Analysis Notes (only a draft, and the first one!)

Aalysis Notes (oly a draft, ad the first oe!) Ali Nesi Mathematics Departmet Istabul Bilgi Uiversity Kuştepe Şişli Istabul Turkey aesi@bilgi.edu.tr Jue 22, 2004

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Cotets 1 Prelimiaries 9 1.1 Biary Operatio........................... 9 1.2 Biary Relatios........................... 10 2 Real Numbers 13 2.1 Axioms for Additio......................... 14 2.2 Axioms for Multiplicatio...................... 16 2.3 Distributivity............................. 17 2.4 Axioms for the Order Relatio................... 19 2.5 Totally Ordered Sets......................... 21 2.6 Completeess Axiom......................... 23 3 Other Number Sets 27 3.1 Natural Numbers ad Iductio.................. 27 3.1.1 Expoetiatio........................ 30 3.1.2 Factorial............................ 31 3.1.3 Sequeces........................... 33 3.2 Itegers ad Ratioal Numbers................... 34 3.2.1 Expoetiatio........................ 35 3.3 Uiqueess of the Real Number System.............. 36 3.4 Complex Numbers.......................... 39 4 Real Vector Spaces 43 5 Metric Spaces 49 5.1 Examples................................ 49 5.2 Defiitio ad Further Examples.................. 50 5.3 Normed Real Vector Spaces ad Baach Spaces ad Algebras.. 52 5.4 Ope Subsets of a Metric Space................... 54 6 Sequeces ad Limits 57 6.1 Defiitio............................... 57 6.2 Examples of Covergece i R ad C................ 60 6.2.1 The Sequece (1/)..................... 61 3

4 CONTENTS 6.2.2 The Sequece (α )..................... 61 6.2.3 The Sequece (α /!)................... 62 6.3 Covergece ad the Order..................... 63 6.4 Covergece ad the Four Operatios............... 64 6.5 More O Sequeces.......................... 67 6.6 Cauchy Sequeces.......................... 68 6.7 Covergece of Real Cauchy Sequeces............... 69 6.8 Covergece of Some Sequeces................... 73 6.8.1 The Sequece ((1 + 1/) )................. 73 6.8.2 The Sequeces (2 1/ ) ad ( 1/ )............. 74 6.8.3 The Sequece (x ).................... 74 6.9 Divergece to Ifiity........................ 76 6.10 Limit Superior ad Iferior..................... 78 6.11 Complete Metric Spaces....................... 79 6.12 Completio of a Metric Space.................... 80 6.13 Supplemetary Problems....................... 82 7 Series 83 7.1 Defiitio ad Examples....................... 83 7.2 Easy Cosequeces of the Defiitio................ 86 7.3 Absolute Covergece........................ 87 7.4 Alteratig Series.......................... 88 7.5 Criteria for Covergece....................... 89 7.6 Supplemetary Problems....................... 92 7.6.1 Midterm of Math 152.................... 93 7.6.2 Solutios of the Midterm of Math 152........... 94 8 Supplemetary Topics 97 8.1 Liouville Numbers.......................... 97 9 Covergece of Fuctios 101 9.1 Poitwise Covergece of a Sequece of Fuctios........ 101 9.2 Uiform Covergece of a Sequece of Fuctios......... 103 9.3 Uiform Covergece of a Series of Fuctios........... 107 9.4 Uiform Covergece ad Metric.................. 109 9.5 Limits of Fuctios.......................... 109 9.6 Covergece of a Family of Fuctios............... 110 9.7 Supplemetary Topics........................ 111 10 Topological Spaces 113 10.1 Defiitio ad Examples....................... 113 10.2 Closed Subsets............................ 115 10.3 Iterior................................ 115 10.4 Closure................................ 116 10.5 Base of a Topology.......................... 116 10.6 Compact Subsets........................... 117

CONTENTS 5 10.7 Covergece ad Limit Poits.................... 119 10.8 Coected Sets............................ 120 11 Cotiuity 121 11.1 Cotiuity o Metric Spaces..................... 121 11.2 Cotiuity o Topological Spaces.................. 123 11.3 Cotiuous Fuctios ad R..................... 127 11.4 Uiform Cotiuity.......................... 128 11.5 Uiform Covergece ad Cotiuity................ 129 11.6 Supplemetary Topics........................ 129 11.6.1 A Cotiuous Curve Coverig [0, 1] 2............ 129 12 Differetiable Fuctios 131 12.1 Defiitio ad Examples....................... 131 12.2 Differetiatio of Complex Fuctios................ 132 12.3 Basic Properties of Differetiable Fuctios............ 132 12.4 Rules of Differetiatio....................... 133 12.5 Relatioship Betwee a Fuctio ad Its Derivative....... 133 12.6 Uiform Covergece ad Differetiatio............. 134 12.7 Secod ad Further Derivatives................... 136 13 Aalytic Fuctios 137 13.1 Power Series.............................. 137 13.2 Taylor Series............................. 138 13.2.1 Calculatig Taylor Polyomials............... 140 13.3 Aalytic Fuctios.......................... 142 13.4 Trascedetal Fuctios...................... 143 13.4.1 Expoetiatio ad Trigoometric Fuctios....... 143 13.4.2 Iverse Trigoometric Fuctios.............. 146 13.4.3 Logarithm........................... 146 13.4.4 Hyperbolic Fuctios.................... 146 13.5 Supplemet.............................. 146 13.5.1 Trigoometric Fuctios................... 146 13.5.2 Series............................. 146 13.6 Notes................................. 147 14 Graph Drawig 149 14.1 Drawig i Cartesia Coordiates................. 149 14.1.1 Asymptotes.......................... 149 14.2 Parametric Equatios........................ 149 14.3 Polar Coordiates.......................... 149 14.4 Geometric Loci............................ 149 14.5 Supplemet.............................. 151 14.5.1 Lipschitz Coditio..................... 151 14.5.2 A Metric O R....................... 151

6 CONTENTS 15 Riema Itegral 155 15.1 Defiitio ad Examples....................... 155 15.2 Fudametal Theorem of Calculus................. 155 15.3 How To Itegrate?.......................... 155 15.3.1 Power Series......................... 155 15.3.2 Trigoometric Fuctios................... 155 15.4 Itegratio of Complex Fuctio.................. 156 15.4.1 Fuctios Defied by Itegratio.............. 156 15.5 Applicatios.............................. 156 15.5.1 Applicatio to Series..................... 156 15.5.2 Stirlig Formula....................... 159 15.5.3 Euler s Γ Fuctio...................... 161 16 Supplemets 165 16.1 Stoe-Weierstrass Theorem..................... 165 17 Fourier Series 167 17.1 Hilbert Spaces............................ 167 17.2 Fourier Series............................. 167 18 Topological Spaces (cotiued) 169 18.1 Product Topology.......................... 169 18.2 Homeomorphisms........................... 170 18.3 Sequeces i Topological Spaces.................. 170 18.4 Sequetial Compactess....................... 171 18.5 Supplemets............................. 171 18.5.1 T 0 -Idetificatio....................... 171 19 Exams 173 19.1 Midterm Math 121 (November 2002)................ 173 19.2 Fial Math 121 (Jauary 2003)................... 177 19.3 Resit of Math 121, February 2003.................. 179 19.4 Correctio of the Resit of Math 121, February 2003........ 181 19.5 Secod Resit of Math 121, March 2003............... 184 19.6 Midterm of Math 152, April 2004.................. 186 19.7 Fial of Math 152, Jue 2004.................... 186

CONTENTS 7 Foreword This is a revolvig ad cotiuously chagig textbook. The preset versio is far from complete. It may cotai several errors, omissios, oversights etc. Do ot circulate ad cosult it oly with great suspicio. The ew versio ca be foud at www.aliesi.org. Chapters 1 through 19 form the first semester of a four semester course.

8 CONTENTS

Chapter 1 Prelimiaries 1.1 Biary Operatio Let X be a set. A biary operatio o X is just a fuctio from X X ito X. The biary operatios are ofte deoted by such symbols as +,,,, etc. The result of applyig the biary relatio to the elemets x ad y of X is deoted as x + y, x y, x y, x y, x y etc. Examples. i. Let X be a set ad let c X be ay fixed elemet. The rule x y = c defies a biary operatio o X. This biary operatio satisfies x y = y x for all x, y X (commutativity) ad x (y z) = (x y) z (associativity). ii. Let X be a set. The rule x y = x defies a biary operatio o X. Uless X 1, this biary operatio is ot commutative. But it is always associative. iii. Let U be a set. Let X := (U) be the set of subsets of U. The rule A B = A B defies a biary operatio o X. This biary operatio is commutative ad associative. Note that A U = U A = A for all A X. Such a elemet is called the idetity elemet of the biary operatio. The rule A B = A B defies aother biary operatio o X, which also commutative ad associative. is the idetity elemet of this biary operatio. Examples i ad ii do ot have idetity elemets uless X = 1. iv. Let U be a set. Let X := (U) be the set of subsets of U. The rule A B = A\B defies a biary operatio o X, which is either associative or commutative i geeral. It does ot have a idetity elemet either, although it has a right idetity elemet, amely. v. Let U be a set. Let X := (U) be the set of subsets of U. The rule A B = (A \ B) (B \ A) defies a biary operatio o X, which is 9

10 CHAPTER 1. PRELIMINARIES commutative ad associative (harder to check) ad which has a idetity elemet. Every elemet i this example has a iverse elemet i the sese that, if e deotes the idetity elemet of X for this operatio, the for every x X there is a y X (amely y = x) such that x y = y x = e. Exercises. i. Let A be a set. Let X be the set of fuctios from A ito A. For f, g X, defie the fuctio f g X by the rule (f g)(a) = f(g(a)) for all a A. Show that this is a biary operatio o X which is associative, ocommutative if A > 1 ad which has a idetity elemet. The idetity elemet (the idetity fuctio) is deoted by Id A ad it is defied by the rule Id A (a) = a for all a A. Show that if A > 1 the ot all elemets of X have iverses. ii. Let A be a set. Let Sym(A) be the set of bijectios from A ito A. For f, g Sym(A), defie the fuctio f g Sym(A) by the rule (f g)(a) = f(g(a)) for all a A (as above). Show that this is a biary operatio o X which is associative, ocommutative if A > 2 ad which has a idetity elemet. Show that every elemet of Sym(A) has a iverse. 1.2 Biary Relatios Let X be a set. A biary relatio o X is just a subset of X X. Let R be a biary relatio o X. Thus R X X. If (x, y) R, we will write xry. If (x, y) R, we will write x Ry. Biary relatios are ofte deoted by such symbols as R, S, T, <, >,,,,,,,,,,, etc. Examples. i. R = X X is a biary relatio o the set X. We have xry for all x, y X. ii. R = is a biary relatio o X. For this relatio, x Ry for all x, y X. iii. Let R := δ(x X) := {(x, x) : x X}. The R is a biary relatio o X. We have xry if ad oly if x = y. iv. The set R := {(x, y) X X : x y} is a biary relatio o X. Thus, for all x, y X, (x, y) R if ad oly if x y.

1.2. BINARY RELATIONS 11 v. Let A ad Y be two set ad B A. Let X be the set of fuctios from A ito Y. For f, g X, set f g if ad oly if f(b) = g(b) for all b B. This is a biary relatio o X. It has the followig properties: Reflexivity. For all f X, f f. Symmetry. For all f, g X, if f g the f f. Trasitivity. For all f, g, h X, if f g ad g h, the g h. A relatio satisfyig the three properties above is called a equivalece relatio. The relatio i Example iii is also a equivalece relatio. Exercises. i. Let A ad Y be two set. Let be a oempty set of subsets of A satisfyig the followig coditio: For all B 1, B 2, there is a B 3 such that B 3 B 1 B 2. Let X be the set of fuctios from A ito Y. For f, g X, set f g if ad oly if there is a B such that f(b) = g(b) for all b B. Show that this is a equivalece relatio o X.

12 CHAPTER 1. PRELIMINARIES

Chapter 2 Real Numbers We will defie the set of real umbers axiomatically. We will supply some umber of axioms (which are by defiitio statemets that we accept without proofs) ad we will state that the set of real umbers is a set that satisfies these axioms. We will either ask or aswer the (importat) questio of the existece of the set of real umbers. This questio is part of Math 111. However we will prove that the set of real umbers is uique i a sese to be made precise (Theorem 3.3.1). We will be iterested just i two (biary) operatios o R, called additio ad multiplicatio. Apart from these two operatios, we will also be iterested i a (biary) relatio <. Our defiitio will take some time, till page 23. Defiitio 2.0.1 A set R together with two biary operatios + ad, two distict costats 0 R ad 1 R ad a biary relatio < is called a set of real umbers if the axioms A1, A2, A3, A4, M1, M2, M3, M4, D, O1, O2, O3, AO, MO ad C (that will be stated i this chapter) hold. The biary operatio + is called additio, the biary operatio is called multiplicatio, the elemet 0 is called zero, the elemet 1 is called oe, the biary relatio < is called the order relatio. For two elemets x, y R, x+y will be called the sum of x ad y; istead of x y we will ofte prefer to write xy. xy will be called the product of x ad y. If x < y, we will say that x is less tha y. We will use the expressios greater tha, ot less tha etc. freely. We also defie the biary relatios, > ad as follows: x y x < y or x = y x > y y < x x y x > y or x = y The fact that 0 1 (which is explicitly stated i the defiitio) is importat ad will be eeded later. Oe caot prove this fact from the rest of the axioms. 13

14 CHAPTER 2. REAL NUMBERS Ideed, the set {0} satisfies all the axioms (take 1 to be 0) that we will state. I fact, if 0 = 1, the there ca be oly oe elemet i R, amely 0, because, for ay x R, x M2 = 1x = 0x A2 = 0. 2.1 Axioms for Additio We start with the axioms that ivolve oly the additio. A1. Additive Associativity. For ay x, y, z R, x + (y + z) = (x + y) + z. The first axiom tells us that whe addig, the paretheses are uecessary. For example, istead of (x + y) + z, we ca just write x + y + z. Similarly, istead of (x + y) + (z + t) or of ((x + y) + z) + t, we ca just write x + y + z + t. Although this fact (that the paretheses are useless) eeds to be prove, we will ot prove it. The iterested reader may look at Bourbaki. A2. Additive Idetity Elemet. For ay x R, x + 0 = 0 + x = x. A3. Additive Iverse Elemet. For ay x R there is a y R such that x + y = y + x = 0. A set together with a biary operatio, say +, ad a elemet deoted 0 that satisfies the above axioms is called a group. Thus (R, +, 0) is a group. Below, ivestigatig the structure (R, +, 0), we will i fact ivestigate oly the properties of a group. Note that the elemet y of A3 depeds o x. Note also that A3 does ot tell us that x + y = y + x for all x, y R, it oly tells us that it is so oly for the specific pair x ad y. We also ote that 0 is the oly elemet that satisfies A2; ideed if 0 1 also satisfies A2, the 0 = 0 + 0 1 = 0 1. We ow prove our first result: Lemma 2.1.1 Give x R, the elemet y as i A3 is uique. Proof: Let x R. Let y be as i A3. Let y 1 satisfy the equatio x + y 1 = 0. We will show that y = y 1, provig more tha the statemet of the lemma. We start: y A2 = y + 0 = y + (x + y 1 ) A1 A3 A2 = (y + x) + y 1 = 0 + y 1 = y 1. Thus y = y 1. Sice, give x, the elemet y that satisfies A3 is uique, we ca ame this elemet as a fuctio of x. We will deote it by x ad call it the additive iverse of x or just mius x. Therefore, we have: x + ( x) = ( x) + x = 0.

2.1. AXIOMS FOR ADDITION 15 Sice, by A2, 0 is its ow iverse, we have 0 = 0. As we have said, the proof of Lemma 2.1.1 proves more, amely the followig: Lemma 2.1.2 If x, y R satisfy x + y = 0, the y = x. We start to prove some umber of well-kow results: Lemma 2.1.3 For all x R, ( x) = x. Proof: It is clear from A3 that if y is the additive iverse of x, the x is the additive iverse of y. Thus x is the additive iverse of x. Hece x = ( x). Lemma 2.1.4 If x, y R satisfy x + y = 0, the x = y. Proof: Follows directly from lemmas 2.1.2 ad 2.1.3. Lemma 2.1.5 For all x, y R, (x + y) = ( y) + ( x). Proof: We compute directly: (x + y) + (( y) + ( x)) = x + y + ( y) + ( x) = x + 0 + ( x) = x + ( x) = 0. Here the first equality is the cosequece of A1 (that states that the paretheses are useless). Thus (x+y)+(( y)+( x)) = 0. By Lemma 2.1.2, ( y)+( x) = (x+y). We should ote that the lemma above does ot state that (x+y) = ( x)+ ( y). Although this equality holds i R, we caot prove it at this stage; to prove it we eed Axiom A4, which is yet to be stated. We ow defie the followig terms: Lemma 2.1.6 For all x, y R, Proof: Left as a exercise. x y := x + ( y) x + y := ( x) + y x y := ( x) + ( y) (x y) = y x ( x + y) = y + x ( x y) = y + x The ext lemma says that we ca simplify from the left. Lemma 2.1.7 (Left Cacellatio) For x, y, z R if x + y = x + z the y = z. Proof: Add x to the left of both parts of the equality x + y = x + z, ad usig associativity, we get y = z. Similarly we have, Lemma 2.1.8 (Right Cacellatio) For x, y, z R if y + x = z + x the y = z. Fially, we state our last axiom that ivolves oly the additio.

16 CHAPTER 2. REAL NUMBERS A4. Commutativity of the Additio. For ay x, y R, x + y = y + x. A set together with a biary operatio, say +, ad a elemet deoted 0 that satisfies the axioms A1, A2, A3 ad A4 is called a commutative or a abelia group. Thus (R, +, 0) is a abelia group. 2.2 Axioms for Multiplicatio Let R = R \ {0}. Sice 1 0, the elemet 1 is a elemet of R. I this subsectio, we will replace the symbols R, + ad 0 of the above subsectio, by R, ad 1 respectively. For example, Axiom A1 will the read M1. Multiplicative Associativity. For ay x, y, z R, x (y z) = (x y) z. As we have said, we will prefer to write x(yz) = (xy)z istead of x (y z) = (x y) z. Axioms A2 ad A3 take the followig form: M2. Multiplicative Idetity Elemet. For ay x R, x1 = 1x = x. M3. Multiplicative Iverse Elemet. For ay x R there is a y R such that xy = yx = 1. We accept M1, M2 ad M3 as axioms. Thus (R,, 1) is a group. All the results of the previous subsectio will remai valid if we do the above replacemets. Of course, the use of the axioms A1, A2, A3 i the proofs must be replaced by M1, M2, M3 respectively. That is what we will do ow: Lemma 2.2.1 Give x R, the elemet y as i M3 is uique. The proof of this lemma ca be traslated from the proof of Lemma 2.1.1 directly: Proof: Let x R. Let y be as i M3. Let y 1 satisfy the equatio xy 1 = 1. We will show that y = y 1, provig more tha the statemet of the lemma. We start: y M2 = y1 = y(xy 1 ) M1 M3 M2 = (yx)y 1 = 1y 1 = y 1. Thus y = y 1. Sice, give x R, the elemet y R that satisfies M3 is uique, we ca ame this elemet as a fuctio of x. We will deote it by x 1 ad call it the multiplicative iverse of x or sometimes x iverse. Therefore, we have: xx 1 = x 1 x = 1. Note that x 1 is defied oly for x 0. The term 0 1 will ever be defied. Sice, by M2, 1 is its ow iverse, we have 1 1 = 1.

