7 moves from a to (3, the continuous way f/7 changes between Ua and Up shows that there are 7 between a and (3 such that x ^C1. 4* This result shows that if A is a countable subset of (0,1) that L)X£A[X} is a first category subset of [0,1]. As a consequence, the set of rational numbers in (0,1) produces irrational Cantor Sets.

10.3

Sums of Cantor Sets

As we saw earlier, in Chapter 8, a very interesting result occurs when adding together two Cantor Ternary Sets; namely C + C = {x + y : x,y € C} = [0,2]. One immediate consequence is that a set which is small in terms of measure and category can sum with itself to be large in both of those terms. In this section we

SUMS OF CANTOR SETS

187

look at some applications and generalizations that come from this addition. These results come from sources such as [38], [45], [54], [56] and [57]. For our first investigation we switch up our point of view. Normally we use Cr to denote a Cantor Set where the proportion removed from the middle of an interval is r. Associated with this r is a value X, the relative length of each subinterval retained. So A = (1 — r)/2. Then for 0 < A < 1 we have a Cantor Set C\ defined by Cx

e[0,l]:x

{l-\)Yiai\i\,

=

where aj £ {0,1}. We want to consider sums of these Cantor Sets, C 7 + C\, in terms of Hausdorff dimension and Lebesgue measure. As we have seen before, dim HausV-^X (CA and so if

In 2 ln(l/7)

In 2

RVA): In 2

+ ln(l/A)

<1,

then d\mHaus(C1

+ C\) < dim/f a u s (C 7 ) + dimHaus(C\)

< 1.

This implies that p(Cj + C\) < 1, and thus the sum of these Cantor Sets cannot have interior. In the other direction, we use the Gap Lemma from [60]. This tells us that if 1

I-27

A >1, 1 - 2A

then C1 + C\ must contain an interval. However, these two inequalities do not cover all of the square [0,1/2] x [0,1/2] (the region between the two graphs below). The sum of Cantor Sets with subscripts from this region is unknown as far as interior is concerned. For most of the points in the region, things do work out nicely where most is in terms of measure and nicely in nonzero measure. This result is from [70].

188

GENERALIZATIONS AND APPLICATIONS

Theorem 10.5 For any 7 with 0 < 7 < 1, there exists a set E7 C (0,1/2) with /x(£' 7 ) = 0 such that ifX^E^ and In 2

M1J1)

+

In 2

RVA)

>

*'

//zen p,(C1 + C\) > 0. In [13], the authors are answering questions concerning nonmeasurable sets. As intuitively stated in their paper, The Cantor middle-third set, when added to itself, gives an entire interval, [0,2]. So certainly there exists a measure zero set that when added to itself gives a nonmeasurable set. These results use the concepts of Bernstein Sets and Transfinite Induction . So there is some background information we must go over before we can proceed to results. A total order on a set S is a binary relation (usually written <) on S with the properties of totality (for all a,b G S either a < b or b < a), antiy symmetry (a < b and b < a implies a = bfor all a.b G S), and transitivity (for all a,b,c G S, if a < b and b < c, then a < c). The set S is well-ordered if it has a total ordering on it and every non-empty subset of S has a least element in this ordering. The set Z + with < is a well-ordered set while IR+ with the same relation is not; (0,1) is a non-empty subset ofW+ with no least element. Ordinal numbers are the sizes one can have for a well-ordered set. These include: the nonnegative integers { 0 , 1 , 2, 3 , 4 , . . . } ; the value to, the first transfinite ordinal and the size of the whole numbers; w + 1 (we will not go into the details here, but U> + 1>1+UJ

fl

= u>), LU + 2, ..., bj\, u>2, ■■■ y OJ^; and more.

EXAMPLE 10.1 The natural numbers N in their usual ordering is a well-ordered set of type to with the property that every number except 1 has a predecessor. Another well-ordering of the natural numbers is 1,3,5, 7 , 9 . . . 2 , 4 , 6 , 8 , . . . . Here, the order type is LO+UJ and neither the numbers 1 nor 2 have a predecessor. Thus order types are not unique. ■

A

EXAMPLE 10.2 In the plane, M2, we usually do not think of ordering elements, comparing ( — 1,2) with ( 2 , - 1 ) . However, working in full generality, there is a way to take ordered pairs (a, b) from A x B and put a total order on things, assuming

SUMS OF CANTOR SETS

189

A and B are themselves totally ordered sets. This is the lexicographic order. We say that (a, b) < (a*,b*) if a < a* or (a = a* and b < b*).

Transfinite Induction is an extension of mathematical induction to sets that can be well-ordered. These sets can be beyond countably infinite, in which case a proof by Transfinite Induction usually has three parts: • The Zero Case: P(0) is true. • The Successor Case: Let a + 1 be the successor for a. We must show that if P(a) is true (or possibly P{0) is true for all ft < a), then P(j3) is true. • The Limit Case: If X is a limit ordinal (a limit ordinal has no predecessor; that is, there is no a such that a + 1 = A), then P(X) follows from P{0), /3 < X. fl

EXAMPLE 10.3 For an example of applying Transfinite Induction we present a result due to Sierpinski: There exists a set X in M? such that for any line £, £ n X contains exactly two points. Start by well-ordering the set of all lines in the plane (there are c, the size of the continuum, of these) in such a way that each line has fewer than c prede­ cessors. Let £a be the line associated with the ordinal number a. To create X, assume that for each (3 < awe have a set Xp with the property that iff < j3 the line £j intersects the set Xp in exactly two points and no line £ intersects Xp in more than two points. Additionally, we insist that (1) iff < /3, then X1 C Xp, and (2) each Xp has less than c many points. Define Ya by Ya =

Up
Then Ya contains less than continuum many points and ya^£ contains at most 2 elements for any line and is exactly two for every j3 < a. If the cardinality ofYa n £a = 2, then let Xa = Ya. If not, since there are less than c points in Ya n £p where /3 < a and continuum many points on a line, we can find a point on £a which is not a point on £p, (3 < a. Adding one or two (however many is necessary) points to Ya will then create our Xa. Either way stage a is complete. We now have a set Xa defined and let X = \JaXa. ■ As stated, we also need the concept of a Bernstein Set. A Bernstein Set is a set B CM. with the property that for any uncountable closed set, S, in the real line both BnS and Bc n S are non-empty. It is not obvious that Bernstein Sets must exist, so we shall go through the construction. In order to do so, we need two lemmas, which we present without proof.

190

GENERALIZATIONS AND APPLICATIONS

Lemma 10.2 Any uncountable closed set in K has cardinality continuum. Lemma 10.3 There are continuum many uncountable closed subsets in K. Now on to the construction. Begin by well-ordering the real line. Call this order­ ing -<. If we use c denote the cardinality of the continuum, via Lemma 10.3 we can order all the uncountable closed subsets ofM. via ordinals smaller than continuum, {Fa : a < c}. By Lemma 10.2 each Fa has cardinality c. Using -< choose the first two elements in FQ, call them XQ and yo- Let X\ and yi be the first two elements of F\ (again using -<) that are different from XQ and yo. Now suppose a < c and that for each /3
-Up
is non-empty since Fa has cardinality c while the cardinality of\Jp
G D x D x Pa x Pa : a < c}

such that for each a < c ca=aa

+ ba and {dp : (3 < a} n {Aa+1 + Aa+i)

=0

SUMS OF CANTOR SETS

191

where Aa = Up . This is the set o/subsums of the series. Also, let oc

rn = 5Z

afc

'

k=n+l

the nth tail of the series. The following facts have been produced several times. The first time seems to be in 1914 [41]. • E is a perfect set. ■ E is a finite union of closed intervals if and only if an < rnfor n sufficiently large. E is an interval if and only ifan < rnfor all n. ,

Ifan>

rnfor n sufficiently large, then E is homeomorphic to the Cantor Set.

Now let us look at the sum of copies ofE. Using the notation from [57], we let ®mS = {si + s2 + s 3 H

h sm : Si e S}.

This is the algebraic sum of S with itself m times. Applying this type of sum to the set E, we have the following theorem. Theorem 10.7 There is a positive integer rnfor which (BmE is a finite union of intervals if and only if limsup — < oo. rn Moreover, the smallest positive integer rnfor which ®mE is a finite union of intervals is the smallest integer m such that an/rn < rnfor all but a finite number of integers n.

