THRY OF FUNC OF CMPLX VARIABLE
THRY OF FUNC OF CMPLX VARIABLE M 361
Popular in Course
Popular in Mathematics (M)
This 70 page Class Notes was uploaded by Reyes Glover on Sunday September 6, 2015. The Class Notes belongs to M 361 at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/181468/m-361-university-of-texas-at-austin in Mathematics (M) at University of Texas at Austin.
Reviews for THRY OF FUNC OF CMPLX VARIABLE
Report this Material
What is Karma?
Karma is the currency of StudySoup.
Date Created: 09/06/15
Introduction to Real Analysis M361K Preface These notes are for the basic real analysis class The more advanced class is M365C They were writtten7 used7 revised and revised again and again over the past ve years The course has been taught 12 times by eight different instructors Contributors to the text include both TA7s and instructors Cody Patterson7 Alistair Windsor7 Tirn Blass7 David Paige7 Louiza Fouli7 Cristina Caputo and Ted Odell The subject is calculus on the real line done the right way The main topics are sequences7 lirnits7 continuity7 the derivative and the Riemann integral it is a challenge to choose the proper amount of preliminary rnaterial before starting with the main topics In early editions we had too much and decided to move some things into an appendix to Chapter 2 at the end of the notes and to let the instructor choose what to cover We also removed much of the topology on R rnateral from Chapter 3 and put it in an appendix In a one semester course we are able to do most problems from Chapter 376 and a selection of certain prelirninary problems from Chapter 2 and the two appendices March 2009 Contents Chapter 1 Introduction Chapter 2 Preliminaries Numbers and Functions 1 The Absolute Value 2 lntervals 3 Functions Chapter 3 Sequences Limits Archimedean Property Convergence Monotone Sequences least upper bound and greatest lower bound Subsequences 6 Cauchy Sequences 7 Decimals r5905 9 Chapter 4 Limits and Continuity 1 Limits 2 Continuous Functions 3 Uniform Continuity Chapter 5 Differentiation 1 Derivatives 2 The Mean Value Theorem Chapter 6 Integration 1 The De nition 2 lntegrable Functions 3 Fundamental Theorems of Calculus 4 Integration Rules Appendix Appendix to Chapter 2 1 The Field Properties 2 The Order Properties 3 The Ordered Field Properties 4 Set Theory 5 Cardinality 6 CONTENTS Mathematical Induction Appendix Appendix to Chapter 3 1 2 3 Open and Closed Sets Compactness Sequential Limits and Closed Sets CHAPTER 1 Introduction Goals The purpose of this course is three fold 1 to provide an introduction to the basic de nitions and theorems of real anal ysis 2 to provide an introduction to writing and discovering proofs of mathematical theorems These proofs will go beyond the mechanical proofs found in your Discrete Mathematics course 3 to experience the joy of mathematics the joy of personal discovery Proofs Hopefully all of you have seen some proofs before A proof is the name that math ematicians give to an explanation that leaves no doubt The level of detail in this explanation depends on the audience for the proof Mathematicians often skip steps in proofs and rely on the reader to ll in the missing steps This can have the ad vantage of focusing the reader on the new ideas in the proof but can easily lead to frustration if the reader is unable to ll in the missing steps More seriously these missing steps can easily conceal mistakes most mistakes in proofs begin with it is clear that In this course we will try to avoid missing any steps in our proofs each statement should follow from a previous one by a simple property of arithmetic by a de nition7 or by a previous theorem7 and this justi cation should be clearly stated Writing clear proofs is a skill in itself Often the shortest proof is not the clearest There is no mechanical process to produce a proof but there are some basic guidelines you should follow The most basic is that every object that appears should be de ned when a variable7 function7 or set appears we should be able to look back and nd a statement de ning that object 1 Let 8 gt 0 be arbitrary 2 Let f 2x 1 3 Let A x E R x13 7 27x12 16 7 4 0 4 By the de nition of continuity there exists a 6 gt 0 such that 1 2 1 INTRODUCTION Always watch out for hidden assumptions In a proof you may want to say Let x E A be arbitrary77 but this does not work if A g the empty set A common error in real analysis is to write lim an or limfz without rst checking whether the Hoe mam limit exists often this is the hardest part The audience for which a proof is intended is what determines how the proof is written Your audience is another student in the class who is clueless as to how to prove the theorem Logic We will avoid using logical notation in our de nitions and statements of theorems Instead we will use their English equivalents Beyond switching to the contrapositive and negating a de nition formal logical manipulation is rarely helpful in proving statements in real analysis You should be familiar with the basic logical operators if P and Q are propositions ie statements that are either true or false then you should understand what is meant by 1 not P 2 P or Q 7 the mathematical use of or is not exclusive so P or Q is true even if both P and Q are true 3 P and Q 4 if P then Q or P implies Q 5 P if and only if Q sometimes written P is equivalent to Q Similarly if P is a predicate that is a statement that becomes a proposition when an object such as a real number is inserted for x then you should understand 1 for all x Pz is true 2 there exists an x such that P is true Simple examples of such a Pz are x gt 077 or 2 is an integer77 You should know the formulae for negating the various operators and quanti ers Most of our theorems will have the form of implications if P then Q77 P is called the hypothesis and Q the conclusion De nition The contrapositive of the implication if P then Q77 is the implication if not Q then not P77 The contrapositive is logically equivalent to the original implication This means that one is valid true if and only if the other is valid Sometimes it is much easier to pass to the contrapostive formulation when proving a theorem De nition The converse of the implication if P then Q is the implication if Q then P The converse is not logically equivalent to the original implication 1 INTRODUCTION 3 De nition A statement that is always true is called a tautology An implication that is a tautology is called a valid argument A statement that is always false is called a contradiction To show that an argument is not valid it suf ces to nd one situation in which the hypotheses are true but the conclusion is false Such a situation is called a counterexample One technique of proof is by contradiction To prove P implies Q77 we might assume that P is true and Q is false and obtain a contradiction This is really proving the contrapositive CHAPTER 2 Preliminaries Numbers and Functions What exactly is a number If you think about it to give a precise answer to this question is surprisingly dif cult As is often the case the word number7 re ects a concept of which we have some intuitive understanding but no concrete de nition We will try to describe exactly what we should expect from a number system These expectations will lead us to conclude that a number is an element of R the collection of real numbers Most of us have probably heard of the real numbers but may not be exactly sure what they are In particular we might ask Why exactly are the real numbers so important ls there some other system that would also suf ce to be our system of numbers What is wrong with just using the natural numbers the integers or the rational numbers To answer this question we will have to pin down the expectations we have on a number system More precisely we will have to decide exactly what properties should hold in a system of numbers It is actually not so dif cult to make progress in this area it only requires a little bit of introspection and memory Think back to when you rst met the idea of a number Probably the very rst purpose of a number in your life is that it allowed you to count things 50 states 32 professional football teams 7 continents 5 golden rings etc Needing to count things leads us to the invention or discovery depending on your point of view of the natural numbers the numbers 1 2345 Mathematicians typically denote the collection of natural numbers by the symbol N Though this collection can be constructed quite rigorously from the standard axioms of mathematics we will assume that we are all familiar with the natural numbers and their basic properties such as the concept of mathematical induction see appendix to Chapter 2 The natural numbers ful ll quite successfully our goal of being able to count The next thing that will expect of our number system is that it should be able to answer questions like the following If the Big Twelve has 12 football teams and the Big Ten has 11 shockingly its true how many teams do the conferences have between them77 In other words we will need to add We will also multiply The natural numbers are already well suited for these tasks Really this should not come as a surprise After all adding is really just a different way of looking at counting ie 6 2 PRELIMINARIES NUMBERS AND FUNCTIONS adding three and ve is the same as taking three dogs and ve cats and counting the total number of animals As we all know multiplication is really repeated addition Having addition naturally leads us to subtraction This is the rst place the natural numbers will fail us Subtracting 7 from 2 is an operation that cannot be performed within N The need for subtraction therefore is one of the reasons that N will not work as our entire number system Thus we are lead to expand to the integers As we all probably know the integers are comprised of the natural numbers the number zero and the negatives of the natural numbers at this point you might protest and say that zero should be included as a natural number as it allows us to count collections which contain no objects in fact many mathematicians do include zero in N but the distinction is of little importance The collection of integers is denoted by Z Again we will assume we know all the basic properties of Z like prime factorization The integers are a very good number system for most purposes but they still have an obvious defect we cannot divide Surely any reasonable number system allows division If you and l have a sandwich and we each want an equal share how much do we each get Needing division we throw in fractions symbols which are comprised of two integers one in the numerator and one in the denominator of course the denominator is not allowed to be zero A fraction will represent the number which results when the numerator is divided by the denominator Things start to get a little bit complicated here We can now have more than one symbol that that stand for 2 the same number E and will both stand for integer 1 Combining all the numbers we have so far gives Q the collection of rational numbers Again we will assume that we are familiar with all its basic properties Before we go on to justify our assertion that Q is not a suf cient number system we have another property to point out Notice that most of our properties so far involve operations among our numbers namely addition subtraction multiplication and division We call these types of properties algebraic ln mathematics the word algebra describes the study of operations The property we are going to discuss next is not algebraic Suppose then that I pick a rational number and you pick another We can easily decide which is bigger Namely is bigger than i where a b c and d are integers if ad is bigger than be assuming b and d both positive we can easily assure both denominators are positive by moving any negative into the numerator Since ad and be are integers we know how to compare them because we know how to compare natural numbers and how to take negatives into account Since we can always compare any two rational numbers in this way we say that Q is totally ordered In retrospect we should have demanded this property of our number system from the beginning Numbers should come with some notion of size Fortunately we got it for free Moreover it is interesting to notice that the our expectation that a number system should include the natural numbers and that it should have certain algebraic properties is enough to lead us to include all of Q We did not to request that our 2 PRELIMINARIES NUMBERS AND FUNCTIONS 7 system be ordered to nd Q The order properties turn