2.3. DISTRIBUTIVITY 17 As we have oticed, the proof of Lemma 2.2.1 proves more, amely the followig: Lemma 2.2.2 If x, y R satisfy xy = 1, the y = x 1. Lemma 2.2.3 For all x R, (x 1 ) 1 = x. Proof: As i Lemma 2.1.3. Lemma 2.2.4 If x, y R satisfy xy = 1, the x = y 1. Proof: As i Lemma 2.1.4. Lemma 2.2.5 For all x, y R, (xy) 1 = y 1 x 1. Proof: As i Lemma 2.1.5. We should ote that the lemma above does ot state that (xy) 1 = x 1 y 1. Although this equality holds i R, we caot prove it at this stage; to prove it we eed Axiom M4. The ext lemma says that we ca simplify from the left ad also from the right. Lemma 2.2.6 (Cacellatio) Let x, y, z R. i. If xy = xz the y = z. ii. If yx = zx the y = z. Proof: Left as a exercise. The lemma above is also valid if either y or z is zero (without x beig zero), but we caot prove it yet. Fially, we state our last axiom that ivolves oly multiplicatio. M4. Commutativity of the Multiplicatio. For ay x, y R, xy = yx. Thus (R,, 1) is a abelia group. Sometimes, oe writes x/y or x y istead of xy 1. 2.3 Distributivity I the first subsectio, we stated the axioms that ivolve oly the additio, ad i the secod subsectio, the axioms that ivolve oly the multiplicatio. Util ow there is o relatioship whatsoever betwee the additio ad the multiplicatio. For the momet they appear to be two idepedet operatios. Cosequetly, at this poit we caot prove ay equality that ivolves both operatio, e.g. the equalities ( 1) 1 = 1 ad ( 1)x = x caot be prove at this stage. Below, we state a axiom that ivolves both additio ad multiplicatio.

18 CHAPTER 2. REAL NUMBERS LD. Left Distributivity. For all x, y, z R, x(y + z) = xy + xz. Sice the multiplicatio is commutative, we also have the partial right distributivity valid if x, y, z 0 ad y + z 0: (y + z)x M4 = x(y + z) D = xy + xz M4 = yx + zx. Lemma 2.3.1 For all x R, x0 = 0. Proof: Sice x0 + 0 A2 = x0 A2 = x(0 + 0) LD = x0 + x0, by Lemma 2.1.7, 0 = x0. I do ot kow whether oe ca prove the right distributivity i its full geerality or the equality 0x = 0 from the axioms above. It seems (but I may be wrog) that we eed to add the right distributivity to the list of our axioms as well. This is what we will do ow: D. Distributivity. For all x, y, z R, x(y + z) = xy + xz ad (y + z)x = yx + zx. Axiom D will be the oly axiom relatig the additio ad the multiplicatio. Lemma 2.3.2 For all x R, 0x = 0. Proof: As i Lemma 2.3.1. It follows that M4 is valid for all x, y R. A set R together with two biary operatios + ad ad costats 0 ad 1 that satisfy the axioms A1, A2, A3, A4, M1, M2, M3, M4 ad D is called a field. Thus (R, +,, 0, 1) is a field. We ivestigate some other cosequeces of distributivity: Lemma 2.3.3 For all x, y R, ( x)y = (xy) x( y) = (xy) ( x)( y) = xy Proof: We compute directly: 0 = 0y A3 = (x + ( x))y = D xy + ( x)y (the first equality is Lemma 2.3.2). Thus ( x)y is the additive iverse of xy ad so ( x)y = (xy). This is the first equality. The others are similar ad are left as exercise. It follows that we ca write xy for ( x)y or x( y). Corollary 2.3.4 For all x R, ( 1)x = x. Corollary 2.3.5 ( 1) 1 = 1.

2.4. AXIOMS FOR THE ORDER RELATION 19 2.4 Axioms for the Order Relatio The first three axioms below ivolve oly the biary relatio iequality <. Very soo we will relate the iequality to the operatios + ad. O1. Trasitivity. For all x, y, z R, if x < y ad y < z the x < z. O2. Irreflexivity. For all x R, it is ot true that x < x, i.e. x x. O3. Total Order. For ay x, y R, either x < y or x = y or y < x. Lemma 2.4.1 For ay x, y R, oly oe of the relatios hold. x < y, x = y, y < x Proof: Assume x < y ad x = y hold. The x < x, cotradictig O2. Assume x < y ad y < x hold. The by O1, x < x, cotradictig O2. Assume x = y ad y < x hold. The x < x, cotradictig O2. Now we state the two axioms that relate the iequality with the two operatios + ad : OA. For all x, y, z R, if x < y the x + z < y + z. OM. For all x, y, z R, if x < y ad 0 < z the xz < yz. A set R together with two biary operatios + ad, two costats 0 ad 1 ad a biary relatio < that satisfies the axioms A1, A2, A3, A4, M1, M2, M3, M4, D, O1, O2, O3, OA ad OM is called a ordered field. Thus (R, +,, 0, 1) is a ordered field. Below, we ivestigate the properties of ordered fields. Lemma 2.4.2 If x < y the y < x. Proof: Addig x y to both sides of the iequality x < y, by OA we get the result. Lemma 2.4.3 If 0 < x < y the 0 < y 1 < x 1. Proof: Left as a exercise. Lemma 2.4.4 If x < 0 ad y < 0 the 0 < xy. Proof: By Lemma 2.4.2, 0 < x ad 0 < y. The, 0 2.3.2 = 0( y) OM < ( x)( y) 2.3.3 = xy. For ay x, we defie x 2 to be xx.

20 CHAPTER 2. REAL NUMBERS Corollary 2.4.5 For ay x R, x 2 0. Proof: Left as a exercise. If x < 0 we say that x is strictly egative, if x > 0 we say that x is strictly positive. If x 0 we say that x is opositive ad if x 0 we say that x is opositive. We let R >0 = {x R : x > 0} R 0 = {x R : x 0} = R> 0 {0} R <0 = {x R : x < 0} = R >0 R 0 = {x R : x 0} = R <0 {0} = R 0 To fiish the axioms of real umbers, there is oe more axiom left. This will be the subject of oe of the later subsectios. Exercises ad Examples. i. The set {0, 1} with the followig additio 0 + 0 = 1 + 1 = 0 0 + 1 = 1 + 0 = 1 ad multiplicatio defied by 0 x = x 0 = 0 for x = 0, 1 ad 1 1 = 1 satisfies all the axioms about additio ad multiplicatio (A1, A2, A3, A4, M1, M2, M3, M4, D), but there is o order relatio o {0, 1} that satisfies the order axioms (O1, O2, O3, OA, OM). ii. The set Z of itegers, with the usual additio, multiplicatio ad the order relatio, satisfies all the axioms except M3. iii. The set Q of ratioal umbers, with the usual additio, multiplicatio ad the order relatio, satisfies all the axioms. We will see that Q does ot satisfy the Completeess Axiom that we will state i the ext subsectio. We assume for the rest that we are i a structure that satisfies the axioms stated util ow. iv. Show that if xy = 0 the either x or y is zero. v. Defie 2 to be 1 + 1. Show that 2x = x + x. vi. Show that the axioms above imply that R is ifiite. vii. Prove that (x + y) 2 = x 2 + 2xy + y 2 ad that x 2 y 2 = (x y)(x + y). (Recall that x 2 was defied to be xx). viii. Prove that if 0 < x < y the x 2 < y 2. ix. Prove that if x 2 = y 2 the either x = y or x = y.

2.5. TOTALLY ORDERED SETS 21 x. Prove that if x < y ad z < t the x + z < y + t. xi. Prove that if 0 < x < y ad 0 < z < t the xz < yt. xii. Prove that if x < y, the x < x+y 2 < y. xiii. Defie x to be max(x, x) (i.e. the largest of the two). For all x, y R, show the followig: a) x 0. b) x = 0 if ad oly if x = 0. c) x = x. d) x + y x + y. e) Coclude from (d) that x y x y. f) Show that x y x y. (Hit: Use (e)). g) Show that xy = x y. The umber x is called the absolute value of x. xiv. Let x, y R. Show that max(x, y) = x + y + x y 2, mi(x, y) = x + y x y. 2 xv. Show that x y x y for ay x, y R. xvi. Show that Example v o page 9 is a commutative group. 2.5 Totally Ordered Sets Let X be a set together with a biary relatio < that satisfies the axioms O1, O2 ad O3. We will call such a relatio < a totally ordered set, or a liearly ordered set, or a chai. Thus R is a totally ordered set. Thus (R, <) is a totally ordered set. We defie the relatios x y, x > y, x y, x y etc. as usual. I a totally ordered set (X, <) oe ca defie itervals as follows (below a ad b are elemets of X): (a, b) = {x X : a < x < b} (a, b] = {x X : a < x b} [a, b) = {x X : a x < b} [a, b] = {x X : a x b} (a, ) = {x X : a < x} [a, ) = {x X : a x} (, a) = {x X : x < a} (, a] = {x X : x a} (, ) = X

22 CHAPTER 2. REAL NUMBERS A elemet M of a totally ordered set X is called maximal if o elemet of X is greater tha M. A elemet m of a poset X is called miimal if o elemet of X is less tha m. Let (X, <) be a totally ordered set. Let A X be a subset of X. A elemet x X is called a upper boud for A if x a for all a A. A elemet a X is called a least upper boud for A if, i) a is a upper boud of A, ad ii) If b is aother upper boud of A, b a. Let X be a poset. A subset of X that has a upper boud is said to be bouded above. The terms lower boud, greatest lower boud ad a set bouded below are defied similarly. Lemma 2.5.1 A subset of a totally ordered set which has a least upper boud (resp. a greatest lower boud) has a uique least upper boud (resp. greatest lower boud). Proof: Let X be a totally ordered set. Let A X be a subset which has a least upper boud, say x. Assume y is also a least upper boud for A. By defiitio y x ad x y. The x = y by O3. Thus if A is a subset of a totally ordered set whose least upper boud exists, the we ca ame this least upper boud as a fuctio of X. We will use the otatio lub(a) or sup(a) for the least upper boud of A. We use the otatios glb(a) ad if(a) for the least upper boud. Examples ad Exercises. i. Let X = (R, <). Let A = (0, 1) R. The ay umber 1 is a upper boud for A. 1 is the oly least upper boud of A. Note that 1 A. ii. Let X = (R, <). Let A = [0, 1] R. The ay umber 1 is a upper boud for A. 1 is the oly least upper boud of A. Note that 1 A. iii. Let X = (R, <). Let A = (0, ) R. The A has o upper boud. But it has a least upper boud. iv. Let X = (R, <) ad A = { x oly least upper boud of A. x+1 : x R ad x 1}. Show that 1 is the v. Let X = (R, <) ad A R ay subset of R. Defie A = { a : a A}. i. Show that if x is a upper boud for A the x is a lower boud for A. ii. Show that if x is the least upper boud for A R the x is the greatest lower boud for A.

2.6. COMPLETENESS AXIOM 23 vi. Let (X, <) be a totally ordered set ad A X. Let B := {x X : x is a upper boud for A}. a) Assume that sup(a) exists. Show that if(b) exists ad if(b) = sup(a). b) Assume that if(b) exists. Show that sup(a) exists ad sup(a) = if(b). vii. Let X be a totally ordered set ad b X. Show that b is the oly least upper boud of (, b], ad also of (, b). viii. O R R defie the relatio as follows (x, y) (x 1, y 1 ) by either y < y 1, or y = y 1 ad x < x 1. a) Show that this is a total order (called lexicographic orderig). b) Does every subset of this liear order which has a upper boud has a least upper boud? ix. O N N defie the relatio as above (lexicographic order). a) Show that this is a total order. b) Does every subset of this liear order which has a upper boud has a least upper boud? x. Fid a poset where the itersectio of two itervals is ot ecessarily a iterval. 2.6 Completeess Axiom We ow state the last axiom for R. This axiom is differet from the others i the sese that all the other axioms were about a property of oe, two or at most three elemets of R. But this oe is a statemet about subsets of R. C. Completeess. Ay oempty subset of R which is bouded above has a least upper boud. It follows from Lemma 2.5.1, that a subset X of R which has a upper boud has a uique least upper boud. We deote it by sup(x) or lub(x). The least upper boud of a set is sometimes called the supremum of the set. Note that the supremum of a set may or may ot be i the set. This completes our list of axioms. From ow o we fix a set R together with two biary operatios + ad, two distict costats 0 R ad 1 R ad a biary relatio < that satisfies the axioms A1, A2, A3, A4, M1, M2, M3, M4, D, O1, O2, O3, AO, MO ad C stated above. (The existece of such a structure is prove i Math 112.)

24 CHAPTER 2. REAL NUMBERS Lemma 2.6.1 Ay oempty subset of R which is bouded below has a greatest lower boud. Proof: Left as a exercise. Use Exercise v, 22 or Exercise i, 24 below. The greatest lower boud of a (oempty) set is deoted by if(x) or glb(x). The greatest lower boud is sometimes called the ifimum of the set. Note that the ifimum of a set may or may ot be i the set. Exercises. i. Show that if X R has a least upper boud the the set X := { x : x X} has a greatest lower boud ad if( X) = sup(x). ii. Suppose X, Y R have least upper bouds. Show that the set X + Y := {x + y : x X, y Y } has a least upper boud ad that sup(x + Y ) = sup(x) + sup(y ). iii. Suppose X, Y R 0 have least upper bouds. Show that the set XY := {xy : x X, y Y } has a least upper boud ad sup(xy ) = sup(x) sup(y ). Does the same equality hold for ay two subsets of R? The completeess axiom makes the differece betwee Q ad R. The equatio x 2 = 2 has o solutio i Q but has a solutio i R. This is what we ow prove. Theorem 2.6.2 Let a R >0. The there is a x R such that x 2 = a. Proof: Replacig a by 1/a if ecessary, we may assume that a 1. Let A = {x R 0 : x 2 a}. For x A, we have x 2 a a 2. It follows that 0 a 2 x 2 = (a x)(a + x), so a x. We proved that a is a upper boud for A. Let b = lub(a). We will show that b 2 = a. Clearly b 1 > 0. Assume first that b 2 < a. Let ɛ = mi( a b2 2b+1, 1) > 0. The (b + ɛ)2 = b 2 + 2bɛ + ɛ 2 b 2 + 2bɛ + ɛ = b 2 + ɛ(2b + 1) b 2 + (a b 2 ) = a. Hece b + ɛ A. But this cotradicts the fact that b is the least upper boud for A. Assume ow that b 2 > a. Let ɛ = mi( b2 a 2b, b) > 0. Now (b ɛ)2 = b 2 2bɛ + ɛ 2 > b 2 2bɛ b 2 (b 2 a) = a. Thus (b ɛ) 2 > a. Let x A be such that b ɛ x b. (There is such a x because b ɛ is ot a upper boud for A). Now we have a < (b ɛ) 2 x 2 a (because b ɛ 0), a cotradictio. It follows that b 2 = a. Remark. Sice every oegative real umber has a square root, the order relatio < ca be defied from + ad as follows: for all x, y R, x < y if ad oly if z (z 0 y = x + z 2 ).

2.6. COMPLETENESS AXIOM 25 Exercises. i. Let A R be a subset satisfyig the followig property: For all a, b A ad x R, if a x b the x A. Show that A is a iterval. ii. Let x 3 mea x x x. Show that for ay x R there is a uique y R such that y 3 = x.

26 CHAPTER 2. REAL NUMBERS

Chapter 3 Other Number Sets 3.1 Natural Numbers ad Iductio We say that a subset X of R is iductive if 0 X ad if for all x X, x + 1 is also i X. For example, the subsets R 0, R, R \ (0, 1) are iductive sets. The set R >0 is ot a iductive set. Lemma 3.1.1 A arbitrary itersectio of iductive subsets is a iductive subset. The itersectio of all the iductive subsets of R is the smallest iductive subset of R. Proof: Trivial. We let N deote the smallest iductive subset of R. Thus N is the itersectio of all the iductive subsets of R. The elemets of N are called atural umbers. Theorem 3.1.2 (Iductio Priciple (1)) Let X be a subset of R. Assume that 0 X ad for ay x R, if x X the x + 1 X. The N X. Proof: The statemet says that X is iductive. Therefore the theorem follows directly from the defiitio of N. Suppose we wat to prove a statemet of the form for all x N, σ(x). For this, it is eough to prove i) σ(0), ii) If σ(x) the σ(x + 1). Ideed, assume we have proved (i) ad (ii). let X := {x R : σ(x)}. By (i), 0 X. By (ii), if x X the x + 1 X. Thus, by the Iductio Priciple, N X. It follows that for all x N, σ(x). Lemma 3.1.3 i. lub(n) = 0, i.e. 0 is the least elemet of N. ii. If x N \ {0} the x 1 N. 27

28 CHAPTER 3. OTHER NUMBER SETS iii. N is closed uder additio ad multiplicatio, i.e. if x, y N the x + y, xy N. iv. If 0 < y < 1 the y N. v. Let x N. If x < y < x + 1, the y N. vi. Let x, y N. If x < y the x + 1 y. vii. Let x, y N. Assume y < x. The x y N. viii. If x, y N ad y < x + 1, the either y = x or y < x. Proof: i. Clearly R 0 is a iductive set. But N is defied to be the smallest iductive subset of R. Thus N R 0. Sice 0 N, it follows that 0 is the least elemet of N. ii. Assume that for x N\{0}, x 1 N. The the set N\{x} is a iductive set (check carefully). Sice N is the smallest iductive set, N N \ {x} ad x N. iii. Let x, y N. We proceed by iductio y to show that x + y N, i.e. lettig σ(y) deote the statemet x+y N, we show that σ(0) holds ad that if σ(y) holds the σ(y + 1) holds. If y = 0 the x + y = x + 0 = x N. Thus σ(0) holds. Assume ow σ(y) holds, i.e. that x+y N. The x+(y +1) = (x+y)+1. Sice x + y N by assumptio, we also have x + (y + 1) N. Thus σ(y + 1) holds also. Therefore σ(y) holds for all N. The proof for the multiplicatio is left as a exercise. iv. The set N \ (0, y] is a iductive set as it ca be show easily. Thus N N \ (0, y] ad y N. v. We proceed by iductio o x. The previous part gives us the case x = 0. Assume the statemet holds for x ad we proceed to show that the statemet holds for x + 1. Let x + 1 < y < (x + 1) + 1, the x < y 1 < x + 1. By the iductive hypothesis, y 1 N. By part (i), either y = 0 or y N. Sice 0 < x + 1 < y, we caot have y = 0. Thus y N. vi. Assume ot. The x < y < x + 1, cotradictig part (v). vii. By iductio o x. If x = 0, the the statemet holds because there is o y N such that y < x. Assume the statemet holds for x. We will show that it holds for x+1. Let y N be such that y < x+1. The either y < x or y = x. Now (x + 1) y = (x y) + 1. I case y < x, the iductio hypothesis gives x y N ad so (x + 1) y N. I case y = x, we have (x + 1) y = 1 N. viii. By (v), y x. Lemma 3.1.4 Ay oempty subset of N has a least elemet. Proof: Let = X N. Assume X does ot have a least elemet. We will prove that X =, which is the same as provig that o elemet of N is i X. We will first show the followig statemet φ() for all : No atural umber m < is i X. Sice there are o atural umbers < 0 the statemet φ holds for 0. Assume φ() holds. If φ( + 1) were false, the would be i X by Lemma 3.1.3.viii ad it would be the smallest elemet of X, a cotradictio. Thus φ() holds for ay. Now if a atural umber were i X, sice φ() holds, would be the smallest elemet of X, a cotradictio.