192

GENERALIZATIONS AND APPLICATIONS

Proof: Let m be a fixed positive integer. We construct a new sequence {c„} by c

= c(q-l)m+2

(q-l)m+l

= ■ • • = Cqm — aq

for q = 1,2,3,.... We can easily see that the ^ cn is a convergent series with 0 < c n + i < cnfor all n, and ®mE is also the set of sub sums of"^2 cn. Let oo k=n+l th

c

be the n tail of Y2 n- Then by the second bullet item above, (BmE is a finite union of intervals if and only if cn < rnfor all but finitely many index values. This inequality is clearly true if rn ^ nq, for some q. If n = mq, then cn = aq and fn — mrq. Thus ®mE is an interval if and only if aq < mrq for all but a finite number of integer values of q. The conclusion of the theorem now follows, tfk With only small changes to the proof we can state this next corollary. Corollary 10.7.1 The smallest integer m such that an/rn < m for all n is the small­ est integer for which ®mE is an interval. The Cantor Ternary set is the set of subsums ofY2 &n, where an = 2/3 n . Then an

rn

2/3"

ZZn+i

2/3fc

•

Thus, again, C + C = (B2C is an interval. Again, let E be the set of subsums of^an. This time we wish to look at sets of the form aE + bE. Letting s = Y2 an> the following are easy to see.

• E=

s-E,

• aE + bE = aE + (—b)E + bsfor any real numbers a and b, • aE + bE = a(E + (b/a)E) for any real numbers a, b with a / 0. The result of these facts is that we can transform aE + bE in such a way that we only really need to study the set E + xE, where 0 < x < 1. Now we come to a new theorem. Theorem 10.8 Assume an+i/an

< x < I for all n. Then

1. E + xE is an interval if and only if (a„ - r n ) / r n _ i < x < rn/(an for all nfor which an > rn.

- rn)

DIFFERENCES OF CANTOR SETS

193

2. E + xE is homeomorphic to the Cantor Set if {an ~ r „ ) / r „ _ i < x < rn/(an

- rn)

for n sufficiently large. Proof: Construct a new sequence {cn} for which C2n-i = an and C2n = xanfor n = 1,2,3,... and note that cn+\ < cn. Then E + xE is the set of sub sums of the series ^2 cn- Again let DO

k=n+l th

the n tail of ^2 cn. As before, E + xE is an interval if and only if cn < fn for all n. Now r^n = (1 + x)rn and r fn for n sufficiently large. Hence E + xE is homeomorphic to the Cantor Set if xan > (1 + x)rn and an > xan + (1 + x)rn for n sufficiently large. 4* 10.4

Differences of Cantor Sets

The difference between two Cantor Sets concerns itself with sets of the form Cr, the so-called middle Cantor Set where in the construction each closed interval, I, is divided into three subintervals and the middle open interval, whose proportional length isr-£(I), is removed. Of use here is the value s, defined by s = (1 —r)/2. This is the proportional length of the intervals kept and allows us to write the construction fo Cr as the invariant of the two contractions T\(x) = sx andTi{x) = sx + (1 — s). Note here that TQ(0) = 0 and T\(1) = 1, showing that the interval removed is, in fact, in the middle. Now given two values a and b, the Cantor Sets Ca and C\, are both measure zero, both uncountable, and both nowhere dense. To separate these, we look at their set differences with themselves Ca-Ca

= {x-y

:x,y e Ca).

A second way to think of this is via translation. For a fixed t, Ca + t = {x + t : x € Co]. Then we can write the difference as Ca-Ca

=

{t:Can(Ca+t)^(D}.

1 94

GENERALIZATIONS AND APPLICATIONS

The first result, straightforward and not needing proof, isthatCa — Ca C [—1,1]. The rest of the results in this section follow the ideas that the "more" that is in a set, the more often that set should intersect itself; and ifa>b, then Ca has less to it than Cb. If the idea of less/more is measure, then we get this theorem which does a small bit of segregating for r values. Theorem 10.9 If s < 1/3, then Cr — Cr is a Cantor Set with measure zero. If s > 1/3, then Cr - Cr = [-1,1]. The next result shows that under the correct circumstances we know exactly how many points of intersection there are. This result holds only for some values of s. Theorem 10.10 For any s £ (0, y/2 - 1) and any natural number n, let

<—"(£)■ Then the intersection ofCr and Cr + tn will contain exactly 2™ points. There are intersections that are singletons. To describe those, we define the set A r = 1Cr — 1, the set difference of Cr and 1 — Cr. We then arrive at a lemma on intersections. Lemma 10.4 Choose s 6 (0,1/2). IfCr fl (Cr + t) is a single point, then t G A r . Continuing, let us define a set A^ as the union of all the images of A r under the transformation n-l

T(x) = snx + ^vls\l

- s),

i=0

where n is any positive integer and V{ is either — 1, 0, or 1. Some things to point out:

■ Afl c [-1,1], ■ Aft has measure zero, and • AR contains all of the points ofCr — Cr for which Cr n (Cr + t) is a finite set of points. Now that we know something about the intersection being finite, we turn to the question of how often that happens. When s is large enough, we would think that the Cantor Sets are big enough for the intersection to not be so sparse. That turns out to be correct. Theorem 10.11 If s > 1/3, then CT D (Cr + t) is a Cantor Set for almost all t 6

[-1.1].

PRODUCTS OF CANTOR SETS

195

We close this section with a result that is not for almost all t, but for all t. It looks at when the difference contains a Cantor Set. Theorem 10.12 Ifs e (V% - 1,1/2). then Ca n (C 0 +1) contains a Cantor Set for allt€ (-1,1). There are more results of this type to be found, using thickness from Chapter 8. See [46]. 10.5

Products of Cantor Sets

Let A and B be sets in Rn and Rm, respectively. The Cartesian product of A and B is the set of ordered pairs for which the first coordinate is an element of A and the second coordinate an element of B. Formally we have, Definition 10.1 Let A C K" and B C M.m. The Cartesian product of A and B is the set given by Ax B = {(a,b) :aeA,beB}. The set Ax B is a subset ofM.n+m. All of these ideas below can be further generalized from R 2 to higher dimensional spaces, but we will stay in the plane. S

EXAMPLE 10.4 Let A = [0,1] C R, the unit interval. Then A x 1/2 is a horizontal line and ■ A x A = {(x,y) : x,y G [0,1]} which is the unit square in the plane.

\M EXAMPLE 10.5 Let C denote the Cantor Set in the line. Then C x C is called the Cantor Dust. We will denote this set in M2 by D. A picture of the construction of D is below. ■

1 96

GENERALIZATIONS AND APPLICATIONS

Let us look at some properties of products and, by doing so, properties of the Cantor Dust. Theorem 10.13 Let A J c l a measure zero set in M2.

both with measure zero. Then their product Ax

B is

Proof: First we note that if [a, b] and [c, d] are intervals in the real line, then [a. b] x [c, d] is a rectangle in K 2 and the area of the rectangle is the value of the (outer) measure of the rectangle; m*([a, b] x [c, d}) = (b — a) ■ (d — c). Now since A and B are measure zero sets, given e > 0 there exists finite collections of open sets {Ui} and < \fe. {Vj} in the line such that A C UUi, B C UVj and J2m*(Ui),Jl'm*(Vj) Then Ax

B c \J(Ui

x Vj)

and

m*{A x B) < J2m*(Ui x Vj) < J2rn*(ui) -J2m*(Vj) Since e is arbitrary, this shows the outer measure of Ax measurable with (Lebesgue) measure zero. 4k

< £-

B is zero and such sets are

Corollary 10.13.1 The Cantor Dust has measure zero. The result above does generalize to measurable sets with positive measure: The Cartesian product of two measurable sets is a measurable set and the measure of the product is equal to the product of the individual measures. We now turn our attention to the various notions of dimension. Many of these results can be found in [30] and are presented without proof. For the first example above, when the product of the unit intervals is the unit square, it is obvious that d i m H o n a ( [ 0 , 1 ] x [0,1]) = dim/, o ,„ s ([0,1]) + dim// a ,, i s ([0, l]); however, this is not the usual result. The best we tend to get is a lower bound. Theorem 10.14 Let A c t " and B C Wn. d\mHaus(A

Then

x B) > dimHaus(A)

+

dimHaus(B).

In his book, Falconer shows a construction of two sets E and F, both with Hausdorff dimension zero, while d\mHaus{E

x F) > 1.

In order to get an upper-bound result, we must replace one of the Hausdorff di­ mensions in the right side with Box-Counting Dimension.

CANTOR TARGET

197

Theorem 10.15 For any sets A C l " and B C WLm dimHaus(A

x B) < d\mHaus(A)

+ dimfcox(jB).

This leads to the following corollary: C R m , ifdimHaus{B) Corollary 10.15.1 For any sets AcRnandB then dmvHaus{A x B) = dmiHaus(A) +dimHaus(B). S

=

d\mbox{B),

EXAMPLE 10.6 For any two Cantor Sets, Cr and Cs, we have dimHaus(Cr

x Cs) = dimHaus(Cr)

+ dim f f a u s (C s ).

