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Introduction to Analysis

Steve Halperin

(with contributions from Elizabeth Hughes)

September 1, 2013

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Contents

Chapter 1 Overview

Chapter 2 The Literature and Language of Mathematics

Chapter 3 Basic Set Theory

Chapter 4 The Real Numbers

Chapter 5 Infinite Sequences

Chapter 6 Continuous Functions of a Real Variable

Chapter 7 Derivatives

Chapter 8 Area and Integrals

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Chapter 1

Overview

1.1 Purpose of the Course

Most of your university courses focus on new material for you to learn, and testyou on how well you have learned and understood it. This course has a differentobjective: the primary purpose is to help students learn how to do something,namely to write proofs of mathematical statements. Mathematical proofs arejust a sequence of statements, each following logically from the next. A commonformat is a sequence of statements of the form:

Since Statement A (the hypothesis) is true therefore Statement B (the con-clusion) is true.

Example 1 1. Since the student is in Math 310, she is registered at the Uni-versity of Maryland.

To see this as a prototype for a mathematical statement we analyze itas follows:

Hypothesis: The student is in Math 310.Conclusion: The student is registered at the University of Maryland.

The conclusion follows logically from the hypothesis because to be in Math310 you must be registered at the University.

2. If every student in Math 310 gets at least 7/10 on a homework then theclass average is at least7/10.

Hypothesis: Every student in Math 310 gets at least 7/10 on a home-work.Conclusion: The class average is at least7/10.

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In this example the conclusion is not immediately obvious from the hy-pothesis. Thus we need to provide a proof, consisting of the followingstatements:

- Let the number of students in the class be n and let xi be the gradeof theith student. (Establishes notation.)- Then by definition the class average isn

i=1 xin

.

- Since eachxi7, thereforeni=1 xin

7 =n

i=1 xin

7nn

=

ni=1(xi 7)

n 0.

Therefore the class average is at least7/10.

3. If you get an A on your next report card you wil l be excused from doingthe dishes for a week.

Hypothesis: you get an A on your next report card.Conclusion: you will be excused from doing the dishes for a week.

In this example the statement is true because it is made by an authorityfigure like a parent. NOTE: In Mathematics, authority figure argumentsarenotallowed!

Everyone, even at an early age, learns what it means for a statement to followlogically from another, and sometimes it will be obvious that the statement iscorrect. So, writing proofs does not require you to learn to think logically. Youalready know how to do that. It does, however, require you to do two otherthings:

1. Learn the mathematical language.

2. Write in clear, precise, unambiguous complete sentences.

Here, then, is the central point on which to focus: A mathematical state-

ment must say exactly what the author means. Even a small change inthe order of words can change the meaning, and while you may feel that thereader will understand what you mean even if you have not said it exactly, thatis simply not acceptable when writing mathematics.

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Example 2 1. The writer means: Every student has a different breed ofdog.The writer says: Every student in her class has a different dog.The writer should say Every student in her class has a dog, and no twostudents have the same breed of dog.

2. The writer means She can find anx for every >0 such that|xxn|< ifn is sufficiently large.The writer says There is an xn such that|xxn| < .// The writershould say In a sequence xn, n = 1, 2, 3, for every > 0 there is anatural numberNsuch that|x xn|< ifnN.

This text will consist mainly of statements and examples, and their proofs, andthe exercises also ask you to prove statements and construct examples. In eachcase you are expected to provide a proof of the statement, or that your exampledoes what it is supposed to.

Thus in homework, tests, and the final exam the essential criterion bywhich your work will be judged is whether it meets the standards

below.

1. Your answers must be a sequence of mathematical statements:

- typed or written legibly in ink;

- clear, precise, and unambiguous;

- without excess verbiage;

- written in complete, grammatical sentences.

2. Each statement must follow logically from the previous ones, with anexplanation as to why.

3. You will get alow grade if your answers do not meet these requirementseven if they give the impression that you understand the material.

4. Your homework answers may rely on facts as outlined below, but youmust in each case indicate which fact you are using and give the explicitreference:

- any statement already proved in class;

- any statement in any earlier exercise;

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- any statement in class that I have labelled a quotable fact.

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Chapter 2

The Literature and

Language of Mathematics

2.1 The Literature of Mathematics

Mathematicians have built up a body of knowledge over several thousand years,expressed in theorems and examples, all validated by rigorous proofs, and thatonce validated are then true forever. This is theliterature of mathematics. Eachnew piece of knowledge depends on what came before, and sometimes it takesmany decades for little pieces to fit together to establish some remarkable newphenomenon.

The classical example is Fermats last theorem! Fermat was a 17th centurynumber theorist, and after he died in 1637 the following statement was foundwritten in the margin of one of his books: I have found the most wonderfulresult, but the margin is too small for me to write down the proof. The simpleassertion was this: Ifa, b, c, n are natural numbers all greater than 1, and if

an + bn =cn,

then n = 2.

In the following three hundred years the search for a proof was one of theholy grails of mathematics. Much of our knowledge in number theory andgeometry was developed in the process. And finally in 1995 Andrew Wiles

(Princeton) published a proof using many of the results which had been estab-lished over the preceding centuries.

Some of you may wonder about the relation of mathematics to the physi-cal world we live in. Now mathematical knowledge itself is validated just byproofs, and so its correctness stands for all time and does not depend in anyway on our physical experience. Nonetheless, much of mathematics is inspired

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by that physical experience, mathematical knowledge is regularly applied in al-most every branch of science and engineering, and indeed many mathematiciansfocus on building mathematical models and computational algorithms that di-rectly reflect/compute physical phenomena. As a Nobel Laureate in Physics,Eugene Wigner, wrote in 1960 in a paper entitled The Unreasonable Effective-ness of Mathematics. . . the mathematical formulation of the physicists oftencrude experience leads in an uncanny number of cases to an amazingly accuratedescription of a large class of phenomena.

2.2 The Language of Mathematics

If theorems and examples are our literature, then the way we express them isour language. There are two central distinguishing features to mathematical

language. First, it is (and must be) utterly without ambiguity. A mathematicalstatement can never admit of more than one interpretation. Second, math-ematical statements and writing is terse and to the point. The result is thatmathematical writing is information-intensive.

Definitions and notation are essential tools in keeping our language unam-biguous and terse and, as with any new language, it is essential that you inter-nalize the definitions and notation rather than simply memorize them. Indeed,when learning to speak or write a new language you need to be able to use thewords spontaneously without having to call up each corresponding English wordand then translate it. In the same way, you need to be able to speak/write math-ematics, not just remember the dictionary of definitions.

You learn a second language most easily by speaking it with others to whomit comes naturally. You learn to drive a car by driving it and to walk by walk-ing. You learn to write/speak mathematics by writing it and presenting it andgetting feedback when you get it right and how to correct it when you dont.The golden rule when writing: never write anything whose meaning isunclear to yourself! You can also use this text to find many detailed exam-ples of how to write a proof correctly.

The language of mathematics consists of assertions about mathematicalobjects. Mathematical objects include the natural numbers, the integers,the rationals, the real numbers, sets, maps, functions and many other things.Mathematics uses symbols to denote objects, but in any specific proof each

symbol must have a precise meaning and the same symbol may neverbe used with two different meanings. In particular, we fix the followingnotation for the entire course:

1. Nwill denote the set of natural numbers, and we write N ={1, 2, 3, . . .}.2. Z will denote the set of integers: Z={0, 1, 2, 3 . . .}.

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3. Q will denote the set of rational numbers: Q={p/q|p, qN andq= 0}.

4. R will denote the set of real numbers.

5. Cwill denote the set of complex numbers.

We also use the sigmanotation for sums:

Ifvi are numbers or vectors (or anything else we know how to add) then wewrite

ni=1

vi = v1+ v2+ vn.

More generally, we also write

kni=k1

vki =vk1+ vk1+1+ vk1+2+ vkn .

Certain words and expressions frequently appear in mathematical statements:

1. Definition: This is a sentence (or sentences) which assigns a specificmeaning to a word or symbol.

2. Let....: This is a statement which defines some notation or terminologyyou will be using.

3. Since: This means because.

4. Such that: This means with the property that.

5. If A then B . This assertion states thst if A is true then B is true.Here A is the hypothesis or assumption (both words mean the samething), andB is the conclusion. This is often denoted by

AB.

It may also be phrased as Assume A; then B, or as B if A, or as Aimplies B.

6. A only if B. This means thatA is not true unless B is true. In otherwords, ifB is not true then A is not true.

7. A if and only if B This means that ifB is true then so is A, and

that ifB is not true then A is not true. In other words,A is true if andonlyB is true. This is often denoted by

AB.

8. The converse to the statement If A is true then B is true is thestatement if B is true then A is true.

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9. The contrapositiveto the statement A is true is the statement Ais not true.

10. Theorem, proposition, lemma: These are names for mathematicalstatements that are going to be proved (see next item for proof.

11. Therefore means as a consequence of what just preceded.

12. For every xS there exists a y such that A is true means thatfor every choice ofx in a given set Sthere exists a y with the propertiesprescribed by A.

13. q.e.d. stands for quid erat demonstrandum, which is Latin for what wasto be proved. It is used at the end of a proof to signal that the proof iscomplete.

Important note:Items 5, 6, and 7 are themselves assertions and they may betrue or false!

Example 3 Here are some illustrative examples:

1. Letn be the least natural number such thatn2 andn is not a prime.2. Ifx = 2 andy = 4 thenx+y = 6. This assertion is true. However, the

assertionx = 2andy = 4ifx + y= 6is false, because ifx= 1 andy = 5thenx + y= 6.

3. Suppose x and y are natural numbers. Then xy = 2 only if x 2 andy2. This is a true statement. But the statementxy = 2 if and only ifx2 andy2 is false.