3.1. NATURAL NUMBERS AND INDUCTION 29 There is a slightly more complicated versio of the iductive priciple that is very ofte used i mathematics: Theorem 3.1.5 (Iductio Priciple (2)) Let X be a subset of N. Assume that for ay x N, The X = N. ( y N (y < x y X)) x X. Proof: Assume ot. The N \ X. Let x be the least elemet of N \ X. Thus y N (y < x y X). But the by hypothesis x X, a cotradictio. How does oe use Iductio Priciple (2) i practice? Suppose we have statemet σ(x) to prove about atural umbers x. Give x N, assumig σ(y) holds for all atural umbers y < x, oe proves that σ(x) holds.this is eough to prove that σ(x) holds for all x N. We immediately give some applicatios of the Iductio Priciples. Theorem 3.1.6 (Archimedea Property) Let ɛ R >0 there is a N such that x < ɛ. ad x R, the Proof: Assume ot, i.e. assume that ɛ x for all N. The the set N is bouded above by x/ɛ. Thus N has a least upper boud, say a. Hece a 1 is ot a upper boud for N. It follows that there is a elemet N that satisfies a 1 <. But this implies a < + 1. Sice + 1 N, this cotradicts the fact that a is a upper boud for N. Lemma 3.1.7 Ay oempty subset of N that has a upper boud cotais its least upper boud. Proof: Let = A N be a oempty subset of N that has a upper boud. Let x be the least upper boud of A. Sice x 1 is ot a upper boud for A, there is a a A such that x 1 < a x. By parts (iv) ad (v) of Lemma 3.1.3.vii a is the largest elemet of A. Theorem 3.1.8 (Itegral Part) For ay x R 0, there is a uique N such that x < + 1}. Proof: Let A = {a N : a x}. The 0 A ad A is bouded above by x. By Lemma 3.1.7, A cotais its least upper boud, say. Thus x < + 1. This proves the existece. Now we prove the uiqueess. Assume m N ad m x < m + 1. If < m, the by Lemma 3.1.3.vii, x < + 1 m x, a cotradictio. Similarly m. Thus = m. Theorem 3.1.9 (Divisio) For ay, m N, m 0 there are uique q, r N such that = mq + r ad 0 r < m.

30 CHAPTER 3. OTHER NUMBER SETS Proof: We first prove the existece. We proceed by iductio o (Iductio Priciple 2). If < m, the take q = 0 ad r =. Assume ow m. By iductio, there are q 1 ad r 1 such that m = mq 1 + r 1 ad 0 r 1 < m. Now = m(q 1 + 1) + r 1. Take q = q 1 + 1 ad r = r 1. This proves the existece. We ow prove the existece. Assume = mq + r = mq 1 + r 1, 0 r < m ad 0 r 1 < m. Assume q 1 > q. The m > m r 1 > r r 1 = mq 1 mq = m(q 1 q) m. This is a cotradictio. Similarly q q 1. Hece q 1 = q. It follows immediately that r = r 1. Exercises. i. Show that for ay N \ {0}, 1 + 3 +... + (2 1) = 2. ii. Show that for ay N\{0}, 1 2 +2 2 +3 2 +...+ 2 = (+1)(2+1)/6. iii. Show that for ay N, 1 1 1 3 + 1 1 3 5 +... + 1 2 1 1 2 + 1 = 2 + 1. iv. Show that for ay N \ {0}, 1 + 3 +... + (2 1) = 2. v. Show that for ay N\{0}, 1 2 +2 2 +3 2 +...+ 2 = (+1)(2+1)/6. vi. Show that for ay N, 1 1 1 3 + 1 1 3 5 +... + 1 2 1 1 2 + 1 = 2 + 1. vii. Show that for ay N \ {0}, 1 2 + 2 2 +... + 2 = (2+1)(+2) 6. 3.1.1 Expoetiatio Let r R. For N, we defie r, th power of r, as follows by iductio o : r 0 = 1 if r 0 r 1 = r r +1 = r r Note that 0 0 is ot defied. We will leave it udefied. Note also that the previous defiitio of r 2 coicides with the oe give above: r 2 = r 1+1 = r 1 r = rr. Propositio 3.1.10 For r R ad N ot both zero, we have, i. (rs) = r s. ii. r r m = r +m. iii. (r ) m = r m.

3.1. NATURAL NUMBERS AND INDUCTION 31 Proof: Left as a exercise. Theorem 3.1.11 i. Let r R 0 ad N \ {0}. The there is a uique s R such that s m = r. ii. Let r R ad let N be odd. The there is a uique s R such that s m = r. Proof: (ii) follows from (i). (i) is proved as i Theorem 2.6.2. Left as a exercise. The umber s is called the m th -root of r. 3.1.2 Factorial For N, we defie! by iductio o : Set 0! = 1, 1! = 1 ad ( + 1)! =!( + 1). This just meas that! = 1 2.... Exercises. i. Show that a set with elemets has! bijectios. ii. Fid a formula that gives the umber of ijectios from a set with elemets ito a set with m elemets. iii. Prove that for N, the set {0, 1,..., 1} has 2 subsets. (Hit: You may proceed by iductio o ). iv. Show that! > 2 for all large eough. v. Show that (x 1) x x 1 for all x > 1. (Hit: By iductio o ). vi. Show that if 0 < x < 1 ad > 0 is a atural umber, the (1 x) 1 x + ( 1) 2 x 2. vii. Show that for ay N \ {0}, 1 3 + 2 3 +... + 3 = (1 + 2 +... + ) 2. viii. Show that for ay N \ {0}, 1 4 + 2 4 +... + 4 = (+1)(63 +9 2 + 1) 30. Choose k. For, k N ad k, defie ( k ) =! k!( k)!.

32 CHAPTER 3. OTHER NUMBER SETS Exercises. ( i. Show that k ( ii. Show that 0 ( iii. Show that 1 ) = ( k ). ) ( ) = = 1. ) =. iv. Show that ( + 1 k + 1 ) ( = k ) ( + k + 1 ( ) v. Deduce that N. (Hit: By iductio o ). k vi. Show that for N ad 0 k, a set with elemets has subsets with k elemets. ). ( k vii. ( Show) that for N ad k N with k, a set with elemets has subsets with k elemets. k ) viii. Show that for x, y R ad N, (x + y) = ( k k=0 ) x k y k. (Hit: By iductio o ). ix. Show that ( ) k=0 = 2 k. x. Show that k=0 ( 1)k ( k ) = 0. xi. Compute (x + y + z) 3 i terms of x, y ad z. xii. Compute (x + y + z) 4 i terms of x, y ad z. xiii. Show that for x > 1 ad N, (x 1) x x 1. xiv. Show that for x < 1, (1 x) 1 x. xv. Show that for N \ {0}, (1 + 1 ) (1 + 1 +1 )+1. (See Theorem 6.8.1).

3.1. NATURAL NUMBERS AND INDUCTION 33 3.1.3 Sequeces Let X be a set. A sequece i X is just a fuctio x : N X. We let x := x() ad deote x by listig its values (x ) rather tha by x. We ca also write x = (x 0, x 1, x 2, x 3,..., x,...). ( If X = R, we speak of a real sequece. For example 1 +1 ) is a real sequece. We ca write this sequece more explicitly by listig its elemets: (1, 1/2, 1/3, 1/4, 1/5,..., 1/( + 1),...). If we write a sequece such as (1/), we will assume implicitly that the sequece starts with = 1 (sice 1/ is udefied for = 0). Thus, with this covetio, the above sequece ( 1 +1 ) may also be deoted by (1/). For ( example, the sequece listed as 1 ( 1) ) starts with = 2 ad its elemets ca be (1/2, 1/6, 1/12, 1/20, 1/30,..., 1/( 1),...). A sequece (x ) is called icreasig or odecreasig, if x x +1 for all N. A sequece (x ) is called strictly icreasig, if x < x +1 for all N. The terms decreasig, strictly decreasig ad oicreasig are defied similarly. Let (x ) be a sequece. Let (k ) be a strictly icreasig sequece of atural umbers. Set y = x k. The we say that the sequece (y ) is a subsequece of the sequece (x ). For example, let x = 1 +1 ad k = 2. The y = x k = x 2 = 1 2+1. Thus the sequece (y ) is (1, 1/3, 1/5, 1/7,..., 1/(2 + 1),...). If we take k 0 = 1 ad k = 2 := 2 2... 2 ( times), the the subsequece (y ) becomes (1/2, 1/3, 1/5, 1/9, 1/17,..., 1/(2 + 1),...). We ow prove a importat cosequece of the Completeess Axiom: Theorem 3.1.12 (Nested Itervals Property) Let (a ) ad (b ) be two real sequeces. Assume that for each, a a +1 b +1 b. The N [a, b ] = [a, b] for some real umbers a ad b. I fact a = sup{a : N} ad b = if {a : N}. Proof: Sice the set {a : N} is bouded above by b 0, it has a least upper boud, say a. Similarly the set {b : N} has a greatest lower boud, say b. I claim that N [a, b ] = [a, b]. If x a, the x a for all. Likewise, if x b, the x b for all. Hece, if x [a, b], the x [a, b ] for all. Coversely, let x N [a, b ]. The a x b for all. Thus x is a upper boud for {a : N} ad a lower boud for {b : N}. Hece a x b.

34 CHAPTER 3. OTHER NUMBER SETS Exercises. i. Ca you fid a icreasig sequece (a ) ad a decreasig sequece (b ) N of real umbers with a < b m for all, m N such that [a, b ) =? ii. Prove Theorem 3.1.12 for R (with closed cubes or balls rather tha closed itervals). iii. Let G be a subset of R cotaiig 1, closed uder multiplicatio ad iversio. Let Seq(R) be the set of sequeces of R. For a = (a ) ad b = (b ) i Seq(R), set a b if ad oly if for some g G, a = gb evetually. Show that is a equivalece relatio o Seq(R). Fid the equivalece classes whe G = {1}, G = {1, 1} ad G = R. 3.2 Itegers ad Ratioal Numbers I N, we ca add ad multiply ay two umbers, but we caot always subtract oe umber from aother. We set Z = { m :, m N}. Now i Z we ca add, multiply ad subtract ay two umbers. The elemets of Z are called itegers. I Z, we ca add, multiply ad subtract ay two umbers, but we caot always divide oe umber to aother. We set Q = {/m :, m Z, m 0}. The elemets of Q are called ratioal umbers. Now i Q we ca add, multiply, subtract ad divide ay two umbers, with the oly exceptio that we caot divide a ratioal umber by0. We will ot go ito further detail about these umber systems. We trust the reader i provig ay elemetary statemet about umbers, for example the decompositio of itegers ito prime factors. The structure (Q, +,, 0, 1, <) satisfies all the axioms A1-4, M1-4, O1-3, OA, OM. But it does ot satisfy the Completeess Axiom C as the followig lemma shows (compare with Theorem 2.6.2). Lemma 3.2.1 There is o q Q such that q 2 = 2. Proof: Assume ot. Let q Q be such that q 2 = 2. Let a, b Z be such that q = a/b. Simplifyig if ecessary, we may choose a ad b so that they are ot both divisible by 2. From (a/b) 2 = q 2 = 2 we get a 2 = 2b 2. Thus a 2 is eve. It follows that a is eve. Let a 1 Z be such that a = 2a 1. Now 4a 2 1 = a 2 = 2b 2 ad 2a 2 1 = b 2. Hece b is eve as well, a cotradictio. Theorem 3.2.2 Q is dese i R, i.e. for ay real umbers r < s, there is a ratioal umber q such that r q s. Proof: By Theorem 3.1.6, there is a atural umber such that 1 < (r s). Now cosider the set A := {m N : m/ < s}. By Theorem 3.1.6 agai, A is a bouded set. By Lemma 3.1.7, A has a maximal elemet, say m. Thus m/ < s ad m+1 s. We compute: s m+1 = m + 1 < s + (r s) = r.

3.2. INTEGERS AND RATIONAL NUMBERS 35 3.2.1 Expoetiatio Let r R. At page 30, we have defied r for N (except for 0 0, which was left udefied). If r 0, we ca exted this defiitio to Z by r = (r ) 1. Note that the previous defiitio of r 1 coicides with the oe give above. Propositio 3.2.3 For r R ad Z ot both zero, we have, i. (rs) = r s. ii. r r m = r +m. iii. (r ) m = r m. Proof: Left as a exercise. If r R 0 ad q Q, we ca also defie r q as follows: Let m N \ {0}, by Theorem 3.1.11, there is a uique s R such that s m = r. Set s = r 1/m. Now for Z ad m N \ {0}, defie r /m to be (r 1/m ). Propositio 3.2.4 For r R 0 ad q, q 1, q 2 Q, we have, i. (rs) q = r q s q. ii. r q 1 r q 2 = r q 1+q 2. iii. (r q1 ) q2 = r q1q2. Proof: Left as a exercise. Exercises i. Show that if q Q is a square i Q, the 2q is ot a square i Q. ii. Let a < b be real umbers. Show that for each N, there are ratioal umbers a < b such that [a, b ] = [a, b]. Hit: See Theorem 3.2.2. iii. Let a < b be real umbers. Show that for each N, there are ratioal umbers a < b such that (a, b ) = [a, b]. Hit: See the exercise above. iv. Let a < b be real umbers. Show that for each N, there are ratioal umbers a < b such that [a, b ] = (a, b). Show that if a ad b are oratioal umbers (a ad b are still ratioal umbers), the we ca ever have [a, b ] = [a, b]. Hit: See the exercise above. v. For each N, let a ad b be such that a +1 < a < b < b +1. Show that (a, b ) is a ope iterval (bouded or ot). vi. Show that for all ratioal umber q > 0, there is a ratioal umber x for which 0 < x 2 < q. vii. a) Show that if x ad y are two oegative ratioal umbers whose sum is 1, the a ax + by b. b) Let q be a ratioal umber such that a q b. Show that there are two oegative ratioal umbers x ad y such that x + y = 1 ad q = ax + by. c) Show that the umbers x ad y of part b are uique.

36 CHAPTER 3. OTHER NUMBER SETS viii. Show that for ay x R, there is a uique Z such that x < +1}. ix. Show that for ay, m Z, m 0 there are uique q, r Z such that = mq + r ad 0 r < m. x. Let A l ad A r be two oempty subsets of Q such that a) A l A r = Q. b) A l A r =. c) Ay elemet of A l is less tha ay elemet of A r. Show that sup(a l ) = if(a r ). 3.3 Uiqueess of the Real Number System We assumed without a proof that there was a structure (R, +,, <, 0, 1) that satisfies all our axioms. Assumig there is really such a structure, ca there be other structures satisfyig the axioms of real umbers? Of course! Just reame the elemets of R ad defie the additio, the multiplicatio ad the order accordigly to get aother structure that satisfies the same axioms. For example, the structure (R, +,, <, 0, 1 ) defied by R = {0} R 0 = (0, 0) 1 = (0, 1) (0, r) + (0, s) = (0, r + s) (0, r) (0, s) = (0, r s) (0, r) < (0, s) r < s satisfies the axioms of R. Oe might argue that this structure we have just defied is ot very differet from the old oe, that all we did was reamig the elemet r of R by (0, r). Ideed... Ad that is all oe ca do as we will soo prove. Note that i the above example, the map f : R R defied by f(r) = (0, r) is a bijectio that has the followig properties: f(0) = 0 f(1) = 1 f(r + s) = f(r) + f(s) f(r s) = f(r) f(s) r < s f(r) < f(s) for all r, s R. We will show that if (R, +,, <, 0, 1 ) is a structure that satisfies all the axioms of real umbers the there is a bijectio f : R R that satisfies the above properties. This meas that R is just R with its elemets reamed: r R is amed f(r). Note also that all the theorems we have proved for (R, +,, <, 0, 1) are also valid for (R, +,, <, 0, 1 ). I particular R has a

3.3. UNIQUENESS OF THE REAL NUMBER SYSTEM 37 smallest iductive set N, cotais a dese subset Q similar to Q i R (Theorem 3.2.2) etc. Theorem 3.3.1 (Uiqueess of the Real Number System) Let (R, +,, <, 0, 1 ) be a structure that satisfies all the axioms of the real umbers. The there is a uique ocostat map f : R R such that for all r, s R, (1) f(r + s) = f(r) + f(s) (f is additive) (2) f(r s) = f(r) f(s) (f is multiplicative) Furthermore such a map must be a bijectio ad must also satisfy (3) r < s f(r) < f(s) (f is order preservig) (4) f(0) = 0 (5) f(1) = 1. Proof: 1. We first prove that a map that satisfies (1) ad (2) must be a bijectio that satisfies (3), (4), (5). 1a. f(0) = 0. A map f that satisfies (1) must sed 0 to 0, because 0 + f(0) = f(0) = f(0 + 0) = f(0) + f(0) ad so by simplifyig we get f(0) = 0. 1b. f( x) = f(x). (Here the sig o the right had side stads for the additive iverse i R for the biary operatio + ). Ideed, 0 = f(0) = f(x + ( x)) = f(x) + f( x) ad so f( x) = f(x). 1c. f is oe to oe. Assume f(x) = 0 for some x R \ {0}. We will get a cotradictio. We compute: f(1) = f(x 1 x) = f(x 1 ) f(x) = f(x 1 ) 0 = 0 ad so, for all r R, f(r) = f(r 1) = f(r) f(1) = f(r) 0 = 0, cotradictig the fact that f is ocostat. Thus f(x) = 0 implies x = 0. We ca ow show that f is oe to oe. Assume f(x) = f(y). The 0 = f(x) f(y) = f(x) + ( f(y)) = f(x) + f( y) = f(x + ( y)) = f(x y). By above x y = 0, ad x = y. Hece f is oe to oe. 1d. f(1) = 1. Sice f(1) = f(1 1) = f(1) f(1), f(1) is either 0 or 1. But sice f(0) = 0 ad f is oe to oe, f(1) = 1. 1e. f(n) = N. We ca show by iductio o that f() N. Thus f(n) N. It is also clear that f(n) is a iductive subset of N. Hece f(n) = N. 1f. f(q) = Q. By 1e ad 1b, f(z) = Z. Let /m Q with, m Z, m 0. The f(m) f(/m) = f(m /m) = f() f(z) = Z. Sice f(m) f(z) ad sice f(m) 0 (because m 0, see 1c), from this we get f(/m) Q. Thus f(q) f(q ). Sice f(z) = Z, it is also easy to show that f(q) = Q. 1g. If x < y i R the f(x) < f(y) i R. Note first that, i R, x y if ad oly if x + z 2 = y for some z R. The same statemet holds i R. Assume x, y R are such that x < y. The there is a z R \ {0} such that y = x + z 2. Therefore f(y) = f(x) + f(z) 2 (here f(z) 2 is the squarig i R, i.e. f(z) 2 = f(z) f(z)). Sice f(z) 0, we get f(x) < f(y).