Specifically, for the Cantor Dust, D, we obtain dim H<XUS{D) = 2 dim# a u s (C) = £ | = 1.261.... . If we use just (Upper) Box-Counting Dimension, then the inequality we get is a reverse of the one we have for Hausdorff dimension. Theorem 10.16 For any two sets AcRn dimbox(A

andB C Mm

x B ) < dimbox(A)

+

dimbox(B).

Proof: It is straightforward to see that if A can be covered by Ns (A) sets of diameter d and B can be covered by N$(B) sets of diameter S, then Ax B can be covered by the NS(A)Ns(B) "squares" formed by the product of these sets. Using the laws for logarithms and some simple algebra we arrive at \n(NS(AxB)) , ]n(NS(A)) ,. hmsup —^ ^-^r—JJ- < hmsup , ; " + limsup ,5^0+ -ln(5) ^0+ -ln(S) s^0+

\n(N5(B)) , -ln(5)

.\/;

which is what we need. 4 10.6

Cantor Target

A slightly different way to take the Cantor Set from M. to R 2 is via the Cantor Target. This is created by rotating the Cantor Set about the origin. Specifically, we can write the target, T, in polar coordinates via T = {(r,0)

:reC,0<e<2Tr}.

This set is used in the investigation of electromagnetic waves through a self-similar aperture. The one result we shall show is determining its Hausdorff dimension.

198

GENERALIZATIONS AND APPLICATIONS

Theorem 10.17 The Cantor Target, T, has Hausdorff dimension given by d\mHaus{T)

= 1 + ——. In 3

Proof: Let f : R 2 —>■ R 2 be the function f{x, y) = (x cos j/, y cos x). This mapping has the property \f(x,y)-f(x',y')\<\(x,y)-(x',y')\ where distance in R 2 is measured by the Euclidean metric. This property d{g(a),g(b)) 0) is called the Lipschitz property and g would be called a Lipschitz function. The target is then T = f(C x [0, 2TT}). This gives us
= =

dim f f a u s / ( C x [0, 2n}) < dimHaus(C x [0, 2TT]) dimHaus C+ dimHaus[0, 27r] = j ^ | + 1.

In the other direction, if f is restricted to the domain [|, 1] x [0, n], then f~x is also a Lipschitz function. NowT is a subset o / / ( C n [ | , l ] x [0, 7r]) and so dimHaus(T)

> = =

10.7

d i m / / a u s / ( ( C n [§, 1]) x [0, TT}) dimifous((Cn[§)l])x[0,7r]) dimHaus(C

n [|, 1]) + dim ffaus ([0, TT]) = £ f + 1. 4

Ana Sets

Recall from Chapter 8 the set difference definition of a Cantor Set:

C = I\\JOi, i>i

where I is a finite closed interval and {Od\i > 1} is a (finite or infinite) collection of disjoint open intervals contained in I. In this section, we introduce a sequence

ANA SETS

199

and show how these sequences can be used to create sets that fit this definition of a Cantor Set. We begin with what we call an Ana Set and then proceed to the Golden Ana Set. The Ana Set is based on a verbal sequence. We begin with the letter "a" in stage 0. In stage 1, we replace the "a " with "ana ". In the subsequent stages every "a" is replaces by "ana" and every "n" is replaced by "ann". This is really just pronouncing each letter as "An a" or "An n". Thefirstfour stages are a ana ana ann ana ana ann ana ana ann ann ana ann ana Pictorially, this can be illustrated by starting with the closed unit interval. At stage k the unit interval is divided into 3fe equal-sized subintervals (corresponding to the number of letters at that stage). The intervals corresponding to an a are kept, and the open intervals corresponding to an n are removed. Ai= A2 = ^ 3 =

A4= Ana Sets (and then Golden Ana Sets) were introduced in [63]. In the article by Pe ([62]), he answers the question, "At stage k, how many a's and n's are there?" There are (3 fc " 1 + l ) / 2 a's and (S^1 - l ) / 2 n's. The Golden Ana Set is created by bringing up the following misconception: Since n is a consonant, then perhaps it should be said as "A n". Thus the Golden Ana sequence starts with "a" and there after every "a" is replaced with "ana" while every "n" is replaced by "a n". Thefirstfour stages here are then a ana ana an ana ana an ana ana an ana an ana He then shows that at stage k the number of a's is a(k) = F(2k ~ 1) and the number of n's is n(k) = F(2k — 2) where F(k) is the kth Fibonacci number. We then have a(k) _ F ( 2 f c - 1 ) , _ l + \/5 njkj ~ F(2k-2) ~*^~ 2 ' the Golden Ratio, hence the name of the set. For the Golden Ana set thefirstfour iterates look like

200

GENERALIZATIONS AND APPLICATIONS

G3=. Cx4 =

Corresponding to each of these constructions, there are three sets which we can then define. These are • The intersection of the Ak,

A = nf=lAk • The limsup of the sets: A = limsupk^,^ Ak = n j ^ (U^ = f c A m ), which is the set of points in infinitely many of the Ak• The liminf of the sets: A = lim inffe^oo Ak = U^L1 (C\™=kAm), the set of points in all but finitely many of the Ak• And the corresponding G = C\Gk, G = lira sup Gk, and G_ = liminf GkThe sets A and G are of the form [0,1] \ (uOj), hence Cantor Sets. The set A is a familiar set, it is the Cantor Ternary set, C. The limsup and liminf are different. Looking at the pictures we can see that later stages reintroduce points that were take out previously; e.g. there will be points in (1/3,2/3) despite this being the first interval removed. In [40], Jones was able to characterize some of the more interesting numbers that are in G. His main result was Theorem 10.18 Let Fj be the j t h Fibonacci number, Lj the j t h Lucas numbers, and 4> represents the Golden Ratio. Then the following numbers are all in the Golden Ana Set: L ■ ' i

-L-, + 1

3+1

£i>

^j+1

1

^

£23 A f p M 2j + l '

Since the Ana Set and the Golden Ana Set are both Cantor Sets, it is natural to want to characterize them in terms of some of the "usual" ideas about Cantor Sets. This was done in [72]. We start with the Golden Ana Set, since the Ana Set coincides with the Cantor Set. Theorem 10.19 The Golden Ana Set is a porous set in R. As with most applications of porosity, the result gives an immediate corollary. Corollary 10.19.1 The Golden Ana Set is both measure zero and nowhere dense. Proof: This is a consequence of the following: For any k 2 <

_ ^

< 3

•f'2fc-2

AVERAGE DISTANCE

201

so each interval I is divided into three pieces and the gap created is at least the same size as the sub-intervals on either side which came from I. That is, given e > 0 there is an m such that 3/F2 m < e < l/-p2m-2 and this means that in the interval (x,x + 3/F 2 m ) there is a gap of length \jFim which is in the complement of G. Thus we get our result. 4b Moving on to limsup's and liminf's we obtain some differing results. Theorem 10.20 The set A, A, G, and G_ are each dense in [0,1]. Let us sketch the proof of this. At each stage in the construction the unit interval is divided into 3n (or F(2k)) pieces. At least one of every three pieces relates to the letter "a" and so contains points that will be in A, A, G, or G. As the number of pieces increase without bounds this means the points in these sets will be dense in the unit interval. The liminf's turn out to be small sets. Somewhat expectedly. Theorem 10.21 The sets A and G are each a-porous sets. Hence they are measure zero and first category. However, when we turn our attention to the sets that come from limsup, these turn Q

out to have some largeness to them. The complements A being a-porous. Hence

Q

and G are small, both

Theorem 10.22 The sets A and G have full measure and are residual in [0,1]. One counterintuitive result on Cantor Sets is how such a sparse set can be added to itself to form a large set (an interval). We close this section with some summation results for Ana and Golden Ana Sets and an open question. Now A C A C A and A + A = [0, 2] so any sum of two of those sets gives an interval. The results are different for Golden Ana Sets. The sum G + G is not an interval. For example, G + G misses all of the points between 11/24 and 1/2. However, using what is called normalized thickness, 7(G) = T ( G ) / ( 1 + T(G)), it has been shown that

G + G + G + G= [0,4]. Nothing is known about either G + G + G or the sum ofG_ and itself. However, since G has positive measure, the famous theorem of Steinhaus proves that G+ G contains an interval.