out to be more important in telling us which potential numbers we should NOT include such as the imaginary number 239 Q comes very close to satisfying everything we want in a number system Unfortu nately it is still lacking Suppose we draw a circle whose diameter is 1 The length around the circle usually called the circumference should certainly be a number lf7 however7 we restrict ourselves to the rational numbers7 this length will not be a number the number is of course usually denoted 7139 and it is not a rational number The same could be said of the length of one of the sides of a square whose area is 2 this number is usually denoted These two examples merely comprise our attempt to give geometric demonstration that Q is lacking as a number system The real more general property that we seek7 called completeness7 is actually quite subtle and has to do with the presence of something like gaps7 in Q the absence of the number V or the number 7139 is an example of such a gap These gaps have to do with something called a Cauchy sequence which we will study in detail in this course One consequence of lling in these gaps is that we are able to perform calculus This7 in turn7 allows us to express all the lengths and areas7 volumes7 etc of geometric objects like the examples above as real numbers One of the fundamental results in mathematics is that the collection of real numbers is the ONLY system of numbers which satis es all of our demands We will formulate our demands precisely throughout the rst three sections of this chapter with the exception of the completeness axiom We are thus forced to admit that the real numbers comprise the only possible choice of a number system To give an exact de nition of the real number is surprisingly complicated ln fact7 the rst rigorous construction of the real numbers was given by Georg Cantor as late as 1873 by comparison7 the rational numbers where constructed in ancient times For our purposes7 we will take it on faith that the real numbers exist as a number system In the appendix to this chapter7 we will state precisely the important prop erties of the real numbers which we will also take on faith and note that these line up with our expectations of the number system except for the completeness axiom whose importance might not be clear at rst keep our geometric example in mind In the next chapter7 we will also describe a way to de ne the real numbers rigorously using decimal expansions there are actually several well known ways Finally7 it is important to realize that the properties given in the appendix which we will call accimns7 together with the completeness axiom7 are the ONLY properties that we assume about R Strictly speaking7 any other statement we want to make must be proven from either from our axioms or from properties we have already assumed about N7 Z7 and Q or7 of course7 some combination of the two In general7 however7 this can get to be a little bit tedious Hence we will allow you to assume all of the basic7 or obvious7 properties of the real numbers Unfortunately7 8 2 PRELIMINARIES NUMBERS AND FUNCTIONS deciding which properties are obvious is a subjective process Therefore if there is any doubt about whether a statement is obvious you should prove it rigorously from the axioms or at least describe how to prove it rigorously Actually the ability to decide when statements are obvious or trivial7 is an important skill in mathematics Possessing this ability can often be a re ection of great mathematical maturity and insight In the appendix to this chapter we will also derive some properties of R that follow from our axioms We will work on some of these in class but thereafter you may consider them known The appendix also contains a discussion of basic set theory and induction 1 The Absolute Value An important property of the real numbers is that we can de ne a size on them which we call the absolute value The de nition is relatively simple and yet it has vast and important consequences De nition Given a real number a E R we de ne the absolute value of a denoted lal to be a if a is nonnegative and 7a if a is negative 21 For a E R 1 lal 2 0 2 lal 0 if and only if a 0 and 3 lal 2 a The following technical observations will be of assistance in some arguments involving the absolute value 22 For all a E R a2 lalz 23 For all ab E R with ab 2 0 we have a2 S b2 if and only if a S b Likewise a lt b if and only if a2 lt b2 The next statement gives two fundamental properties of the absolute value 24 Let ab E R then 1 W MW and 2 labl S W W Hint One can prove these by laboriously checking all the cases eg a gt 0 b S 0 but in each case an elegant proof is obtained by using our observations to eliminate the absolute value and then proceeding using the properties of arithmetic The second inequality above is perhaps the most important inequality in all of anal ysis It is called the triangle inequality The remaining results in this section are important consequences of the triangle in equality 25 Let abc E R Then we have 2 INTERVALS 9 lt1 ia7bi 2 llal 7 WI and lt2 ia7ci ia7biib7cl You have probably measured the length of something using a yardstick You might place one end at zero and see where the other end lies to get the length In a tight place you might place the yardstick and nd one end at 7 and other at 13 and conclude the length is 13 7 7 6 lf I told you one end was at z and the other was at y what would be the length Well y 7 z ify gt z and z 7y if z gt y In short it would be lx 7 So we can regard lx 7 yl as the distance between the numbers z and y Note this works for all real numbers even if one or both is negative Note that this explains why see 252 we call 242 the triangle inequality 251 is often called the reverse triangle inequality 26 Let x 8 E R with 8 gt 0 Then we have the following two statements 1 S 8 if and only if 76 S x S 8 The double inequality 76 S x S 8 means 78 xand a 2 lfaERlx7alSeifandonlyifa78 x aa Remark By a similar proof the same properties hold with 3 replaced by lt 2 Intervals lntervals are a very important type of subset of R Loosely speaking they are sets which consist of all the numbers between two xed numbers called the endpoints We also informally allow the endpoints to be ioo Depending on whether the endpoints are nite and whether we include them in our sets we arrive at 9 different types of intervals in R De nition An interval is a set which falls into one of the following 9 categories assume ab E R with a lt b We apply the word bounded7 if both the endpoints are nite Otherwise we use the word unbounded7 1 Bounded open intervals are sets of the form abz Raltltb 2 Bounded closed interval are sets of the form 11 xeanSsgb 3 There are two type of half open bounded intervals One type is sets of the form ab Ra zltb 4 The other is sets of the form abz Raltx b 10 2 PRELIMINARIES NUMBERS AND FUNCTIONS 5 There are also two types of unbounded open intervals not equal to R One type is sets of the form aoo x6 Ralt 6 The other is sets of the form 7007b x E R s lt b 7 There are two types of unbounded closed intervals not equal to R One type is sets of the form aoo x6 Ra 8 The other is sets of the form 7007bl x E R x S b 9 The whole real line R foo7 00 is an interval We count R as being open7 closed7 and unbounded Some authors include the empty set7 Q7 and single points7 a for some a E R7 as intervals To distinguish these special sets7 people often call them degenerate intervals7 whereas sets of the above would be nondegenerate invervals7 We will reserve the word interval7 for the nondegenerate case That is7 in our language7 an interval is not allowed to be g or 1 De nition The closure of an interval I7 denoted T is the union of and it endpoints Thus De nition The interior of an interval I7 denoted 1quot is I rninus its endpoints Thus 07W 07W 07W 0 bl a 00 la 00 07 00 7007bo 7007blo 7007b R R 3 Functions Even more basic to mathematics than the concept of a number is the concept of a function Roughly speaking7 a function f from a set A to a set E is a rule that assigns 3 FUNCTIONS 11 to each x E A an element f E B In this case we write f A a B What do we mean by rule Let7s try to be more precise De nition A function f from A to B denoted by f A a B is a subset f of the Cartesian product A gtlt B ab a E Ab E B satisfying 1 for each a E A there exists b E B such that 11 E f 2 for all a E A and for all bb E B if 11 E f and ab E f then b b We could combine the two hypotheses into a single statement for each a E A there exists a unique b E B such that 11 E f Rather than writing 11 E f it is customary to write fa b Technically then a function from A to B is just a special subset of A gtlt B Mathemati cians however rarely think of functions in this way The idea of a rule is intuitively more accurate whereas the formal de nition is just a way to make it precise It is very important to realize that a function is not the same thing as a formula Many beginning students in mathematics think that in order to nd a function they need to nd a formula using variables This is not the case It is perfectly reasonable to de ne a function by saying something like De ne a function from the set of real numbers to the set 01 by assigning the value 1 to all rational numbers and the value 0 to all irrational numbers Since every number has been given a value and no number has been given more then one our rule gives a function This function f R a 0 1 would probably be more commonly described by saying that for z E R 1 if z E Q f 96 7 0 if s E RQ but either description would work Notice that it would be impossible to nd what most people would call a formula7 to describe this function De nition Let f A a B be a function The set A is called the domain of f and B is called the co domain The range of f denoted by fA is fA fa i 06A Most of the functions we consider will be of the form f R a R or f A a R where A Q R 12 2 PRELIMINARIES NUMBERS AND FUNCTIONS A function is sometimes called a mapping or a transformation lf fa b we might say f maps a to b77 or f sends a to b77 Some functions have certain speci c properties that we shall name De nition Let f A a B 1 f is onto or surjective if fA B That is7 f is onto if7 for every b 6 B7 there exists some a E A such that fa b 2 f is 171 or oneto one or injective if for all 11702 6 A7 fa1 fa2 implies a1 12 That is7 f is 171 if7 for all 11702 E A such that al 31 12 we haVe 01 7 02 3 f is a bijection or 171 correspondence if it is 171 and onto This is equivalent to for all b E B there exists a unique 1 E A with fa b Note it is usually simpler to show that a function is a bijection by showing it is 171 and onto separately If a function is a bijection then you can reverse it77 to obtain a function going the other way The following theorem makes this precise Theorem 27 Let f A a B be a bijection Then there exists a bijection g B a A satisfying 1 for all a 6 A7 gfa a 2 for all b 6 B7 fgb b Furthermore7 this function g is unique if 91 and 92 are bijections satisfying 1 and 2 then 91 92 Would it be enough to assume 91 and 92 both satisfy 1 The bijection g is called the inverse function of f and is usually denoted by f l Do not confuse this with 1f De nition Let f A a B Let D Q A7 and C Q B 1 The image of D under f7 denoted fD7 is fD 1 i 96 E D 2 The pre image7 of C under f7 denoted fquotlC7 is f 1O a e A fa e O The set f 1C always exists even when the function f 1 does not exist The context of a problem will tell you which f l 7 is being used We are using the symbol f 1 in two different ways When the function f 1 does exist then there are two different ways of reading f 1C it can be read as the direct image of the set C under the function f 1 or it can be read as the inverse image of C under the function f These give the same set and thus there is no ambiguity7 in this case 3 FUNCTIONS 13 De nition Let f A a B and g B a C The composition 9 o f A a C is de ned by g 0 MW 900 28 Let f A a B and g B a C be two bijections Then 9 o f A a C is also a bijection Caution If f and g are functions7 it is not true in general that f o g g o f In fact7 these two compositions may have completely different domains and CO domainsl Note If f A a B is a bijection then f 1 o f A a A is the identity map on A and fof l B a B is the identity map on B The identity map on the set S is the map z d S a S de ned by z dx z for all z E S CHAPTER 3 Sequences 1 Limits Our basic object for investigating real analysis will be the sequence De nition A sequence in R is a function f N a R Example fn n2 and gn both de ne sequences in R We typically do not use functional notation to