38 CHAPTER 3. OTHER NUMBER SETS ad 1h. f is oto. Let r R \ Q. Cosider the sets L r = {q Q : q < r} = Q (, r) R r = {q Q : r < q} = Q (r, ). The sets L r ad R r partitio Q. Sice f : Q Q is a order preservig bijectio, f 1 (L r) ad f 1 (L r) partitio Q ito two oempty covex subsets (of Q) ad every elemet of f 1 (L r) is strictly less tha every elemet of f 1 (R r). The sup(f 1 (L r)) = if(f 1 (R r) (Exercise x, page 36). Let r be this umber. Thus r = sup(f 1 (L r)) = if(f 1 (R r). We claim that f(r) = r. If f(r) < r, the (because Q is dese i R) there is a q Q such that f(r) < q < r. Thus q L r. Let q Q be such that f(q) = q. Thus q f 1 (L r). So q < r, hece q = f(q) < f(r), a cotradictio. Thus f(r) r. Similarly f(r) r. Therefore f(r) = r. 2. Existece. We ow prove the existece of the map f. We defie our map f first o N. We let f(0) = 0 ad assumig f() has bee defied for N, we defie f( + 1) to be f() + 1. Thus by defiitio, f(0) = 0 f( + 1) = f() + 1 By iductio o, oe ca prove that for all N, f() N. Also f(n) is clearly a iductive subset of R. Hece f(n) = N. We claim that f : N R is oe to oe. Assumig f() = f(m) for, m N, we will show that = m. We proceed by iductio o. If = 0, the 0 = f(0) = f() = f(m), therefore, by the very defiitio of f, m caot be of the form k + 1 for some k N, i.e. m = 0. If = k + 1 for some k N, the f(k) + 1 = f(k + 1) = f() = f(m), therefore m 0 ad so m = l + 1 for some l N. Now f(k) + 1 = f(m) = f(l + 1) = f(l) + 1 ad so f(k) = f(l). By iductio k = l ad = k + 1 = l + 1 = m, provig the f : N R is oe to oe. We will deote the image of N uder f by, i.e. we let f() =. The proofs that for, m N, f( + m) = f() + f(m) ad f( m) = f() f(m) are easy as well ad are left as a exercise. From the additivity of f, it follows that f must preserve the order o N. Now cosider the subset Q := {a/b : a N ad b N \ {0} of R. Here a/b stads for a b 1 ad b 1 deotes the multiplicative iverse of b R with respect to. We exted f : N R to Q by settig f(/m) = f()/f(m) Q R. We should first check that this map is welldefied, meaig that /m = p/q for, m, p, q N should imply f()/f(m) = f(p)/f(q). Ideed, if /m = p/q for, m, p, q N, the q = mp, so f()f(q) = f(q) = f(mp) = f(m)f(p) ad f()/f(m) = f(p)/f(q). Thus f : Q Q R. We still deote by f this exteded map.

3.4. COMPLEX NUMBERS 39 f : Q Q is certaily oto: If a/b Q with a, b N, the f() = a ad f(m) = b for some, m N ad so f(/m) = f()/f(m) = a/b. It is also easy to check that f : Q R is oe to oe, additive, multiplicative ad order preservig. We ow exted f to R. Let r R. Cosider the disjoit sets ad L r = {q Q : q r} R r = {q Q : r < q}. Note that the sets L r ad R r partitio Q ad they do defie r, as r is both sup L r ad if R r. Cosider the sets f(l r ) ad f(r r ). Sice f : Q R is order preservig, ay elemet of f(l r ) is strictly less tha ay elemet of f(r r ). Also, sice f(q) = Q, the sets f(l r ) ad f(r r ) partitio Q. Therefore sup f(l r ) = if f(r r ). We let f(r) = sup f(l r ) = if f(r r ). It is a matter of writig to show that f : R R satisfies all the required properties. 3. Uiqueess. Assume f : R R ad g : R R are such maps. The g 1 f : R R is such a map as well. Therefore it is eough to show that a ocostat map f : R R satisfyig (1) ad (2) (therefore also (3), (4) ad (5)) is Id R. Oe ca show quite easily (as above) that f is idetity o Q. Sice Q is dese i R, it follows that f = Id R. 3.4 Complex Numbers Let C = R R. O C we defie two operatios called additio ad multiplicatio as follows: Additio : (x, y) + (z, t) = (x + z, y + t) Multiplicatio : (x, y)(z, t) = (xz yt, xt + yt) It is easy to check that the first ie axioms about the additio ad multiplicatio of real umbers do hold i C: A1, A2, A3, A4 (with O C = (0, 0) as the additive idetity ad ( x, y) as the additive iverse of (x, y)), M1, M2, M3, M4 (with 1 C := (1, 0) as the multiplicative idetity ad x y ( x 2 +y, 2 x 2 +y ) as the multiplicative iverse of the ozero elemet (x, y)) 2 ad D hold. It ca be checked that o order satisfyig the axioms O1, O2, O3, OA ad OM ca be defied o C (See Exercise x, page 42). Thus C is a field which is ot a ordered field. The set C together with the additio ad multiplicatio is called the field of complex umbers, each elemet of C is called a complex umber.

40 CHAPTER 3. OTHER NUMBER SETS The followig property is easy to check ( ) (x, y) = (x, 0) + (0, y) = (x, 0) + (y, 0)(0, 1). Let us cosider the map α : R C give by α(r) = (r, 0). followig hold: α is oe to oe α(r + s) = α(r) + α(s) α(0) = 0 C α( r) = α(r) α(rs) = α(r)α(s) α(1) = 1 C α(r 1 ) = α(r) 1 The the Thus the map α trasports the structure (R, +,, 0, 1) oto the substructure (α(r), +,, 0 C, 1 C ) of (C, +,, 0 C, 1 C ); i other words, the oly differece betwee the two structures (R, +,, 0, 1) ad (α(r), +,, 0 C, 1 C ) is the ames of the objects, what is called r i the first oe is called (r, 0) i the secod. The property ( ) above ow ca be writte as ( ) (x, y) = α(x) + α(y)(0, 1) We also ote the followig: α(r)(x, y) = (rx, ry). From ow o, we will idetify R ad its image α(r) via the map α, i.e. we will let r = α(r) = (r, 0). Although r (r, 0) = α(r), we will make the idetificatio r = α(r) for the sake of otatioal simplicity. I particular, we idetify 0 ad 0 C, as well as 1 with 1 C. I this way we will see R as a subset of C. Thus the two properties above are ow writte as ( ) (x, y) = x + y(0, 1) Let i = (0, 1). Note that r(x, y) = (rx, ry) ad that for ay (x, y) C, we have i 2 = ii = (0, 1)(0, 1) = ( 1, 0) = 1, ( ) (x, y) = x + yi From ow o, we will represet a complex umber z as x + iy (or as x + yi) for x, y R rather tha as the pair (x, y). Note that, give z C, the real umbers x, y for which z = x + iy are uique (this would ot have bee so if we assumed x ad y were i C rather tha i R). The additio ad the multiplicatio of complex umbers with this otatio become: (x + yi) + (z + ti) = (x + z) + (y + t)i (x + yi)(z + ti) = (xz yt) + (xt + yz)i

3.4. COMPLEX NUMBERS 41 As is the custom, we will write x yi istead of x + ( y)i. The iverse of a ozero complex umber is give by the formula (x + yi) 1 = x x 2 + y 2 y x 2 + y 2 i. Cojugatio. We cosider the followig map : C C give by x + yi = x yi (here x, y R). For every α, β C, the followig properties are easy to verify: α + β = α + β αβ = α β α 1 = α 1 Also, α = α if ad oly if α R. The complex umber a is called the cojugate of α. Norm. If α = x + yi with x, y R the it is easy to check that αa = x 2 + y 2 R 0. Thus we ca take its square root. We defie Hece α := αα = x 2 + y 2. α 2 = x 2 + y 2. The oegative real umber α is called the orm idexorm of a complex umber of α. Note that if α R C, the the orm of α is equal to the absolute value of α, so that the two meaigs that we have give to the otatio α coicide. For all α, β C, we have the followig properties: Lemma 3.4.1 For all α, β C, i. α + β α + β, ii. α β α β. P 1 α 0 ad α = 0 if ad oly if α = 0 P 2 αβ = α β Proof: i. O the oe had, α + β 2 = (α + β)α + β = (α + β)(α + β) = αa + αβ + aβ + ββ = α 2 + αβ + aβ + β 2. O the other had, ( α + β ) 2 = α 2 + 2 α β + β 2. Thus we eed to prove that αβ + αβ 2 α β, or that αβ + αβ 2 ααββ. Settig γ = αβ, this meas that it is eough to prove that γ + γ 2 γγ for all γ C. Set γ = x + yi where x, y R. The γ + γ = 2x ad γγ = x 2 + y 2. Thus the statemet γ + γ 2 γγ for all γ C is equivalet to the statemet

42 CHAPTER 3. OTHER NUMBER SETS x x 2 + y 2 for all x, y R. We will prove this last iequality. Sice x 2 x 2 + y 2, takig the square roots of both sides, we ifer x x 2 + y 2. Thus x x x 2 + y 2. ii. From part (i) we have α = β + (α β) β + α β. Thus α β α β. Agai from part (i), we have β = α+(β α) α + β α = α + α β. Thus β α α β. These two iequalities mea exactly that α β α β. Exercises. i. Show that ( 3 2 + 1 2 i)3 = i. ii. Show that ( 2 2 + 2 2 i)2 = i. iii. Show that for ay complex umber α there is a polyomial p(x) = ax 2 + bx + c R[X] such that p(α) = 0. (Note: a, b ad c should be real umbers). iv. Show that for every α C there is a β C such that β 2 = α. v. Let α, β, γ C. Assume that ot both α ad β are zero. Show that the equatio αx 2 + βx + γ = 0 has a solutio i C. Show that this equatio has at most two solutios i C. (Hit: Recall the quadratic formula ad its proof). vi. Show that for every α C ad every N \ {0}, there is a β C such that β 2 = α. vii. Let a 0, a 1,..., a R. Show that if α C is a solutio of a 0 + a 1 x +... + a x the α is also a solutio of this equatio. viii. Show that there is a bijectio betwee {α C : α > 1} ad {α C : 0 < α < 1}. ix. For α, β C, defie the distace, d(α, β) by d(α, β) = α β. Show the followig: i. d(α, β) = 0 if ad oly if α = β, ii. d(α, β) = d(β, α), iii. d(α, β) d(α, γ) + d(γ, β). x. Show that o order satisfyig the axioms O1, O2, O3, OA ad OM ca be defied o C (Hit: See Corollary 2.4.5).

Chapter 4 Real Vector Spaces Let R deote the cartesia product of copies of R, i.e. R = {(r 1,..., r ) : r i R for i = 1,..., }. The elemets of R will be called vectors. We will deote a vector by x, v, a etc. To deote the vectors we will use the followig covetio as a rule: x = (x 1,..., x ), v = (v 1,..., v ), a = (a 1,..., a ). The vector (0,..., 0) will be deoted by 0. Give v = (v 1,..., v ), we let v = ( v 1,..., v ). Give v = (v 1,..., v ) ad w = (w 1,..., w ), we defie the sum v + w of v ad w as v + w = (v 1 + w 1,..., v + w ). Give v = (v 1,..., v ) ad r R, we defie the scalar multiplicatio of r R with the vector v as r v = (rv 1,..., rv ). With these defiitios, lettig V = R, the followig hold: A1. Additive Associativity. For ay x, y, z V, x + ( y + z) = ( x + y) + z. A2. Additive Idetity Elemet. There is a vector 0 such that for ay x V, x + 0 = 0 + x = x. A3. Additive Iverse Elemet. For ay x V there is a y V (amely x) such that x + y = y + x = 0. A4. Commutativity of the Additio. For ay x, y V, x + y = y + x. B1. Associativity of the Scalar Multiplicatio. For ay r, s R ad x V, r(s x) = (rs) x. 43

44 CHAPTER 4. REAL VECTOR SPACES B2. Distributivity of the Scalar Multiplicatio 1. For ay r, s R ad x V, (r + s) x = r x + s x. B3. Distributivity of the Scalar Multiplicatio 2. For ay r R ad x, y V, r( x + y) = r x + r y. B4. Idetity. For ay x V, 1 x = x. Note that the properties A1, A2, A3, A4 are the same as i sectio 2.1. Thus (R, +, 0) is a commutative group. A set V o which a biary operatio + ad a scalar multiplicatio R V V that seds a pair (r, v) of R V to a elemet of V (deoted by r v) satisfyig the properties A1-A4 ad B1-B4 is called a vector space over R or a real vector space. More precisely, a real vector space is a triple (V, +, R V V ) satisfyig the properties above. If i the above axioms R is replaced by a field F, the resultig structure is called a vector space over F. Note that ay vector space over R is also a vector space over the field Q of ratioal umbers. But i this moograph, we will oly eed real vector spaces. As with the real umbers, oe ca show that the elemet 0 of a vector space satisfyig A2 is uique. Oe ca also show that, give x V, the elemet y that satisfies A3 is uique. We set y = x. Of course all the cosequeces of Axioms A1-A4 ivestigated i Sectio 2.1 hold. For example, we have (x) = ( x). Examples. i. Let V = {(x, y, z, t) R 4 : x 2y + 3t = 0}. The V is a vector space with the usual compoetwise additio ad scalar multiplicatio. ii. R itself is a real vector space. iii. The set C of complex umbers is a vector space over R. iv. The sigleto set {a} is a vector space over R if we defie a + a = a ad ra = a for ay r R. We will deote a by 0 of course. v. Let R[x] be the set of real polyomials i x, i.e. R[x] = {a 0 + a 1 x +... + a x : a i R}. (We will avoid the mathematical defiitio of the polyomials. Keep i mid oly the fact two polyomials a 0 + a 1 x +... + a x ad b 0 + b 1 x +... + b m x m are equal if ad oly if = m ad a i = b i for all i = 0,...,. The precise mathematical defiitio of polyomials is give i Math 211). We defie the additio of two polyomials p(x) = a 0 + a 1 x +... + a x ad q(x) = b 0 + b 1 x +... + b m x m

45 as p(x) + q(x) = (a 0 + b 0 ) + (a 1 + b 1 )x +... + (a k + b k )x k where k = max(, m) ad a i = 0 i case i > ad b j = 0 i case m > j. For example (1 2x 2 + x 3 ) + (3 + x + 2x 2 + 3x 5 + x 6 ) = 4 + x + x 3 + 3x 5 + x 6. Ad we defie the scalar multiplicatio of a real umber r with a polyomial a 0 + a 1 x +... + a x as r(a 0 + a 1 x +... + a x ) = ra 0 + ra 1 x +... + ra x. With these defiitios, it is easy to check that R[x] satisfies A1-A4 ad B1-B4, thus is a real vector space. 2. Cosider the set R(x) := {p/q : p R[x], q R[x] \ {0}} with the covetio that p/q = p 1 /q 1 if ad oly if pq 1 = p 1 q. We defie the additio ad scalar multiplicatio as expected: p q + p 1 q 1 = pq 1 + p 1 q qq 1 r p q = rp q for p, q, p 1, q 1 R[x] ad r R. With these defiitios, it is easy to check that R(x) satisfies A1-A4 ad B1-B4, thus is a real vector space. (You may wat to check that R(x) is a field). vi. Let X be ay set ad V ay real vector space. Let Fuc(X, V ) be the set of all fuctios from the set X ito V. For two fuctios f, g Fuc(X, V ) ad a real umber r R, we defie the fuctios f + g, rf Fuc(X, V ) as follows (f + g)(x) = f(x) + g(x), (rf)(x) = r f(x), for all x X. With these defiitios the set Fuc(X, V ) becomes a real vector space. The elemet 0 of Fuc(X, V ) correspods to the fuctio that seds every elemet of X to 0 V. Give f Fuc(X, V ), the elemet f Fuc(X, V ) that satisfies A3 is defied by ( f)(x) = f(x). vii. Let V be a vector space. Cosider the set Seq(V ) of sequeces from V. Thus by defiitio a elemet v Seq(V ) is of the form (v ) N where v V for all N. For two sequeces v = (v ) ad w = (w ) ad a real umber r, set v + w = (v + w ) ad rv = (rv ). The Seq(V ) becomes a vector space. The zero elemet 0 of Seq(V ) is the zero sequece that cosists of zero vectors ad (v ) = ( v ).