10.8 Average Distance In [37], the authors introduce a new way to distinguish self-similar sets from one another and use Cantor Sets in many of their examples. Unlike the last section, this method can work for sets where an interval is divided into more than three sections and more than two of the subintervals are retained. For example, they introduce two

202

GENERALIZATIONS AND APPLICATIONS

generalized Cantor Sets. In the first, called M\, is created using an iterative process where one starts with the closed interval [0,1], divides it into 16 equal subintervals, and then keeps only subintervals numbered 1, 5, 11, and 16. The second, Mi, uses the same division of the unit interval, but discards all but pieces 7, 2, 75, and 16. Thus Mi contains larger gaps than M\. However, in terms of measure (measure zero), category (nowhere dense), cardinality (uncountably many points), and dimen­ — dim.Box(Mi) = 1/2), they both agree. Thus the need for a sion (dimtfaus(Mi) new indicator of size. We begin with the usual Cantor construction, Cr, where the unit interval is divided into three pieces and the middle one removed, and r refers to the proportion removed. Now fix an r. At each stage, Kn, in the construction of Cr there are 2™ closed subintervals and we can compute the distance between points, T>(Kn). The average distance for Cr is then the limit of this calculation. Definition 10.2 Let CT be a Cantor Set. The average distance between points for C, is [CK- W \X ~ v\dx dy r -u i V(Cr) = lim V{Kn) = lim ■ " * " * * " ' / ' = L±i n->oo TWOO JJK axay r +6 K Thus the usual Cantor Set has average distance between points given by V{C) = \ . With r fixed, the Cantor Set Cr is guaranteed to be a measure zero set. There are Cantor Sets with positive measure, the so-called Fat Cantor Sets. What we need for a fat set is a variable ratio r^ at stage k and r^ —>• 0. In this instance, we will fix r and at stage k remove a middle interval of proportion r . We will write this as C\.k. This type of set has an average distance formula. Theorem 10.23 The average distance between points of the set Crk is 3(2 - r) As we have seen with Cantor Dust, it is possible to take these ideas into higher dimensions. Our notation here will be that C\ = C r C l and C " + 1 = CT x C™ C M™+1. Before we can continue we must define distance as these higher dimensions admit different generalizations of the Euclidean distance. We are using the taxicab metric; recall, in R 2 x = {x\, x^) and y = (y\, 1/2) and d(x,y)

= \x2 - xi\ + Ij/2 ~V\\-

Then it turns out that Theorem 10.24 IfC™ is the n-dimensional

v{cr;

Cantor Dust, then

'r + V r+3

NON-AVERAGING SETS

203

As for the motivating examples Mi and M2, the average distance for these can be computed using results on self-similar sets. This is Corollary 9 in [37], using the fact that the similitudes are linear function on M with a fixed contraction ratio. Theorem 10.25 If {Ti, T 2 , . . . , T/v} is a set of contractions on [0,1] satisfying the Open Set Condition with the added property that Ti = ax + bi, then for the invariant K we obtain T,(K\ 1

;

=

2

^l
2

~

b

i)

N -Nr

This leads us to T>(M\) = 25/63 andT>(Mi) = 29/63. This gives us quantitative evidence that the points in M% are more spread out than the points in M\. 10.9

Non-Averaging Sets

Let S be a set of real numbers. We say that S is non-averaging iffor any two points a,b £ S, their arithmetic mean, (a + b)/2, is not an element of S. For a fixed natural number n, it is easy to come up with a non-averaging set containing n elements. An active problem in number theory is as follows: Estimate the maximum number v(n) of elements in a non-averaging subset of {1,2,3,..., n}. It is currently known that cnl/1° < u(n) < cn2/s, where c is some constant, but the exact fractional exponent is not known. The Cantor Set is almost a non-averaging set. Keeping in mind the base three expansion of points in the set, for x,y E C we can write x — Y2ai/3l and y = ^2 bi/y where a,i, bi is either a 0 or a 2. If k represents the first natural number where k, then in base 3 the expansion of z is z = O.ci . . . Cfc_il000 ..., which is the same as z = 0.ci...c f c _i0222... giving z 6 C. This point, having a terminating expansion, is a endpoint of a con­ tiguous interval. Thus we can get the following example \9 EXAMPLE 10.7 Let C* be the Cantor Set with the endpoints of the contiguous intervals removed. Then C* is an example of an uncountable non-averaging set. ■ This is extended further by Foran in [31]. In this paper he constructs a nonaveraging set which in one way is very large (it has Hausdorff dimension 1), while in many ways is small (strongly porous, hence measure zero and nowhere dense).

204

GENERALIZATIONS AND APPLICATIONS

To prove the dimension is large he using the theorem below. Theorem 10.26 Let {nk} be a subsequence of the natural numbers and let n(m) be the number of j < m which are not in the subsequence. Let f be a function which takes each finite sequence of digits into a digit. Suppose E is the set of x such that if x = Yl fli/101, then for each k, ank = f(a,i, a-2,..., ank-\). Suppose there is a number a and a natural number N such that for each m > N, n(m) > ma — N. Then dimHaus(E) > a andUa(E) > 0. Notice that not only does this say the Hausdorff dimension is 1, but it also says the Hausdorjf measure is positive. This is a special circumstance as there are many sets with dimension 1, but measure 0. Now we give Foran's construction. It is Cantor-like in that blocks of decimals are made 0 representing intervals being removed. Let S = {Np} — {10p} be our subsequence of natural numbers and note that Np+i > Np + 2. Let E(S) be the set of all x G [0,1] such that if

where a* = 0 , 1 , . . . , 9 and s(j) = j(j + l ) / 2 and bj = 0 , 1 , , . . . , s(j) — 1 and the a,j and bj satisfy for each x G E{S), i. bn + bn+l ■ ios = YZ~=\h)when

n

= Nv>

2. an+i = 0 when n ^ Np, 3. bn+2 = 0 when n = Np. then E(S) is a non-averaging set. By item 2 the elements ofE(S) can be averaged by doing so to the blocks bn of their decimals. By item 1, ifx, y,z G E(S) and z = (x + y)/2, then x = y = z follows from the fact that each of their block are equal. By item 3, E(S) is a porous set. For ifx = Y.h3l®~s(i)> n = Np, a = £™ + 1 bjlO'8^, then x G [a, a + 10"S("+2)] and (a + KT S ("+ 2 ), a + Kr^™ + 1 )) n E = 0. Hence the porosity of E at x is at least lim n->oo

jQ-s(n+l) _ lQ-s(n+2) -,——r = 1. 10_s(n+1)

Now fix m and suppose n, K are such that n(n + l)/2 > m > (n — l)n/2 and s{10K) <m< s(10K+1). Then n(M > m — n — 3(10 i f + 1 — l ) / 9 — 2>K since at most n of the a,j are fixed by item 2 and at most 3(10^ + 1 — 1) + 2>K ai are fixed by items] and3 Thusn(m) > m - v / 2 m - 3 ( 1 0 \ / 2 m / 9 + l/21n(2m)). Leta G (0,1); then n(m) — am > m(l — a) — \/2m — 3(10\/2m/9 + l/21n(2m)). Since the right side of the inequality approaches infinity, there is an N such that for m > N, n(m) — am > —N. Thus the set E(S) satisfies the hypotheses of the theorem for each a G (0,1) and hence d\ra.Haus{E{S)) = 1.

CANTOR SERIES AND CANTOR SETS

205

10.10 Cantor Series and Cantor Sets As stated Chapter 1, Cantor worked in several areas of mathematics including in­ cluding set theory, number theory, and trigonometric sequences. Let us introduce a new topic here, the Cantor Series. The idea behind series representations is to take an x £ (0,1) and un(x) so that oo n=l

where each un(x), n = 1, 2 , . . . is a function of x which is one-to one. An example such a series (not a Cantor Series) is as follows: Define un(x) as un(x) = en(x) ■ 10 _ n with en(x) a non-negative integer such that 0 < en(x) < 9 for each n and there are infinitely many n such that en(x) ^ 0. Definition 10.3 Let B — {bj} be a sequence of integers where bj > 2 for all j and let ej = ej(x) be integers satisfying 0 < ej < bj - 1, Then if ej ^ Ofor infinitely many j , then

for j = 1,2,

oo

.

. &1&2 * • -&j

3 — i-

is the Cantor Series of x. We note here that these x will be irrational numbers in (0,1). From this series idea we can then define a Cantor Set Ea by I

oo

Ea = I x — y

—-;—-—— : 0 < a,i < 6j_i, ai + 0 unless a,_i = 0

with the understanding that oo ^ 0. This can then bring us back over to continued fractions. For each Cantor Set E there is a corresponding a that has a regular continued fraction representation given by 1 a & i - l &2-1+T6 3 - 1 + " .