discuss sequences instead we write things like n2f 1 or il If we say Consider the sequence 1nf 177 we are referring to the sequence whose value at n is 1 Sequences like general functions can be de ned without using an explicit formula Example We can de ne a sequence anf 1 by 1 nth digit to right of the decimal point for 7139 Thus 10311 14159 Example We can de ne a sequence recursively that is using the previous members of the sequence to de ne the next element A famous example of a recursively de ned sequence is the Fibonacci sequence given by 111 121 an2an1anforn21 Find 131415 A sequence is not to be confused with a set For example 1 1 1 1 but 111 denotes f N a R with fn 1 for all n This is an example of a constant sequence A sequence is really an in nite ordered list of real numbers which may repeat Remark Sometimes we will write consider the sequence 1nf 477 Of course this is given by fn 131 for n E N so it is a sequence even though for convenience or whatever reason we chose to start the sequence at n 4 lntuitively a sequence 1f 1 converges to a limit L if the terms get closer and closer to L as 71 gets larger and larger Now we need to make this idea precise so we can use it in our future discussions First note that 10 00 should converge to 0 under our yet to be formulated 15 16 3 SEQUENCES de nition But it is not true that each term is closer to 0 than the previous term So the rough intuitive de nition needs clari cation The following de nition makes this intuition precise De nition A sequence 171 is said to converge to L E R if for all 8 gt 0 there exists N E N so that for all n 2 N7 lan 7 Ll lt 8 L is called a limit of the sequence anff1 lfthere exists an L E R such that anff1 converges to L then we say 1011 converges or un il is a convergent sequence Note It is important to realize that the N that you choose will typically depend on 8 Typically we expect that N will get larger as 8 gets closer to 0 This de nition seems to have been rst published by Bernard Bolzano7 a Czech math ematician7 in 1816 It is the notion of limit that distinguishes analysiscalculus from7 say7 algebra This de nition came about 150 years after the creation of calculus due independently to Newton and Leibniz The fact that it will probably take you some time to understand and become happy with it is therefore no surprise The old guys were pretty sharp and still struggled with the notion Lemma 31 Let a 2 0 be a real number Prove that if for every 8 gt 0 we have that a lt 8 then a 0 32 Prove that if anff1 converges to L E R and 1011 converges to M E R then L M Hint argue by contradication We have just shown if a sequence has a limit then that limit is unique Thus it makes sense to talk about the limit of a sequence and to write lim an L Hoe Caution Before we write lim an we must know that can has a limit Hoe 33 Negate the de nition of lim an L to give an explicit de nition of 1011 does not converge to L77 TOO Remark We can write anff1 does not converge to L as un il 74gt L Caution We should not write simian 31 L to mean un il does not converge to L unless we know that lim an exists Hoe De nition A sequence can is said to diverge if it does not converge to L for any L E R We will distinguish two special types of divergence 1 A sequence can is said to diverge to 00 if for all M E R there exists N E N such that for all n 2 N we have an 2 M In an abuse of notation we often write lim an 00 Hoe 2 A sequence can is said to diverge to foo if for all M E R there exists N E N such that for all n 2 N we have an S M In an abuse of notation we often write 1320a foo 2 ARCHIMEDEAN PROPERTY 17 00 34 Prove that a sequence annL set converges to L E R if and only if for all 8 gt 0 the nEN laniLl 28 is nite This is saying that the number of terms can which are more than 8 distance from L must be nite This has a couple of interesting corollaries 1 if we change nitely many terms of a sequence we do not alter its lirniting behaviour if the sequence originally converged to L then the altered sequence still converges to L7 and if the original sequence diverged then so does the altered sequence 2 if we remove a nite number of terms from a sequence then we do not alter its lirniting behaviour if the sequence originally converged to L then the altered sequence sequence still converges to L7 and if the original sequence diverged then so does the altered sequence 35 Prove using the de nition of a limit that lirn 71L 0 Hoe Let7s examine the proof here Your proof should look something like this PROOF We will show lirn 0 Let 8 gt 0 be arbitrary but xed We must nd 00 NENsothat ifnZNthen iOl lt8Wthh is the same asilta ChooseNEN sothatlt8ThenifnZN lt8 D We have used here two things First the basic properties of order which we will assume you know Secondly7 we have used that given 8 gt 0 there exists N E N with lt 8 or in other forrn7 lt N This is called the Archirnedian property 2 Archimedean Property R has the Archimedean Property7 which says that for every x there exists an n E N with z lt n This seerningly innocuous property has many consequences 1 For every 8 gt 0 there exists an n E N such that l 0 lt 7 lt 8 n 2 for every x E R there exists an m E Z such that m S x lt m 17 3 for every x E R and n E N there exists an m E Z such that m 1 m i zlt 71 71 18 3 SEQUENCES 36 Using these we can prove that every real number can be approximated arbitrarily well by a rational number We say that Q is dense in R 1 Prove that for every x E R and every 8 gt 0 there exists m E Q such that V L m 0 S x 7 7 lt 8 n 2 Prove that for every x E R and every 8 gt 0 there exists E Q such that m 76 lt z 7 7 S 0 n Similarly7 the irrational numbers are dense in R 37 Prove 1 For all g E Q 0 the number x2 is irrational 2 Prove that for every x E R and every 8 gt 0 there exists m E Q such that 0 x72Elt6 n 3 Prove that for every x E R and every 8 gt 0 there exists m E Q such that n 78ltx72E 0 n 3 Convergence 38 Prove using the de nition of a limit that a lim 1 7 17 b lim 1 L271 n21 nam2n23 239 39 Prove or disprove To disprove you need only give a counterexample 1 If lim an L then lim lanl 3 4 lfangcngbn for alln ENand liman limbnLthen limcnL 2 If lim lanl lLl then lim an L lf lim lanl 0 then lim an 0 This is normally referred to as the Squeeze Theorem for sequences De nition A sequence anff1 is called bounded if the set an n E N is contained in a bounded interval Note Thus an is bounded if and only if there exists K 2 0 with lanl S K for all n E N 310 Prove that if anff1 is a convergent sequence with lim an L then 10211 is bounded ls the converse true We can now justify our use of the word diverges77 in the situation that a sequence goes to ioo 3 CONVERGENCE 19 311 Show that if an diverges to 00 then an diverges Show that if an diverges to 700 it also diverges 312 Let bnn1oo be a convergent sequence with lirn bn M with M 31 0 Prove that there exists an N E N such that for all n 2 N we have lbnl gt The following are often called the Limit Laws77 for sequences 313 Let anff1 and bn 1L be sequences and c E R be an arbitrary real number We can de ne new sequences 0 1011 cuL baff 7 and an bnf1 lf bn 31 0 for all n E N then we can de ne Prove that 1 If anff1 is a convergent sequence and c E R then c anff1 is a convergent sequence and lirn c an c lirn an Hoe Hoe 2 If anff1 and b9211L are convergent sequences then an bn 1 is a con vergent sequence and lirn an b lirn an lirn bn Hoe Hoe Hoe 3 If anff1 and b9211L are convergent sequences then an bnff1 is a conver gent sequence and lirn an bn lirn an lirn b Laoo Laoo Lace 4 If anff1 and b9211L are convergent sequences with bn 31 0 for all n E N and lirn bn 31 0 then Hoe h all limHoe an H00 be hm Hoe be 39 Hint for 3 The problem in 3 is that we have two quantities changing sirnultane ously To deal with this we use a very common trick in analysis we add and subtract additional terrns7 which does not affect the value7 and then group terms so that each term is a product of things we can control Let L lirn an and M lirn b We can Hoe Hoe write anbnaLManbnaLMLbnaLbn anaLbnLbnaM Hint for 4 Given 3 it suf ces to prove explain why that 1 1 lirn a Lace bn lirn bn 39 Laoe 20 3 SEQUENCES Let M lim bn and notice that Hoe i L i 7 M M bn M 7 lenl MZQ if lbnl gt Now use Problem 312 314 Suppose a 3 an lt b for all n E N Prove that if lim an L7 then L E 11 315 Let anff1 and bn 1L be convergent sequences with a lim an and b lim b Prove that if an 7 bn i for all n E N then a b Would this theorem still be true if7 instead of equality7 you had an 7 bn lt i What if an 7 bn 2 4 Monotone Sequences least upper bound and greatest lower bound In general it can be dif cult to show that a sequence converges For certain classes of sequences checking for convergence can be much easier De nition Let an be a sequence We say 1 an is increasing if for all n 6 N7 an 3 an 2 an is decreasing if for all n 6 N7 an gt an 3 an is monotone if it is increasing or decreasing Before proceeding we need some new de nitions De nition Let A Q R and z E R 1 z is called an upper bound for A if7 for all a 6 A7 we have a S x 2 z is called an lower bound for A if7 for all a 6 A7 we have x S a 3 The set A is called bounded above if there exists an upper bound for A The set A is called bounded below if there exists a lower bound for A The set A is called bounded if it is both bounded above and bounded below 4 z is called a maximum of A if z E A and z is an upper bound for A 5 z is called a minimum of A if z E A and z is a lower bound for A In 4 we write x maxA min A Note there is only one maximum or minimum7 if any 316 For each of the following subsets of R ltagtAza bAaER0lta 17 cAa R2 a dAaERaZlt27 1 Find all lower bounds for A and all upper bounds for A7 4 MONOTONE SEQUENCES LEAST UPPER BOUND AND GREATEST LOWER BOUND 21 2 Find minA and maxA if they exist 3 Discuss whether A is bounded De nition Let A Q R and z E R 1 z is called a least upper bound of A or suprernurn of A if a z is an upper bound for A b if y is an upper bound for A then x S y 2 z is called a greatest lower bound of A or in mum of A if a z is a lower bound for A b if y is a lower bound for A then y S x 317 Let A Q R Prove that if z is a supremum of A and y is a supremum of A then s y Hence if the set A has a supremum then that supremum is unique and we can speak of the supremum of A and write sup A Similarly if the set A has an in mum then that in mum is unique and we can speak of the in mum of A and write inf A 318 Find ian and sup A if they exist for each of the following subsets of R The above de nitions could also be made in Q or in any other subset of R Of course the answers to the exercises would depend upon the universe in question For example if A is the set of all positive irrationals then ian would not exist inside of the universe R Q The Completeness Axiom R is complete That is if A Q R A 31 g and A is bounded above then supA exists Though supA need not be in the set A there are elements of A arbitrarily close to sup A 319 Let A Q R be non empty and bounded above Prove that if s supA then for every 8 gt 0 there exists an a E A with s 7 8 lt a S 5 What would the equivalent property of ian be 320 Let A Q R be bounded above and non empty Let s sup A Prove that there exists a sequence can Q A with liman s Prove that in addition the sequence V L an can be chosen to be increasing ie an 3 can for all n E N What would be the analogous results for t inf A if A is bounded below 22 3 SEQUENCES 321 Let A Q R De ne 7A7z A 1 Prove that z is an upper bound for A if and only if is is a lower bound for 2 Prove that z supA if and only if 7x inf 7A 3 Prove that if A is non ernpty and bounded below then ian exists 322 Prove or disprove 1 If A7 B Q R are nonernpty sets such that for every 1 E A and for every b E B we have a S b then supA and inf B exist and supA S infB 2 If A7 B Q R are nonernpty sets such that for every 1 E A and for every b E B we have a lt b7 then supA and inf B exist and supA lt infB 323 Let A7 B C R be non ernpty7 bounded sets We de ne A B a b a E A and b E B Prove that supA B supA supB and infA B infA infB 324 Let f R a R and g R a R and A Q R Assume that fA and gA are bounded Prove that supf 9A S sup fA SUP9A Give examples where one has 77 and also lt77 State the analogous results for inf 325 Prove the following 1 If an is increasing and unbounded then an diverges to 00 2 If an is decreasing and unbounded then an diverges to foo 3 Let an be increasing and bounded Let L supan n E N Then lirn an L 4 Let an be decreasing and bounded Let L infan n E N Then lirn an L Hoe 326 Let A C R be a non ernpty7 bounded set Let Oz supA and