46 CHAPTER 4. REAL VECTOR SPACES viii. Let V be a vector space. Cosider the set Seq f (V ) of sequeces from V which are zero after a while. Thus by defiitio a elemet v Seq(V ) is of the form (v ) N where v V for all N ad for which v = 0 for large eough, i.e. for N for certai N (that depeds o v). For two such sequeces v = (v ) ad w = (w ) ad a real umber r, set v + w = (v + w ) ad rv = (rv ). The Seq f (V ) becomes a vector space. The zero elemet 0 is the zero sequece of zero vectors ad (v ) = ( v ). Exercises. Suggestio: First do Exercise i. i. Let V be a real vector space. Let W be a subset of V. Suppose that W is closed uder additio ad scalar multiplicatio, i.e. suppose that w + w W for all w, w W ad rw W for all r R ad w W. Show that W is a vector space. Such a subset is called a subspace of V. ii. Are the followig sets vector space uder the usual additio ad scalar multiplicatio? {(x, y, z) R 3 : x + y + z = 0} {(x, y, z) R 3 : x + y + z = 1} {(x, y, z) R 3 : xyz = 0} {(x, y, z) R 3 : x 2 + y 2 + z 2 = 0} {(x, y, z) R 3 : x 3 + y 3 + z 3 = 0} {(x, y, z) R 3 : xyz 0} {(x, y, z) R 3 : x Q} iii. Which of the followig are ot vector spaces over R (with the compoetwise additio ad scalar multiplicatio) ad why? V 1 = {(x, y, z) R 3 : xy 0} V 2 = {(x, y, z) R 3 : 3x 2y + z = 0} V 3 = {(x, y, z) R 3 : xyz Q} V 4 = {(x, y) R 3 : x + y 0} V 5 = {(x, y) R 2 : x 2 + y 2 = 0} V 6 = {(x, y) C 2 : x 2 + y 2 = 0} iv. Are the followig sets vector space uder the usual additio of fuctios ad scalar multiplicatio of a fuctio with a real umber? {f Fuc(X, R) : f(x) = 0} where x is a fixed elemet of X {f Fuc(R, R) : f(x) = 0 for ay x (0, 1)} {f Fuc(R, R) : f(0) 0} v. Let V ad W be two real vector spaces. Cosider the set Hom(V, W ) of fuctios f : V W such that f(v 1 + v 2 ) = f(v 1 ) + f(v 2 ) ad f(rv) = rf(v) for all v, v 1, v 2 V, r R Show that Hom(V, W ) is vector space uder the usual additio of fuctios ad scalar multiplicatio (see Example vi, page 45).

vi. Show that R(x) is a field. Fid all orders o R(x) that makes it a ordered field. 47

48 CHAPTER 4. REAL VECTOR SPACES

Chapter 5 Metric Spaces 5.1 Examples. Real Numbers ad Absolute Value. Let us cosider the fuctio d : R R R 0 defied by d(x, y) = x y. This fuctio has the followig properties: For all x, y, z R, i. d(x, y) = 0 if ad oly if x = y, ii. d(x, y) = d(y, x), iii. d(x, y) d(x, z) + d(z, y). We will see several examples of sets X together with a fuctio d : X X R 0 satisfyig the three properties above. Euclidea Spaces. For x = (x 1,..., x ) R ad y = (y 1,..., y ) R defie d( x, y) = x 1 y 1 2 +... + x y 2. We will show that d satisfies the properties i, ii, iii stated above. The first two are immediate. It will take us some time to prove the third equality. For two vectors x = (x 1,..., x ) ad y = (y 1,..., y ), defie their scalar product as x y = x i y i. i=1 We have for all x, y, z R ad α, β R, C1. (α x + β y) z = α x z + β y z C2. x y = y x C3. x(α y + β z) = α x y + β x z (C1 ad C2 are direct cosequeces of the defiitio. C3 follows from C1 ad C2.) Note that x x 0, so we ca take its square root. Let x = ( x x) 1/2. It is 49

50 CHAPTER 5. METRIC SPACES easy to show from the defiitio that, for all x R ad α R, we have C4. x 0,ad x = 0 if ad oly if x = 0 C5. α x = α x Now we show that for all x R we have C6. x y x y Ideed, let z = α x β y where α = x y ad β = x 2. Use z 2 0 to prove C6. Details: 0 (α x β y)(α x β y) = α 2 x x 2αβ x y + β 2 y y = α 2 x 2 2αβ x y + β 2 y 2 = ( x y) 2 x 2 2( x y) 2 x 2 + x 4 y 2 = ( x y) 2 x 2 + x 4 y 2. Hece ( x y) 2 x 2 x 4 y 2. Simplifyig we get ( x y) 2 x 2 y 2. Takig the square roots we get x y x y. This proves C6. Now we prove C7. x + y x + y For this we use C6 ad the fact that x + y 2 = x 2 + 2 x y + y 2. Details: x + y 2 = ( x + y)( x + y) = x 2 + 2 x y + y 2 x 2 + 2 x y + y 2 A6 x 2 + 2 x y + y 2 = ( x + y ) 2. Thus x + y 2 ( x + y ) 2. Takig the square roots we fially obtai C7. We ow remark that d( x, y) = x y 1/2 ad we prove the third property (iii) above: d( x, z) = x z 1/2 1/2 C7 = ( ( x y)+( y z) ) = ( x y + y z ) 1/2 x y 1/2 + y z 1/2 = d( x, y) + d( y, z). Thus we have proved the followig result. Theorem 5.1.1 For x = (x 1,..., x ) R ad y = (y 1,..., y ) R defie d( x, y) = x 1 y 1 2 +... + x y 2. The the couple (R, d) satisfies the properties i, ii, iii above. 5.2 Defiitio ad Further Examples A set X together with a fuctio d : X X R 0 is said to be a metric space if for all x, y, z X, MS1. d(x, y) = 0 if ad oly if x = y. MS2. d(x, y) = d(y, x). MS3. (Triagular Iequality.) d(x, y) d(x, z) + d(z, y). The fuctio d is called a metric or a distace fuctio o X. We will give several examples of metric spaces.

5.2. DEFINITION AND FURTHER EXAMPLES 51 Euclidea Metric. By Theorem 5.1.1, (R, d) where d( x, y) = x 1 y 1 2 +... + x y 2 is a metric space, called the Euclidea metric space, or sometimes the usual metric o R. Metric o a Product. The followig theorem will be prove much later (Lemma 14.5.4), whe we will give a sese to the term x r for x R 0 ad r R \ {0}. I ay evet, the theorem below makes sese for p N eve at this poit. Theorem 5.2.1 For i = 1,...,, let (X i, δ i ) be a metric space. Let X = X 1... X. Let p 1 be ay real umber. For x, y X let d p (x, y) = ( i=1 δ i(x i, y i ) p ) 1/p. The (X, d p ) is a metric space. If we take X i = R ad p = 2 i the theorem above with d i (x, y) = x y, we get the Euclidea metric o R. Uless stated otherwise, whe we speak about the product of fiitely metric spaces, we will always take p = 2. Iduced Metric. Ay subset of a metric space is a metric space with the same metric, called the iduced metric. More precisely we defie the iduced metric as follows: Let (X, d) be a metric space. Let Y X. The (Y, d Y Y ) is a metric space. We say that the metric of Y is iduced from that of X. Sup Metric. For i = 1,...,, let (X i, d i ) be a metric space ad let X = X 1... X. For x, y X, set d (x, y) = max{d i (x i, y i ) : i = 1,..., }. The (X, d ) is a metric space as it ca be checked easily. Discrete Metric. Let X be ay set. For x, y X, set d(x, y) = { 1 if x y 0 if x = y The (X, d) is a metric space. This metric o the set X is called the discrete metric. p-adic Metric. Let X = Z ad p a positive (prime) iteger. For distict x, y Z, set d(x, y) = 1/p if p divides x y but p 1 does ot divide x y. Set d(x, x) = 0. The (X, d) satisfies the properties i, ii ad iii stated above. I fact it satisfies the stroger triagular iequality d(x, y) max{d(x, z), d(z, y)}. Such a metric is called ultrametric. Eve further d(x, y) = max{d(x, z), d(z, y)} if d(x, z) d(y, z). It follows that ay triagle i this space is isosceles.

52 CHAPTER 5. METRIC SPACES Metric o Sequeces. Let X be ay set. Cosider the set Seq(X) of sequeces from X. Thus a elemet x Seq(X) is of the form (x ) N where x X for all N. For two distict sequeces x = (x ) ad y = (y ), set d(x, y) = 1/2 if is the first atural umber where x y. The (Seq(X), d) is a metric space. Note that, i this metric, the maximum distace betwee two sequeces is 1. We ed this subsectio with a simple but useful calculatio. Propositio 5.2.2 I a metric space (X, d), for all x, y, z X, we have d(x, z) d(z, y) d(x, y). Proof: We have to show that d(x, z) d(z, y) d(x, y) ad d(z, y) d(x, z) d(x, y). We have d(x, z) d(x, y) + d(y, z) = d(x, y) + d(z, y), so the first iequality follows. The secod oe is similar ad is left as a exercise. Exercises. i. Show that the discrete metric is ultrametric. ii. Show that the metric o the set of sequeces defied above is a ultrametric. 5.3 Normed Real Vector Spaces ad Baach Spaces ad Algebras Most of what oe ca say about R ca be said mot à mot about R ad C, ad most of what oe ca say about R ad C ca be said more geerally (mot à mot agai) for Baach spaces ad algebras, cocepts that geeralize R, R, C ad C ad that will be defied i this sectio. Sice the proofs are exactly the same, we prefer treat the most geeral case. Readers who will be psychologically affected by this geerality may prefer to read this treatise by cosiderig oly the cases of R ad R, ad sometimes C. A ormed real vector space is a vector space V together with a map : V R such that NVS1. For all v V, v 0. NVS2. For all v V, v = 0 if ad oly if v = 0. NVS3. For all v V ad r R, rv = r v. NVS4. For all v, w V, v + w v + w. Examples ad Exercises. i. The Euclidea space R with the usual orm is a ormed space.

5.3. NORMED REAL VECTOR SPACES AND BANACH SPACES AND ALGEBRAS53 ii. Cosider the set ω R of real sequeces (r ) whose terms are all zero except for fiitely may of them. It is clear that ω R is a real vector space. Defie (r ) = =0 r2. The reader should show that this turs ω R ito a ormed vector space. iii. Let X be a set ad (V, ) a ormed vector space. Cosider the set B(X, V ) of fuctios from X ito Y which are bouded. Thus, a fuctio f : X V is i B(X, V ) if ad oly if there is a real umber M such that for all x X, f(x) < M. Show that B(X, V ) is a vector space. For f B(X, V ), defie f to be the miimum of the umbers M as above. Thus f = sup{ f(x) : x X}. Show that the vector space B(X, V ) together with defied as above is a ormed vector space. Suppose V = R. Show that if f, g B(X, R), the f g B(X, R) ad that f g = f g. iv. Let X be a set ad (V, ) a ormed vector space. Cosider the set F(X, V ) of fuctios from X ito Y. Show that F(X, V ) is a vector space. For f F(X, V ), defie f = if{sup{ f(x) : x X}, 1}. Show that the vector space F(X, V ) together with defied as above is a ormed vector space. Suppose V = R. Is it true that if f, g B(X, R), the f g F(X, R) ad that f g = f g. Propositio 5.3.1 Let (V, ) be a ormed real vector space. The map d(v, w) = v w defies a metric o V. Proof: Easy. A ormed real metric space which is complete with respect to the metric defied above is called a Baach space. Defie Baach algebra. Problem. Suppose V is a ormed real vector space ad W V a subspace. Ca you make V/W ito a ormed vector space i a atural way? Exercises. i. Show that if V is a Baach space, the so is B(X, V ). ii. Show that if V is a Baach space, the so is F(X, V ).

54 CHAPTER 5. METRIC SPACES 5.4 Ope Subsets of a Metric Space Let (X, d) be a metric space, a X ad r R. The the ope ball with ceter a ad radius r is defied as B(a, r) = {x X : d(a, x) < r}. The circle with ceter a ad radius r is defied as B(a, r) = {x X : d(a, x) = r}. A arbitrary uio of ope balls i a metric space is called o ope subset. Propositio 5.4.1 A subset U of a metric space is ope if ad oly if for every a U there is a r R >0 such that B(a, r) U. Proof: ( ) Let a U. Sice U is a uio of ope balls, there is a ope ball i U that cotais a. Set a B(b, s) U. Let r = s d(a, b). The r > 0. We claim that B(a, r) B(b, s). Ideed, let x B(a, r). The d(x, b) d(x, a) + d(a, b) < r + d(a, b) = r + (s r) = s. Thus x B(b, s). This proves the claim. Now B(a, r) B(b, s) U. ( ) Suppose that for ay a U, there is a r a R >0 such that B(a, r a ) U. Sice r a > 0, a B(a, r a ). It follows that U = a U B(a, r a). Hece U, beig a uio of ope balls, is ope. Lemma 5.4.2 I a metric space, the itersectio of two ope balls is a ope subset. Proof: Let B(a, r) ad B(b, s) be two ope subsets of a metric space (X, d). To show that B(a, r) B(b, s) is ope, we will use Propositio 5.4.1. Let c B(a, r) B(b, s). The r d(c, a) > 0 ad s d(c, b) > 0. Let ɛ = mi(r d(c, a), s d(c, b)). The ɛ > 0. We claim that B(c, ɛ) B(a, r) B(b, s). Let x B(c, ɛ). The d(a, x) d(a, c) + d(c, x) < d(a, c) + ɛ d(a, c) + (r d(a, c)) = r. Thus x B(a, r). Similarly x B(b, s). Propositio 5.4.3 Let (X, d) be a metric space. The, i. ad X are ope subsets. ii. A arbitrary uio of ope subsets is ope. iii. A fiite itersectio of ope subsets is a ope subset. Proof: i. If a X, the the ball with ceter a ad radius 0 is the emptyset. Hece the emptyset is ope (beig a ope ball). The whole space X is the uio of all the ope balls of radius, i.e. X = x XB(x, 1). Thus X is ope. ii. This is clear by defiitio of a ope subset.

5.4. OPEN SUBSETS OF A METRIC SPACE 55 iii. Let U ad V be two ope subsets. Thus U ad V are uios of ope balls. Write U = a A B(a, r a) ad V = b B B(b, s b) where A ad B are subsets of X ad r a ad s b are real umbers. Now U V = B(a, r a ) B(b, s b ) = (B(a, r a ) B(b, s b )). a A b B a A, b B Thus, by part ii, it is eough to show that B(a, r a ) B(b, s b ) is ope. But this is the cotet of Lemma 5.4.2. Exercises. i. Describe B( 0, 1) i the three metric spaces (R, d 1 ), (R, d 2 ) ad (R, d ) defied above. ii. Cosider R with oe of the distaces d p (p 1) or d defied above. Show that ay ope subset of R is a coutable uio of ope balls. iii. A isometry betwee two metric spaces (X, d 1 ) ad (Y, d 2 ) is a bijectio f : X Y such that d 1 (x 1, x 2 ) = d 2 (f(x 1 ), f(x 2 )). Fid all isometries of R ito R. iv. Fid all isometries of R 2 ito R 2. v. Show that the metric spaces (R, d p ) ad (R, d ) defied above all have the same ope subsets. vi. Let X = {1/2 : N} R. Cosider X with the atural metric (the iduced metric). Show that ay subset of X is ope. vii. Let X = {1/2 : N} {0} R. Cosider X with the atural metric (the iduced metric). Show that ope subsets of X are the cofiite subsets 1 of X ad the oes that do ot cotai 0. viii. Show that a closed ball B(a, r) = {x R : d(a, x) r} is ot ope i the Euclidea metric uless r < 0. d(x,y) 1+d(x,y) ix. Let (X, d) be a metric space. Show that d 1 (x, y) = Show that the ope subsets i both metrics are the same. is a metric. x. Show that R caot be the uio of two oempty ope subsets (for the usual metric). xi. Show that R caot be the uio of two oempty ope subsets (for the usual metric). 1 A subset Y of X is called cofiite if X \ Y is fiite.

56 CHAPTER 5. METRIC SPACES

Chapter 6 Sequeces ad Limits 6.1 Defiitio Let (X, d) be a metric space, (x ) N a sequece i X ad x X. We say that the sequece (x ) N coverges to x if for ay ɛ > 0 there is a N N such that d(x, x) < ɛ for all atural umbers > N. We the write lim x = x. We call x the limit of the sequece (x ). Whe the limit of a sequece (x ) exists we say that the sequece (x ) is coverget or that it coverges, otherwise we say that the sequece is diverget or that it diverges. Note that, above, we used the symbol without defiig it. Above we oly defied lim x x = x as if it were oe sigle word. We will ever defie the symbol, there is o such a object i mathematics. Examples ad Exercises. i. A sequece (x ) N where x = x for all N coverges to x. Such a sequece is called a costat sequece. ii. A sequece (x ) N where x = x for all greater tha a certai coverges to x. Such a sequece is called a evetually costat sequece. iii. If we delete fiitely may elemets from a sequece or add fiitely may elemets to a sequece, its divergece ad covergece remais ualtered, ad its limit (if it exists) does ot chage. iv. Let x =. metric of R. The the sequece (x ) does ot coverge i the usual v. Let x = ( 1). The the sequece (x ) does ot coverge i the usual metric of R. 57

58 CHAPTER 6. SEQUENCES AND LIMITS vi. Let X be a metric space. Let a 1,..., a k X. Let (x ) be a sequece of X such that for each, x = a i for some i = 1,... k. Show that the sequece (x ) coverges if ad oly if it is evetually costat. vii. Let x = 0 if is ot a power of 2 ad x 2 = 1. The the sequece (x ) does ot coverge i the usual metric of R. viii. Cosider Z with the 2-adic metric ad set x = 2. The the sequece (x ) coverges to 0. ix. Cosider a set X with its discrete metric. The a sequece (x ) coverges to some elemet i X if ad oly if the sequece (x ) is evetually costat. x. Show that if i Theorem 3.1.12, lim (b a ) = 0, the N [a, b ] is a sigleto set. Remarks. i. I the defiitio of covergece, ɛ should be thought as a small (but positive) real umber. The iteger N depeds o ɛ. The smaller ɛ is, the larger N should be. ii. I the defiitio of covergece, we could replace d(x, x) < ɛ by d(x, x) ɛ, i.e. the sequece (x ) N coverges to x if ad oly if for ay ɛ > 0 there is a N N such that d(x, x) ɛ for all atural umbers > N. iii. I the defiitio of covergece, sice Q is dese i R, we could take ɛ to rage over Q istead of over R. iv. The covergece of a sequece depeds strogly o the metric. It is possible that a sequeces coverges to some elemet for a certai metric, but that this same sequece diverges for some other metric. v. The otatio lim x x = x suggests that the limit of a sequece is uique whe it exists. This is ideed the case ad will be prove right ow. Theorem 6.1.1 I a metric space, the limit of a sequece, if it exists, is uique. Proof: Let (X, d) be a metric space, (x ) a sequece from X ad a, b X. Suppose that a ad b are limits of the sequece (x ). We will show that a = b by showig that d(a, b) < ɛ for ay ɛ > 0. Let ɛ > 0. Sice a is a limit of (x ) there is a N such that d(x, a) < ɛ/2 for all > N. Similarly there is a N 1 such that d(x, b) < ɛ/2 for all > N 1. Let N = max(n, N 1 ) + 1. The d(x N, a) < ɛ/2 ad d(x N, b) < ɛ/2. Hece d(a, b) d(a, x N ) + d(x N, b) < ɛ/2 + ɛ/2 = ɛ. Lemma 6.1.2 I a metric space, lim x = x if ad oly if lim d(x, x) = 0.

6.1. DEFINITION 59 Proof: This is a triviality. A subset of a metric space is said to be bouded if it is cotaied i a ball. For example Z is ot bouded i R (with its usual metric). But the set {1/2 : N} is bouded. A sequece (x ) is called bouded if the set {x : N} is bouded. Theorem 6.1.3 A coverget sequece is bouded. Proof: Let (x ) be a sequece covergig to some a. Take ɛ = 1 i the defiitio of the covergece. Thus there exists a N such that d(x, a) < 1 for ay > N. Let r = max(d(x 1, a),..., d(x N, a), 1) + 1. The the sequece (x ) is etirely i B(a, r). Exercises i. Show that lim 1/ = 0 i the usual metric of R. ii. For a atural umber we defie ν() = sup{m N : 2 m }. Show that lim ν()/ = 0. iii. Let x = 0 if is ot a power of 2 ad x 2 = 1/. Show that the sequece (x ) coverges to 0 i the usual metric of R. iv. Show that a sequece of itegers coverges i the usual metric of R if ad oly if it is evetually costat. v. Let x = ( ) k=0 ( 1) k 2 k k. Does the sequece (x ) coverge? vi. Let (x ) be a real sequece. Assume that there is a r > 0 such that for all m, x x m > r. Show that the sequece (x ) diverges. vii. Let (X, d) be a metric space, (x ) a sequece i X ad x X. Show that lim x = x if ad oly if lim d(x, x) = 0. viii. Let (a ) be a sequece of oegative real umbers. Suppose that the sequece (a 2 ) coverges to a. Show that the sequece (a ) coverges to a. Does this hold for sequeces which are ot oegative? ix. Let X be the set of sequeces of zeroes ad oes. For two distict elemets x = (x ) ad y = (y ) of X defie d(x, y) = 1/2 if is the least atural umber for which x y. Let d(x, x) = 0. The (X, d) is a metric space. Let χ be the elemet of X whose first coordiates are 1, the rest is 0. Show that lim χ exists. Fid the limit. x. Let X be the set of sequeces of zeroes ad oes with oly fiitely may oes. Cosider X as a metric space with the metric defied above. Let χ be the elemet of X cosidered above. Note that χ X. Show that lim χ does ot exist i X (but it exists i X as the above exercise shows).