4. This example highlights the importance of getting the order right,with two similar statements:

(a) For each student in this class there is a date on which that studentwas born.

(b) For each date there is a student in the class who was born on thatdate.

The first statement is true and the second is false.

5. A natural numbern is odd if and only if it has the formn= 2k+ 1 forsome natural numberk.

6. For every rational number there is an integer which is larger. (Here youfirst have to pick the rational numberp/q and only then can you pick aninteger which works (egp2).

7. For every positive rational number, a, there is a natural number n suchthat 1/n < a. Here again you first have to picka which will necessarilyHave the forma= p/q withp, qN. Then once you know whata is youcan setn= q+ 1.

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Exercise 1 1. For each statement below, the symbols are used either cor-rectly or incorrectly. Replace the incorrect statements by coorectones.

(a) Ifx0 thenxN.

(b) Ifp, qR thenp/qcan be constructed such thatp/qQ.

(c) IfxN, yZ, andzC,thenx,y, zR.

2. Describe the error in each of the following statements.

(a) Ifp, qN, wherep2 andq6, thenp + q is even.

(b) Laptops are popular among college students. Therefore at least onestudent at Maryland has a laptop.

(c) It is true thatU= 6 for everyx, yQsuch thatx + y < U.

3. Does (a) or (b) correctly add such that to the statement: every studenthas a height h?

(a) Every student has a heighth such thatha wherea is the height ofthe tallest student.

(b) Every student has a heighth such that every student is in a class.4. Does the statement: SupposeA is a natural number. Recall the definition

of A is that A is a prime number correctly specify that A is a primenumber? If not, change the statement to do so.

5. Is the fol lowing statement true or false? Provide a proof for your answer.Suppose A and B are natural numbers, and that A = 5 if and only ifB= 2. IfB= 2, thenA= 7.

6. What is the contrapositive of the following statements?

(a) A= 8 ifB= 9.(b) A= 8 ifB

= 9.

7. What is the converse of the following statements?

(a) B= 9 ifA= 8.

(b) A= 8 ifB= 9.

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8. The Proposition below is missing a hypothesis. Add that to the statementand then change the proof so it is correct.Proposition: The student is reading a book.Proof: Let the student be reading from a collection of pages.Define a book to be a collection of pages such that the pages are boundtogether and published.Therefore the student is reading a book.q.e.d.

9. The proof of this proposition is incomplete. Add in a step to make itcomplete. Then prove the converse of the proposition. Proposition: IfxNis an odd number, then it is not an even number. Proof: An odd numberis a number of the form2n 1withnN.. LetxNbe an odd number.An even number has the form2n, withnN. Therefore x is not an evennumber. q.e.d.

10. What is the hypothesis and what is the conclusion in the following asser-tions?

i) The cube of an odd natural number is odd.ii) For every >0 there is somepNsuch that1/p.

11. What is the contrapositive of the statement: For every > 0 there is apoint(x, y)in the plane whose distance from the origin is between/2and.

12. What is the converse of the following statement? Supposen, mN. Ifn > m then for somek

N, km > n. Is the original statement true? Is

the converse true? provide proofs that show your answers are correct.

2.3 Proofs

Recall that a mathematical proof is a sequence of statements each followinglogically from the previous ones, and which together demonstrate the truth ofan assertion. These statements must satisfy the following criteria

- Some may define notation or terminology.- Each of the other statements must follow logically from the preceding onesand the hypotheses, and must explain why!- The last statement will be the conclusion.

Tips for Writing Proofs from Elizabeth (a former Math 310 student):

1. Learning to write proofs is not like learning how to multiply matrices, andit can take mental effort and time.

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2. Remain calm; if you put enough time into this class, you will figure outhow to write proofs.

3. Give yourself plenty of time to work on a proof.

4. Before you start the proof, look back at the given definitions and lemmasthat you think might be useful for this proof. Make sure you fully under-stand the definitions and lemmas and see if you can find a way to connectthem back to what you are trying to prove.

5. Restate definitions in your proofs, so you are less likely to use a definitionincorrectly.

6. Working in groups is fine, but make sure you try each exercise thoroughlyon your own first. You will not have your group to help when it comes

time to take the test.7. Never hesitate to ask Professor Halperin a question when you are stuck:

either in class, in his office, or by email. He will welcome your interest: Iknow!

8. Most importantly: This class is designed to teach you how to write proofs.While it is important you understand the concepts, you must be able to usewhat you learn to complete a correctly written proof in order to succeed.

There are two standard techniques which are often used to prove statements:

1. Proof by contradiction. Here you suppose that the assertion to beproved is false and use that assumption to show that a fact already known

to be true is false. It follows that the assertion cannot be fals, and so itmust be true. The proof of the next Proposition is an example of proofby contradiction.

Proposition 1 There are infinitely many prime numbers.

Proof: Suppose there were only finitely many prime numbers, so thatthere was a largest prime numberp.Letn = (

k=pk=1 k) + 1.

Thenn > p and nis not divisible by any natural number mp.Butnmust be divisible by some prime number q.Therefore there must be a prime number q > p.

This contradicts the hypothesis that p was the largest prime number.Therefore there must be infinitely many primes. q.e.d.

Remark: This proof was discovered by Euclid about 300 BC.

2. Proof by induction. This is based on the following

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Induction Principle: Suppose given a sequence of statements S(n), onefor each n

N. Suppose that

i)S(1) is true, and

ii) wheneverS(n) is true then also S(n + 1) is true.

ThenS(n) is true for all nN.

The induction principle describes a basic property which we treat as anaxiom about N.

The proof of the next Proposition is an example of proof by induction:

Proposition 2 For everynN,k=nk=1

k3 =

n(n + 1)

2

2.

Proof: We prove this by induction on n, with S(n) the equation above.

When n = 1 both sides of the equation equal 1 and so they equal eachother.

Suppose now by induction that S(n) is true for some n.

Then for this n,k=nk=1

k3 =

n(n + 1)

2

2.

We have to show that S(n + 1) is true; i.e., that

k=n+1k=1

k3 =

(n + 1)(n + 2)

2

2.

Butk=n+1k=1

k3 =

k=nk=1

k3 + (n + 1)3

and(n + 1)(n + 2)

2

2=

(n + 1)2(n2 + 4n + 4)

4 =

n(n + 1)

2

2+ (n + 1)3.

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Thus, since we have assumed S(n) is true it follows that

(n + 1)(n + 2)

2

2

=k=nk=1

k3 + (n + 1)3 =k=n+1k=1

k3.

ThereforeS(n+1) is true, and the proposition follows by induction. q.e.d.

There is another form of the induction principle we shall also use, asillustrated in the next Proposition:

Proposition 3 Suppose given a sequence of statementsS(n), one for eachnN, and that

i) S(1) is true, and

ii) wheneverS(k) is true forkn then also S(n + 1) is true.

ThenS(n) is true for allnN.

Proof: Let T(n) be the statement: S(k) is true for all kn.

SinceS(1) is true, so is T(1).Suppose by induction that T(n) is true.ThenS(k) is true for all kn.Therefore by hypothesis, S(n + 1) is true.SinceS(k) is true for k

n as well, it is true for k

n + 1.

Thus T(n + 1) is true.Now it follows from the induction principle that T(n) is true for all n.ThereforeS(n) is true for all n. q.e.d.

The principle of induction can also be used for construction, and we illus-trate this by the following example.

Example 4 Construction by induction

An infinite sequence of integersnkQ, k1is constructed by inductionby requiring that:

n1= 1, and

nk+1=

i=ki=1

ni2

, ifi=k

i=1ni is even;

k, ifi=k

i=1ni is odd.

Exercise 2 1. The proofs below are incomplete. Complete them to correctproofs.

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(a) Proposition: There are infinitely many even natural numbers.Proof: An even natural number has the form2nwithn

N. Suppose

there is a largest even natural number r. There there must be somepNwithp > r. Lets= 2p. Therefore there are infinitely many evennumbers. q.e.d.

(b) Proposition: IfkNandk2 is even thenk2 is divisible by4.Proof: The square of an odd natural number is odd. Thereforek mustbe even and so k2 is divisible by4. q.e.d.

2. Is (a) or (b) a proof that the following statement is wrong? Statement:All ducks are red.

(a) There exists a yellow duck, A. Therefore not all ducks are red.

(b) We have not seen all ducks. Therefore some may not be red.

3. Correct these sloppy attempts at proofs:

(a) Show that the following statement is false: If the sum of an odd num-ber of natural numbers is odd then each of those natural numbers isodd.Proof: Recall that2 is a natural number. Notice that2 + 2 + 2 = 6.But 6 is not odd despite being the sum of an odd number of naturalnumbers. Also,4 is a natural number and is not odd. q.e.d.

(b) Prove the following statement: Ifn2 is odd thenn is an odd naturalnumber.Proof: Supposen2 is odd wheren is any even natural number. Then

it must be true thatn2 =n(n) =n1+ + nn = i = 1nni. Howeverany even number plus another even number must be even. Thereforen must be any odd natural number. Therefore ifn2 is odd thenn isan odd natural number. q.e.d.

4. Prove or disprove the fol lowing statement: For every x Q there is auniquenNwhich is the cloural number to x.

2.4 Counterexamples

Counterexamplesare examples which demonstrate that some assertions arefalse. Thus the example asked for in Exercise 1.3 is a counterexample. There

is a much deeper example from the theory of equations. You learned in schoolthat the quadratic equation

ax2 + bx + c= 0

has two solutions, x= (b b2 4ac)/2a. One might ask whether the solu-tions to higher order equations can also be expressed in formulas that only use

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nth roots, for some natural numbers n, and indeed this is true for cubic andquartic equations.