In the paper [81] several properties of these sets and continued fractions are established. Also worth pointing out, this paper links these ideas to applications in the study of the quasi-crystal spectrum, high energy physics, and E-infinity theory. For example, they state that the fractal dimension of spacetime can be expressed as 4+ . ,

l

l

=[4:4,4,4,...].

206

GENERALIZATIONS AND APPLICATIONS

First, let us look at a new, recursive way to describe the continued fraction sion of an x 6 (0,1). Let T : (0,1) —> (0,1) be defined by T(x)

= x

where |_VXJ is the integer part ofl/x.

Then we can write 1

Tk~la where T°a = a and Tka

expan­

6fc - 1 +

Tka'

T(Tk-1a).

=

As we recall from Chapter 4, a simple continued fraction [oo : a\,a,2,Q>3, ■ ■ •] has a,i}, which in rational number form is written —, convergents Ci ~ [CLQ : ai,a^,..., Pi is an integer and qi a positive integer. Then the values of the numerators and denominators satisfy the equations Pi

=

qi

=

diPi-i+Pi-2,

algi_i+ft_2

for i — 2, 3, 4, 5 , . . . with initial values Po = a0,

p\ = aia.0 + 1,

<7o = 1,

qi

=ai.

We can now write a in terms of the transformation T as a

pk + qk +

Tapk_i Tkaqk^

for any k. The main result in [81] is the following theorem on the Hausdorff dimension of these Cantor Sets. Theorem 10.27 For any integer sequence {bn} with bi > 2, let a and T^a be de­ fined as above. Then the Hausdorff dimension of the Cantor Set Ea is An(a-Ta-T2a---Tk-1a) a i m f f a u s Ea = hm mf ■ fe-voo — ln(6i • 62 • • • bk) Although a little off topic, something we have covered already is the construction of continuous, nowhere differentiable functions. In [77], Wen gives another such construction that is accessible to first year calculus students and uses a Cantor Se­ ries.

LIOUVILLE NUMBERS AND IRRATIONALITY EXPONENTS

10.11

207

Liouville Numbers and Irrationality Exponents

A number is algebraic if it is the root of a polynomial with integer coefficients. Thus 3, 5/2, and \f2 are all algebraic numbers. If a number is not algebraic, then it is called transcendental. As easy as examples of algebraic numbers were, examples of transcendental numbers were difficult to find. In 1844, Joseph Liouville established Liouville numbers and became the first to prove the existence of transcendental num­ bers. Definition 10.4 A real number x is called a Liouville number if x is irrational and for each positive integer n there exists integers p and q such that <

1

where q > 1. Some definitions call for x to be a real number, but no rational number can be a Liouville number. This is easy to see. Suppose there exists integers c, d with d > 0 so that x = c/d. Choose a positive integer n so that 2n~l > d. Then for any integers p, q with q > 1 and p/q ^ c/d we have cq — pd 1 ~ ~ dq dq

1 1 > r 2 ~q " q n l

Thus a rational x cannot be a Liouville number. In [58] we see a proof that every Liouville number is a transcendental number. Since we have looked at various notions of size in Chapter 3, we will take a moment to look at the set L of Liouville numbers in terms of category and measure. Theorem 10.28 The set of Liouville numbers is second category set, yet of measure zero. Proof: To begin with, since every point in L is irrational we will write it as

L = Qcn (n„G„) where the Gn are open sets made by the following union of open intervals: Gn = U~ 2 U~-co (p/q - l/qn,p/q

+

l/qn).

Obviously all rational numbers are in each Gn hence each Gn is dense in the line. Each Gn is also an open set. Thus G% is nowhere dense. The complement of L is then Q c U (UG£), a first category set. To look at measure, we will show that for any positive integers m and n we can contain L n (—m,m) is a sequence of intervals whose length adds to less than (Am + l)/(n — 2). Thus the total length can be as small as we desire.

208

GENERALIZATIONS AND APPLICATIONS

ForfixedTO,n L n (-TO, m)GGnn

(-TO, m) C U~ 2 U™J_m, (p/q - l/qn,p/q

+

l/qn).

This means L n (—TO,TO)/ias an open cover using intervals whose lengths total to

E ^ £r=-m, 2 /«" = E ~ 2(2^9 + l)(2/g») < E ~ 2(4mg + g)(l/g") = (4TO + 1) £ ~

2

l / 9 n _ 1 < (4™ + 1) /i°° ^1_™ dx = ^ ± 1 .

Tn« gives us L fl (—m, m) liaj measure zero, hence L is measure zero. 4 77?e irrationality measure of a number is a measurement of how the value can be approximated by a rational number. Definition 10.5 For any real number x the irrationality exponent of x, /x(x) is the supremum of all real numbers p such that 1 P < — q qn has infinitely many rational number solutions, p/q. For a rational number, x, the irrationality measure is p(x) = 1. The ThueSiegel-Roth Theorem states that if x is irrational, but algebraic, then p(x) — 2. Louiville numbers are those numbers whose irrationality exponent is infinite. In [12], Bugeaud applied this idea to the Cantor Set to answer the question, "How close can irrational elements of Cantor Sets be approximated by rational number in Cantor's Set?" We present his answer, without proof, in the next theorem. Theorem 10.29 Let T be a real number with r > 2. Then the Cantor Set, C, contains uncountably many elements whose irrationality exponent is equal to r. In fact, for X > Oifwesetno(T,X) = 1 + max{0, |_(ln(2/A))/lnrJ}, then the real number

n>n ( )fr,A)

is an element ofC with p{rjT^\) = r ([-\ is the greatest integer function). Although we are not presenting the proof of this, we will note that it makes use of the so-called Folding Lemma ofMendes France [32] which deals with continued fractions using only the numbers 0 and 1. These relate to the peaks and valleys one gets when folding a sheet of paper, hence the name. Regardless, this furthers the connections between the topics in this book.

SETS OF SUMS OF CONVERGENT ALTERNATING SERIES

10.12

209

Sets of Sums of Convergent Alternating Series

Let an be a sequence of numbers where an = 0 or 1, and let {bn} be a fixed sequence of real numbers. We wish to look at all the possible series we can create via the product of(—l)a" and bn; that is OO

71=1

and the collections

5=JE(-l)a"^:K}G{0,l}N|. It is known that ifbn < J^'kLn+i ^k, then S is just a closed interval. The results in [22] show that under some circumstances the set S will be a generalization of a Cantor Set. There are several assumptions we will place on the situations. To begin with, S depends on the choice of bn, but not the sign of bn- Therefore we may assume bn > Ofor all n. If ^ bn = oo and lim^^oo bn = 0, then the Calculus II result on rearranging series shows that S — R. Hence we just need to investigate absolutely convergent series so assume oo

Y2 bn < oo. 71 =

1

One property of absolutely convergent series is that the sum is unchanged if the terms of the series are reordered. Thus our last assumption is h > b2 > b3 > ■ ■ ■ > bn > ■ ■ ■ > 0,

the series of terms is nonincreasing. Let us now relate this to a Cantor-like Set. We begin with the usual construction: At each step in this process we have an interval I and from that interval we remove a middle subinterval J so that £(J)/£(I) = 1/3, where £(■) refers to the length of an interval. Our first generalization, creating a new Cantor Set, was done by insisting the middle interval removed have ratio £(J)/£(I) = r, where 0 < r < 1. We will denote this Cantor Set by Cr. Now, rather than keep r constant, we use the sequence {r„}. For explanation's sense, let us label the two subintervals left after J is removed. If I = [a, b] and J — [c, d], then the left/right intervals remaining will be [a, c] and [d, b], respectively. This, again, then gives us Kn where each Kn consists of2n disjoint, closed intervals and CTn = C\Kn. The Ternary Cantor Set has rn constant at 1/3 and we have seen before that if rn —> Owe have a Fat Cantor Set, one with positive measure. In this generalization, we loosen the restriction on rn to let it be negative. So we will assume rn £ ( — 1,1) for all n G N. We now need an additional requirement: There is an increasing sequence of integers n^ such that rnk > 0.