60 CHAPTER 6. SEQUENCES AND LIMITS xi. Let X = Q with the 3-adic distace. Let x = 1 + 3 + 3 2 +... + 3. Show that (x ) is coverget i Q. xii. Let X = Z with the 3-adic distace. Let x = 1 + 3 + 3 2 +... + 3 X. Show that (x ) is ot coverget i Z (but it is i Q). xiii. Let A be a bouded subset of a metric space (X, d). Show that for ay b X there is a ball cetered at b that cotais A. xiv. Let A ad B bouded subsets of R with its usual Euclidea metric. Let ad A + B = {a + b : a A, b B} AB = {ab : a A, b B}. Show that A + B ad AB are bouded subsets of R. xv. Let A be a bouded subset of R (with its usual metric) that does ot cotai 0. Let A 1 = {a 1 ; a A}. a) Show that A 1 is ot ecessarily bouded. b) Assume there is a r > 0 such that B(0, r) A =. Show that A 1 is bouded. c) Coversely show that if A 1 is bouded the there is a r > 0 such that B(0, r) A =. xvi. Let (x ) be a sequece of R (with its usual metric). Show that (x ) is bouded if ad oly if there is a real umber r such that x r for all N. xvii. Show that ay real umber is the limit of a ratioal sequece. 6.2 Examples of Covergece i R ad C The most importat cases for us are the case of R ad C with their usual metric d(x, y) = x y. Traslatig the defiitio of covergece of a sequece to this case, we obtai the followig special case: A sequece (x ) of real or complex umbers coverges to x R if ad oly if for ay ɛ > 0 there is a N N such that x x < ɛ for all atural umbers > N. We kow that the limit x, whe it exists, is uique. I this subsectio, we prove the covergece ad divergece of some importat sequeces: (1/) =1, (α ) ad α /! for ay r. I the meatime we prove a criterio for covergece (Sadwich Lemma). The followig result will be hady. Theorem 6.2.1 Let α C. The lim α = 0 if ad if lim α = 0. Proof: This is a immediate cosequece of the defiitio.

6.2. EXAMPLES OF CONVERGENCE IN R AND C 61 6.2.1 The Sequece (1/) Let us start by showig that lim 1/ = 0. Lemma 6.2.2 lim 1/ = 0. Proof: Let ɛ > 0 be fixed. By Archimedea Property of the real umbers (Theorem 3.1.6), there is a atural umber N such that Nɛ > 1. The 1/N < ɛ. Now for all > N, 1/ 0 = 1/ < 1/N < ɛ. Exercises. i. Prove that lim ( ii. Prove that lim ( iii. Prove that lim ( iv. Prove that lim ( +3 2 + 5 +3 ) = 0. 2 + 5) = 0. +3 2 + 5 ) 1 3+2 = 0. ) 2 1 2 3 +3 2 + 5 = 0. 6.2.2 The Sequece (α ) Propositio 6.2.3 If r ( 1, 1) the the sequece (r ) coverges to 0. If r ( 1, 1] the sequece (r ) diverges. Proof: We first claim that if s > 1, the for all atural umbers, (1+s) 1+s. We proceed by iductio o to prove the claim. The claim clearly holds for = 0. We ow assume the claim holds for ad prove it for + 1. Sice 1 + s > 0, (1 + s) +1 = (1 + s) (1 + s) (1 + s)(1 + s) = 1 + (1 + )s + s 2 1 + (1 + )s. This proves the claim. We ow retur to the proof of the propositio. Assume r ( 1, 1). Sice for r = 0 the statemet is clear, we may assume that r 0. Let ɛ > 0 be a real umber. We have to show that there exists a atural umber N such that for all > N, r = r 0 < ɛ. We may therefore assume that r 0. Thus r > 0. Let s = 1/r 1. Note that r = 1 1+s. It is also easy to check that s > 0. Thus by the claim (1 + s) 1 + s. Let N > 0 be a atural umber such that 1/s Nɛ (Theorem 3.1.6). Now for all > N, r = r = ( ) 1 1 1 + s 1 + s < 1 1 + Ns < 1 Ns ɛ. This proves the first part of the propositio. Claim. Let r > 1. The the sequece (r ) is ubouded. Proof of the Claim. By the first part of the propositio the sequece (1/r ) coverges to 0. Hece for all ɛ > 0 there is a N such that for all

62 CHAPTER 6. SEQUENCES AND LIMITS > N, 1/r < ɛ, i.e. r > 1/ɛ. Thus (r ) is ubouded. This proves the claim. Now by Theorem 6.1.3, (r ) is diverget if r > 1. The case r 1 is left as a exercise. Corollary 6.2.4 If r > 1 the the sequece (r ) is ubouded. Corollary 6.2.5 Let α C. The the sequece (α ) coverges to 0 if α < 1, coverges to 1 if α = 1, diverges otherwise. Proof: If α < 1 the the result follows from Theorem 6.2.1 ad Propositio 6.2.3. If α = 1 the the statemet is clear. Assume α > 1. The the sequece ( α ) is ubouded by Corollary 6.2.4. By Theorem 6.1.3, (α ) is diverget. Exercises. i. Fid lim ( 1 ). ii. Fid lim ( 1 2 + 1 ). 6.2.3 The Sequece (α /!) Propositio 6.2.6 For ay r R, the sequece (r /!) coverges to 0. Proof: We may assume that r > 0. Let ɛ > 0. Let x = r /!. Let N be a atural umber such that r/n < 1 (Theorem 3.1.6). We claim that for each atural umber k, ( ) k r x N +k x N. N + 1 We prove this by iductio o k. The claim is clear for k = 0. Assume it holds for k. The x N +k+1 = rn +k+1 (N + k + 1)! = rn +k (N + k)! r N + k + 1 = x r N +k N + k + 1 ( ) k ( ) k+1 r r r x N N + 1 N + k + 1 x N. N + 1 This proves the claim. Sice 0 < r r N +1 < 1, by Propositio 6.2.3, the sequece (( N +1 )k ) k coverges r to 0. Therefore there exists a N 1 such that for k N 1, ( N +1 )k < ɛ/x N. Thus for > N + N 1, by lettig k = N 1, we see that x < ɛ. This proves the propositio. Corollary 6.2.7 For ay α C, the sequece (α /!) coverges to 0. Proof: Follows from Theorem 6.2.1 ad Propositio 6.2.6.

6.3. CONVERGENCE AND THE ORDER 63 6.3 Covergece ad the Order We first prove a result that will show that a sequece that is squeezed betwee two sequeces covergig to the same umber coverges. Lemma 6.3.1 (Sadwich Lemma) Let (x ), (y ) ad (z ) be three real sequeces. Suppose that lim x ad lim z exist ad are equal. Suppose also that x y z evetually (i.e. for > N for some N ). The lim y exists ad lim x = lim y = lim z. Proof: Let a be the commo limit of (x ) ad (z ). Let ɛ > 0. Sice lim x = a there is a N 1 such that for > N 1, x a < ɛ/3. Sice lim z = a there is a N 2 such that for > N 2, z a < ɛ/3. Let N = max(n, N 1, N 2 ). The for > N, we have y a y z + z a (z x ) + z a (z a) + (a x ) + z a < ɛ/3 + ɛ/3 + ɛ/3 = ɛ. Corollary 6.3.2 lim 1/ 2 = 0. Proof: Follows from Lemmas 6.2.2 ad 6.3.1. Secod proof: Let ɛ > 0. Let N > 1 be such that 1/2 < Nɛ. The for all > N, 1/ 2 0 = 1/ 2 1/2 < 1/2N < ɛ. Corollary 6.3.3 Let (x ) be a covergig sequeces ad r R. Assume that x r. The lim x r. Note that we could well have x > r ad lim x = r, for example take x = r + 1/. Corollary 6.3.4 Let (x ) ad (y ) be two covergig sequeces. that x y for almost all. The lim x lim y. Suppose Exercises. i. Fid lim ( 1 ). ii. Show that lim ( 1 2 + 1 ) = 0. iii. Show that lim ( 2 ) = 1/2. iv. Let (x ) be a coverget sequece i C (or a Baach space X), r > 0 ad assume that x r for all. Show that lim x r. v. Let (x ) be a coverget complex sequece. Show that lim x = lim x. (Solutio: Let x be the limit. Sice a b a b for all a, b C, x x x x. Thus it is eough to show that the limit of the right had side is 0, which is obvious.)

64 CHAPTER 6. SEQUENCES AND LIMITS 6.4 Covergece ad the Four Operatios I this subsectio we study the behavior of the covergece of real sequeces uder several operatios. We first show that a covergig sequece of R that does ot coverge to a elemet a is bouded away from a. We will eed this result later i this subsectio. Propositio 6.4.1 Let X be a metric space ad a X. Let (x ) be a covergig sequece of X that does ot coverge a. The there is a r > 0 such that B(a, r) {x : N} is fiite. Proof: Let b be the limit of (x ). Let r = d(a, b)/2. The r > 0. Sice b is the limit of (x ), there exists a N such that d(x, b) < r for all > N. Thus, for > N, d(x, a) d(x, b) d(a, b) = d(x, b) 2r = 2r d(x, b) > r. (We used Lemma 5.2.2 i the first iequality). Hece B(a, r) {x : N} {x 1,..., x N }. Now we aalyze the relatioship betwee the four operatios o R ad the covergece of sequeces. Theorem 6.4.2 Let (x ) ad (y ) be sequeces covergig to a ad b respectively. The the sequeces (x + y ) ad (x y ) coverge to a + b ad ab respectively. I particular, for r R the sequece (rx ) coverges to ra. Proof: (+). Let ɛ > 0. Sice the sequece (x ) coverges to a, there is a N such that x a < ɛ/2. Sice the sequece (y ) coverges to b, there is a N 1 such that x a < ɛ/2. Let N = max(n, N 1 ). The for > N, (x + y ) (a + b) = (x a) + (y b) x a + y b < ɛ/2 + ɛ/2 = ɛ. Thus (x + y ) coverges to a + b. ( ). Assume first b = 0. We have to show that lim x y = 0. Let ɛ > 0. By Exercise xvi, page 60, there exists a r > 0 such that x r for all. Sice (y ) coverges to 0, there exists a N such that for all > N, y = y 0 < ɛ r. Now for > N, we have x y 0 = x y = x y < r ɛ r = ɛ. Thus the sequece (x y ) coverges to 0. Assume ext b 0. Let ɛ > 0. Sice b 0 ad (x ) coverges to a,, there exists a N such that x a ɛ 2 b for all > N. Sice (x ) is a coverget sequece, it is bouded by Theorem 6.1.3. By Exercise xvi, page 60, there is a M > 0 such that x < M for all N. Sice (y ) coverges to b, there exists a N 1 such that for > N 1, y b < ɛ 2M. Let N = max(n, N 1 ). For > N, we have x y ab = x (y b) + (x a)b x (y b) + (x a)b x y b + x a b M ɛ 2M + ɛ 2 b b = ɛ. Thus the sequece (x y ) coverges to ab. For the last part: Take y = r for all apply the above part. Corollary 6.4.3 Let (x i ) i ad (y i ) i be sequeces covergig to a ad b respectively. The the sequece (x i y i ) i coverges to a b.

6.4. CONVERGENCE AND THE FOUR OPERATIONS 65 Corollary 6.4.4 Let (x i ) i be a sequece covergig to a. Let k N\{0}. The the sequece (x k i ) i coverges to a k. Corollary 6.4.5 Let (x i ) i ad (y i ) i be sequeces covergig to a ad b respectively. Let α, β C. The the sequece (αx i + βy i ) i coverges to αa βb. Corollary 6.4.6 Let (x i ) i be a sequece covergig to a. Let a, a 1,..., a k C. The the sequece (a +a 1 x i +...+a k x k i ) i coverges to a +a 1 a+...+a k a k. If a, a 1,..., a k C, the a term of the form a 0 +a 1 x+...+a k x k is called a (complex) polyomial. The umbers a, a 1,..., a k are called its coefficiets. If a k 0, the the atural umber k is called the degree of the polyomial p(x). A polyomial p(x) = a 0 + a 1 x +... + a k x k gives rise to a fuctio from C ito C whose value at α C is p(r) := a 0 + a 1 α +... + a k α k. Thus the corollary above ca be traslated as follows: Corollary 6.4.7 Let (x ) be a sequece covergig to α. Let p(x) R[x]. The the sequece (p(x )) coverges to p(α). I geeral, if the sequece (x ) is coverget ad x 0 for all, the the sequece (x 1 ) is ot ecessarily coverget. This is what happes if we take x = 1/. However if the limit of (x ) is ot zero, the the covergece of the sequece (x 1 ) is guarateed as the followig theorem shows: Theorem 6.4.8 Let (x ) be a sequece of C covergig to a ozero umber α. Assume x 0 for all. The the sequece (x 1 ) coverges to α 1. Proof: Let ɛ > 0. By Propositio 6.4.1, there is a r > 0 such that B(0, r) {x : N} is fiite. Let N be such that x r for all > N. Sice a ad r are ozero ad sice the sequece (x ) coverges to α, there exists a N 1 such that x α < ɛ for all > N 1. Let N = max(n, N 1 ). Now for > N we have, x 1 a 1 = x a x a < ɛ a r a x a r a r = ɛ. Thus the sequece (x 1 ) coverges to α 1. Corollary 6.4.9 Let (x ) ad (y ) be sequeces covergig to α ad β respectively. Assume β 0 ad that y 0 for all. The the sequece (x /y ) coverges to α/β. Corollary 6.4.10 Let p(x) ad q(x) 0 be two polyomials of degree d ad e respectively. The the sequece (p()/q()) coverges if ad oly if e f. If e > f, the the limit is zero. If e = f the the limit is α/β where α ad β are the leadig coefficiets of p(x) ad q(x) respectively. Corollary 6.4.11 Assume lim a exists ad is ozero. The the sequece (a /a +1 ) coverges to 1.

66 CHAPTER 6. SEQUENCES AND LIMITS The sequece (a /a +1 ) may ot coverge if lim a = 0. For example, choose { 1/ if is eve a = 1/ 2 if is odd Clearly lim a = 0, but a = a +1 (+1) 2 +1 2 if is eve if is odd co- Ad the subsequece (+1)2 diverges to, although the subsequece +1 2 verges to 0. Corollary 6.4.12 Let (x ) be a sequece covergig to α. Let p(x), q(x) C[x]. Assume that q(a) 0. The the sequece (p(x )/q(x ) coverges to p(α)/q(α). Exercises. i. Fid the followig limits ad prove your result usig oly the defiitio of covergece. a. lim 2 5 5+2 b. lim 2 5 +2 c. lim 2 5 2 +2 ii. Let 1 < r 1. Show that the sequece (r /) coverges to 0. iii. Show that +3 a. lim 2 +3+2 = 0. ( +3 b. lim + 5) = 0. 2 c. lim ( d. lim ( +3 2 + 5 ) 1 3+2 = 0. ) 2 1 2 2 3 1 3 5 = 0. iv. Show that if a, b > 0 the lim (a + b ) 1 = max(a, b). v. Let (a ) ad (b ) be two real sequeces that coverge to two differet umbers i the usual metric. Show that {a : N} {b : N} is fiite. vi. Show that the set of coverget sequeces of R (i the usual metric) is a real vector space. (For the additio of sequeces ad scalar multiplicatio of a sequece by a real umber, see Example vii, page 45).

6.5. MORE ON SEQUENCES 67 vii. Show that the set of sequeces of R (i the usual metric) that coverges 0 is a real vector space. viii. Show that the set of bouded sequeces of R (i the usual metric) is a real vector space. 6.5 More O Sequeces Lemma 6.5.1 A subsequece of a coverget sequece is also coverget; furthermore the limits are equal. Proof: Let (x ) coverge to x ad let (y i ) i = (x i ) i be a subsequece of (x ). We will show that lim i y i = x also. Let ɛ > 0. Sice lim x = x, there is a N such that for all i > N, d(x i, x) < ɛ. Sice the sequece ( i ) i is strictly icreasig, i i (oe ca prove this by iductio o i). Thus for all i > N, i > N also ad hece d(y i, x) = d(x i, x) < ɛ. Limit i the Topological Laguage. We ca traslate the defiitio of a limit of a sequece i a laguage that ivolves oly ope subsets (rather tha the metric). This is what we will do ow: Theorem 6.5.2 Let (X, d) be a metric space, (x ) a sequece from X ad x X. The lim x = x if ad oly if for ay ope subset U cotaiig x, the sequece (x ) is evetually i U, i.e. there is a N such that x U for all > N. Proof: ( ) Let U be a ope subset cotaiig x. By Propositio 5.4.1, there is a ɛ > 0 such that B(x, ɛ) U. Sice lim x = x, there is a N such that for all > N, d(x, x) < ɛ, i.e. x B(x, ɛ) U. ( ) Let ɛ > 0. The B(x, ɛ) is ope. Thus, by hypothesis, there is a N such that for all > N, x B(x, ɛ), i.e. d(x, x) < ɛ. Hece lim x = x by defiitio of limits. Exercises. i. Let (a ) be a coverget sequece of real umbers. a. Does the sequece (a 2 ) coverge ecessarily? b. Assume a 0. Does the sequece (a /a +1 ) coverge ecessarily? ii. Let x,m = +m. Fid lim lim m x,m ad lim m lim x,m. iii. Let (x ) ad (y ) be two sequeces. Defie { x/2 if is eve z = if is odd y 1 2 Show that (z ) coverges if ad oly if (x ) ad (y ) both coverge to the same umber.