However, it is not true for all fifth degree equations, as was shown by Abeland Ruffini in 1824. The easiest way to show this is by an explicit counterex-ample and a very simple one was proved about a century ago by a famousalgebraist, Emil Artin, who showed that roots of the equation x5 x 1 = 0cannot be expressed in this way.

Exercise 3 Do the following:

1. Show by a counterexample that the following statement is false: Ifa, bNthena3 + b3 is the cube of a natural number.

2. Prove by contradiction that2 is not the square of a rational number.

3. Prove by induction that the sum of an odd number of natural numbers isodd if each of the natural numbers is odd. What is the converse statement?Show by a counterexample that the converse is false.

4. Prove that if n is an odd natural number then n2 is odd. What is theconverse statement? Decide if it is true or false and prove your answer.

5. What is the converse of the statement A implies B? Construct an ex-ample where a statement is false but the converse is true.

6. What is the converse of the statement A is true if and only if B is true.?What is the differnce in meaning between the original statement and theconverse?

7. Construct by induction an infinite sequence of natural numbers (xn)n1such that forn2, xn >

n1i=1 xi.

8. Construct an example to show that if in each of two school classes theaverage GPA of the boys is bigger than that of the girls it may not bethe case that when the classes are combined this is still true: i.e., in thecombined classes the average GPA of the girls may be bigger than that ofthe boys.

9. Show by counterexample that the fol lowing statement is false: For everyp, qN, p2 + q2 >(p + q)2 50p 51q.

10. Show by counterexample that the following statement is false: If s >39, t > s + 5, andu >33, thent + u >82 for all t,u,sN.

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Chapter 3

Basic Set Theory

3.1 Sets

As we move out of the familiar territory of the plane we need a language thatpermits us to speak and write precisely, to formulate clear statements, and toprovide proofs and examples that are clear and unambiguous.

The basic language we start with is that of sets and maps. Of course, as wego forward in the course we shall add to this vocabulary, but for the moment,lets just get used to playing with sets and maps.

Definition 1 Sets

A set is any specified collection of things, abstract or concrete. The thingsare called the elements of the set. If x is an element in a set S we say xbelongsto S, and denote this byxS.

Note: To define a set we must specify its elements.

Definition 2

1. A finite set is a set with only finitely many elements. In this case|S|denotes the number of elements inS.

2. A subset of a set S is a set W all of whose elements are elements ofS. We denote this byW S. We often use the following notation for asubsetW S:

W ={xS| },where specifies which elements inSare inW.

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3. The product of sets S and T is the set S T whose elements are theordered pairs(x, y) withx

S andy

T. We often write this as

S T ={(x, y)| xS andyT}.

4. The set with no elements. It is called the empty set and is denoted by.

Definition 3 Families of sets

1. A family of sets is a setmathcalSwhose elements are other sets. Thisis ometimes denoted bymathcalS={S}. If eachS is a subset of a setS then we sayS is a family of subsets ofS.

2. IfS is a family of sets then their union,

S and their intersection,

S are the sets defined byx

S if and only if xS for someS,

andx

S if and only if xS for everyS.

Example 5 Sets

1. N is the set of natural numbers. For eachnNletSnNbe the subsetof all natural numbers except forn. ThenmathcalS={Sn| nN} is afamily of subsets ofN. Moreover,

nSn = Nbecause every natural number is in someSn.

On the other hand, n

Sn =

because any natural numberk is not in one of theSn: namely, k /Sk.2. The set, R , whose elements are all the real numbers.

3. A line in the plane is a subset ofR2, and is also an element in

L.

4. The set, R2, whose elements are all the points in the plane:

R2 ={(x, y)| xR andyR}= R R

5. The set whose elements are all the cathedrals in France.

6. The set of all maps from a setS to a setT.

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7. The setSwhose elements are all the subsets of a setS.

8. The subsetT Swhose elements are all the subsets ofSwith exactly twoelements.

9. The setL whose elements are all the lines in the plane.10. The subset ofL whose elements are all the vertical lines in the plane.

3.2 Maps

Definition 4 A map between sets, usually written: ST, is a rule whichassigns to each elementxS asingle specified element(x)T.

If: STis a map between sets then theimage ofis the subsetI mTdefined byIm={(x)| xS}.

Note: To define a map we must specify three things: the two sets S and T,and the rule - see the example below.

Example 6 Three maps are defined as follows:

1.: NN; (n) =n2

2. : Z

N; (n) =n2

3.: NZ; (n) =n2

In this example all three maps are different because the two sets are different ineach case even though the formula is the same!

There are three important classes of maps of sets:

Definition 5

1. A map : S T isonto ifIm isallofT (i.e., Im= T). In otherwords, is onto if and only if every elementy can be written in the formy= (x) for at least onex

S.

2. A map : S T is 1-1 if it maps different elements in S to differentelements inT. In other words, is 1-1 if and only if for everyx1, x2S,

(x1) =(x2)x1= x2.

3. A bijection is a map : STof sets which is both onto and 1-1.

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Caution: Do not confuse the difference between the definition of a set mapand the definition of a 1-1 set map. A map : S

T associates to x

S a

single element (x)T. Thus for any map : ST,

x1= x2(x1) =(x2).

A 1-1 map : ST satisfies

x1= x2(x1) =(x2).

Thus : S T is 1-1 means that any element in yI mcan be trackedback to a single element xT such that (x) =y.Example 7

1. The maps, , in Example 6 are respectively 1-1, but not onto, neither1-1 or onto, and 1-1 but not onto.

2. The map : NNgiven by

(m) =

1, m= 1, 2,m 1, m3

is onto, but not 1-1.

3. The map : ZNgiven by

(k) =

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3. S is the set of students enrolled in the University of Maryland on Oct. 1,2012 and : S

N is the rule which assigns to each student the least

number of inches which is greater than or equal to their height.

4. S is the set of students enrolled in the University of Maryland on Oct. 1,2012,C is the set of all countries that existed prior to Oct. 1, 2012, andis the rule which assigns to each student the country in which they wereborn.

5. : NNassigns to eachnNthenth integer in the decimal expansionof.

Remark: A map from a set, S, to R is often called a real valued function onS, and the formulas you are used to from calculus usually define real valuedfunctions on R. However, it is interesting toi note that not every real valued

function on R has a formula.

Lemma 1 Suppose: ST, : T W, and: W Uare maps of sets.Then

1. ( ) = ( ).

2. is a bijection if and only if there is a set map : T S such that = idS and = idT.

3. The map in ii) is the only map which satisfies = idS and =idT.

Proof:

1. Both sides evaluated at any xSgive (((x)).

Therefore both sides are the same map.

2. Suppose is a bijection.

Lety be any element in T.Because is onto, there is an element xS such that (x) =y.Because is 1-1, there is only one such element x.Thus we may define a set map by setting (y) =x.By construction, = idS.

On the other hand, every element yT has the form y = (x) for somexS.It follows that (y) =(((x)).

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But we constructed so that ((x) =x.Thus

(y) =(x) =y.

Therefore, = idT.

In the other direction, suppose there is a map : T S such that = idS and = idT.Then any yT satisfies y = ((y)).Therefore, is onto.Similarly, supposex1= x2.Thenx1= (x1) =(x2) =x2.Therefore is 1-1.

3. Suppose 1 is any map such that 1

= idS and

1= idT.

Then by Lemma 1 i),

= idS = (1 ) ) =1 ( ) =1 idT =1.

q.e.d.

Definition 7 The map constructed in Lemma 1 ii) is called the inverse of and is denoted1.

Exercise 4 Sets and maps

1. Determine if each of the following is a map:

(a) : NN, defined by

(x) =

1, 2 x= 1,x + 2 x >1.

(3.1)

(b) : RR, defined by

(x) =

1 x= 1,x + 2 x= 1. (3.2)

(c) : CCdefined by:

omega(x) =

5 x= 1, 2, 3x + 2 x= 1, 2, 3. (3.3)

2. Below are several attempted proofs showing that the map : n n2 inExample 5.1 is 1-1. Choose the proof that is correctly written and say whatis wrong (grammar, logic, incompleteness) with the other attempts.

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(a) Recall a map : S R is 1-1 if and only if for every x, y S:(x) = (y) if and only ifx =y. Recall that every natural numberhas a unique square. Thus we must prove that if k, n N and ifk2 =n2 thenk= n. We do this by contradiction. If this is not truethen eitherk > n ork < n. Ifk > n thenk = n +m withmN.It follows thatk2 =n2 + 2nm+m2 > n2, which is a contradiction.The same argument applies ifk < n; just switch the roles of k andn. q.e.d.

(b) Recall that a map : SR is 1-1 if and only if: for everyx, yS:(x) =(y) ifx = y. Recall that every natural number has a uniquesquare. Therefore ifk =n thenk2 =n2. Therefore the map is 1-1.q.e.d.

(c) Define a map : NNhave the rule(x) =n2 fornN. Notice(n2)1/2 =n2/2 =n. q.e.d.

(d) Recall a map : S R is 1-1 if and only if for every x, y S:(x) = (y) if and only if x = y. Notice k2 = n2 k = n.Therefore k2 = n2 if and only if k = n. Therefore the map is 1-1.q.e.d.

3. What is||?4. IfS andTare finite sets show that|S T|=|S||T|.5. IfS is a finite set show that|S| = 2|S|.6. Prove that:

(a) for any setS, idSis a bijection.

(b) The composite of onto set maps is onto.

(c) The composite of 1-1 set maps is 1-1.

(d) The inverse of a bijection and the composite of two bijections arebijections.

7. Suppose: S

T is a map of sets.

(a) IfT is finite and the map is 1-1 show thatS is finite and that|S| |T|.

(b) If S is finite and the map is onto show that|T| is finite and that|S| |T|.

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8. Suppose: ST is a setmap and thatS andTare finite sets with thesame number of elements. Show that the following three conditions areequivalent:

(a) is a bijection.