B infA7 and let anff1 C A be a convergent sequence7 with a lirn an Prove that B S a S a Hoe Notice that a need not be equal to either of 04 or 67 even if an is strictly increasing or strictly decreasing For exarnple7 let A 071 so 04 1 and B 07 and let an Then an is strictly decreasing and a lirn an lf7 instead7 Hoe an 7 2 then an would be strictly increasing and a lirn an In both Hoe cases7 a 31 04 and a 31 6 Come up with a different example where an is rnonotonic ie7 increasing or decreasing but its limit is not 04 or B 5 SUBSEQUENCES 23 327 If A C R is a non empty7 bounded set and B C A7 prove infA S infB S supB S supA 328 If A7 B C R are both non empty7 bounded sets7 prove supA U B maxsupAsupB 329 We analyze the important sequence 7quot ff17 where r E R 1 Prove that if 0 S r lt 1 then r f1 converges to 0 Hint Show the sequence is decreasing Proceed by contradiction suppose L infan n E N gt 07 nd can with L 3 an lt r lL7 and hence produce a contradiction 2 Prove that if r gt 1 then 7 311 diverges to 00 3 Use Problem 39 to complete the proof of the following fact The sequence 7 converges if and only if 71 lt r S 1 If r 1 then lim 7 1 l 71ltrlt1thenlimr 0 n 00 The following example illustrates how powerful monotonicity is 330 Let a1 2 and let an be generated by the recursive formula 1 2 an1 5an Tn for n 2 1 1 Prove that 2 lt can for all n E N Hint Assuming an gt 0 turn an gt 2 into an equivalent condition on a quadratic polynomial Proceed by induction 2 Prove that an is decreasing and hence converges 3 Now take limits on both sides of the recursive formula using Problem 313 and use the fact that lim an lim an Hoe Hoe to nd lim an Hoe Let x E R5 What can you say about the sequence given by 11 1 and an an for n 21 5 Subsequences If we have a sequence an7 a subsequence is a new sequence formed by skipping possibly in nitely many terms in an n1 De nition A sequence b10211 is a subsequence of 6 if there exists a strictly increasing sequence of natural numbers 711 lt 712 lt so that for all k 6 N7 bk am 24 3 SEQUENCES Example 17 17 17 is a subsequence of 17 71 17 71 In fact any sequence of i17s is a subsequence of 17 71 17 71 In particular we see that a subsequence of a divergent sequence may be convergent Example g is a subsequence of Both 6 and converge to 0 331 Prove or disprove lf bn is a subsequence of an and lim an L7 then lim bn L Now we can play a fun game to see whether we can nd subsequences with better properties than the original sequence 332 Prove that every sequence of real numbers has a monotone subsequence Hint Consider two cases The rst is the case in which every subsequence has a minimum element In this case we can extract an increasing subsequence 333 Let zn f be an increasing sequence Suppose that there exists a subse quence of that converges to a point z E R Prove that also converges to x Combining this with our earlier work on monotone sequences7 Problem 3257 we get a very useful result 334 Prove that every bounded sequence of real numbers has a convergent subse quence 335 Let an and bn be sequences such that an1 3 an lt bn S bn1 for all n 6 N7 and lim bn 7 an 0 If we de ne In an7bn then I1 3 I2 3 I3 3 1 Prove that there exists a p E R such that p E In for all n E N 2 Prove that if q E In for all n E N then q p 6 Cauchy Sequences The problem with the de nition of a convergent sequence is that it requires us to know or guess what the limit is in order to prove a sequence converges We saw when looking at bounded monotone sequences that sometimes it is possible to show that a sequence converges without knowing the limit We now give another de nition that will address the problem De nition A sequence an is called a Cauchy sequence if for all 8 gt 0 there exists N E N such that for all 771771 2 N we have an 7 am lt 8 336 Negate the de nition of Cauchy sequence to give an explicit de nition of L4 00 77 ann1 is not a Cauchy sequence 7 DECIMALS 25 This will turn out to be very useful for proving that a sequence diverges 337 Prove that every convergent sequence is Cauchy This is true much more generally than just for R It is particularly useful in the contrapositive form if 171 is not Cauchy then an f 1 diverges 338 Prove that every Cauchy sequence is bounded ls the converse true 339 Let an be a Cauchy sequence and let am be a convergent subsequence Prove that an is convergent and lirn an klirn am Hoe Aux Again this statement does not depend on special properties of R However7 Problern 339 together with what we know about sequences in R proves the converse of Problem 337 is true for sequences in R 340 Let an be Cauchy sequence in R Prove that an is convergent A Cauchy sequence in R is thus the same as a convergent sequence in R The advan tage is that the de nition of a Cauchy sequence makes no reference to the limit it is an intrinsic property of the sequence This fact does not hold in every universe7 ls every Cauchy sequence in Q convergent to some point in Q 341 Prove or give a counterexarnple If a sequence of real numbers 95 has the property that for all 8 gt 07 there exists N E N such that for all n 2 N we have lxn 7 znl lt 87 then is a convergent sequence How is this different from the de nition of a Cauchy sequence 7 Decimals 342 The way that many people think of real numbers is as decirnal expansions Of course the only decirnal expansions that we can easily write down either terrninate 10125 Z 054 8 50 or become periodic 03333 03 0142857142857 0142857 Write down a decimal that is not periodic in such a way that the pattern is clear It is an interesting fact that a decimal expansion which either terrninates or becomes periodic represents a rational number Unfortunately some numbers have two decimal expansions For example we could write 1 1 7 01249 8 All the numbers that have multiple decirnal expansions are rational However7 not all rational numbers have multiple expansions 26 3 SEQUENCES Shortly we will prove that 7139 has a decimal expansion even though not all the digits are known as of this writing the rst 1241100000000 decimal digits are known They were calculated by the laboratory of Yasurnasa Kanada at the University of Tokyo Chao Lu of China holds the Guinness Book of Records record for reciting digits of 7T he recited 67890 digits Suppose s E 01 For a nite sequence 711 71 with n E 0 1 9 we can de ne a closed subin terval of 01 by 711 712 M71 7 711 712 M71 71k 1 n n i 7 7 77 k 10 100 101 l Jr10k 10 100 101 1 10k IO 1 2 3 I4 5 6 I7 8 9 0 01 02 03 04 05 06 07 08 09 1 FIGURE 1 01 is the union of the intervals 1019 In this case x E 2 First we notice that 071l U01i and hence there exists n1 6 0 1 9 such that z 6 1m 02 021 022 023 024 025 026 027 028 029 03 FIGURE 2 Since z E 2 we subdivide 2 Now we suppose that we have 711 71 with n E 0 1 9 such that s E Imwm Notice that In1nk U01n1nki and hence there exists 71k 6 01 9 such that z 6 my nknk1 This process de nes an in nite sequence 7117127133 and the decimal representation of z is then nlngng In general if z E R we let no be the largest integer less than or equal to x Then the decimal expansion for z is 710111712713 where 0711712713 is the decimal expansion for z 7 no 6 0 1 7 DECIMALS 27 343 1 For each k E N de ne M E Q by k 71071171k Prove that klim xk x H00 2 Explain how a real number x may have two decimal expansions 3 Suppose my 6 R and suppose ak and bk are decimal expansions for z and y respectively In addition7 assume neither ak nor bk ends with a constant sequence of 97s see the previous question Show that z lt y if and only if there exists a k E N with a0a1a2 ak lt b0b1b2 bk 4 Suppose my 6 R and suppose ak and bk are decimal expansions for z and y respectively Describe and prove how to nd a decimal expansion for z y 5 Prove that a decimal expansion is eventually periodic if and only if it comes from a rational number 344 Let dn 1 be an arbitrary sequence with dn 6 017quot 79 for each n E N De ne a sequence 1011 by wlw d an0d1d2dn210 k1 Prove that anff1 converges to a real number Hence7 give another proof that for any 17 b E R with a lt b there exists a rational number x 6 Lab CHAPTER 4 Limits and Continuity 1 Limits The intuitive idea of a number L E R being the limit of a function f as x approaches a point p is that for all x close enough to p the value of the function is as close as we like to L The limit should not depend on the value of f at p but only on the value of f at points z near p Indeed for a limit to exist at p it is not even necessary that f be de ned at p but only that f be de ned at points z near p De nition Let I Q R be an interval f I 7 R a function p E I and L E R We say that L is a limit of f as x approaches p if for every 8 gt 0 there exists a 6 gt 0 such that for all z E I if 0 lt lx 7pl lt 6then 7 Ll lt 8 When proving that L is a limit of f as x approaches p we are given an arbitrary 8 gt 0 and have to nd a 6 gt 0 exactly as we had to nd a N E N when proving that L was the limit of a sequence In practice this means that we seek to estimate lf 7 Ll from above making it lt 8 using lx 7pl lt 6 This de nition gives you no way of nding a limit L Exactly as for sequences Problem 32 we have to show that limits are in fact unique 41 Let I Q R be an interval f I 7 R a function and p E I Suppose L and M are both limits of f as x approaches p Show that L M This shows that if a limit of f as x approaches p exists then it is unique Now we can talk about the limit of f as x approaches p and write limfz L map Note When we write limfz L we are making two assertions the limit of f as x map approaches p exists and its value is L Exactly as with sequences we must take care never to write limfx until after we have shown that the limit exists map If the point p can be approached from both sides by points of I that is ifp E I then we can de ne the lefthand limit and righthand limit of f as x approaches p De nition Let I Q R be an interval f I 7 R a function p E I and L E R We say that L is a righthand limit of f as x approaches p if for every 8 gt 0 there exists a6 gt 0 such that I pp6 31 Q and for all z E I ifp lt z ltp6then lf 7 Ll lt 8 We say that L is a lefthand limit of f as x approaches p if for 29 3O 4 LIMITS AND CONTINUITY I I I I I I I I I I I I I I I I I I f 17 b a R I Y I I I I I I I I I I I I I r 6 b FIGURE 1 Showing that the limit of f 17 a R as x approaches p is L a possible choice of 6 gt 0 for the given 8 gt 0 every 8 gt 0 there exists a 6 gt 07 such that I p 7 67p 31 I and for all z E I if p76ltxltpthenIfx7LIlt8 Similarly left hand and right hand limits are unique why and are denoted by lim f map L and limIfx L respectively map 42 Let I Q R7 f I a R a function7 and p E I Prove that limfz L if and map only if lim f lim f L map 417 43 Let I Q R be an interval and p E I Give the negation of the de nition of limfx L map Note The negation of limfx L77 is not limf 31 L77 since that implies map map the existence of the limit In words we would phrase the negation as fx does not approach L as x approaches p There are two possibilities limfz exists but map limfz 31 L7 or f has no limit as x approaches p map 44 Let f R a R be given by f 3 for all z E R Let p E R be arbitrary Prove that limfz 3 map 1 LIMITS 31 Remark In this case the 6 that you get does not depend on 8 for any p For a function f de ned on an interval this only occurs when f is constant 45 Let f R 7 R be given by f z for all z E R Let p E R Prove limfz p map Remark Now we see that 6 depends on 8 and that 6 goes to 0 as 8 goes to 0 However the choice of 6 still does not depend on the choice of p 46 Let f R 7 R be given by f 3x 7 5 for all z E R Let p E R Prove limfz 3p 7 5 map 47 Let f R 7 R be given by fx 2 for all z E R Let p E R Prove limfz p2 map Hint Start the proof by writing down what needs to be shown The goal is always to estimate 7 102 from above by some function of 6 that goes to 0 as 6 goes to 0 By the difference of squares formula If 7 102 l Pl l 7 pl The de nition of the limit gives us the estimate Ix 7 pl lt 6 However since our choice of 6 is only allowed to depend on 8 and p but not on x we must estimate lx pl by some quantity independent of x Since we know that lx 713 lt 6 we can say lt lpl 6 and hence we can estimate l96pl S M W lt mm 5 and hence W 7p2l lt mm 55 Now if we x an 8 gt 0 then we can always choose 6 gt 0 such that 2lpl 56 lt 8 If we complete the square or use the quadratic formula then we nd an optimal choice for 6 However it is important to remember that we dont need to nd the best 6 we just need to nd a 6 gt 0 that works If we knew 6 S 1 then we could estimate lx pl lt 2lpl 1 and hence If 7102 lt 2lpl 16 From this it is