210

GENERALIZATIONS AND APPLICATIONS

Our new construction is as before with the exceptions that we now say £(J)/t(I)

= \rn\

and, ifrn < 0 then the remaining left/right intervals will be (again, respectively) [a, d] and [c, b\. Still we have Kn, although this means Kn and Kn+\ Crn = C\Kn. We now have the following result:

can be the same set, and

Theorem 10.30 Let {rn} be a sequence of numbers in (—1,1) such that there is a subsequence {rHk} which is positive. Then one of the following is true: 1. C is a Cantor-like Set. 2. C can be written as a closure of a countable union of disjoint closed intervals. The proof will be omitted. It is straightforward, based on what we already know about a Cantor Set when rn > 0. There is a question of uniqueness here. Is it possible for two sequences rn and rn such that CTn = Cfn ? The answer is yes, for example with

r„ = {1/7,1/3,1/7,1/3,...} and fn = {-3/7, -1/5,1/3, -3/7, -1/5,1/3,...}. Our attention turns back to the set S. The statements of the results and their proofs are quite long and detailed, but can be found in [22]. Theorem 10.31 Let {bn} be a nonincreasing and positive sequence satisfying oc

b

^2

n>

bk

'

k=n + l

Then the corresponding set of sums S where S= lreR:r

= Y2{-l)a"bnforsome{an}

€ {0,1} N I

is Cantor-like. More precisely, the starting interval I of the construction is I

J2bk'Y.bk fc=l fc=l

and the positive sequence {r„} is defined by bn - 2^k=n+l b

T,T=n k

"•

n = 1,2,3,.

THE MONTY HALL PROBLEM

211

Conversely, if a > 0, I = [—a, a], and {rn}, rn e (0,1) then there is a correspond­ ing Cantor-like set C. For this C there exists a unique decreasing positive sequence {bn} satisfying bn > E f c l n + i bk Siven h

2 whose set of sums is C. If we reverse the requirement on the tails of the sum ofbn, then we end up with an interval rather than a Cantor-like set. Theorem 10.32 Let {bn} be a nonincreasing and positive sequence

satisfying

oo

bn < ^2

bk

k=n+l

-

Then the corresponding set of sums S where S = I r G E : r = ] T ( - l ) a " & n for some {an}

e {0,1}N I

is the interval oo

oo

■ Hfefc>J2

h

fe=l fc=l

More results can be found in various papers by Dindos and by Menon. T. Saldt has also published many papers on this topic, but in Slovak.

10.13

The Monty Hall Problem

As we have stated many times before, one of the more intriguing properties of the Cantor Set is its pervasiveness. It often shows up where it is least expected. In this section, we see how a variation of the Cantor Set appears in a generalization of a famous recreational math problem. First, the original problem. The Monty Hall Problem You are a contestant on the game show Let's Make a Deal. The host, Monty Hall, shows you three giant boxes. Behind one of the boxes is a brand new car, the other two contain "zonks," gag prizes worth nothing. You do not know which box hides the car. You choose one of the boxes to be yours. Monty then reveals one of the boxes you did not choose and shows you the zonk inside that box. Now you are given a choice. Keep the box you originally chose, or switch to the other box whose contents are unknown. The question is, is it better to keep your box or change prizes?

212

GENERALIZATIONS AND APPLICATIONS

The answer, which surprises many including mathematicians and statisticians, is you should always switch boxes. The probability of getting the car is doubled by switching. We shall not go into explicit detail, although the interested reader may like [66]. The heuristic argument is easy enough to follow. When the contestant makes his/her first choice, Box 1, Box 2, or Box 3, there is no extra information and we are looking at a random guess. Thus the probability of choosing the box with the car is 1/3. The fact that Monty reveals a non-winning box does not change this. So when there are two boxes left, the probability that the other box contains the car is 1 — 1/3 = 2/3, hence the probability of winning is doubled when you switch. There are many articles written on this topic and variations on the game. A par­ ticularly good one for undergraduates is [67]. We will look at a problem from [59]. In a TV contest, the host randomly hides a single prize in one of several boxes. The contestant chooses a box and then the host - who knows where the prize is hidden - picks a box that is not the one the contestant chose, opens it, and shows the contestant and the audience that the box does not contain the prize. The empty box is removed and the contestant is given the choice of remaining with the chosen box or picking a new box. If the contestant chooses a new box, the remaining boxes, including the one previously held by the contestant, are jumbled randomly. This process continues until there are only two boxes left: the one the contestant holds and one other. The contestant is then offered one last chance to change. The question, as always, is what are the probabilities of winning associated with each box? If the contestant always stays with the box originally chosen, then the probability of having the winning box is 1/n. On the other end of the spectrum, if the contestant holds onto the original box until the last possible moment, and then switches, the probability of having the winning box is 1 — 1/n = — - . Of course, when n = 3 this is the Monty Hall Problem and we have the previous probabilities ofl/3and2/3. Let us now look at the probabilities associated with other strategies. During the course of the game, there are n — 2 offers to switch boxes. This means 2™~2 differ­ ent strategies which can be represented by a finite sequence of decreasing positive integers {n, a,k, flfc-i, • • •, a,2, ai} where n is the number of boxes at the start of the game, k changes were made, and a,i denotes that a change of boxes was made when there were ai boxes with which one is left to switch. For example, {5, 3,1} means the game started with 5 boxes and changes were made twice, once when there were 3 boxes left, and once when there was 1 box left (note: this 1 box, and the previ­ ous 3 boxes, are the number of choices of boxes with which to switch; this means if a i = 1 there are two boxes, one the contestant is holding, the other with which the contestant can swap) giving 02 — 3 and a\ — 1. Two items worth pointing out in our notation • cik 7^ n — 1. One box is held by the contestant, one box has already been revealed to be empty, thus the number of boxes from which the contestant can choose is at most n — 2.

THE MONTY HALL PROBLEM

213

■ n > cik > fflfe-i > ■ ■ ■ > ci2 > di > I. The sequence is actually strictly decreasing. Now we turn our attention to computing probabilities. If the first switch is made when there are ak boxes left for choosing, then let pk denote the probability of choos­ ing correctly among the ak boxes. The value ofpk is given by \

nj

ak

this product meaning that the wrong box was chosen at first and the right box chosen at the switch. In the same vein, pk-i

= (1 -ak)

. flfc-i

This last expression can be rewritten by substituting in the formula for pk to arrive at Pfe-i = 1 1 1 1 • Ofe-i flfe-iafc ak-\akn Continuing this process we arrive at the probability of obtaining the winning box by switching when there is one box left from which to choose. This value is p\ and is given by 1 1 (-l)fc_1 (-l) f c pi = + • ■ • + -i '+ —i 1 . a^ai ■ ■ ■ akn a\ a\a-2 aia'2---ak So, in our example {5,3,1} we have the contestant's probability of winning is 1 1 1 11 1 1-3 1-3-5 15 Surprisingly, this can be worked in the backwards direction. Given a rational number pi, we can determine the values ofn and the a». This is due to the following theorem. Pi

Theorem 10.33 Any rational number p/q € (0,1] has a unique representation q

a\

a^a-i

a\ai---a,k

where a* are positive integers such that 1 < ai < a2 < • • • < a.fc-i < afc — 1. Proof: The interval (0,1] can be written at the union of intervals of the form {{n 1) _ 1 , n - 1 ] . Thus any value a £ (0,1] belongs to exactly one of the intervals ((n l ) - 1 , n - 1 ] . And so 1 / 1 a = n Ai \n

1 ^ * 1 n+lj n

Ai n{n+\)

214

GENERALIZATIONS AND APPLICATIONS

where Ai G [0,1). If we say ai = \\/{n + 1), we get that a = (1 — a\)/n and cti £ (0, (n + 1) _ 1 ). Doing as we did before, this time with (0, (n + l ) " 1 ) , we find ot\ = (1 — a2)/mfor some m > n, and this leads us to 1 n

1 nm

-OL2-

We summarize the recursive process as 1

with ao = a,

ati-i

and OL\

1 -

aj-iftj,

where \x\ is the greatest integer function. Now, when a is an irrational number, then ai must be irrational for all i and the process never terminates. If a is a rational number, then the process is a modified Euclidean algorithm with a strictly decreasing sequence of remainders, hence the process must terminate. In applying the Euclidean algorithm to our rational number we create a sequence of dividends a\. a,2, ■ ■ ■, a^ with the property that ak-i < a*; — 1. This, plus the fact that 1 1 1 1 < < gives us the uniqueness of expansion that we needed. 4* As a consequence of this proof we now have the fact that for any number a G (0,1] we have 1 1 (-l)*-1 a= + •■• + + ••• , ai aia,2 aia 2 ---a/ c where 1 < ai < a 2 < 03 < • • •. This the referred to as the Pierce expansion of the number. We will denote this expansion by < a\, 02, ^3, • • • >• For a quick example where the sequence is infinite, let us look at < 1,2,3,4,... >. This represents the infinite series 1 1 1 1 lT~2!+3!~4!+'"' which is the Taylor Series expansion of f(x) = 1 — ex evaluated at x = —1. Thus 1 1 l!~2!

+

1 1 3!~4!

+

1 '"~ ~e' i

Incidentally, this proves that e is an irrational number. For this application, we go back to the historical origins of the Cantor Set. The set answered the question over the existence of a non-empty, perfect, nowhere dense set. Similar to the Cantor Ternary Set consisting ofpoints in [0,1] that have a ternary expansion that has no 1 's, we leave something out of the Pierce expansion.