(b) is onto.

(c) is 1-1.

9. Construct set maps and from a set S to a set T such that is 1-1but not onto and is onto but not 1-1.

10. IfS andTare finite sets show that there are|T||S| elements in the set ofmaps fromSto T. How many elements are there in the subset of bijectionsofSto itself?

3.3 Equivalence Relations

A relation between two sets, SandTis a generalization of the idea of a map:

Definition 8 Arelation,R, between a setSand a setTis a subsetRSTof the product ofS with T. If (x, y)R we say that the relation relates y tox, or that y isrelated to x, and we write xRy. If S=T we say that R is arelation in S.

Note: If : S T is a map, then{(x, (x))| x S} is an example of arelation. Relations corresponding to set maps are those subsets ofS T thatsatisfy the following property: every x

S is related to a single y

T. By

contrast,any subsetRS Tis a relation between SandT.Exercise 5 Relations

1. Identify whether which of the following subsets R SxT is a relationdetermined by a map:

(a) S is all ofR andT is all ofZ, andR={(2, 3), (3, 4), (4, 5)}.(b) S= N, T =R andR={(n, 1/n)|nN}.(c) S= /bn, T =NandR={(n2, 1/n)|nN}.

2. Show that the relation{(x2, x)| x R} R2 is not the relation deter-mined by a map.

3. Which are the integersnNsuch that the relation{(xn, x)|x R} R2 corresponds to a map from R to R? Prove thecorrectness of your answer.

4. IfSandTare finite sets, how many relations are there betweenSandT?Compare this with the number of maps fromS to T- see Exercise 4.

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Perhaps you have heard the expression,

You cant see the forest for the trees.

The speaker is accusing the audience that they are so bound up in the details ofthe trees that they dont come to grips with the more global properties of theforests. And in fact, sometimes instead of looking at individual trees, we wantto look at the forests as individuals. Similarly, instead of considering individualpeople we might want to talk about cities, which are collections of people. Wemight want to describe the properties of species, which are collections of ani-mals, rather than the distinctions between individual animals.

This leads to a very important mathematical idea, which we formalize in thefollowing way:

Definition 9 A partition of a setS is a family of non-empty subsetsSiSsuch that every elementxSbelongs to exactly one subsetSi.Every partition{Si} of a set Sdetermines a specific relation in S S; namely,we set xRy if and only ifx and y are in the same subset Si. This is called therelation of the partition.

Example 9

1. The single set S is a partition of a non-empty setS. The correspondingrelation isxRy for everyx, yS;i.e., R= SxS.

2. The family of subsetsSx, xSof a non-empty setS is a partition ofS,since everyx belongs to a singleSx. The corresponding relation iisxRy

if and only ifx= y.

3. A partition of the set,S, of sophomore students at the University of Mary-land is defined by: For each possible GPA, , S is the set of studentswhose GPA is.

Lemma 2 The relation of a partition{Si} of a set S satisfies the followingproperties:

1. For everyxS, xRx. (The relation isreflexive.)

2. IfxRy then also yRx. (The relation issymmetric.)

3. IfxRy andyRz thenxRz. (The relation is transitive.)

Proof::

1. The relation is reflexive becausex and x belong to the same Si !

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2. IfxRy then by definition x andy are in the sameSi and so y Rx.Therefore the relation is symmetric.

3. We have to show that ifxRy andy Rz then xRz.

By hypothesis, x belongs to some (unique) SiSincexRy it follows thatySi as well.SinceyRz it follows that also zSi.Thus by definition xRz and the relation is transitive.

q.e.d.

Definition 10 Anequivalence relation in a setS is a reflexive, symmetric,and transitive relation. Equivalence relations are denoted byx

y.

Lemma 2 states that the relation of a partition is an equivalence relation.Conversely we have

Proposition 4 Every equivalence relation,, in a set S is the relation of aunique partition ofS.

Proof:Fix an equivalence relation,, in S.

For each xSdefine a subset S(x)S by

S(x) ={yS| yx}.

Now let

{Si

}denote the family of distinct subsets ofSsuch that each Si = S(x)

for somexS.

We will prove three things:

1. This is a partition ofS.

2.is the relation of the partition.3. This is the only partition withas ite relation.

Step 1. This is a partition ofS.

By reflexivity each xS(x) and so the Si are not empty and every x in S

belongs to some Si.

Thus to show this is a partition we have to show that any element of S canbelong to only one subset Si.

In other words we have to show that ifyS(x) andyS(z) thenS(x) =S(z).

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We first prove that for any xS, ifuS(x) then

S(u) =S(x).

In fact, since uS(x) we have ux, and ifvS(u) we have vu.Thus vux, and since the relation is transitive, vx.It follows thatvS(x); i.e. S(u)S(x).On the other hand, since u x and the relation is symmetric, x u; i.e.,xS(u).By what we have just shown this implies that S(x)S(u).SinceS(u)S(x) andS(x)S(u) we have proved that S(u) =S(x).

Finally, if y S(x) and y S(z), then by the equality we just provedS(x) =S(y) =S(z).

Therefore{Si} is a partition and Step 1 is proved.Step 2. The equivalence relationis the relation of the partition.

Denote the relation of the partition by R, so that xRy if and only ifx andy are in the same subset Si of the partition.

Then it follows from Lemma 2 that:

xRyx, yS(w), somewS, S(x) =S(w) =S(y)xS(y)xy.

Thus R =.

Step 3. The partition{Si} is the unique partition which has as its equiva-lence relation.

Let{T} be a partition ofSwithas its equivalence relation.

IfxSthen the subsetT in the partition containing x will satisfy

yx if and only if yT.

Thus T = S(x).It follows that every T is one of the S(x).Since every element ofS is in some T it follows that every S(x) is one of theT.

Therefore the partition{T}is the partition{Si}. q.e.d.Exercise 6 Equivalence relations

1. Provide a complete proof for the statement in Example 9.3.

2. Below ia an attempted formal proof for Lemma 2. State everything that iswrong with it.

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Proof: Let Si be a partition of a set S. Recall a partition of a set P isa family of non-empty subsets P

iP such that every element x

PP

belongs to exactly one subsetPi. ForxSi, xRx because x and x belongto the same uniqueSi. ThereforeSi is symmetric. IfxRy, thenx andyare in the sameSi soy Rx is also true. ThereforeSi is reflexive. Since wesuppose x, y Si if xRy, then if yRz we need z Si. Therefore ifxRyandyRx, thenxRz becausex, y, andz z are all in the sameSi. ThereforeSi is transitive. Since Si is reflexive, symmetric and transitive, it is anequivalence relation. q.e.d.

3. Two UM students are related if they both take at least one class in common.Is this an equivalence relation?

4. Two UM students are related if they take all their classes in common. Isthis an equivalence relation?

5. Supposef: STis a map between two sets. Say x and y in S are relatediff(x) =f(y). Is this an equivalence relation?

6. SupposeS=mi=1Si is a partition of a set withn elements for some nat-ural numbern.

(a) If eachSi has the same number of elements, k,show thatk dividesnand thatn/k= m.

(b) If eachSi hasi elements findn.

(c) If eachSi hasi3 elements, findn.

(d) If eachSi has an odd number of elements show thatm is even if andonly ifn is even.

7. LetQ be the set of all subsets of Q. Thus the elements ofQ are thesubsets SQ. Define a relation

Q Qby settingST if

(a) for everyaSthere is somebTsuch thatba, and(b) for everycTthere is somedSsuch thatdc.

Then

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(a) Show that is an equivalence relation.

(b) Show that ifSis an equivalence class of subsets thenSS

S

is an element inS.(c) Show that if S Q is not equivalent to Q then for some rational

number, a, a > x for allxS.(d) IfaQare the sets

S={bQ| b < a} and T ={bQ| ba}

equivalent?

8. LetTS denote the set of maps from a setS to a setT. Define a relationRTS TS by setting

f Rg

iff(x) =g(x)except for finitely many pointsxS(depending, of course,onf andg). Show this is an equivalence relation.

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Chapter 4

The Real Numbers

The bedrock of analysis is an understanding of the real numbers. Notice thedifference between the rational numbers and the real numbers: we can describethe rational numbers explicitly as quotients of one integer by another. No suchsimple expression is available for the reals. We may have an intuitive under-standing of these as the points on the real number line, as distances, or as(possibly) infinite decimals, but we cannot use intuitive understandings tomake rigorous proofs.

The way we solve this problem is by connecting the reals to the rationalsin a way that allows us to extend properties of the rationals to the reals in arigorous way. Our first step is to identify the basic properties of the rationals.

4.1 Properties of the Rationals

Here are the two fundamental concepts for the rationals with which we are fa-miliar:

1. Algebra: The operations of addition, subtraction, multiplication anddivision.

2. Order: Ifband a are rational numbers then

a < bb a= m/n for some m, nN.

Notation: We use a > bto mean the same thing as b < a.

The basic properties of the ordering in the rationals are contained in thenext lemma. While they are utterly and totally familiar we shall give formalproofs to provide more examples of what a proof looks like.

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Lemma 3

1. For each rational numberaQexactly one of the following three possi-bilities is true:

(a) a >0

(b) a= 0

(c) a 0 anda < b.

(d) Ifa < b thena >b.

3. IfaQ thena >0a >1/n for somenN.

Proof:: We prove each statement separately.

1. Suppose aQ.By definition a is the quotient of two integers.Thus exactly one of the following three possibilities is true:

(a) a= m/n with m, nN, or(b) a= 0, or

(c) a=m/nwith m, nN.In the first case a 0 = m/n with m, nN and so a >0.In the second case, a = 0.In the third case 0 a =a= m/n with m, nN and so 0> a, whichis the same as a a.