easy to choose a 6 gt 0 such that both 6 S 1 and 2lpl16 lt 8 Note Regardless of how we produce our choice of 6 it depends on the point p For a xed 8 we see that the 6 required gets smaller and smaller as lpl gets bigger and bigger How could you tell this from looking at the graph of f 48 Using the de nition of the limit of a function ie the 8 6 de nition prove that lin2x2 7 z 1 7 49 Let I C R be an interval and let f I 7 R with p E I Assume there exist a lt b and 6 gt 0 such that for all z E I if lx7pl lt 6 then f E ab Prove that if L lim f exists then L E ab map 32 4 LIMITS AND CONTINUITY 410 Let f R a R be given by 0 z is irrational f 96 1 z is rational Prove that lirnfz does not exist for any p E R map 411 Let f R a R be given by 0 z is irrational Wt s z is rational Prove that lirnfz exists for only one value of p map 412 Let f ab a R and let p E ab Assume lirn L exists with L gt 0 Prove map that there exists 6 gt 0 such that if z E ab and 0 lt lx ipl lt 6 then f gt L2 The next problem relates the limit of a function as x approaches p to the limits of sequences 413 Let I Q R be an interval f I a R a function p E I and L E R Prove that lirnfx L if and only if for every sequence Q I p with lirn xn p map Hoe we have lirn fzn L Hoe Hint To prove if for every sequence C I a with lirn xn p we have Hoe lirn fxn L77 you should switch to the contrapositive Hoe 414 State the contrapositive of the sequential characterization Problern 413 of limits ie get a new if and only if statement by negating both sides This statement is quite useful for proving that a function has no limit as x approaches p 415 Which of these limits if any exist Prove your answer l 1 lin sing 2 lin s s1n The following are the analogues of the limit laws for sequences Problern 313 416 Let I Q R be an interval p E I and f I a R and g I a R functions satisfying lirnfx L lirngx M map map Let c E R Prove that 1 lirnc f c L 2 CONTINUOUS FUNCTIONS 33 2 gym gz L M lt3 gm we L M 4 lfgz0forz6IandM0theniEn There are two ways to prove these statements one is to use the de nition of limit directly and the other is to use the sequential characterization of the limit Problem 413 and our earlier limit theorems for sequences Problem 313 Remark The condition g 74 0 for z E I is stronger than necessary It ensures that the function is de ned for all z E I If limgz 74 0 then there exists an map interval J Q I with p E j and g 74 0 for all z E J 417 Let I C R be an interval and let p E I Let f g and h be functions on I p such that g 3 fx 3 h for all z E I Prove that if limgx map lim h L E R then lim f L This is the Squeeze Theorem for functions map map 2 Continuous Functions De nition Let I Q R be an interval f I a R a function and p E I We say that f is continuous at p if for every 8 gt 0 there exists a 6 gt 0 such that if z E I and 196 7191 lt 5 then lfW fpl lt 8 lf f is continuous at all p E I it is called continuous If S Q I and f is continuous at all p E S it is called continuous on S 418 Let f I a R where I is an interval and let p E I Negate the de nition of f is continuous at p 419 Let f R a R be given by 1 0 S s S 1 fa T 0 otherwise Find all points p E R at which f is continuous Justify your answer 420 Let f R a R be given by 0 z is irrational 1 z is rational Find all points p E R at which f is continuous Justify your answer 421 Let f R a R be given by 0 z is irrational z z is rational 34 4 LIMITS AND CONTINUITY Find all points p E R at which f is continuous Justify your answer The de nition of f being continuous at p is very similar to the de nition of the limit of f as x approaches p The following theorem makes the connection explicit This is probably the de nition of continuity that you saw in your Calculus class 422 Let I C R be an interval and p E I Prove that f is continuous at p if and only if limfz fp map Since we have a sequential characterization of the limit of f as x approaches p Problem 413 we can obviously produce a sequential characterization of continuity 423 Let I Q R be an interval f I a R a function and p E I Prove that f is continuous at p if and only if for every sequence Q I with lim zn p we have Hoe 11m f 9 f P Hoe This is slightly different from a direct application of Problem 413 since we have every sequence Q I rather than every sequence Q I ls the theorem still true if we replace every sequence Q I77 with every sequence Q I p 424 Give the contrapositive to the sequential characterization of continuity Prob lem 423 The limit laws for functions Problem 416 can be reinterpreted in terms of continuous functions 425 Let I Q R be an interval f I a R and g I a R functions and p E I Assume that f and g are continuous at p Let c E R Prove f g is continuous at p c f is continuous at p 3 f g is continuous at p lf 9z 31 0 for z E I then is continuous at p 426 Prove that every polynomial function is continuous on R There is one more operation which we can perform with functions that has no direct analogue for sequences 427 Let I J Q R be intervals f I a R and g J a R functions with fI Q J Let p E I Prove that if f is continuous at p and g is continuous at fp then 9 o f is continuous at p This can be proved either directly from the de nition or by repeated application of Problem 423 One of the most important theorems about continuous functions on intervals is the lntermediate Value Theorem 2 CONTINUOUS FUNCTIONS 35 428 lf f is a continuous function on ab with a lt b and fa lt y lt fb or fa gt y gt fb then there exists p E ab with fp y Hint Suppose fa lt y lt fb and let E x E ab fx lt Let p sup E The point p can be written as the limit of a sequence xn E E and as the limit of a sequence x E ab E Hence prove that fp y 429 Let I Q R be any interval and f I a R a non constant continuous function Prove that fI is an interval Remark First prove that it suf ces to show that given any two points Cd E fI the entire interval between them is contained in fI In general we cannot say any more about the interval fI De nition We say that a function f I a R is bounded if the set fI is bounded Thus f is bounded if and only if there exists K E R such that S K for all z E I 430 Give an example of a continuous function f 01 a R that is not bounded Thus the continuous image of a bounded interval may be unbounded 431 Give an example of a continuous function f 01 a R such that f01 is a closed and bounded interval Thus the continuous image of an open interval may be a closed interval However in the special case of a continuous function on a closed and bounded interval we can say a lot more 432 Let f be a continuous function on ab with a lt b Show that f is bounded Hint Proceed by contradiction Suppose that f is not bounded above and construct a sequence 0311 with xn E ab for all n E N such that lim fan 00 400 Now apply the sequential characterization of continuity Problem 423 to obtain a contradiction 433 Let f be a continuous function on ab with a lt b Show that there exist zmM E ab such that fm 3 fx 3 fM for all z E ab We say f achieves its maximum and minimum value Hint Construct a sequence 0302 with xn E ab for all n E N such that hm supf i E Now apply the sequential characterization of continuity Problem 423 to nd a point p E Ml with fp supf t 96 6 fall 36 4 LIMITS AND CONTINUITY 434 Give an example of f 01 7 R that is bounded continuous and has neither a maximum nor a minimum Can you do the same for f 01 7 R 3 Uniform Continuity Sometimes we encounter a property stronger than continuity Recall that f I 7 R is continuous on I if for all p E I and for all 8 gt 0 there exists a 6 gt 0 such that for all x E I if Ix 7pI lt 6 then 7 fpI lt 8 For a continuous function the 6 generally depends upon both 8 and the point p as previous exercises have illustrated lf we remove the dependence on p we have uniform continuity De nition Let I Q R be an interval and f I 7 R a function We say that f is uniformly continuous on I if for all 8 gt 0 there exists a 6 gt 0 such that for all xy E I if Ix 7yI lt 6 then 7 lt 8 435 Suppose I Q R is an interval Prove that if f I 7 R is uniformly continuous on I then f is continuous on I 436 Let fx mxc for some 7710 E R Prove that fx is uniformly continuous on R 437 Negate the de nition of uniform continuity 438 Let f R 7 R be given by fx x2 Prove that f is not uniformly continuous on R Hint Fix an 8 gt 0 and show that no matter what 6 gt 0 is chosen you can always choose xy E R such that Ix 7 yI lt 6 and Ix2 7sz 2 8 439 Prove that if I is a bounded interval and f I 7 R is uniformly continuous then f is bounded This together with Problem 430 shows that we can have continuous functions on 0 1 that are not uniformly continuous As in the previous section the case of functions on closed and bounded intervals is very different 440 Let f IabI 7 R be a continuous function Prove that f is uniformly continuous Hint Suppose that f is not uniformly continuous Then there exists an 8 gt 0 such that for every 6 gt 0 there exists xy E IabI such that Ix7yI lt 6 but 7fyI 2 8 Show that there exists two sequences xn C IabI which both converge to the same point p E IabI but such that 7 2 8 Hence conclude that f is not continuous at p CHAPTER 5 Differentiation 1 Derivatives De nition Let I C R be an interval7 f I a R be a function7 and p 6 I0 The function f is said to be differentiable at p if hm we 7 fp zap 7 p If f is differentiable at p then we de ne the derivative of f at p7 denoted f p7 by W hm we 7 fp map zip exists If S Q I7 f is said to be differentiable on S if f is differentiable at z for all z E S 51 1 Let c E R be arbitrary and f R a R be de ned by f c Prove that f is differentiable on R and that f 0 for all z E R 2 Let f R a R be de ned by f x Prove that f is differentiable on R and that f 17 for all z E R 3 Let f R a R be de ned by fx x2 7 1 Prove that f is differentiable on R and nd f for z E R 4 Let f R a R be de ned by f Prove that f is not differentiable at 0 We can relate the notion of f being differentiable at p with our earlier notion of continuity 52 Let I Q R be an interval7 f I a R a function7 and p E I Prove that f is differentiable at p if and only if there exists a function b I a R that is continuous at p such that f96 M 96 7 WW Moreover7 if there exists a function b I a R that is continuous at p such that f96 M 96 7 WW then f p 451 Hint This is really just a very useful restatement of the de nition 37 38 5 DIFFERENTIATION Now many of our results about differentiability of f will follow from our rules for continuous functions Problem 425 applied to b 53 Let I Q R be an interval f I a R a function and p E I Prove that if f is differentiable at p then f is continuous at p In particular using Problem 52 the usual rules of differentiation come from Problem 425 by some simple algebra 54 Let I Q R be an interval f I a R a function g I a R a function 0 E R and p E I If f is differentiable at p and g is differentiable at p then 1 c f is differentiable at p and C f p 0 f p 2 f g is differentiable at p and f g f p Mp 3 f g is differentiable at p and f 9W fp 9 p f p 91 4 if gp 31 0 then g is differentiable at p and f gpf pfpg p GNP 2H t 9 MM Hint You can prove these either directly from the de nition or by writing f fp x 7 p and g 91 x 7 p and using Problem 52 You should try both ways 55 Let 132 a0 I am I an l be a polynomial Prove that for all z Pz a1 2a2x nanznil There is one more standard differentiation rule the Chain Rule 56 Let IJ Q R be intervals f I a R and g J a R functions with fI Q J Let p E I Prove that if f is differentiable at p and g is differentiable at fp then 9 o f is differentiable at p and g o f p g fp f p Hint Since 9 is differentiable at fp there exists a function 7 J a R such that 996 gfp x fp WW for all z E J Replace x by f and then use the fact that fx 7fp zip Perhaps the most important applications of differentiation are in optimization and in estimation In optimization we try to nd the maximum or minimum value of a given function often subject to one or more constraints 2 THE MEAN VALUE THEOREM 39 2 The Mean Value Theorem De nition Let I Q R be an interval and f I a R a function We say p E I is a local maximum if there exists a 6 gt 0 such that for all s E I with 1x 7 pl lt 6 f S fp We say p E I is a local minimum if there exists a 6 gt 0 such that for all z E I with 1x 710 lt 6 f 2 fp We begin with a lemma that relates local maxima and minima with differentiation 57 Let I Q R be an interval and f I a R a function Prove that ifp E I is either a local maximum or a local minimum and f is differentiable at p then f p 0 Hint Assume p is a local maximum and compute the signs of ml ifwdlml ifl maer 7 p map 7 p If f is