THE MONTY HALL PROBLEM

fl

215

EXAMPLE 10.8 Let C be the set of real numbers in (0,1] whose Pierce expansion contains no odd integers. In [59] we see that this is a perfect and nowhere dense set. We also note that it is measure zero, due to the fact that

n—1

N

'

Additionally, this set is uncountable as there is a one-to-one correspondence between the element ofC and (0,1] given by /ai

a2 a 3

\

■

CHAPTER 11

EPILOGUE

We have how seen what began as a search for a pathological example; an example of a perfect, nowhere dense set is now so much more. The Cantor Set is useful in multiple areas, including topology, analysis, abstract algebra, and fractal geometry. It lends itself to pathological examples in studies such as probability theory, excep­ tional (e.g., measure, category, dimension) sets, and generalized continuity. Still, there are many different directions for new research by professors, graduate stu­ dents, and undergraduates. This text serves as a window into some of the areas one can search for new results. Of course this book contains just a small amount of the work done on this topic. As of May 2013, MathSciNet® lists over 2600 papers and books containing the keyword "Cantor Set."

The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

217

REFERENCES

1. Steven Astels, "Thickness measures for Cantor sets," Electron. Res. Announc. Am. Math. Soc. 5(1999), 108-111. 2. George Bachman, "Introduction to p-adic Numbers and Valuations Theory," Academic Press, 1964. 3. In Soo Baek, "Dimensions of the perturbed Cantor set," Real Anal. Exchange, 19(1), 1993/94, 269-273. 4. In Soo Baek, "Hausdorff measure on perturbed Cantor sets," Real Anal. Exchange, 20(2), 617-621. 5. Michael Barnsley, "Fractals Everywhere," Academic Press, 1988. 6. C.L. Belna, G.T Cargo, M.J. Evans, andP.D. Humke, "Analogues of the Denjoy-YoungSaks Theorem," Trans. AMS, 271 (1982), 253-260. 7. A.S. Besicovitch and S.J. Taylor, "On the complementary intervals of a linear closed set of zero Lebesgue measure," J. London Math. Soc, 29(1954), 449-459. 8. Duane Boes, Richard Darst, and Paul Erdos, "Fat, symmetric, irrational Cantor sets," Am. Math. Monthly 88 (1981), no. 5, 340-341. 9. Z.I. Borevich and I.R. Shafarevich, "Number Theory," Academic Press, 1966. 10. Andrew M. Bruckner, "Differentiation of Real Functions," American Mathematical So­ ciety, 1994. 11. Andrew M. Bruckner, Judith B. Bruckner, and Brian S. Thomson, "Elementary Real Analysis," Prentice-Hall, 2001. The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

219

220

REFERENCES

12. Yann Bugeaud, "Diophantine approximation and Cantor sets," Math. Ann., 341 (2008), 677-684. 13. Krzystof Ciesielski, Hajrudin Fejzic, and Chris Freiling, "Measure zero sets with nonmeasurable sums," Real Anal. Exchange, 27 (2001/02), 783-793. 14. W. A. Coppel, "An interesting Cantor set," Am. Math. Monthly, 90 (1983), 456-460. 15. Sylvain Crovisier and Michal Rams, "IFS attractors and Cantor sets," Topology Apps., 153 (2006), 1849-1859. 16. T.W. Cusick, "Sums and products of continued fractions," Proc. AMS, 27 (1971), 35-38. 17. T.W. Cusick, "On M. Hall's continued fraction theorem," Proc. AMS, 38 (1973), 253254. 18. Richard B. Darst, "Some Cantor sets and Cantor functions," Math. Mag., 45 (1972), 2-7. 19. Joseph Warren Dauben, "Georg Cantor-His Mathematics and Philosophy of the Infi­ nite," Princeton University Press, 1990. 20. A. Denjoy, "Sur Une Propriete des Series Trigonometrigues," Verlag v.d.G.V. der Wis-en Natuur. Afd., 1920. 21. Keith Devlin, "The Joy of Sets," Springer-Verlag, 1993. 22. Martin Dindos, "Generalized Cantor sets and sets of sums of convergent alternating series," J. Appl. Anal, 7 (2001), 131-150. 23. Bohuslav Divis, "On the sums of continued fractions," Ada Arith., 22 (1973), 157-173. 24. Josef Dobos, "The standard Cantor function is subadditive," Proc. AMS, 124 (1996), no. 11, 3425-3426. 25. E.P. Dolzenko, "Boundary properties of arbitrary functions (in Russian)," Izv. Ada. Nauk SSSR Ser. Mat., 31 (1967), 3-14. 26. Gerald Edgar, "Measure, Topology, and Fractal Geometry," Springer-Verlag, 1990. 27. M.J. Evans, PD. Humke, and K. Saxe, "A characterization of asymmetrically porous symmetric Cantor sets" Proc. AMS, 112 (1994), 805-810. 28. M.J. Evans, "Some theorems whose a-porous exceptional sets are not a -symmetrically porous," Real Anal. Exchange, 17(1991/92), 809-814. 29. Kenneth Falconer, "The Geometry of Fractal Sets," Cambridge University Press, 1985. 30. Kenneth Falconer, "Fractal Geometry, Mathematical Foundations and Applications," J. Wiley & Sons, 1990. 31. James Foran, "Non-averaging sets, dimension, and porosity," Canad. Math. Bull., 29 (1986), 60-63. 32. Michel Mendes France, "Sur les fractions continues limitees ," Ada Arith., 23 (1973), 207-215. 33. Russell A. Gordon, "The Integrals of Lebesgue, Denjoy, Perron, and Henstock," AMS, 1994. 34. Fernando Q. Gouvea, "Was Cantor surprised?" Am. Math. Monthly 118 (2011), no. 3, 198-209. 35. Marshall Hall, Jr., "On the sum and product of continued fractions," Ann. Math., 48 (1947), 966-993.

REFERENCES

221

36. Paul R. Halmos, "Naive Set Theory" Springer-Verlag, 1974. 37. Gregg Hartvigsen, Christopher C. Leary, and Dennis A. Ruppe, "Fractals, average dis­ tance, and the Cantor set," Fractals, 18 (2010) 327-341. 38. Kathryn E. Hare, Franklin Mendivil, and Leandro Zuberman, "The sizes of rearrange­ ments of Cantor sets," Canadian Math. Bull., 56 (2013), no. 2, 354-365. 39. Brian R. Hunt, Ittai Kan, and James A. Yorke, "When Cantor sets intersect thickly," Trans. Am. Math. Soc, 339 (1993), no. 2, 869-888. 40. Corey Jones, (private communication) 41. K. Kakeya, "On the partial sums of an infinite series," Tohoky Sci, Rep., 3 (1914), 159164. 42. Hidefumi Katsuura, "Continuous nowhere-differentiable Functions-an application of contraction mappings," Am. Math. Monthly, 98. no. 5 (1990), 411-416. 43. John L. Kelley, "General Topology," Van Nostrand, 1975. 44. Neal Koblitz, "p-adic Numbers, p-adic Analysis, and Zeta-Functions (2nd ed.)," Springer-Verlag, 1984. 45. Roger Kraft, "What's the difference between Cantor sets?," Am. Math. Monthly, 101 (1994), 640-650. 46. Roger Kraft, "Intersections of Thick Cantor Sets," Mem. Am. Math. Soc, 97 (1992), no. 468. 47. Steven G. Krantz, "An Episodic History ofMathematics: Mathematical Culture Through Problem Solving" MAA, 2010. 48. Timothy Lance and Edward Thomas, "Arcs with positive measure and a space-filling curve," Am. Math. Monthly 98 (1991), no. 2, 124-127. 49. R. Daniel Mauldin, "The Scottish Book: Mathematics from the Scottish Cafe," Birkhauser, 1982. 50. Mark McClure, "Self-similar structure in Hilbert's space filing curve," Math. Mag., 76 (2003), no. 1, 40-47. 51. Mark McClure, "The Hausdorffdimension of Hilbert's coordinate functions." Real Anal. Exchange 24 (1998/99), no. 2, 875-883. 52. P.K. Menon, "On a class of a perfect set," Bull. Am. Math. Soc, 54(1948), 706-711. 53. Ioana Mihaila, "The rationals of the Cantor set," College Math J., 35 (2004), 251-255. 54. Pedro Mendes, "Sum of Cantor sets: self-similarity and measure," Proc AMS, 127 (1999), 3305-3308. 55. Sheldon Newhouse, "The abundance of wild hyperbolic sets and nonsmooth stable sets for diffeomorphisms," Inst. Hautes Etudes Sci. Publ. Math., (1979), no. 50, 101-151. 56. J.E. Nymann, "The Sum of the Cantor Set with Itself," I'Enseign. Math., 39(1993), 177178. 57. J.E. Nymann, "Linear combinations of Cantor sets," Colloq. Math., LXVIII(1995) 259264. 58. John C. Oxtoby, "Measure and Category," Springer-Verlag, 1987. 59. Jaume Paradis, Pelegri Viader, and Liuis Bibiloni, "A mathematical excursion: From the three-door problem to a Cantor-type set," Am. Math. Monthly, 106 (1999), 241-251.