(b) Since (a + c) (a + b) =c b it follows that

(a+c)(a+b) =m/n for some m, nNcb= m/n for some m, nN.

Thereforea + b < a + c if and only if b < c.

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(c) Suppose c >0 and a < b.Thenc = p/qand b

a= m/n for somep,q,m, n

N. Thus

cb ca= c(b a) =pm/qn.SincepmN and q nN it follows that ca < cb.

(d) Suppose a, b are rationals and that a < b.Thenb a= m/n for some m, nN.Thusa (b) = (b a) =m/n.Therefore, by definition,a >b.

3. Suppose a >0.Thena = p/qfor some p, qN.Thus a1/q >1/(q+ 1).Thereforea >1/nwith n = q+ 1.Conversely, for any nN, 1/n > 0.Thus ifa >1/nthen a >1/n > 0 and so a >0.

q.e.d.

4.2 Introducing the Reals

Recall that while we may have an intuitive understanding of the real numbersas the points on the x-axis of the plane, or as distances, or as (possibly) infinitedecimals, we cannot use intuitive understandings to make rigorous proofs.Thus we introduce the reals and list certain elementary properties as axioms,from which all the other properties will be deduced. In this way we rigorously

extend properties of the rationals to the reals.

Property One: The real numbers are a set, denoted by R, containing therational numbers.

Property Two: The operations of addition, subtraction, multiplicationand division are defined for real numbers, coinciding with the old operations inthe rationals, and with the same properties.

Property Three: The ordering of the rationals extends to an ordering ofthe real numbers such that

1. For each real numberx exactly one of the following three possibilities istrue:

(a) x >0, in which case we say x is positive.

(b) x= 0.

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(c) x 0x >1/n for some nN.

2. Ifx >0 thenx >1/2k for somekN.3. If x,y,z,w are any three real numbers, then:

(a) x + y < x + z if and only ify < z .

(b) x >0 if and only ifx 0.(c) zx < zy ifz >0 andx < y .

(d) Ifx < y thenx >y.

Proof:: We prove each statement separately.1. Suppose xR.

If x > 0 it follows from Property Three above that x > 1/n for somenN.

Conversely, suppose x >1/nfor some nN.Thenx >1/n > 0.Now it follows from Property Three thatx >0.

2. By the first assertion, 1/n < x for some n

N.

We show by induction on pN that for p1,p

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and so the inequality follows by induction on p.In particular,

x >1/n > (1/2)n.

3. Suppose x,y, z,wR.(a) Since (x + z) (x + y) =z y it follows from Property Three that

x + y < x + z if and only if y < z.

(b) Ifx >0 then 0 (x) =x >0.Therefore, by Lemma 4.2a,x 0.We show by contradiction that 1/x >0.First, suppose 1/x 0.Since the product of positives is positive (Property Three) it wouldfollow that

1 = (1/x)x >0.But since 1> 0 this would imply 0 = 1 + (1)> 0, which is false.It follows by contradiction that it is not true that 1/x 0.

(c) Since x < y Property Three states that y x >0.Sincez >0 by hypothesis and the product of positive real numbersis positive (Property Three), it follows that

zy zx = z(y x)> 0.

Now it follows from Property Three thatzy > zx.

(d) Ifx < y then Property Three, y x >0.Therefore, (x) (y) =y x >0.Therefore, by Lemma 4.3b,y y.

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q.e.d.

The preceding Lemma is fairly elementary. It simply establishes propertiesyou have been taking for granted for years. Henceforth we will use these prop-erties without having to refer to Lemma 4 for justification.

The next result has more substance. It shows that every positive real numbercan be approximated arbitrarily well by rational numbers.

Proposition 5 Ifx any real number, and ifk is any natural number, then thereare rational numbersa, b such that

a < x < b

andb a 0.By Property Three 1/n < x < mfor some n, mN.Setq= 2n + 2k.Then

1/q < x < m = mq/q.

Therefore there is a largest natural number r such that

r/q < x.

In particular,(r+ 1)/qx,

and so(r+ 2)/q > x.

Therefore,r/q < x 0.In either case there is a rational number c such that c + x >0.

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Therefore by what we have just proved, there are rational numbers a, b suchthat

a < c + x < b and b a z such that

c > a + b.

5. Show that the product of a positive real number with a negative real numberis negative.

6. Letx andy be positive real numbers. Show that ifx >1 then (i) xy > y,(ii) 0< 1/x

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8. Ifa < bare non-negative real numbers show that for anypN, bpap p(b

a)bp1. (Hint: use the factoring

bp ap = (b a)p1i=0

aibp1i.)

9. Ifa, b are non-negative real numbers show that for anypN, bp > ap ifand only if b > a. (Hint: Show that if b > a then bp > ap. Then showthat ifb < a thenbp < ap. Then note that ifb= a thenbp =ap. Finally,use the fact that exactly one of the three possibilities, a < b,a > b, a = bmust be true.)

10. Show by induction that for anyx, yR and anybN

(x + y)n =ni=0

n!i!(n i)! x

iyni.

4.3 Absolute value

Definition 11 The absolute value of x R, denoted by|x|, is the non-negtative real number defined by:

|x|= x if x0 and |x|=x if x 0,

thenx= y.

Proof:: We prove this by contradiction.Suppose the assertion is false.Then for some x=yR we would have|x y|< for every >0.But since x y= 0,|x y|> 0.Thus by Property Three for the reals, for somenN, 1/n 0, since 1/n > 0.q.e.d.

Remark: Lemma 5 illustrates the difference between algebra and analysis. Inalgebra we show equality directly. In analysis we sometimes show two numbersare the same by showing that their difference is arbitrarily small.

Lemma 6 For any real numbersx andy:

1.|x| x and|x| x.

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2.|xy|=|x||y|.

3.|x + y| |x| + |y| (The triangle inequality).4.||x| |y|| |x y|.

Proof:: We prove each statement separately.

1. Ifx0 then|x|= x.Otherwisex

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q.e.d.

Important Remark about proofs: The proof of Proposition 5 illustratesan important point. Sometimes in a proof we need to distinguish multiplecases and handle each one separately.

Exercise 8 Absolute value

1. If x is a positive real number show that for some > 0, then y R ispositive if|x y|< .

2. Ifx, zR show that for every >0 there is a >0 such that ifyRsatisfies|y x|< then|zy zx|< .

3. If x R show that for every > 0 there is a > 0 such that if y Rsatisfies

|y

x|< then

|y2

x2

|< . (Hint: Usey2

x2 = (y

x)(y +x).)

4. IfxR andx= 0 show that for every >0 there is a >0 such that ifyR satisfies|y x|< theny= 0 and|1/y 1/x|< .

4.4 Bounds

Definition 12 Bounds

1. A subsetSR is bounded above if for somebR,xb for all xS.

In this case we write

Sb or bS,and sayb is anupper bound forS.

2. A subsetSR is bounded below if for someaR,xa for all xS.

In this case we writeSa or aS,

and saya is a lower bound forS.

3. A subset S R is bounded if it is both bounded above and boundedbelow.

Lemma 7 SupposeSandTare non-void subsets ofR. Then every element ofS is a lower bound for T if and only if every element of T is an upper boundforS. In this case we write

ST or T S.

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Proof:: The statement that every element ofSis a lower bound forT is true ifand only ifx

T for every x

S.

This is equivalent to the statement:

xy for all xS and all yT .But this is equivalent to saying that every yT is an upper bound for S.q.e.d.

Example 10 Bounds

1. Any real numberx1 is a lower bound forN.2. The rationals are not bounded below or above.

3. The setSof real numbers whose squares are less than2 is bounded aboveby2 and below by

2.

In fact, ifx >2 thenx2 =xx >2x >4 and so x /S.Also, if x 2 and x2 = xx = (x)(x) > 4 and again,x /S.Thus2S2.

Exercise 9 Bounds

1. Show that the set of real numbers of the form x/y with|x| >|y| > 0 arenot bounded above or below.

2. Show that the set of real numbers of the form x/y with|y| >|x| > 0 isbounded and find explicit upper and lower bounds.

3. SupposeST R are non-void sets. IfST show thatS has exactlyone element.

4. More generally, supposeS R andT R are non-void sets. IfS Tshow thatS Tcan have at most one element. IfS T is non void showthat its unique element is an upper bound forSand a lower bound forT.

4.5 Sup and Inf

In everything we have done so far there is nothing to suggest that there are realnumbers which are not rationals! All we have assumed about the real numbersat this point is listed in Properties One, Two, and Three at the start of Section4.2.

We correct this with one final property, still as an axiom, which is the fun-damental property of the real numbers and which is the key fact whichmakes analysis possible.

Property Four: Let Sbe a non empty subset ofR which is bounded above.ThenShas an upper bound, b with the following property:

Ify is any upper bound for S thenby.

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Lemma 8 The upper boundb in Property Four is unique.

Proof:: Suppose c satisfies the same condition as b.Thenc is an upper bound for S.Thereforebc.Butb is an upper bound for S.Thereforecb.Thereforeb = c.q.e.d.

Definition 13 The real number b in Property Four is called the least upperbound forSand is denoted byb=sup(S).

Important Remark: If S is bounded above, sometimes sup(S) is in S andsometimes it is not! For example if

S={xR| x1}or if

S={xR| x

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2. Let S ={x R| x = 1/n for some n N}. ThenS is bounded belowby 0. Moreever, if x is any positive real number then by Property Threethere is some natural numbern such that1/n < x. Thusx is not a lowerbound for S and so 0 is the greatest lower bound: inf(S) = 0. Thus inthis example, inf(S) isnotan element ofS.

4.6 Powers

We begin by recalling the definiton of integral powers of a real number x, whichwe do by induction.

Definition 15 Let x be a non-zero real number. If n N then xn is definedinductively by

x1 =x and xn+1 =x

xn.