68 CHAPTER 6. SEQUENCES AND LIMITS iv. Let (x ) be a real sequece ad (y ) a complex sequece. Let y C. Assume that lim x = 0 ad y y x for all. Show that lim y = y. v. Let (x ) be a coverget sequece of real umbers. Let y = x 1 +... + x. Show that lim y exists ad is equal to lim x. 6.6 Cauchy Sequeces The elemets of a coverget sequece, sice they approach to a fixed elemet, should also approach to each other. A sequece that has this property is called a Cauchy sequece. We formalize this as follows: A sequece (x ) is called a Cauchy sequece if for ay ɛ > 0 there is a N such that d(x, x m ) < ɛ for all, m > N. We formalize this by writig lim,m d(x, x m ) = 0. We first prove what we have said i a mathematical way: Theorem 6.6.1 A coverget sequece is a Cauchy sequece. Proof: Let (x ) be a sequece covergig to x. Let ɛ > 0. Sice lim x = x, there is a N such that d(x, x) < ɛ/2 for all > N. Now if, m > N, the d(x, x m ) d(x, x) + d(x, x m ) < ɛ/2 + ɛ/2 = ɛ. O the other had, a Cauchy sequece eed ot be coverget (see Exercises x, page 59 ad xii, page 60). But we have the followig: Propositio 6.6.2 A Cauchy sequece that has a coverget subsequece is coverget. Furthermore the limits are the same. Proof: Let (x ) be a Cauchy sequece. Let (y i ) i = (x i ) i be a coverget subsequece of (x ). Say lim i y i = x. We will show that lim x = x. Let ɛ > 0. Sice (x ) is a Cauchy sequece, there is a N such that if, m > N, the d(x, x m ) < ɛ/2. Also, sice lim i y i = x, there is a N 1 such that if i > N 1 the d(y i, x) < ɛ/2. Let N = max(n, N 1 ). The for all i > N, we have i i > N (because the sequece ( i ) i is strictly icreasig) ad so d(x i, x) d(x i, y i ) + d(y i, x) = d(x i, x i ) + d(y i, x) < ɛ/2 + ɛ/2 = ɛ. Propositio 6.6.3 A Cauchy sequece is bouded. Proof: Let (x ) be Cauchy sequece. Take ɛ = 1 i the defiitio. Thus there exists a N such that for all, m > N, d(x, x m ) < 1. I particular, for all > N, d(x N+1, x ) < 1. Let r = max(1, d(x 0, x N+1 ) + 1,..., d(x N, x N+1 ) + 1). The {x : N} B(x N+1, r).

6.7. CONVERGENCE OF REAL CAUCHY SEQUENCES 69 Exercises. i. Let (a ) ad (b ) be two Cauchy sequeces such that the set {a : N} {b : N} is ifiite. Show that lim a = lim b. ii. Give a example of each of the followig, or argue that such a request is impossible. a) A Cauchy sequece that is ot mootoe. b) A mootoe sequece that is ot Cauchy. c) A Cauchy sequece with a diverget subsequece. d) A ubouded sequece with a Cauchy subsequece. iii. Let X 1,..., X m be metric spaces. Let X = X 1... X be the product space. Let (x ) be a sequece i X. Set x = (x 1,..., x m ). The (x ) is Cauchy if ad oly if x i ) is Cauchy for all i = 1,..., m. iv. Let x 1 = 1, x 2 = 2 ad x = (x 1 + x 2 )/2 for > 2. a. Show that 1 x 2 for all. (Hit: By iductio o ). b. Show that x x +1 = 1/2 1 for all. (Hit: By iductio o ). c. Show that if m > the x x m < 1/2 2 for all. (Hit: Use part c). d. Show that (x ) is a Cauchy sequece. e. Fid its limit. Note that the sequece is formed by takig the arithmetic mea of the previous two terms. It ca be guessed that the limit is at the two thirds of the way from x1 = 1 to x2 = 2, i.e. it is 5/3. This ca be show by some elemetary liear algebra. But there is a easier way: Note that part b ca be sharpeed ito x x +1 = ( 1) /2 1, thus x +1 1 = x +1 x 1 = (x +1 x ) +... + (x 2 x 1 ) = 1 1/2 + 1/4... + ( 1) 1/2 1. It follows that the limit x, that we kow it exists from the previous questio, is equal to the ifiite sum x = 1 + 1 1/2 + 1/4... + ( 1) 1/2 1 +... Now just remarque that 2x = 2 + 2-1 + 1/2-1/4 +... = 3 + (1/2-1/4 + 1/8...) = 3 + (2-x) ad so x = 5/3. v. We say that a sequece (x ) is cotractive if there is a costat c, 0 < c < 1, such that x +2 x +1 c x +1 x for all. Show that every cotractive sequece is coverget. Hit: x +1 x c x 1 x 0. Thus we ca estimate x x m as is doe above ad show that such a sequece is Cauchy. 6.7 Covergece of Real Cauchy Sequeces The purpose of this subsectio is to prove the followig importat theorem: Theorem 6.7.1 Every real Cauchy sequece is coverget.

70 CHAPTER 6. SEQUENCES AND LIMITS Recall that a sequece (x ) of real umbers is called odecreasig (resp. oicreasig) if x x m (resp. x x m ) for all m. A oicreasig or odecreasig sequece is called a mootoe sequece. To prove this theorem, we first prove that every mootoe ad bouded sequece coverges. Theorem 6.7.2 (Mootoe Covergece Theorem) A mootoe ad bouded sequece of R coverges. Proof: Let (x ) be a icreasig sequece of real umbers which is bouded above. The the set {x : N} has a least upper boud l. We claim that lim x x = l. Let ɛ > 0. Sice l ɛ is ot a upper boud, there is a N such that l ɛ < x N. Sice the sequece (x ) is icreasig, l ɛ < x for all N. Sice l is a upper boud, l ɛ < x l for all N. Thus l x ɛ for all N. The secod part follows easily from the first. Theorem 6.7.3 Every real sequece has either a icreasig or a oicreasig subsequece. Proof: We will choose a icreasig sequece ( i ) i such that give ay i, either x j > x i for all j > i or either x j x i for all j > i. Let = 0. Cosider the sets G 0 = { > 0 : x > x 0 } ad L 0 = { > 0 : x x 0 }. Either G 0 or L 0 is ifiite. If G 0 is ifiite, cosider oly the subsequece that cosists of x 0 ad the x s for which G 0, deletig the rest. If G 0 is fiite, cosider oly the subsequece that cosists of x 0 ad the x s for which L 0, deletig the rest. Reamig the elemets, assume that (y ) is this ew sequece. I this ew sequece y 0 has the followig property: Either y > y 0 for all > 0 or y y 0 for all > 0. Now cosider y 1 ad the sets ad G 1 = { > 1 : y > y 1 } L 1 = { > 1 : y y 1 }. Either G 1 or L 1 is ifiite. If G 1 is ifiite, cosider oly the subsequece that cosists of y 0, y 1 ad the y s for which G 1, deletig the rest. If G 1 is fiite, cosider oly the subsequece that cosists of y 0, y 1 ad the y s for which L 1, deletig the rest. Reamig the elemets, assume that (y ) is this ew sequece. I this ew sequece y 0 has the followig property: Either y > y 0 for all > 0 or y y 0

6.7. CONVERGENCE OF REAL CAUCHY SEQUENCES 71 for all > 0 ad y 1 has the followig property: Either y > y 1 for all > 1 or y y 1 for all > 1. Next we cosider y 2 ad the sets ad G 2 = { > 2 : y > y 2 } L 2 = { > 2 : y y 2 }. Either G 2 or L 2 is ifiite. If G 2 is ifiite, cosider oly the subsequece that cosists of y 0, y 1, y 2 ad the y s for which G 2, deletig the rest. If G 2 is fiite, cosider oly the subsequece that cosists of y 0, y 1, y 2 ad the y s for which L 2, deletig the rest. Cotiuig this way, we fid a subsequece, say (z ), of the origial sequece (x ) such that for all, either for all m >, x m > x or for all m >, x m x. We call red if the first case occurs, i.e. if for all m >, x m > x. We call gree if the secod case occurs, i.e. if for all m >, x m x. Thus we gave a color, either red or gree, to each atural umber. If there are ifiitely may red umbers, the the sequece (z ) is red is icreasig. If there are ifiitely may gree umbers, the the sequece (z ) is gree is oicreasig. Proof of Theorem 6.7.1. Let (x ) be a real Cauchy sequece. By Propositio 6.6.3, (x ) is bouded. By Theorem 6.7.3, (x ) has a mootoe subsequece, say (y ). The subsequece (y ) is still bouded. By Theorem 6.7.2, (y ) coverges. By Propositio 6.6.2, (x ) coverges. Theorem 6.7.1 ca also be prove as a cosequece of the followig importat result: Theorem 6.7.4 (Bolzao-Weierstrass Theorem) Every bouded sequece i R has a coverget subsequece. Proof: Let (x m ) m be a bouded sequece i R. Choose a, b R such that x m [a, b] for all m. We will choose a Cauchy subsequece (x mi ). Let m 0 = 0. Separate the iterval [a, b] ito two parts [a, a+b a+b 2 ] ad [ 2, b]. Oe of these two smaller itervals cotais x m for ifiitely may m. Let m 1 to be the smallest idex ot equal to m 0 which is i this smaller iterval that cotais x m for ifiitely may m. Now do the same with this smaller iterval. Thus the sequece y i = x mi has the property that for all i > j, d(y i, y j ) (b a)/2 j. By the Theorem of Nested Itervals 3.1.12, the itersectio of these itervals of legth (b a)/2 j is oempty. If y is i the itersectio, the lim i y i = y. Note that this theorem is false i geeral metric space: I a discrete metric space a sequece of distict elemets does ot have a Cauchy subsequece at all.

72 CHAPTER 6. SEQUENCES AND LIMITS Secod Proof of Theorem 6.7.1. Let (x ) be a Cauchy sequece i R. By Propositio 6.6.3, (x ) is bouded. By Theorem 6.7.4, (x ) has a coverget subsequece. By Propositio 6.6.2, (x ) coverges. Exercises. i. Let x = 1/1 2 + 1/2 2 + 1/3 2 +... + 1/ 2. a. Show that for all itegers 1, x 2 1/ < 2. b. Coclude that the sequece (x ) coverges. c. Show that for a iteger large eough, 2 2. d. Coclude that the sequece (x ) is bouded above by 1 + 1/4 + 1/9 + 1/8 (= 107/72). We will see later that this sequece coverges to π 2 /6. ii. Show that the sequece 2, 2 2, 2 2 2 coverges. Fid its limit. iii. Discuss the covergece of the sequece a, a a, a a a for aır 0. iv. Show that if k 2 is a iteger, the sequece (x ) where x = 1/1 k + 1/2 k + 1/3 k +... + 1/ k is Cauchy ad hece coverges. v. Let x 0 = x ad x +1 = 1 4 x. a) Assume that the sequece coverges. Show that the limit is either 2 + 3 or 2 3. b) Show that if 2 3 x 2 + 3 the 2 3 x +1 2 + 3. c) Assume x [2 2, 2 + 2]. Show that (x ) is decreasig. Coclude that lim x = 2 2. d) Discuss the covergece ad the divergece of (x ) for other values of x. vi. Discuss the covergece of the sequece (x ) defied by x 0 x +1 = 4 1/x i terms of x. = x ad vii. Let x 1 = x ad x +1 = 2x. a) Write the first four terms of x whe x = 2. b) Show that if the sequece (x ) coverges, the, it must coverge to either 0 or 2. c) Discuss the covergece of the sequece (x ) accordig to the values of x. viii. Let x 1 = x ad x +1 = x /2 + 1/x. Discuss the covergece of the sequece (x ) accordig to the values of x.

6.8. CONVERGENCE OF SOME SEQUENCES 73 ix. (Existece of Square Roots) Let x 1 = 2 ad x +1 = (x + 2/x )/2. Show that x 2 2 for all. Use this to prove that the sequece (x ) is decreasig. Show that lim x = 2. Modify the sequece so that it coverges to c. x. Prove the Bolzao-Weierstrass Theorem for R. Deduce that every Cauchy sequece of R coverges. 6.8 Covergece of Some Sequeces 6.8.1 The Sequece ((1 + 1/) ) Propositio 6.8.1 The sequece ((1 + 1/) ) coverges to some real umber betwee 2 ad 3. Proof: We will show that the sequece ((1+1/) ) is bouded ad icreasig. Theorem 6.7.2 will the prove the theorem. We proceed i several steps. Claim 1. For all atural umbers > 0 ad all real umbers x > 1, (1 + x) 1 + x. I particular if y < 1 the (1 y) 1 y For = 0, this is clear. (Note that we eed x 1 for this). Assume the iequality holds for. The we have (1 + x) +1 = (1 + x) (1 + x) (1 + x)(1 + x) = 1 + ( + 1)x + x 2 1 + ( + 1)x. (The secod step is by the iductio hypothesis ad for this step we also eed the fact that 1 + x > 0, which we kow it holds). Claim 2. 2 (1 + 1/) 3 for all atural umbers > 0. Replacig x by 1/ i the first claim, we get 2 (1 + 1/) for all atural umbers > 0. Now we show that (1 + 1/) 3, for all atural umbers > 0. We compute carefully. For 1 we have, ) (1 + 1/) = ( i=1 i = 1 + 1 i = 1 + = 1 + i=1 (1 1 < 1 + 1 i=1 1 2... i 1 i=1 2 i 1 = 1 + 1 1 2 1 1 2 = 1 + 2(1 1 2 ) < 3 ( 1)...( i+1) 1 i=1 1 2... i i i=1 1... i+1 1 1 2... i )... (1 i 1 ) 1 1 2... i Claim 3. N \ {0}, (1 + 1 ) (1 + 1 +1 )+1. Sice (1 + 1 +1 )+1 = (1 + 1 1 + 1 +1 )+1 = (1 + 1 1 (+1) )+1, we have to show that (1 + 1 1 (+1) )+1 (1 + 1 ). By settig a = 1 + 1/, this is equivalet to the statemet (a 1 (+1) )+1 a, i.e. to the statemet

74 CHAPTER 6. SEQUENCES AND LIMITS 1 (1 a(+1) )+1 1/a, i.e. to the statemet (1 1 1, if 1, (1 1 (+1) ) +1 1 1 2 +1 = +1 Now apply Theorem 6.7.2. Exercises. (+1) ) +1 2. This proves Claim 3. +1. By Claim i. We have see i this paragraph that the sequece give ((1 + 1/) ) coverges to a real umber > 2. Let e be this limit. Do the followig sequeces coverge? If so fid their limit.. a) lim (1 + +1) 1 ( b) lim 1 + 1. 2) ( c) lim 1 + 1 2. ) ( d) lim 1 + 1 3. 2) ( e) lim 1 + 1 2. 3) 6.8.2 The Sequeces (2 1/ ) ad ( 1/ ) Propositio 6.8.2 lim 2 1/ = 1. Proof: Clearly 2 1/ > 1. It is also clear that the sequece (2 1/ ) is decreasig. Therefore it has a limit, say l. The l = lim 2 1/ = lim 2 1/ = l 1/2. Hece l = 1. Propositio 6.8.3 lim 1/ = 1. Proof: Clearly 1/ 1. We will show that the sequece is oicreasig for 3, i.e. we will show that ( + 1) 1/(+1) 1/, which is equivalet to ( + 1) +1 ad to (1 + 1/). But we have see i the proof of Theorem 6.8.1 that (1 + 1/) 3 for all. Thus (1 + 1/) for 3. Therefore lim 1/ exists, say l. We have, 1 l = lim (2) 1/2 = lim 2 1/2 1/2 = 1 l. (We used Propositio 6.8.2 i the last equality). Thus l = 1. 6.8.3 The Sequece (x ) Propositio 6.8.4 lim x = 0 for x < 1. The sequece diverges otherwise. Proof: Sice lim +1 = 1, there is a N such that x < +1 for all > N. Now for > N, ( + 1) x +1 = ( + 1) x x < x. Hece the sequece (x ) is decreasig after a while. Therefore the sequece coverges, x say to a. Assume a 0. The by Corollary 6.4.11, 1 = lim (+1)x = +1 1 lim +1 x = 1/x, a cotradictio. Thus lim x = 0. The rest is left as a exercise.

6.8. CONVERGENCE OF SOME SEQUENCES 75 Exercises. i. Let (x ) be a sequece such that x 0 for every N ad x +1 /x r for some fixed r (0, 1). Show that lim x = 0. ii. Let x 1 = 1, x 2 = 2 ad for > 2. a) Show that 1 x 2 for all. x = x 1 + x 2 2 b) Show that x x +1 = 1/2 1 for all. c) Show that if m > the x x m < 1/2 2 for all. d) Show that (x ) is a Cauchy sequece. e) Fid its limit. Aswer: Note that the sequece is formed by takig the arithmetic mea of the previous two terms. It ca be guessed that the limit is at the two thirds of the way from x 1 = 1 to x 2 = 2, i.e. it is 5/3. This ca be show by some elemetary liear algebra. But there is a easier way: Note that part b ca be sharpeed ito x x +1 = ( 1) /2 1, thus x +1 1 = x +1 x 1 = (x +1 x ) +... + (x 2 x 1 ) = 1 1/2+1/4...+( 1) 1 /2 1. It follows that the limit x, that we kow it exists from the previous questio, is equal to lim 2 1/2+1/4...+ ( 1) 1 /2 1. Now we just remarque that 2x = 2 lim 2 1/2 + 1/4... + ( 1) 1 /2 1 = lim 4 1 + 1/2... + ( 1) 1 /2 2 = lim 4 (1 1/2... + ( 1) /2 2 ) = lim 5 (2 1/2... + ( 1) /2 2 ) = 5 x, so 3x = 5 ad x = 5/3. iii. Let x 0 = 1, x +1 = 1 + 1/x. Show that (x ) is coverget. iv. For k N ad x R, discuss the covergece of the sequece ( k x ). v. Let f 0 = f 1 = 1, f +2 = f + f +1. Show that the sequece ( f +1 f ) is 2 coverget. (It coverges to 1+ ). 5 vi. We say that a sequece (x ) is cotractive if there is a costat c (0, 1) such that x +2 x +1 c x +1 x for all. Show that every cotractive sequece is coverget. Hit: Note first that x +1 x c x 1 x 0. Now estimate x x m ad show that such a sequece is Cauchy. vii. Let 0 < a 0 = a < b 0 = b. Defie a +1 = (2a b )/(a + b ), b +1 = (a + b )/2. Show that (a ) ad (b ) are both coverget. Show that their limits are equal. (Hit: Use mea iequalities.) viii. Let a 0 = a, b 0 = b, c 0 = c. Defie a +1 = (b + c )/2, b +1 = (c + a )/2, c +1 = (a + b )/2. Show that (a ), (b ) ad (c ) are coverget. Compute their limits.(hit: First take a = b = 0, c = 1.)