differentiable at p then these limits must be equal With this observation we can prove Rolle7s Theorem A version of the theorem was rst stated by Indian astronomer Bhaskara in the 12th century however the rst proof seems to be due to Michel Rolle in 1691 58 Let a lt b Prove that if f ab a R is continuous f is differentiable on ab and fa fb then there exists a point p E ab with f p 0 Hint If f is not constant it has either a maximum value or a minimum value at some 0 E ab An immediate consequence of Rolle7s Theorem is the very important Mean Value Theorem The Mean Value Theorem is used extensively in estimation 59 Let a lt b Prove that if f ab a R is continuous and f is differentiable on ab then there exists 0 E ab with f b i f a 7 f c 7 b 7 a Hint Construct a linear function lx with la fa lb fb and consider 996 f96 96 We now give some consequences of the Mean Value Theorem 510 Let a lt b Prove that if f ab a R is continuous f is differentiable on ab and f p 0 for all p E ab then f is a constant 511 Let a lt I f ab a R be continuous and f differentiable on ab Prove that 1 if f 2 0 for allx E ab then f is increasing on ab ie ifa S x lt y S b then f95 S y 5 DIFFERENTIATION 2 if f S 0 for all z 6 Lab then f is decreasing on 177 ie ifa S x lt y 3 5731811 95 2 y CHAPTER 6 Integration 1 The De nition Our nal chapter is the other half of calculus the de nite integral Again let7s recall a familiar problem to motivate our de nition Problem Let f ab a 0 00 be continuous Find the area of the region bounded by z 1x b y f and y 0 Draw a few pictures of such f7s eg f x2 on 02 Geometry does not give us a formula for this area unless f is quite nice eg f 3 Our approach for nding this area will be to use very thin rectangles to approximate the area and then use a limiting process to obtain the area It looks complicated so be sure to draw some pictures to help you understand the notation In this chapter we de ne f dz commonly called the de nite Riemann in tegral of f over ab You should recall from calculus the short way77 of computing this 2 22 zdx 27l 1 21 2 2 This comes from the fundamental theorem of calculus which we shall prove We will de ne dx so that when f 2 0 and f is continuous this number is the area of the region bounded by y fx z a z b and the x aXis De nition Let a lt b Let f ab a R be a bounded function A partition P of ab is an ordered nite set ax0lt1ltltznb If P and Q are two partitions of ab we say that Q re nes P if Q Q P Let P x0x1xn be a partition of ab For 1 S 239 S n we set Mif P supf zi1 S x S supfq iih il llf P inffx xi1 S x S inf W217 Al ii 7 M71 41 42 6 INTEGRATION We de ne the upper Riemann sum of f with respect to P denoted Uf P by UMHZMMHamp i1 and we de ne the lower Riemann sum of f with respect to P denoted Lf P by mezmmmw 61 Let f m z and g 2 for z E 01 Let n E N and let Pn 0 lial be a partition of 01 Draw a picture illustrating LfP4 UfP4 LgP4 and UgP4 Can you nd expressions for LfPL UfPL L9Pk7and UK97fk7 62 Let f ab a R be bounded and let P be a partition of ab Let m inf fab and M supfab 1 Prove mltbeagt Leap UfP Mew 2 Prove that if Q re nes P then W P LfQ U f Q s U f P Hint First prove that it suf ces to show this when Q P 1 Then prove 2 in this case De nition Let f 1 b a R be bounded We de ne the upper Riemann integral of f denoted Uf by Uf infUf P P is a partition of ab We de ne the lower Riemann integral of f denoted Lf by Lf supLf P P is a partition of ab Why do these exist We say f is Riemann integrable if Lf Uf In this case we call the common value the de nite Riemann integral of f over the interval ab which we denote Remember that the in mum or supremum of a set may not be a member of that set 63 Show that for f and g as in Problem 61 for all n E N UfPL 31 Uf Lf Pn 31 Lf and similarly for 9 We begin by seeing that not every bounded function is integrable Later we will prove that every continuous function is integrable 64 Let f 01 a R be given by i 0 z 0lQ n1x MH Q 2 INTEGRABLE FUNCTIONS 43 Prove or disprove the statement f is integrable 65 Let f 11 a R be bounded and let P and Q be partitions of 11 Prove that Lf7P S Uf7Q Hint Consider the partition R P U Q 66 Let f 11 a R be bounded Show that Lf S Uf 67 Let f 01 a R be given by fx z for all z E 01 Let Pn be the partition of 01 given by 1 Find LfP and UfP 3 Show f is integrable on 01 and nd fol f Remark Notice for my partitions P Q we have Uf P 31 LfQ even though Uf Lf 2 Integrable Functions The de nition of integrability can be dif cult to use directly so we are fortunate to have this next problem 68 Let f ab a R be bounded Show that f is integrable on ab if and only if for all 8 gt 0 there exists a partition P 0 xn of the interval 11 such that W P 7 L021 2040213 7 mm PM lt e i1 69 Let f ab a R be an increasing function Prove that f is integrable on 11 Hint For an increasing function we know explicitly what Mif P and mif P are 610 Let f ab a R be continuous Prove that f is integrable on 11 Hint Use Problern 440 to conclude that f is uniformly continuous Let 8 gt 0 be arbitrary and choose 6 gt 0 so that if Ly E 11 with x 7y lt 6 then 7 lt 17 Let P be any partition with each Aiz lt 6 and use Problern 68 611 Let f and g be integrable functions on 1 b Let c E R be an arbitrary constant Prove that 1 c f is integrable on ab and abcfcabf 44 6 INTEGRATION 2 f g is integrable on ab and b b b f9 f g 913 S and hence conclude that Uf gP M 97P 2 Lf P L97P Use prove the equation Look at f z and g 17 x on the interval I 01 Then supfI 1 supgI and inffI 0 infgI notice that neither f nor 9 has a mam39mum or a minimum on I The function f fx g x17 z 1 and is therefore constant on I Thus supf gI 1 lt 1 1 supfI supgI and inff gI 1 gt 0 0 inffI infgI 612 Let f and g be integrable on ab with f S g for all z E ab Prove that fff fjg 613 Let a lt c lt b and let f ab 7 R Hint Show 7P Mi97P S Uf P UgP Similarly show that Problem 68 to conclude integrability Then 1 Assume f is integrable on ab Prove that f is integrable on 10 and cb 2 Assume that f is integrable on 10 and on cb Prove that f is integrable onlab and abffcbf De nition lf f is integrable on ab we de ne fif 7 We de ne fjf 0 614 Let f be integrable on an interval containing a b and c Prove that no matter what their order 17 c b f f f 615 Let f ab 7 R be integrable Prove that 1 lfl is integrable on ab Hint Prove that Milfl P 7 llm P S Mif P 7 mif P 2 lfffl grim 616 Prove that if f and g are integrable on ab then f g is integrable on ab Hint Since f and g are integrable we may de ne Lf 7supwltzgti s z s b Lg suplgzl a S x S b 3 FUNDAMENTAL THEOREMS OF CALCULUS 45 Use the fact that to conclude MM 9713 milf 9713 S Lf MKQJJ milt97P Mif7P miltf7P Lg Now use Problem 68 3 Fundamental Theorems of Calculus Finally we reach the Fundamental Theorem of Calculus which relates integration and differentiation It is commonly broken into two different theorems 617 Prove the rst Fundamental Theorem of Calculus Let f be integrable on 17 De ne a function F 17 7 R by Fzf F is uniformly continuous on 17 and if f is continuous at c 6 Lab then F c f C Hint For uniform continuity show that if a S x S y S b then Fltygt 7Fltzgt yr Now estimate f from above and below in terms of y 7 x For F c fc show that M7110 1 mlf7fcl s 7 c s 7 c c Use the fact that f is continuous at c to conclude that the right hand side can be made arbitrarily small by taking z suf ciently close to c 618 Prove the second Fundamental Theorem of Calculus If f is differentiable on 17 and f is integrable on 17 then f fb 7 fa Hint If P xmxl 7x is any partition of 17 then7 by Problem 597 for each 1 S 239 S n we can nd ti 6 xi47m so that 46 6 INTEGRATION 4 Integration Rules Once we have the fundamental theorems ofcalculus we can prove two ofthe important rules of integration integration by parts7 and integration by substitution These turn out to be reinterpretations of the product rule from differentiation and the chain rule respectively 619 Prove that if f and g are functions that are differentiable on 17 and both f and g are integrable on 17 then b b f969 96fb9bfa9a 90496 Hint By Problem 54 f g is differentiable and f g f g f g 620 Suppose that u is differentiable on 07d and u is continuous on 07d Let a uc and b Suppose that f is continuous on uc7 Prove that Abfcdf0uu Hint Let f Use Problem 617 to show that F is differentiable and use Problem 56 to show that F o u is differentiable Now apply Problem 618 Remark Notice we dont need to assume that ucd 17 Appendix to Chapter 2 1 The Field Properties Mathematicians study many different types of mathematical objects You may have heard of groups rings topological spaces smooth manifolds vector spaces Banach spaces af ne varieties elliptic curves etc One of the objects which mathematicians study is called a eld In the introduction to the chapter we mentioned several algebraic properties of R The crucial algebraic properties of R can be summarized by saying that R is a eld Notice that all the eld properties listed below would certainly be demanded of any number system As we mentioned above we will take all the properties on faith Hence will call them axioms in mathematics an axiom is a basic statement which is accepted without proof for example the statement which says There exists a set77 is a basic axiom of mathematics AXIOM 1 There exists a set R which contains Q We may de ne two operations on R called addition and multiplication which extend normal addition and multiplication of rational numbers When we say that addition on R extends addition on Q we mean that adding two real numbers which happen to be rational would be the same as the normal addition of rational numbers and likewise for multiplication We will use all the standard notations regarding operations among numbers For example a b is the sum of ab E R As always we write the symbol 7 between two real numbers which are the same and the symbol 79 between two which are not AXIOM 2 Addition of real numbers is commutative For every 1 b E R ab ba AXIOM 3 Addition of real numbers is associative For every 1 b c E R 1 b c a b c AXIOM 4 The real number zero is an additive identity For each a E R 0 a a AXIOM 5 Every real number has an additive inverse For every 1 E R there is a number b E R such that a b 0 We mentioned above that there are many obvious facts about the real numbers that strictly speaking must be proven from the axioms The following is an example as are most of the exercises in this section 47 48 APPENDIX TO CHAPTER 2 A21 For every 1 E R the additive inverse of a is unique That is if b and c are real numbers which satisfy a b 0 and a c 0 we may conclude that b c The previous problem justi es us saying THE additive inverse of a E R rather then AN additive inverse As usual we will use the symbol 7a for the additive inverse of a Notice that strictly speaking 7a is not the same symbol as 71 a that is the number negative 1 times the number a That the two symbols represent the same number will be one the obvious facts we prove below We can also now de ne subtraction If a and b are natural numbers then a 7 b is de ned to be a 7b in words a 7 b is the the sum of a and the additive inverse of b AXIOM 6 Multiplication of real numbers is commutative For every 1 b E R ab ba AXIOM 7 Multiplication of real numbers is associative For every 1 b c E R abc abc AXIOM 8 The number one is a multiplicative identity For every 1 E R a 1 a AXIOM 9 Every real number besides zero has a multiplicative inverse For every 1 E R a 31 0 there is a number b such that ab 1 We have a result about multiplicative inverses analogous to the one we had for additive inverse A22 For every 1 E R other than zero the multiplicative inverse of a is unique That is if b and c are real numbers which satisfy ab 1 and ac 1 we may conclude that b 0 Again we are now justi ed in referring to THE multiplicative inverse of a which we will denote by 1 1 We de ne division in a similar manner as subtraction 11 is de ned to be 1194 that is 11 is de ned to be the product of a and the multiplicative inverse of b AXIOM 10 Multiplication and addition satisfy the distributive property For every abc E R abc abac These algebraic properties of the real numbers are very important but they are not unique to R Q would satisfy all of these axioms and so is also a eld In general there exist many different elds The collection of complex numbers C with usual notions of addition and subtraction is a eld For a prime number p you may be familiar with the collection of numbers modulo p often denoted Z17 It too is a eld and in contrast to Q R and C has nitely many elements We will thus need more properties of R to describe it