Then we setx0 = 1 andxn = 1/xn.

Our next step is to extend the definition to rational powers of a positive realnumber,x. To do this we first need to prove:

Proposition 7 LetSbe the set of positive real numbers. Then for any naturalnumbernNthe map

SS, xxn

is a bijection.

Proof:: We first show the map is injective. For this, letx, ybe any two positivenumbers. Ify > x we have the formula

yn xn = (y x)n1i=0

yixn1i

It follows thatyn > xn in this case. Similarly, ifx > y it follows that xn > yn

Thus ifxn = yn it cannot be true that y > x or x > y and so it must betrue thatx = y. This proves that our map is injective.

Now we prove that our map is surjective. We need to show that if y is apositive real number andnN then there is a positive real number x such thatxn =y.

For this define a setT by

T ={xR|xn > y}.

Observe that T= . In fact, since y > 0, it follows that y + 1 > 1 and so(y+ 1)n y + 1 > y. Thusy+ 1T. On the other hand, 0T and so T isbounded below. By Proposition 6 T has a greatest lower bound, inf(T). We

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setx = inf(T) and show that xn =y.

However, exactly one of the following three possibilities must hold:

xn < y, xn > y, or xn =y.

We prove our assertion by excluding the first two possibilities. Our strategy isas follows.

1. Ifxn < y we construct a positive real number c such thatxn < cn < y.

But if such a c exists then forzT,

c

n

< y < z

n

and so c is a lower bound for T and x would not be the greatest lowerbound.

2. Ifxn < y we construct a positive real number b such that y < bn < xn.But then bT and b < x, so x would not be a lower bound for T.

In summary,if we can constructcandbthen the possibilitiesxn < y, xn >y, are excluded and so it follows that xn =y. It remains to construct c and b.

To construct c choosekN so that 1/k < x and so that

1/k < y xn2nxn1

.

Then set c = x + 1/k.

Sincex < c, xn < cn. Moreover, by Exercise 7.8,

cn xn = (ni=0

n!

i!(n i)! xi(1/k)ni) xn = 1/k

n1i=0

n!

i!(n i)! xi(1/k)ni1.

Sinceo

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To construct b choose kN so that 0< 1/k < x and

1/k

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Definition 17 Letx be any positive real number and leta be any rational num-ber. Then

xa = (x1/n)m,

wherea= m/n andmZandnN.

Exercise 10 Sup, inf, and powers

1. LetSbe a subset ofR which is bounded above. Then the following condi-tions on a real numberb are equivalent:

(a) b= sup(S).

(b) For each >0 there is an elementxSsuch thatb < xb.(c) For each >0 there is an elementxSsuch that0b x < .

2. Show that ifa, bare rational numbers andx is a positive real number thenxaxb =xa+b.

3. Show that ifa, bare rational numbers andx is a positive real number then(xa)b =xab.

4. Show that ifx >1 is a real number and ifa < b are rational numbers then0< xa < xb. (Hint: Writeb = a + cwithc >0. Then writec = m/n withm, n natural numbers, and use the argument in the proof of injectivity inproposition 7 to show thatxc >1.

5. IfkNandx >0 is a real number, show by induction onk that

xk

x if x 1.

6. Identify what is wrong with the following attempt to prove the assertion inthe previous exercise:Proof: LetS(k) be the statementxk x ifx

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(a) Use Exercise 7.3 to findm < nNsuch thatx < m/n 2. Explain why this impliesthat some integerp satisfies10ka < p < 10kb.

(c) Conclude thatp/10k Sand thatx < p/10k < x.(d) Use Exercise 10.1) to conclude thatx= sup(S).

4.7 Constructing the real numbers

Call a setSof rationals that is bounded above full if

yS andz < y implies thatzS.

Then simply definethe real numbers to be the subsets of the rationals that arebounded above and full, and identify each rational p/qwith the set of all ratio-

nalsp/q. Intuitively we have just filled in the holes among the rationals.IfS and T are sets of rationals that are bounded above and full, then we

think of these sets as real numbers x and y and define x < y if S T, andx > y ifS T. Then set x + y to be the set

S+ T ={p/q+ m/n| p/qS and m/nT},

and define the other algebraic operations in a similar way.

Of course, we then need to prove that all these definitions are well-defined,that all algebraic rules are still true, that the properties above for the orderhold, and that the operations and order for the rationals remain as before. And

then, of course, we need to prove Properties One through Four. All this is longand boring, so instead we will simply take for granted the existence of the realnumbers satisfying those properties.

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Chapter 5

Infinite Sequences

5.1 Convergent sequences

Definition 18 Let S be any set, and let k be any integer. An infinite se-quence (xi)ik of elements in S is a choice of elements xi S, with i Zand ik. We say the sequence begins with xk. For anypk, the infinitesequence(xi)ip is called the pth tail of the original sequence.

Remark: The set of elements actually appearing in an infinite sequence maybe finite, as illustrated by the infinite sequence 1, , 0, 1, 0, 1, 0, . . . in which onlytwo integers, 0 and 1 appear.

In the world of applications we never need to know the exact value of areal number - we only need to know it within a given tolerance. Specs for anyengineering design always specify heights, lengths, weights etc. up to so manyfractions of an inch or so many millimeters, or so many fractions of a gram.

Infinite sequences (xi) of real numbers are therefore a fundamental tool forapplications because they can be used to approximate a real number x to withina given tolerance. Intuitively that means that the error,|xi x|, gets arbitrar-ily small as i gets larger. In other words, if we want to approximatex withina given tolerance we may simply use one of the xias long asiis sufficiently large.

Infinite sequences are also a fundamental tool in analysis, and for this weneed as always to formalize the intuitive idea above:

Definition 19 An infinite sequence (xi)ik of real numbers converges toxR if for all >0 there is some integerpk such that

|xi x|< if ip.

In this case we say the infinite sequence is convergent to xand that x is the

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limit of the sequence. We denote this by

limixi = x.

Lemma 10 If an infinite sequence(xi)of real numbers converges, it convergesto a unique real numberx.

Proof:Suppose the sequence (xi) converges to both x and y .Then for any >0 there is some p such that for ip both|xi x|< /2 and|xi y|< /2.It follows that

|x y|=|x xi+ xi y| |xi x| + |xi y|< .

Since this is true for all , x = y.

q.e.d.

Definition 20 If (xi) is a convergent sequence, the uniquex to which it con-verges is called the limit of the sequence and is denoted by limi(xi).

Example 12

1. The infinite sequence 1, 2, 3, . . . of natural numbers. This sequence doesnot converge.

2. The infinite sequence 0, 1, 1, 2, 2, 3, 3, . . . which lists all the integers.This sequence does not converge.

3. The infinite sequence (xn)n1 defined by xn = 1/n. This sequence con-verges to 0.

4. The infinite sequence(xn)n1 defined inductively byx1= 1, and

xn =

n1i=1

2xi .

This sequences does not converge.

5. The infinite sequence (xn)n1 defined by: xn is the nth largest primenumber. This sequence does not converge.

6. The infinite sequence(xn)n1defined by: xn = 1+(1/2)n

. This sequenceconverges to 1.

Exercise 11 ConvergenceAs always, provide complete proofs for your answers.

1. In each of the examples above supply proofs for the statements about con-vergence.

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2. Suppose (xn)n1 is an infinite sequence of real numbers converging to x.Define the sequence(y

n)n1

byyn

= x2n

. Is this sequence convergent, andif so, what is its limit?

3. Suppose (xn)n1 is an infinite sequence of real numbers. Suppose thesequenceyn =x3n converges to y and that the sequencezn =x3n+2 con-verges toz. Ify=z show that the original sequence(xn)is not convergent.

4. Show that if x is any real number then there is an infinite sequence ofrational numbers converging to x.

5. Suppose (xn)n1 is an infinite sequence of real numbers converging to x.Define a sequence (yn)n1 by yn = xn+1 xn. Prove that the sequence(yn)n1 is convergent, and find its limit.

5.2 Bounded sequences

Definition 21

1. An infinite sequence, (xi)ik, of real numbers is bounded above if forsome real numberb,

xib for all ik.

2. An infinite sequence, (xi)ik, of real numbers is bounded below if forsome real numbera,

axi for all ik.

3. An infinite sequence isboundedif it is both bounded above and boundedbelow.

Definition 22 1. An infinite sequence,(xi)ik, of real numbers is increas-ing if eachxixi+1.

2. An infinite sequence, (xi)ik, of real numbers is decreasing if eachxixi+1.

Lemma 11

1. An increasing sequence, (xi)ik, which is bounded above, converges to

limi(xi) =sup

{xi

|i

k

}.

2. A decreasing sequence, (yi)ik, which is bounded below, converges to

limi(yi) =inf{yi| ik}.

Proof:

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1. Set S={xi| ik}.ThenSis bounded above.Fix any >0.By the definition of sup(S), there is some xpsuch that 0< sup(S)xp < .Moreover, because each xiS, for each i,

xisup(S).Therefore, since the sequence is increasing, for any ip,

xpxisup(S).It follows that for ip

|xi

sup(S)

|= sup(S)

xi

sup(S)

xp < .

Since was any positive real number,the sequence converges to sup(S).

2. Set T ={yi| ik}.ThenTis bounded below.SetT ={yi| ik}.SinceTis bounded below,Tis bounded above and

inf(T) =sup(T).Moreover, the sequence (yi)ik is increasing.

Therefore by the first part of the lemma, the sequence (yi)ik convergesto sup(T) =inf(T).Therefore, for every >0 there is somepk such that for ip,

|yi inf(T)|=| yi (inf(T))< .It follows that the sequence (yi)ik converges to inf(T).

q.e.d.