222

REFERENCES

60. J. Palis and F. Takens, "Hyperbolicity and Sensitive Chaotic Dynamics at Homoclinic Bifurcations," Cambridge University Press, 1993. 61. M. Pavone, "The Cantor Set and a Geometric Construction," I'Enseign. Math., 35 (1989), 41-49. 62. Joseph L. Pe, "Ana's golden fractal," Fractals 11 (2003), no. 4, 309-313. 63. Clifford Pickover, "Wonders of Numbers," Oxford Univ. Press, 2001. 64. Hui Rao, Huo-Jun Ruan, and Ya-Min Yang, "Gap sequence, Lipschitz equivalence, and box dimension offractal sets," Nonlinearity, 21 (2008), 1339-1347. 65. Ira Rosenholtz, "Another proof that any compact metric space is the continuous image of the Cantor set," Am. Math. Monthly, 83 (1976) 646-647. 66. Jason Rosenhouse, "The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain Teaser," Oxford University Press, 2009. 67. Jeffrey Rosenthal, "Monty Hall, Monty fall, Monty crawl," Math Horizons, 16 (2008), 5 -7. 68. H.L. Royden, "Real Analysis, 2nd ed.," Macmillan Publishing, 1968. 69. Alan H. Schoenfeld, "Continuous surjections from Cantor sets to compact metric spaces," Proc. AMS, 46 (1974), 141-142. 70. Boris Solomyak, "On the measure of arithmetic sums of Cantor sets," Indag. Math., N.S. 8(1997), 133-141. 71. Claude Tricot, "Douze definitions de la densite logartihmique," C.R. Acad. Sci. Paris, 342 (1981), 549-552. 72. Robert W. Vallin, "Sums and sizes of Ana sets," (to appear) 73. Robert W. Vallin, "Newhouse thickness and porosity of Cantor sets," Real Anal. Ex­ change, 27(2001/02), 349-358. 74. Robert W. Vallin, "Symmetric porosity, dimension, and derivates," Real Anal. Exchange, 17 (1991/92), 314-321. 75. A.C.M. Van Rooij and W.M. Schikhoff, "A Second Course on Real Functions," Cam­ bridge University Press, 1982. 76. VS. Vladimirov, I.V Volovich, and E.I. Zelenov, "p-adic Analysis and Mathematical Physics," World Scientific, 1994. 77. Lu Wen, "A nowhere differentiable continuous function," Am. Math. Monthly, 107 (2000), 450-453. 78. R.F. Williams, "How big is the intersection of two thick Cantor sets," Contemp. Math., 117, Am. Math. Soc, (1991) 163 - 175, 79. Grace C. Young and W. H. Young, "Discontinuous functions continuous with respect to every straight line," The Q. J. of Pure Appl. Math., 41 (1910), 87-95. 80. Ludek Zaji'cek, "Porosity and a-porosity," Real Anal. Exchange, 13 (1987-88), 314-350. 81. Ting Zhong, Jing-Jing Zhang, and Liang Tang, "A class of Cantor sets associated with the regular continued fractions," Comput. Math. Appl, 61 (2011) 2251-2255.

Index

Ta set, 8, 172 Qs set, 8, 172 affine transformation, 106 algebraic numbers, 27 Ana set definition, 198 Golden, 199 Archimedean Property, 54 Baire Category Theorem, 33 Bernstein set, 188 Besicovitch-Taylor dimension, 137 Bolzano-Weierstrass Theorem, 103 Borel sets, 172 boundary, 8 box counting dimension definition, 128 lower, 128 upper, 128 Cantor Cantor Cantor Cantor Cantor

dimension, 183 dust, 195, 202 Function, 161 Series, 205 Set

Ternary, 17 construction, 17, 59, 149 derivation, 150 differences, 193 fat, 38, 49, 185 generalzied, 181 in base 3, 20 p-adic, 80, 86 perturbed, 183 products, 195 sums, 152, 186 thickness, 150 Cantor target, 197 Cantor, Georg, 2 cardinality, 22 Cartesian product, 26 Cauchy sequence, 99 closed set, 8, 30 continued fraction definition, 52 simple, 52 Continuum Hypothesis, 29 Contraction Mapping Theorem, 108 convergence in a topological space, 30 convergence of a function

The Elements of Cantor Sets -With Applications, First Edition. By Robert W. Vallin Copyright © 2013 John Wiley & Sons, Inc.

224

INDEX

pointwise, 158 uniform, 159 convergent?, 56 Darboux integral, 42 Darboux sum, 42 dimension Besicovitch-Tayior, 137 box counting, 127, 128 Cantor, 183 Hausdorff', 132 inductive, 124 modified box-counting, 137 packing, 136 diophantine equation definition, 60 linear, 60 division ring, 71 equivalence class, 23 equivalence relation, 23 Euclidean Algorithm, 61 fat Cantor Set, 38, 49, 185 Fibonacci number, 199 field, 71 first category; 32 fixed point, 107 function Baire class one, 171 bijection, 9 characteristic, 9, 40 continuous, 11, 31 Darboux, 173 decreasing, 29 definition, 9 homeomorphism, 35 increasing, 29 inverse, 9 Lebesgue integral of, 45 linearly continuous, 177 Lipschitz, 10, 14, 15, 141 measurable, 40 restriction, 9 Riemann integral of, 42 separately continuous, 177 simple, 43 Golden Ana set, 199 golden ratio, 56, 199 group abelian, 69 definition, 68 Hausdorff s—measure, 132

Hausdorff s—outer measure, 132 Hausdorff dimension, 132, 154, 182 Hausdorff metric, 105 homeomorphism, 35 inductive dimension, 124 infimum, 120 Integral Lebesgue, 45 Riemann, 42 integral domain, 70 interior, 8 Intermediate Value Property, 173 Iterated Function System, 110 Koch Curve, 111 Lebesgue integral, 44, 45 Lebesgue measure, 132 limit infimum function, 122 of a sequence, 121 limit point, 101 limit supremum function, 122 of a sequence, 121 linearly continuous function, 177 Liouville numbers, 207 Lipschitz function, 10, 14, 15, 141 lower bound, 119 Lucas number, 199 Maclaurin series, 64 mass distribution, 134 Mass Distribution Principle, 135 measurable function, 40 measurable set, 37 metric p—adic, 73 complete metric space, 100 definition, 12 Hausdorff, 105 space, 13, 92 modified box-counting dimension, 137 Monty Hall Problem, 211 non-averaging sets, 203 norm, 73 open ball, 94 open cover, 36 open set, 8 Open Set Condition, 128, 133, 182 ordered derivation, 150 outer measure, 36

p-adic Cantor Set, 80, 86 integer, 77 numbers, 72 rational, 82 packing dimension, 136 partition, 41 perturbed Cantor Set, 183 Pigeon-Hole Principle, 103 porosity definition, 144 generalized, 154 right, 144 porosity index, 145 porous set, 144 quadratic irrational, 53 relation definition, 22 equivalence, 23 Riemann integral, 42 Riemann sum, 41 ring, 70 second category, 32 self-similar, 91 sequence bounded, 100 Cauchy, 99 definition, 10 in a metric space, 97 set Ta, 8, 172 Qs, 8, 172 bounded, 8, 95 closed, 8, 30, 94, 102 closure, 101 compact, 103 countable, 24 definition, 5 dense, 32 equivalent sets, 22 finite, 24 infinite, 24 intervals, 8 measurable, 37 nowhere dense, 32 open, 8, 29, 94 perfect, 102 porous, 144 small, 21 symmetric, 146 totally bounded, 95 uncountable, 24

very porous, 156 Sierpinski Triangle, 110, 141 similarity dimension, 127 simple continued fraction, 52 simple function, 43 space-filling curve, 168 subsequence, 102 supremum, 120 symmetric porosity, 147 thickness, 150 Thomae's Function, 34 topological space, 29 topology basis for, 123 definition, 29, 123 transcendental numbers, 27 transfinite induction, 188 ultra-metric, 74 uniformly Cauchy, 162 upper bound, 119 valuation, 73