uniquely The following relatively simple question might help you to better understand the axioms 2 THE ORDER PROPERTIES 49 A23 Which axioms would still be satis ed if R were replaced with Q with Z with N We will now give some more basic properties about the real numbers which follow from these axioms For our rst result we will see that the multiplication operation on R still boils down to repeated addition as long as one of the numbers is a natural number A24 Multiplication of real numbers by natural numbers is just repeated addition That is if a E R and n E N no is the same as the number which results when a is added to itself 71 times Hint Use induction As promised we will show that the additive inverse of a real number is just that number multiplied by 71 A25 For every 1 E R the product of a and 71 is the additive inverse of a That is 7a 71a We can also de ne integral powers of real numbers in the usual way De nition Let a E R If n is a natural number we de ne a to be the product of a with itself 71 times If a is de ned to be the product of 1 1 with itself 71 times We also de ne a0 to be 1 We have the following basic properties of powers A26 lf ab E R and 77171 E Z then 1 Wm am WW 2 am ama and 3 abm ambm Hint These properties are by no means automatic They must be proven by careful reasoning from the axioms 2 The Order Properties In the previous section we saw the algebraic or eld properties of R In this one we will study the order properties As mentioned in the introduction a set is ordered if we have a rule which tells us given two elements of the set which is bigger AXIOM 11 The real numbers come equipped with an order which extends the order on Q By extends7 we mean that if a and b are rational numbers then a is less than b according to the order on Q if and only if a is less than b according to the order on R 50 APPENDIX TO CHAPTER 2 As usual we denote the order by S AXIOM 12 The order is re exive For every 1 E R a S a AXIOM 13 The order is transitive For every abc E R such that a S b and b S c we have a S c AXIOM 14 The order is antisymmetric For every 1 b E R such that a S b and b S a we have a b AXIOM 15 The order is a total order For every 11 E R either a S b or b S a R is by no means the only set that comes with an order In fact an order can be de ned on any set and many sets like for example the set consisting of all the months in the year have an obvious order Actually there are many different ways to de ne an order on R but there is only one order that will satisfy all the axioms we will list and have listed We will also use the symbols lt gt and 2 with their usual meanings ie a lt b means a S b and a 31 b To make our words precise we will pronounce a S b as a is less than if7 and a lt b as a is strictly less than b77 with similar phrasing for 2 and gt Note then that a is less than b77 includes the possibility that a b This is only a convention but it is one used by many mathematicians Again we have many basic and obvious properties A27 R satis es the trichotomy property if 11 E R then exactly one of the following holds 1 altb 2 agtb or 3 11 As expected a number which is strictly greater than zero is called positive whereas a number which is either positive or zero in other words a number that is greater then zero is called nonnegative We use the terms negative and nonpositive similarly though nonpositive is typically used with less frequency 3 The Ordered Field Properties In this section we will discuss how the algebraic eld properties of R interact with the order properties again in ways that if you think about them should work in any system of numbers AXIOM 16 The order is preserved under addition by a xed number If abc E R andagbthenac bo AXIOM 17 The product oftwo nonnegative numbers is again nonnegative lf 11 E R 0gaand0 bthen0 ab 3 THE ORDERED FIELD PROPERTIES 51 To say that R satis es these additional properties is to say that it is an ordered eld Notice that being an ordered eld is much more restrictive than being a eld and having an order The eld properties and the order properties must also interact in the right way as described by the previous two axioms For example7 although there many orders on the set of complex numbers7 7 there is no order which makes it into an ordered eld this is not too dif cult to prove and we will do so below Demanding that our numbers form an ordered eld tells us that we cannot include imaginary numbers or complex numbers in our number system It also turns out that there is no order on Z17 which makes it into an ordered eld Nevertheless7 R is not the only ordered eld Q is an ordered eld and there are many others We will need one additional property7 called the completeness axiom7 to uniquely de ne R As we mentioned above7 the completeness axiom is signi cantly deeper than the others and we will need to develop several new concepts in the next chapter before we can describe it Again we have many basic properties that follow from the axioms As always7 be careful not to use any facts other than the axioms and other facts we have proven The next result shows that we may multiply inequalities by 71 as long as we are willing to reverse the sign A28 Let a7b E R If a S b then 7b S 7a More generally we may multiply an inequality by a real number7 but7 as expected7 we must reverse the sign if the number is negative A29 Suppose a7b E R and a S b lf 0 E R is nonnegative7 then ac S be lf 0 is nonpositive then be 3 ac Of course the same result holds for strict inequalities unless c 0 by a similar proof A210 Given any number a E R7 there is a number which is strictly larger Hint Finding a number is not dif cult7 but prove rigorously that it is larger A211 lfaER7 a220with a20if and only ifa0 We will not have occasion to use the next two results7 but they are of general interest and they provide insights into ordered elds A212 Suppose we have a eld F that is7 F satis es all the properties which we gave for R in the section on the eld properties In addition suppose there is an element 239 E F which satis es 2392 71 Then there is no order on F which makes it into an order eld that is7 no order which will satisfy all the properties given for R in this chapter Hint Suppose F does indeed satisfy all the properties we have given so far for R How does 239 compare to zero A213 There is no order on C which makes it an ordered eld 52 APPENDIX TO CHAPTER 2 This last result is important because C does satisfy the completeness axiom Thus there are elds other then R which satisfy the completeness axiom7 but no other ordered elds which satisfy it Notice that Q is an ordered eld which is not complete and C is a completed eld which is not ordered All of the assumed properties of R are now in place except completeness The re maining statements of this chapter must therefore be proven from our axioms 4 Set Theory We will not try and de ne what we mean by a set Surprisingly this is actually quite complicated and there is a branch of mathematical logic called set theory that deals with this De nition The basic set theory concepts we expect you to be familiar with are 1 g is the empty set7 ie the set with no elements 2 A Q B means that every element of A is an element of B7 or for all z 6 A7 z E B It is read A is a subset of BF7 3 A B means that A and B have the same elements Another way of saying this is z E A if and only if z E B 4 A B x z E A and z E B A Bis read as A intersect B77 and is called the intersection of A and B 5 A U B x z E A or z E B A U Bis read as A union B77 and is called the union ofA and B lfx E AUB then x E A or z E B It could be in both 6 A and B are called disjoint sets if A B g A and B are disjoint if they have no elements in common It turns out that the operations of union and intersection satisfy distributive laws77 reminiscent of those that hold for the real numbers Theorem A214 Suppose that A7 B7 and C are sets Then the following identities hold De nition 1 AB x z E A and z B AB is read as A minus B77 and is called the set theoretic difference of A and BF7 2 AAB AB U B A and is called the symmetric difference of A and B 3 Ac x z A is called the complement of A 5 CARDINALITY 53 Caution To be honest Ac is not really a set since we have not said what z is other than it is not in A When we use A0 we must have a universal set U in mind The universal set is often unspeci ed and is simply inferred from the context Then we can write Ac U A which is unambiguous Theorem A215 Let A and B be sets Then the following are true A Bc A0 U B0 A U B0 Ac BC AABAUBA B Statements 2 and 3 are called De Morgan7s Laws De nition If A and B are sets then the Cartesian product of A and B is de ned to be AgtltBaba Aandb B where a b denotes the ordered pair R2 R gtlt R is the Cartesian plane De nition We can de ne unions and intersections of large collections of sets If I is a set called the index set and for all 239 E I A is a set then we de ne UA x there exists 239 E I such that z 6 Ai ieI Ai x for allz39EI z 6A ieI Theorem A216 If I is a set and for allz E I A is a set then 1 U ADC H A ieI 61 2 ADC U A ieI ieI These are the De Morgan7s laws for unions and intersections over arbitrary index sets 5 Cardinality Having de ned function77 we can proceed to some fun stuff concerning counting the number of elements in a set How can you count an in nite set How can you determine if two sets even in nite ones have the same number of elements or as we shall say the same cardinality Your rst notions of size may be shattered Read on De nition If A and B are sets we write lAl lBl if there exists a 171 onto function f A a B We read lAl lBl as the cardinality of A equals the cardinality of B Note By Theorem 27 in Chapter 2 lAl lBl lBl 54 APPENDIX TO CHAPTER 2 lntuitively lAl lBl means that both sets have the same number of elements This is not startling for nite sets It is no surprise that labcl l17273l However this de nition can lead to non intuitive results We can have A Q B7 A 31 B yet lAl lBl how7 De nition A is nite if A Q or if there exists n E N with lAl H1727 We then say lAl 0 or lAl n accordingly A is in nite if A is not nite A is countabe in nite if lAl A is countable if A is nite or countably in nite Are all in nite sets also countably in nite A217 Prove that a set A is a countably in nite if and only if we can write A 117027 where 11 31 17 if i 7 j b countably in nite if and only if A is in nite and we can write A 117 127 c countable if and only if A Q or we can write A 117027 Deduce that if B Q A and A is countable7 then B is countable A218 Let lAl lBl and lBl Prove that lAl A219 Prove that 51 1N1 H b N Z1 c Niizeltzgtoi 7476787l Hint Try to make an in nite list of all rationals in 01 Now try to make a list of all rationals gt 1 d N lQl A220 If A is countable and B is countable prove that A gtlt B is countable Hint You want to construct a list of all elements in A gtlt B see 117 Can you make an in nite matrix of these elements starting with 11 12 13 b1 52 53 Can you take this matrix and make a list as in 117c Our next problem is due to G Cantor It is a famous result which shook the math ematical world and has found its way into numerous popular mathscience books Cantor went insane The problems solution relies on the decimal representation of a 5 CARDINALITY 55 real number In turn this actually involves the notion of convergence of a sequence of reals which we address in chapter 3 But you can use it here 13 333 means that 13 limH00 xn where xn 333 n entries Beware of this fact Some numbers have 2 decimal representations7 eg7 1 1000 999 This can only happen to numbers which can be represented as decimals with 9 repeating forever from some point A221 a Prove that 071 is not countable Hint If it were countable then we can list 071 117027 a37 Write each al as a decimal to get an in nite matrix as the following example illustrates a1 013974 12 0000002 13 055556 a4 0345587 a5 09871236 Can you nd a decimal in 07 1 that is not on this list Can you describe an algorithm for producing such a number Could a 05 be equal to al Could a 054 be equal to al or a2 b Show that 1071H H07 c lfa lt b show that l01l labl H071 Ha7bll De nition z E R is irrational if z Q Thus R Q is the set of all irrational numbers Can you prove that irrationals exist The next problem shows much more A222 a Prove that if A and B are countable then A U B is countable b Prove that R Q is uncountable ie7 not countable c Prove that if a lt b then a7b R Q 74 0 This problem shows that every open interval contains irrationals In the next chapter we will show that it contains rationals as well d Prove that if I is countable and