Exercise 12

1. If 0 < x < 1 show that the sequence (xr)r0 converges to 0. (Hint: UseExercise 6.3 to help show that if inf{xn |n N} > 0 then there is arational number a such that 0 < a < inf{x

n

|n N} < 1. Then showthat0 =inf{xn |nN}, and explain why this proves the result.)2. Suppose (xi) and (yi) are infinite sequences of real numbers converging

respectively to x andy.

(a) Show that(xi+ yi) converges to x + y.

(b) Show that(xiyi) converges to xy.

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(c) If y= 0 show that for some n N, yi= 0 for i n. In this caseprove that the infinite sequence(1/y

i)in

converges to 1/y.

3. Show that convergent sequences are bounded. (Hint: Ifxn converges to xthen for someN, x 1< xn < x + 1 ifnN. Then show that the entiresequence is bounded.

4. Which of the fol lowing infinite sequences are bounded, which are increas-ing, which are decreasing, and which converge to a limit? If the sequenceis convergent, find the limit. As always prove your answers.

(a) 1, 1, 1, 1, 1, 1, . . ..(b) xq =x

1/q wherex(0, ) is fixed. (Here the answers may dependonx.)

(c) xm= cos(m).5. Show that if xi is a convergent sequence then so is the sequence x

2i and

find its limit. Is the converse true?

6. Supposexi is a convergent sequence. Show that the sequence yi = x2i isconvergent with the same limit. Is the converse true?

7. If(xi)is an infinite sequence converging tox, show that|xi| is convergent,and find its limit.

8. Set xn = (sin(a/n))/n, where a is a fixed arbitrary real number. Showthat(xn) is a convergent sequence.

5.3 The Intersection Theorem

We now come to our first major result, which is also the first one which reallyuses Property Four about the reals.

Theorem 1 Let

SmSm+1Sm+2 Sk

be an infinite sequence of bounded subsets ofR. Set

ai = inf(Si) and bi = sup(Si).

Then

1. (ak) is an increasing sequence convergent to a= sup{ak};2. (bk) is a decreasing sequence convergent to b= inf{bk};3. (bk ak is a decreasing sequence of non-negative numbers convergent to

b a, andb a0;

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4.

k [a

k, bk] = [a, b].

Proof:By definition, ak is a lower bound for Sk and bk is an upper bound forSk.SinceSkSk+1 it follows thatak is a lower bound for Sk+1andbk is an upperbound forSk+1.Butak+1is the greatest lower bound forSk+1and bk+1is the least upper boundforSk+1.It follows that for each k

akbk, akak+1, and bk+1bk.Therefore (ak)km is an increasing sequence bounded above by bm, and

(bk)km is a decreasing sequence bounded below by am.In particular,(bk ak) is a decreasing sequence of non-negative numbers.

Now Lemma 11 states that (ai) converges toa = sup{ak} and (bj) convergesto b = inf{bk}.Therefore

limk(bk ak) =b a.Moreover, since each (bk ak)0, b a0.

It remains to show that

k

[ak, bk] = [a, b].

But ifx[a, b] then for all k ,akaxbbk,

and so x[ak, bk]. Thusx

k[ak, bk]; i.e.,k

[ak, bk][a, b].

On the other hand, suppose xk[ak, bk].Then for all k , akxbk, and so

a= sup{ak} xinf{bk}= b.Thus x[a, b], and

k

[ak, bk][a, b].

This completes the proof of the theorem.q.e.d.

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Corollary 1 to Theorem 1 Let (Sk

)km

be as in Theorem 1, and supposelimk(bk ak) = 0. Then

k

[ak, bk]

is a single point c. I f (xk) is an infinite sequence with xk [ak, bk] then thesequence (xk) converges to c.

Proof:It follows from Theorem 1 that b a= 0; i.e., b= a.Denote this point by c.Then [a, b] ={c}, and so

k

[ak, bk] ={c}.

Finally, let >0 be any positive number.Then choosenso thatbk ak < for kn.Sincec[ak, bk] and xk[ak, bk] it follows that ifkn, then

|xk c| .

Thus the sequence (xk) converges toc.q.e.d.

With this corollary we may resolve the question: how can you tell from asequence of real numbers whether or not it is convergent?

Proposition 8 A sequence (xi) of real numbers converges if and only if for

each >0 there is some integerk such that|xi xj |< for all i, jk.Proof:Suppose the sequence converges to x.Then for any 0 there is some integerk such that

|xi xj|< if i, jk.In particular this implies that

|xi xk|< if ik.

Thus ifik,xk xi < and xi xk < ,

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and soxk

< xi< x

k.

Next, define a decreasing sequence of subsets ofR by setting

Sn ={xi| in}.

It follows from the inequalities above that some Sk is bounded.Since anySn differs from Sk by only finitely many points, each Sn is bounded.Therefore we may apply Theorem 1.

The inequalities above show that:

xk is a lower bound for Sk

andxk+ is an upper bound for Sk.

Therefore, in the notation of the Theorem,

xk akbkxk+ .

It follows thatbk ak2.

By Theorem 1 the sequence (bn an) is decreasing tob a.It follows thatb a2 for all >0.Thus b = a.Since xn Sn [an, bn] the Corollary implies that the sequence xn convergesto b = a.q.e.d.

Exercise 13

1. Construct a sequence of intervalsI(n)of lengthl(n)such that the sequence(l(n))converges to zero, and such that each(I(n)andI(n+1) have a pointin common, but such that

n

I(n) =.

2. LetSn be the set of rational numbersa satisfying

21/n < a

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5.4 Subsequences

Definition 23 Asubsequenceof a sequence(xn)nk is a sequence of the form(yi = xni)ir, in whichnr < nr+1< nr+2 . . . is a sequence of integers.

Exercise 14 Subsequences

1. Construct a sequence which has two convergent subsequences convergingto different limits.

2. Construct a sequence which for eachnNhas a subsequence convergingto n.

3. Show that any subsequence of a convergent sequence converges to the samelimit.

4. Show that a sequence which is not bounded above has an increasing subse-quence which is not bounded above.

5. Let (xn) be a sequence such that the set S ={xn} is finite. Show thatthere is a subsequence(xni) such that for all i, (xni) = (xn1).

Theorem 2 Every bounded sequence contains a convergent subsequence.

Proof:Let (xn)np be a bounded sequence.Then the set Sof thexn is contained in an interval [a, b] with a < b.Our first step is to construct by induction on k a decreasing sequence of intervals

[ak, bk][ak+1, bk+1]

with the following two properties:

1. xn[ak, bk] for infinitely many integers n, and2. bk ak = ba2k1 .For this set [a1, b1] = [a, b].

Then suppose by induction that the intervals [ai, bi] are constructed for ik.By construction [ak, bk] contains xn for infinitely many n.Suppose [ak, (ak+ bk)/2] contains xn for infinitely many n. Then set

ak+1= ak and bk+1=ak+ bk

2 .

Otherwise [(ak+ bk)/2, bk] must contain xn for infintely many n.In this case set

ak+1= ak+ bk

2 and bk+1= bk.

By our induction hypothesis bk ak = (b a)/2k1, and therefore

bk+1 ak+1= bk ak2

=b a

2k .

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This completes the inductive construction.

Our second step is use induction on k to construct the subsequence (xnk) sothat for all k

xnk[ak, bk].First, set xn1 =xp.Then suppose by induction that the xni are constructed for ik.Since [ak+1, bk+1] containsxnfor infinitely manynthere is a least integerr > nksuch that xr[ak+1, bk+1].Setxnk+1=xr. This completes the inductive construction of the subsequence.

It remains to show that this subsequence is convergent.But ifik then

xni[ak, bk].Therefore ifi, jk, then

|xni xnj | bk ak = b a2k1

.

Now choose any >0. By Lemma 4.2 there is some nN such that1

2n 0 such that

ifxD and if0

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Fix >0. Choose 1> 0 and 2> 0 so that for xD ,

|f(x) u1|< /2 i f 0

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Review of Contrapositives: Contrapositives were defined in Chapter Two:

The contrapositiveof the statement A is true is the statement A is false.

In analysis we frequently prove a statement is true by showing that the con-trapositive leads to a contradiction, and for this it is often helpful to restate thecontrapositive in a more useful form. Here is a typical example:Statement A: limxcf(x) =u.

So what does it mean for Statement A to be false?

If Statement A is true then for every > 0 there is a > 0 satisfying therequirements of the definition above.Therefore to say Statement A is false is precisely to say there must be some

>0 so that no >0 will work for that particular .

Now what does it mean for to work for that ?It means precisely that for every xD such that 0 0 will work for that means that for every >0 thereis some xD such that

0 0 and every > 0 there is some x D suchthat 0

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1. If limxcf(x) = u then limnf(xn) = u whenever (xn) is a sequence ofpoints in D different from c but convergent to c.

2. If limnf(xn) = u whenever (xn) is a sequence of points in D differentfromc but convergent to c, then limxcf(x) =u.

For the first statement our hypothesis is limxcf(x) =u.Then if (xn) is any sequence of points in D different from but converging to c,we need to prove that

limnf(xn) =u.

In other words, for any >0 we need to show that for some k ,

|f(xn) u|< if nk.

Fix some >0.We are supposing f(x) hasu as limit when x approaches c.It follows that for some >0,

|f(x) u|< if 00 there is some x

D such that:

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This defines a sequence (xn) of points in D different from c.Moreover, for any real number > 0 there is some k

N such that 1/k <

(Property Three for the real numbers).Therefore, fornk

|xn c|< 1/n1/k < .Thus the sequence (xn) converges toc.

Now our hypothesis implies that f(xn) converges to u.But we constructed the sequence so that

|f(xn) u|

for all n, which This contradicts our hypothesis.

It follows the contrapositive cannot be correct and therefore that limxcf(x) =u.q.e.d.

Exercise 15 Limits

1. Ifa < b are points in an intervalD, show that [a, b]D.2. Suppose f is a real-valued function defined in an interval (a, b) and that

for someuR, f(x) < u for all x(a, b). If limxbf(x) exists, showthat

limxbf(x)u.

3. For each of the following choices of D, f, and c, decide if limxcf(x)exists and when it does find the limit.

(a) D= (1, 0), c= 0, f(x) = 1, xD.

(b) D= [0, 1), c= 0, andf(x) = 0, xD.

(c) D= (1, 1), c= 0, and

f(x) =

1, x(1, 0],0, x(0, 1).

4. Supposef andg are real-valued functions in an intervalD. IfcD andif limxcf(x) and limxcg(x) exist, show that

limxc(f(x) + g(x)) =limxcf(x) + limxcg(x), and

limxcf(x)g(x) = (limxcf(x))(limxcg(x)).

In particular conclude that these limits exist.

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5. Supposef is a real-valued function in an intervalD . Suppose further thatfor somec

D and some >0, f(x)

= 0 if0

|x

c|< . Show that

limxc1

f(x)

exists if limxcf(x) exists and limxcf(x)= 0.6. Do the following limits exist?

limx0sin(1

x) and limx0xsin(

1

x).

Hint: you may use the following facts:

(a)1sin(y)1 for allyR.(b) sin(k2 ) = (1)k+1.

7. Define a real-valued functionf in the intervalR by

f(x) =

0, x

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Lemma 13 Supposefis a continuous function in an intervalD. Then

1. For all > 0 and any x D there is a > 0 such that if z D and|z x|< then

|f(z) f(x)|< .2. If(xn)is a sequence of points inD converging to xD then the sequence

(f(xn)) converges to f(x).

Proof:

1. Since f is continuous, limzxf(z) =f(x). In particular, the limit exists.Fix any >0.

By the definition of limit, there is some > 0 such that if z D and00.Choose >0 so that

|f(z) f(x)|< if zD and |z x|< .Since the sequence (xn) in our hypothesis converges to xD , it followsthat for some r ,

|xn

x

|< if n

r.

Thus fornr,|f(xn) f(x)|< .

Therefore (f(xn)) converges to f(x).

q.e.d.

Exercise 16 Continuous functions

1. Show that ifR the constant functionR is continuous.2. Iff andg are continuous functions in an intervalD, show that the func-

tionsf+ g andf g are continuous.

3. If f is a continuous function in an interval D, and if f(x)= 0 for allxD, show that 1/f is continuous.

4. Show that polynomials are continuous functions inR.

5. Show that sin x and cos x are continuous functions. (You may use thetrigonometric formulae forsin(x + y) andcos(x + y).)

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6. If f is a continuous function in an interval D and if g is a continuousfunction inR, show that the compositeg

f is continuous.

7. If f is a continuous function defined in an intervalD, show that the re-striction off to a second interval contained inD is continuous.

8. Iffis a continuous function defined in[a, b)and iflimxbf(x) =u, showthat a continuous functiong: [a, b]R is defined by

g(x) =

f(x), x[a, b),

u, x= b.

9. Supposefis a continuous function in an intervalD.

(a) Show that the function|f| defined by|f|(x) =|f(x)| is continuous.

(b) Show that the functionsf+ andf defined by

f+(x) =

f(x), f(x)0,

0, f(x)< 0. and

f(x) =

f(x), f(x)0,0, f(x)> 0.

are continuous. (Hint: Consider1/2(f+ |f|).)

(c) Show thatf=f+ f.Recall that a continuous function is a real-valued function whose value at a givenpoint is the limit of the values at points approaching the given point. Continu-ous functions in a closed interval automatically satisfy a stronger condition:

Proposition 10 Supposefis a continuous function defined in a closed interval[a, b]. Then for all >0 there is some >0 such that

|f(y) f(x)|< if x, y[a, b] and |x y|< .Proof:We assume the Proposition is false and derive a contradiction.If the Proposition is false there is some >0 such that for every >0 there isa pair of points x, y[a, b] satisfying

|f(y) f(x)| and |x y|< .In particular, we may find sequences (xn) and (yn) of points in [a, b] such that

|f(yn) f(xn)| and |xn yn|< 1n

.

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Since these sequences are bounded it follows from Theorem 2 that the sequence(x

n) contains a convergent subsequence (x

ni).

In the same way the sequence (yni) contains a convergent subsequence (ynik ).Now Exercise 14.3 asserts that the subsequence (xnik ) is convergent.

Setlimk(xnik ) =x and limk(ynik ) =y.

Since the xn) and the (yn are in [a, b], it follows that x[a, b] and y[a, b].Since|xn yn|< 1/nit follows that

limk|ynik xnik |= 0,and so

limk(ynik ) =limk(xnik ).

Sincefis continuous it follows from Lemma 13.2

limk(f(ynik ) =limk(f(xnik )).

But this is impossible because

|f(yn) f(xn)| for all n.This is the desired contradiction.q.e.d.

6.4 Continuous functions preserve intervals

In this section we establish an important and fundamental property of contin-uous functions.

Theorem 3 Suppose f is a continuous function defined in a closed interval[a, b]. Then the image offis a closed interval:

f([a, b]) = [c, d]

for some real numberscd.

Consequences of Theorem 3.

1. Maximum and minimum values. Since [c, d] is the image off theremust be points u, v[a, b] such that

f(u) =c and f(v) =d.

Thusf(u)f(x)f(v)

for all x[a, b].

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2. Intermediate values. Since [c, d] is the image of f it follows that forevery y

[c, d] there is some x

[a, b] such that

f(x) =y.

Important Remarks: Let fbe a continuous function defined in the interval[a, b].

1. Then the Theorem states that

(a) The set of values off is bounded.

(b) f has a minimum and maximum value.

(c) Every number between the minimum and maximum value is a valueoff.

2. However, the points u, vwhere fattains its minimum and maximum valuesmay be anywhere inside the interval [a, b], and it may well happen thatv < u!

Example 15

1. [a, b] = [1, 1] andf(x) =

x, x0,2x, x0.

Here the image off is the interval [2, 0] andf attains its minimum at1 and its maximum at0.

2. f is defined in(0, 1)byf(x) = 1/x. Here the values off are not boundedbelow but are bounded above by1; however, fdoes not attain a maximiumvalue since

sup{f(x)}= 1and1 is not a value off.

While the properties above are consequences of Theorem 3 we actually have toprove them first and then use them to prove the theorem! We do this now.

Proposition 11 Supposefis a continuous function defined in a closed interval[a, b]. Then for someu, v[a, b],

f(u)

f(x)

f(v)

for allx[a, b].Proof:Set S={f(x)|x[a, b]}.Then for each n there is some ynS such that

0sup(S) yn < 1n

.

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In particular, limn(yn) =sup(S).

On the other hand, becauseynSwe may writeyn = f(xn) for somexn[a, b].Then (xn) is a bounded sequence.Therefore by Theorem 2 there is a subsequence xnk converging to some pointv.Since each xnk[a, b], it follows that v[a, b].Now it follows from Lemma 13.2 that the sequence f(xnk) converges to f(v).

Butf(xnk) is a subsequence of the convergent sequence f(xnk).Therefore by Exercise 14.3, this subsequence has the same limit.Therefore

f(v) =limkf(xnk) =limnf(xn) =limnyn = supS.

In other words, for all x[a, b],f(x)f(v).

The proof of the existence ofu is identical.q.e.d.

Definition 29 The pointsu, v in Proposition 10 are called respectively an ab-solute minimum point and an absolute maximum point forf.

Proposition 12 Iffis a continuous function defined in an intervalD and iff(x)< f(y) for somex, y D , then for anyw withf(x)< w < f(y) there issomezD withf(z) =w.

Proof:We prove the Proposition when x < y. (The proof when y < x is simi-lar.)First observe that since x, yD it follows that [x, y]D.

We proceed by three steps.

Step One: Iff(u)< w for some u[x, y) then for some >0,

f(s)< w if s[u, u + ).

Proof of Step One: Since f is continuous, limsuf(s) =f(u).Thus ifw f(u)> 0, it follows that for some >0

f(s) f(u) |f(s) f(u)|< w f(u) ifs[u, u + ).

It follows thatf(s)< w ifs[u, u + ).

Step Two: Let S be the set of points t [x, y] such that f(t) < w. Thensup(S)< y.

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Thus f([a, b]) = [f(u), f(v)].q.e.d.

Exercise 17

In these exercises you may assume the standard properties of sin x and cosx.

1. Letfbe any real-valued function defined in an intervalD.

(a) Show that if f attains a maximum then that maximum is sup(S)whereS={f(x)| xD}.

(b) Iffattains a minimum value, what is it?

2. (a) Use Exercise 16.5 to show thatsin(1/x) is a continuous function in

(0, ).(b) Show that this function attains a maximium and a minimum value.

(c) Show that it is not the restriction of a continuous function defined in[0, ].

3. Show that (1x)sin(1/x) is a continuous function in (0, 1] which doesnot attain a maximum or minimum value.

4. Supply the proof for the casev < u in Proposition 11.

5. Supply the proof for the casey < x in Proposition 12.

6.5 Exponential functionsSuppose x >1 is real number. For for every rational number a, a real number,xa is constructed in Exercise 10, with the following properties:

1. x0 = 1 and x1 =x.

2. Ifa, bQ anda < bthen0< xa < xb.

3. xaxb =xa+b for any a, bQ.4. (xa)b =xab for any a, b

Q.

We now extend the construction to all real exponents to define a function

expx : RRas follows: For any yR, set

S(y) ={aQ|ay} and E(y) ={xa | aS(y)}.

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IfaQ satisfies a < y then a is a lower bound for S(y).Thus it follows from Exercise 10 that xa is a lower bound for