for all 239 6 I7 Ai is a countable set then U Ai is countable 161 So irrationals do exist Does this proof give you any explicit number in R Q We have not de ned lAl 3 Bl yet A223 Give a de nition for lAl S Your de nition should satisfy 56 APPENDIX TO CHAPTER 2 3 W S W b lAl 3 Bl and lBl S 0 implies that lAl S Bonus Prove also that c lAl 3 Bl and lBl S lAl iplies that lAl De nition lAl lt lBl if lAl 3 Bl and lAl 31 A224 Prove that for all sets A7 lAl lt lPAl so no largest cardinal number77 exists Hint Show there does not exist a function f A a 73A which is onto by assuming such an f exists and considering B E 73A where B a E A a fa Note 73A denotes the set of all subsets of A Thus 7317 2 7 17 27 6 Mathematical Induction THE THEOREM OE MATHEMATICAL INDUCTION Let P1P2P3 be a list of statements each of which is either true or false Suppose that i P1 is true ii For all n E N if Pn is true thert Pn 1 is true Then for all n E N Pn is true A225 Prove this theorem Hint Suppose it were not true Choose no to be the smallest integer so that Pn0 is false A226 Use mathematical induction to establish the following Make sure in your proof to precisely state what you are taking Pn to be a For allnEN7 12n b For allnEN121L227 t2 M2 c For all n 6 N7 ifn 2 4 then 2 lt 71 Note 71 1 2 3 n 7 1 71 This is called 71 factorial77 Appendix to Chapter 3 1 Open and Closed Sets Continuing our discussion of R we turn to what is called topology This is crucial material for our future discussion of limits and continuity De nition Let 8 gt 0 The interval a 7 67a 8 is said to be an open interval centered at a of radius 8 A31 Let a lt b Show that a7b is an open interval of radius 8 for some 8 gt 0 What is the center What is 6 De nitions Let S Q R a S is open if for all a E S there exists 8 gt 0 with a 7 67a 8 Q S b S is closed if CS R S is open QUICK QUESTION ls every S Q R either open or closed Can you justify your answer A32 Prove that every open interval is an open set and every closed interval is a closed set A33 Classify as open7 closed7 both or neither aroma onWMQeHKDWMUR gHimem De nitions Let S C R a z E intS if there exists 8 gt 0 with x 767z8 Q S b x E bdS iffor alle gt 07 x767z8 S7 Q and x7676 CS 7amp0 Note int77 is short for interior and bd is short for boundary A34 For each S nd intS and bdS awn awn Q wk wean mamem A35 Prove the following S Q R a intS Q S and intS is an open set 57 58 APPENDIX TO CHAPTER 3 b S is open gt S intS c S is open gt S bdS Q d S is closed gt S Q bdS A36 Prove the following a If I is a set and for all 2 6 I7 Ai is an open set7 then U Ai is open 23961 b If I is any set and for all 2 6 I7 E is a closed set then H E is closed 23961 V L c If n E N and Ai is an open set for each 2 S n then H Ai is open 21 d If n E N and Ai is a closed set for each 2 S n then U Ai is closed 21 A37 Show by example that c and d in 217 cannot be extended to in nite intersections unions De nitions Let S Q R7 z E R a z is an accumulation point of S if for all 8 gt 07 y E R 0 lt lx 7 yl lt 8 S 7Q V b S x z is an accumulation point of S c z is an isolated point of S if z E S S d 5 S o 8 Note S is called the closure of S A38 Let S Q R Prove the following a z E S is an isolated point of S if and only if there exists 8 gt 0 such that 767z 6 Sx LetSQR b Let x E R Prove that z E S if and only if for all 8 gt 07 x 7 87 8 S is in nite A39 For each set S below nd S 7 S and all isolated points of S a R M c Q d 01 e Qmlto1gt fgtltRQgtmlto1gt g i n e N A310 Prove the following S Q R a S is closed if and only if S Q S S is closed 7 S is closed if and only if S S b c d If F 2 S and F is closed then F 2 S 2 COMPACTNESS 59 2 Compactness Our next topic in topology is compactness The de nition is quite abstract and will take effort to absorb We will later prove that a continuous function on a compact domain achieves both a maximum and a minimum value 7 quite a useful thing in applications De nitions Let S Q R a Let Alha be a family of open sets Alha is an open cover for S if S Q U A id 139 For example 7171 1 n E Z is an open cover of R Question For all z E Q let 81 gt 0 ls 7 81x 8x z E Q necessarily an open cover of R b Let AlLEI be an open cover for S A subcover of this open cover is any collection Alha where IO Q I such that U A Q S i610 c S is compact if every open cover of S admits a nite subcover ie whenever AlLEI is a family of open sets such that S Q U A then there exists a nite id setFQIsothatSQ UAi ieF This is a very abstract de nition that requires study and time to absorb Note that the de nition requires that every open cover of S admits a nite subcover To show S is not compact you only need construct me open cover without a nite subcover Compactness plays a key role in analysis and topology A311 Which of the following sets are compact a 123 b V c 01 d 01 e R A312 Let S Q R be compact Prove that a S is bounded b S is closed Hint Assume not in each case and produce an open cover without a nite subcover A313 Prove that 01 is compact Hint Let Alha be any open cover of 01 Let B x E 01 0x can be covered by a nite subcover of A id Then 0 E B so B 31 0 Let x supB Show z E B Show z 1 A314 Let K Q R be compact and let F Q K be closed Prove that F is compact Hint lf A id covers F then AlLEI U covers K 60 APPENDIX TO CHAPTER 3 A315 Let K Q R be closed and bounded a Prove minK and maXK both exist if K 31 0 b Prove that K is compact Note From 312 and 315 we see that K Q R is compact gt K is closed and bounded A316 Let 1 Q 2 Q 3 Q be a nested sequence of closed7 bounded and nonempty sets in R Then n1 Hint Assume it is empty Then RO 1n 3010le n1 n1 A317 Let K Q R be compact and in nite Prove that K 31 0 Hint Assume K Q A318 Let A Q R be bounded and in nite Prove that A 31 0 3 Sequential Limits and Closed Sets De nition Let A Q R A is sequentially closed if whenever anff1 is a sequence in A converging to a limit 17 then a E A A319 If A Q R is closed then it is sequentially closed A320 If A Q R is sequentially closed then it is closed A321 If A Q R then A is closed if and only if it is sequentially closed CHAPTER 4 Limits and continuity Our rst goal is to de ne and understand limmaa x L Here 1 D 7 R where D Q R We want the de nition to mean roughly7 as x gets close to a then x is close to L Perhaps a more careful description is as x gets closer and closer to a with x 344 017 then x gets close to7 and could equal7 L What does this mean It is still vague It took mathematicians hundreds of years to learn the best way to precisely de ne this First note that x has to be able to get arbitrarily close to a without being equal to a Thus we will need to have a E D It may be that a D If a E D then the value a has no affect whatsoever on the limit DEFINITION Let D g R a e D and L e R Let f D 7 R Then limmaa x L if for every 6 gt 0 there exists 6 gt 0 so that ifx 6 D7 lx7al lt 6 and x 344 athen l x7Ll lt 8 inlogical form V6gt0 3 6gt OVx 6 D7 0lt lx7al lt6 l x7Ll lt 6 We read limmha x L as the limit of x as x approaches a is L 41 Give the logical form of the negation of limmha x L assuming 1 D 7 R and a E D 42 a Let f R 7 R be given by x 3 for all x E R Let a E R Prove that IimH mg 3 b Let f R 7 R be given by x x for all x E R Let a E R Prove limmha x a c Let f R 7 R be given by x 73x 4 Let a E R Prove that limmaa x 73a 4 d Let f R 7 R be given by x 2x2 7 3x 7 Prove that limmLA x 71 Hint 7 71ll2x2 7 3x 7 5 l2x 7 5 lx 1 Restrict 6 g 1 so tht lx 7 71 lt 6 1x 6 720 2x 6 740 2x75 6 7975 l2x75l lt 9 Ifsgt 0is given nd 7 so that if 6 min1 then lx 7 71 lt 6 l2x 7 5 lx 1 lt 8 e Let f R 7 R be given by x 73x2 x 7 2 and prove that limmag x 3 29 30 4 LIMITS AND CONTINUITY 43 Let f R 7 R be given by 0 x is irrational fz 1 x 1s rational Let x E R Prove that limmaa x does not exist for all a E R 44 Let f D 7 R a E D and L7 M E R Assume limmha x L and limmna x M Prove that L M 45 Let f D 7 R and a E D Let L E R Prove that limmaa x L ltgt for every sequence an Q D a with an 7 a we have an 7 L Hint For lt77 prove the contrapositive 46 State the contrapositive of 45 Note In calculus 4OSCD when you considered limits you required that 3 6 gt 0 so that a 7 60 l 8 a Q D Our de nition is more general in that we only require a E D In calculus you considered 1 sided limits Our de nition includes that as well If f at7 a 8 7 R for some 6 gt 0 then limmha x L means in calculus language that limmna x L or the right hand limit of x at a is equal to L 47 Which of these limits if any exist Prove your answer I I l I I l a 13 s1nm b limoxsin m 48 Limit Theorems Let f D 7 R and g D 7 R and let a E D and LM E R Assume that limmha x L and limmhagx M Let c E R Prove that a lim c x cL x7ia bi 1fmgmLM c lim xgx LM d Ifga 0formeDandM 0thenim Continuous functions DEFINITION Let f D 7 R and let a E D Then 1 is continuous at a if for all 6 gt 0 there exists 6 gt 0 so that ifx E D and lx 7 at lt 6 then 7 al lt 8 in logical form V gt0 3 6gt0Vx7 x E D and lx7al lt6 l x7 al lt8 QUESTION Identify the difference between this and limmha x at77 If f is continuous at all a E D it is called continuous If S Q D and f is continuous at all a E S it is called continuous on S 4 LIMITS AND CONTINUITY 31 49 Let f R a R be given by 1 0 lt x lt 1 m 0 otherwise At what points is 1 continuous ln calculus you learned that a continuous function is one whose graph you can draw without taking your pencil off the paper This is true if the function is continuous on an interval But it is not an accurate description of continuity at a point or even on the domain D if D is not an interval 410 At which points is 1 continuous a f 01 a R x 0 if x is irrational and x 1 if x is rational h f R a R x x if x is rational and x 0 if x is irrational c f RHR xx2x71 d fDHR x Dxx7 1 411 Let f D HR and let a E D Prove a if a is an isolated point of D then f is continuous at a b if a E D then f is continuous at a if and only if x a 412 let 1 01 U 2 a R be given by x x2 for x E 07 1 U At what points is 1 continuous 413 Let f D a R and let a E D Prove f is continuous at a if and only if for every sequence an in D with an a a we have an a a 414 Give the contrapositive to 413 415 Let f D HR 9 D a R and let a E D Assume that f is continuous at a and g is continuous at a Let c E R Prove a f g is continuous at a b of is continuous at a c fg is continuous at a d If gx 7 0 for x E D then is continuous at a 416 Prove that every polynomial is continuous on R 417 Let DE Q R f D a E and g E a R Let a E D Prove that if f is continuous at a and g is continuous at a then 9 o f is continuous at a DEFINITION f D a R is bounded if D is bounded Thus 1 is bounded ltgt 3 K lt 00 such that V x 6 D7 K 32 4 LIMITS AND CONTINUITY 418 Give an example of a continuous function f 07 1 a R that is not bounded a graph will suffice Can you nd a formula for such an 1 419 Let D Q R be compact and let 1 D a R be continuous Prove that fD is compact Hint Show fD is closed and bounded 420 Let D Q R be compact and let 1 D a R be continuous Show 3 071 E D with fa x fb for all z E D Thus 1 takes on a maximum and a minimum ualue 421 Give an example of f 01 a R which is bounded7 continuous and has neither a max7 nor a min Can you do the same for f 07 1 a R 422 Let f 071 a R be continuous and let fa lt 0 lt fb Prove that for some m E 071 0 Hint Let x supy 6 11 y 0 423 Let f 11 a R be continuous and suppose 0 lies between at and fb Show x c for some z E 071 424 Let f 071 a R be continuous Then i7 b is a closed and bounded interval Uniform Continuity lffDARiscontinuousonDthenVaEDV6gt036gt0Vz D7 lziallt6i lfm 7 fal lt 8 In general 6 depends upon both 6 and the point a as previous exercises have illustrated If we remove the dependence on a we have uniform continuity DEFINITION Let f D gt R f is uniformly continuous on D ifV 6 gt 0 3 6 gt 0 V x73 6 D lx yl lt 5 5 We fyl lt 6 425 Let f D a R be uniformly continuous Show that f is continuous 426 Let Pm be a polynomial of degree 1 Prove that Pm is uniformly continuous on R 427 Negate the de nition of uniform continuity 428 Let fz m2 Prove that f is not uniformly continuous on R 429 Let K Q R be a compact set and let 1 K a R be continuous Prove that f is uniformly continuous Hint Let 6 gt O Vm E K 3 6m gt 0 so that ify E K and lmiyl lt 6m then lfz7fyl lt 82 Choose a nite subcover of K from the open cover 7 67quot z 57quot z E Let 6 be the smallest radius in this nite subcover 4 LIMITS AND CONTINUITY 33 430 Give an example of f D a R which is continuous and such that there exists a Cauchy sequence Q D with being divergent 431 Let f D a R be uniformly continuous and let Q D be Cauchy Prove that is also Cauchy in R This page is blank 439 LIMITS AND CONTINUITY