76 CHAPTER 6. SEQUENCES AND LIMITS 6.9 Divergece to Ifiity Let (x ) be a real sequece. We say that the sequece (x ) diverges to ifiity if for all real umbers A there is a N such that for all > N, x > A. Here, A should be regarded as a very large umber, as large as it ca be. I this case oe writes lim x =. Similarly we say that the sequece (x ) diverges to mius ifiity if for all real umbers A there is a N such that for all > N, x < A. Here, A should be thought as a very small egative umber. I this case, we write lim x = If a icreasig sequece (a ) is bouded the lim x R (Theorem 6.7.2), if it is ubouded, the lim x = (Exercise vi, page 77). Similarly for decreasig sequeces. Note that util ow, we have ot defied ifiity, ad we will ever really defie it. Above, we oly defied the phrase lim x = as if it were oe sigle etity. Now we let ad to be two ew distict symbols. We will ever give ay meaig to these symbols. Their oly property is that they are ew symbols. We let R := R {, }, we exted the operatios ad the order relatio defied o R to R partially as follows: < r < for all r R r + = + r = for all r (, ] r + ( ) = r = + r = for all r [, )} ( ) = r = r = for all r (0, ] r = r = for all r [, 0) r( ) = ( )r = for all r (0, ] r( ) = ( )r = for all r [, 0) r/ = r/( ) = 0 for all r R ± /r = ± (1/r) for all r R r = for all r (1, ] r = 0 for all r ( 1, 1) r = 0 for all r (1, ] r = for all r (0, 1) r = for all r ( 1, 0) Note that the followig terms are ot defied:

6.9. DIVERGENCE TO INFINITY 77 ( ) + + ( ) / /( ) ( )/ ( )/ /0 /0 0 ( ) 0 0 (±1) Theorem 6.9.1 Let (x ) ad (y ) be two sequeces i R such that lim x R ad lim y R. Let be ay of the five basic arithmetic operatios (additio, substractio, multiplicatio, divisio ad power). If lim x lim y is defied, the lim x y = lim x lim y. Proof: Left as a exercise. Exercises. i. Show that the sequeces () ad ( 2 ) diverge to ifiity. ii. Show that the sequece (( 1) ) does ot coverge or diverge either to or to. iii. Show that lim (3/2 + 1/) =. iv. Fid the followig limits ad prove your result usig oly the defiitio: a) lim 2 2 5 +2. b) lim 2 2 5 +2. 2 c) lim 3 5 2 +2. 2 d) lim 2 1 2 1. ( e) lim 2 + 1. ) v. Show that if lim x = the lim x =. vi. Show that a icreasig ad ubouded sequece coverges to. vii. Let p(x) ad q(x) 0 be two polyomials of degree d ad e respectively ad with leadig coefficiets a ad b respectively. The the sequece (p()/q()) diverges to ifiity if ad oly if e < f ad a/b > 0. Ad the sequece (p()/q()) diverges to if ad oly if e < f ad a/b < 0.

78 CHAPTER 6. SEQUENCES AND LIMITS viii. Show that if lim x = ad lim y = a R the lim x + y =. ix. Show that if lim x = ad lim y = the lim x +y =. x. Show that if lim x = the lim x 1 = 0. xi. Show that there are sequeces (x ) such that lim x = 0 but lim x 1 ±. xii. Show that if lim x = ad lim y > 0 the lim x y =. xiii. Show that if lim x = ad lim y = the lim x y =. xiv. Let x = 1 + 1/2 +... + 1/. Show that lim x =. (See Theorem 7.1.3). 6.10 Limit Superior ad Iferior Let (a ) be a sequece of real umbers. We let lim a = lim sup{a, a +1,...} R. Sice the sequece (sup{a, a +1,...}) is decreasig, lim a is either a real umber or is ±. lim a is called the limit superior of the sequece (a ). We also defie lim a = lim if{a, a +1,...} R. Sice the sequece if{a, a +1,...} is icreasig, lim a is either a real umber or is ±. lim a is called the limit iferior of the sequece (a ). We always have lim a lim a. Lemma 6.10.1 i. lim a if ad oly if the sequece (a ) has a upper boud. ii. lim a = if ad oly if the sequece (a ) is ubouded by above. iii. If lim a = r R the for all ɛ > 0 there is a N such that for all > N, a < r + ɛ. Proof: (i) ad (ii) are clear We prove (iii). Let ɛ > 0. Let N be such that for all > N, sup{a, a +1,...} r < ɛ. Thus ɛ < sup{a N+1, a N+2,...} r < ɛ ad so sup{a N+1, a N+2,...} < r + ɛ. Thus a < r + ɛ for all > N. We leave to the reader the task of statig ad provig the aalogue of the above lemma for the limit iferior.

6.11. COMPLETE METRIC SPACES 79 Theorem 6.10.2 Let (a ) be a sequece of real umbers. The lim a exists if ad oly if lim a ad lim a are real umbers ad are equal. I this case, lim a = lim a = lim a. Proof: This follows from Lemma 6.10.1.iii ad its aalogue for the limit iferior. Exercises. i. Suppose (u ) ad (v ) are positive sequeces. Let U = lim u ad V = lim v. Assume that if oe of U ad V is, the the other is ozero. The L = lim u v = UV. 6.11 Complete Metric Spaces A metric space is called complete if its Cauchy sequeces coverge. Thus Q is ot complete. Theorem 6.7.1 says that R is complete with its usual metric. We ca geeralize this to the Euclidea spaces. Theorem 6.11.1 The Euclidea space R is complete. To prove this we oly eed to prove the followig. Theorem 6.11.2 Let x k = (x k,1,..., x k, ) R with R cosidered with its Euclidea metric. The i) (x k ) k is a Cauchy sequece if ad oly if (x k,i ) k is a Cauchy sequece for all i = 1,...,. ii) (x k ) k is coverget if ad oly if (x k,i ) k is coverget for all i = 1,...,. Ad i this case lim x k = ( lim x k,1,..., lim x k,). k k k Proof: i. Assume (x k ) k is a Cauchy sequece. Let i = 1,...,. Let ɛ > 0. Let N be such that for all k, l > N, d(x k, x l ) < ɛ. The for all k, l > N, x k,i x l,i (x k,1 x l,1 ) 2 +... (x k, x l, ) 2 = d(x k, x l ) < ɛ. Thus (x k,i ) k is a Cauchy sequece. Coversely, assume that (x k,i ) k is a Cauchy sequece for all i = 1,..., m. Let ɛ > 0. Let N i be such that for all k, l > N i, x k,i x l,i < ɛ/. Let N = max(n 1,..., N ). Now for all k, l > N, d(x k, x l ) = (x k,1 x l,1 ) 2 +... (x k,1 x l,1 ) 2 x k,1 x l,1 +... + x k, x l, < ɛ/ +... + ɛ/ = ɛ. ii. The proof is similar ad we leave it as a exercise. Oe ca geeralize this theorem easily:

80 CHAPTER 6. SEQUENCES AND LIMITS Theorem 6.11.3 For i = 1,...,, let (X i, d i ) be a metric space. Cosider X = X 1... X with the product metric. Let x k = (x k,1,..., x k, ) X. The i) (x k ) k is a Cauchy sequece if ad oly if (x k,i ) k is a Cauchy sequece for all i = 1,...,. ii) (x k ) k is coverget if ad oly if (x k,i ) k is coverget for all i = 1,...,. Ad i this case lim x k = ( lim x k,1,..., lim x k,). k k k Proof: Left as a exercise. Exercises. i. Show that lim x = 0 if 0 x < 1. Corollary 6.11.4 Let (x ) ad (y ) be two covergig sequeces i R m. The (x + y ) ad (x y ) are covergig sequeces ad lim (x + y ) = lim x + lim y ad lim (x y ) = lim x lim y. Corollary 6.11.5 Let (x ) be a covergig sequece i R m. The ay subsequece of (x ) is coverget ad it coverges to the same limit. Corollary 6.11.6 Product of fiitely may complete metric spaces is complete. Corollary 6.11.7 The metric space of complex umbers C is complete. Exercises. i. Let X be ay set. Show that the space Seq(X) defied o page 52 is complete. 6.12 Completio of a Metric Space I this subsectio, give a metric space X, we will fid the smallest complete metric space cotaiig X. Let (X, d) be a metric space. Cosider the set C(X) of Cauchy sequeces of X. O C(X) defie the followig equivalece relatio: (x ) (y ) lim d(x, y ) = 0. Lemma 6.12.1 The relatio is a equivalece relatio o the set C(X).

6.12. COMPLETION OF A METRIC SPACE 81 Proof: Trivial. Set X = C(X)/. For (x ) C(X), let (x ) X be its class. We will tur the set X ito a metric space. To do this we eed the followig. Lemma 6.12.2 i. Let (x ) ad (y ) be two Cauchy sequeces i X. The lim d(x, y ) exists. ii. Let (x ), (x ), (y ), (y ) be four Cauchy sequeces i X. If (x ) (x ) ad (y ) (y ), the lim d(x, y ) = lim d(x, y ). Proof: We first ote the followig: implyig Exchagig ad m, we get d(x, y ) d(x, x m ) + d(x m, y m ) + d(y m, y ), d(x, y ) d(x m, y m ) d(x, x m ) + d(y m, y ). d(x, y ) d(x m, y m ) d(x, x m ) + d(y m, y ). i. It is eough to show that the real sequece (d(x, y )) is Cauchy. Let ɛ > 0. Let N be large eough so that for, m > N, d(x, x m ) < ɛ/2 ad d(y m, y ) < ɛ/2. By the above iequality for, m > N, d(x, y ) d(x m, y m ) > ɛ.. ii. By assumptio lim d(x, x ) = lim d(y, y ) = 0. So the secod part also follows from the iequality prove i the begiig. The lemma above says that we are allowed to defie a map d from X X ito R 0 by the rule, d((x ), (y ) ) = lim d(x, y ). Lemma 6.12.3 (X, d) is a metric space. Proof: Suppose d((x ), (y ) ) = 0, i.e. lim d(x, y ) = 0. By defiitio (x ) = (y ). Clearly d((x ), (y ) ) = d((y ), (x ) ). It remais to prove the triagular iequality: d((x ), (y ) ) = lim d(x, y ) lim (d(x, z ) + d(x, z )) = lim d(x, z ) + lim d(x, z ) = d((x ), (z ) ) + d((z ), (y ) ). Lemma 6.12.4 The map i that seds x X to the class (x) of the costat sequece (x) is a cotiuous embeddig of X ito X. Furthermore d(i(x), i(y)) = d(x, y).

82 CHAPTER 6. SEQUENCES AND LIMITS Proof: If (x) = (y), the lim d(x, y) = 0, i.e. d(x, y) = 0, hece x = y, provig that i is oe to oe. For the last equality: d(i(x), i(y)) = lim d(x, y) = d(x, y). Let us ow show that i is cotiuous. Let (x k ) k be a sequece from X covergig to a. Let us show that the class of costat sequeces (x k ) coverge to the class of the costat sequece (a) whe k goes to ifiity. By Lemma 6.1.2, it is eough to show that lim d((x k ), (a) ) = 0. We have: lim k d((x k ), (a) ) = lim k d(x k, a) = 0. This proves it. Lemma 6.12.5 (X, d) is a complete metric space. Proof: From ow o we idetify X with its image i(x) ad we assume that X is a subset of X. Lemma 6.12.6 X is dese i X. Lemma 6.12.7 Let Y be a metric space. Let f : X Y be a cotiuous map. The there is a uique cotiuous extesio f : X Y of f. Lemma 6.12.8 Uiqueess of X. 6.13 Supplemetary Problems i. Oe tosses a coi. Each head adds oe poit, each tail adds two poits. Let p() be the probability of reachig the iteger. Show that lim p() = 2/3. ii. Fiboacci Sequece. Let f 0 = 0, f 1 = 1 ad defie f +2 = f +1 + f. The sequece (f ) is called the Fiboacci sequece. Show that lim f /f +1 = 2 1+ 5. iii. Let (a ) be a bouded real sequece with the property that every covergig subsequece coverges to the same limit. Show that the sequece coverges. iv. Let (a ) be a bouded real sequece. Let S = {x R : x < a : for ifiitely may }. Show that there exists a subsequece covergig to sup(s). Deduce the Bolzao-Weierstrass Theorem from this.

Chapter 7 Series 7.1 Defiitio ad Examples DO IT IN BANACH SPACES Let (a i ) i be a sequece of complex umbers. (For most of what follows we ca also assume that (a i ) i is a sequece i a ormed vector space). Let s := i=0 a i. Assume that the sequece (s ) coverges i C. The we write a i = lim i=0 a i i=0 ad we say that the series i=0 a i coverges to lim i=0 a i or that its sum is lim i=0 a i. Else, we say that the series i=0 a i is diverget. The umbers s := i=0 a i are called the partial sums of the series i=0 a i. Although we defied a series for complex umbers, the reader is welcome to cosider oly real series, i.e. series where a i R for all i. If i=0 a i is a real series ad lim i=0 a i =, we say that the series i=0 a i diverges to ad we write i=0 a i =. Similarly for. Wheever we write i=0 a i =, we will assume that a i R eve if this is ot explicitly stated. Examples. i. If a i = 1 for all i, the i=0 a i = lim i=0 a i = lim =. I geeral, if (a i ) i is a costat sequece of real umbers, the the series i=0 a i coverges if ad oly if a i = 0. If the sequece (a i ) i is a real costat ozero sequece, the i=0 a i diverges to ± depedig o the sig of a i. ii. If a i = ( 1) i for all i, the i=0 a i does ot exist (is ot eve ± ). 83

84 CHAPTER 7. SERIES Theorem 7.1.1 The set of sequece (a i ) i such that i a i coverges form a vector space. Furthermore we have, (αa i + βb i ) = α i i a i + β b i. The equality persists if i a i ad i b i are i R {, } if the operatios i the equality are allowed. Proof: Easy. Remarks. i. Regroupig the terms of a series may chage the covergece of the series. For example cosider the series i=0 a i where a i = ( 1) i. If we let b i = a 2i + a 2i+1, the =0 b i = 0 because each b i = 0. Thus a diverget sequece may tur ito a coverget sequece. O the other had regroupig the terms of a coverget series does ot chage its covergece ad the sums are equal. For example, the terms of the alteratig series i=1 ( 1)i /i that we will show is coverget ca be regrouped two by two to get that the series i=1 1/(2i + 1)(2i + 2) coverges to the same limit. As we have see we caot dissociate a series without alterig its sum (uless the terms are positive). ii. A rearragemet of the terms of the series may chage the covergece of the series (uless the terms are positive, see Theorem 7.3.2). Cosider the series ( 1) i /i = 1 1/2 + 1/3 1/4 + 1/5.... i We will see later that this series coverges to a ozero umber. Now shuffle the terms as follows: (1 1/2 1/4)+(1/3 1/6 1/8)+...+(1/(2+1) 1/2(2+1) 1/2(2+2))+... Regroupig a positive term with the ext term which follows it, we get 1/2 1/4 + 1/6 1/8 +... + 1/2(2 + 1) 1/2(2 + 2)) +... ad this is half of the iitial series. iii. O the other had, if i=0 a i coverges, the if we regroup the terms without chagig the order, the the sum (a 1 + a 2 +... + a 1 ) + (a 1+1 +... + a 2 ) + (a 2+1 +... + a 3 ) +...,

7.1. DEFINITION AND EXAMPLES 85 remais ualtered. This may be called the associativity of series (see Exercise v, page 86). But we caot dissociate: For example (1 + ( 1)) + (1 + ( 1)) +... = 0, but is ot coverget. 1 + ( 1) + 1 + ( 1) +... iv. The orderig of the terms of the series is importat ad i some cases the sum may chage. For example, summig first the terms with eve subscripts a 2i ad the the terms with odd subscripts a 2i+1 ad the addig these two sums may chage the sum. We give a example i page 89. The series i=0 ri is called geometric series. This will be our first example of a coverget series. Theorem 7.1.2 (Geometric Series) Let α C. The series i=0 αi is coverget if ad oly if α < 1. I this case i=0 α i = 1 1 α. Proof: If α 1, the lim α 0, so that i=0 αi caot be coverget. Assume α < 1. Let s = i=0 αi. The (1 α)s = s αs = 1 α +1 ad so s = 1 α+1 1 α. Therefore by Propositio 6.2.3, i=1 αi = 1 1 α. The series i=1 1/i is called harmoic series. Theorem 7.1.3 (Harmoic Series) i=1 1/i =. Proof: For ay atural umber k ad for the 2 k 1 may atural umbers i [2 k 1, 2 k ), we have 1/i > 1/2 k, so that 2 k 1 1 i=2k 1 1/i > 2k 1 = 1/2. It 2 k follows that s 2 1 = 2 1 i=1 1/i = 2 k 1 k=1 i=2 1/i > k 1 k=1 1/2 = /2, so that the sequece s diverges to ifiity. Exercises. i. Show that the series i=1 1/ii coverges. Fid a upper boud for the sum. ii. Show that the series i=1 1/2i ad i=0 1/(2i + 1) diverge. iii. Show that the sum of the reciprocals of atural umbers whose decimal expasio cotais at least a zero 1/10 +... + 1/90 + 1/100 + 1/101 +... + 1/109 + 1/110 + 1/120 +... diverges.

86 CHAPTER 7. SERIES iv. Let k N be 2. Show that the series i=1 1/i coverges. (Hit: See Exercise iv, page 72). v. Suppose i=0 a i coverges. Let ( k ) k be a strictly icreasig sequece of atural umbers with 0 = 0. Set b k = a k +... + a k+1 1. Show that k=0 b k coverges to i=0 a i. 7.2 Easy Cosequeces of the Defiitio Propositio 7.2.1 If i=0 a i coverges the lim i a i = 0. Proof: Let ɛ > 0. Sice i=0 a i coverges, the partial sums s = i=0 a i form a Cauchy sequece. Thus there exists a N such that for all, m > N, s s m < ɛ. Takig m = +1, we see that for all > N, a +1 = s s +1 < ɛ. Thus for all > N + 1, a < ɛ. This shows that lim a = 0. Corollary 7.2.2 If lim i a i 0 the the series i=0 a i caot coverge. Sice the covergece of a series is othig else but the covergece of the sequece of partial sums, we ca use the completeess of R: Propositio 7.2.3 (Cauchy s Criterio) A sequece i=0 a i coverges if ad oly if for all ɛ > 0, there exists a N such that for all > m > N, i=m a i < ɛ. Proof: The coditio just expresses the fact that the sequece of partial sums ( i=0 a i) is Cauchy. Theorem 7.2.4 If a i 0 ad if the partial sums i=0 a i are bouded the i=0 a i coverges. Proof: Follows from Theorem 6.7.2. Corollary 7.2.5 (Compariso Test) Let 0 a i b i. a. If i=0 b i coverges, the i=0 a i coverges as well. Furthermore i=0 a i i=0 b i. b. If i=0 a i = the i=0 b i =. Corollary 7.2.6 Suppose that u, v > 0 ad u +1 /u v +1 /v evetually. If v coverges the u coverges. Hece if u diverges the v diverges. Proof: Let 0 be a witess for the evetuality. Let > 0. Multiply the iequalities from 0 to 1 to get u /u 0 v /v 0, i.e. u rv where r = u 0 v 0. Clearly v coverges if ad oly if rv coverges. Now apply the compariso test (Corollary 7.2.5).

7.3. ABSOLUTE CONVERGENCE 87 Exercises. i. Cauchy Codesatio Test. Let (x i ) i be a oicreasig, oegative real sequece. Show that i=0 x i coverges if ad oly if i=0 2i x 2 i coverges. [A, page 53] ii. Let a > 0, b > 0. Show that if lim i b i /a i = 0 ad i=0 a i coverges the i=0 b i coverges as well. iii. Let a i > 0, b i > 0. Show that if lim i b i /a i = ad i=0 a i = the i=0 b i =. 7.3 Absolute Covergece If i=0 a i coverges, the we say that the series i=0 a i coverges absolutely. Theorem 7.3.1 (Cauchy) A absolutely coverget series is coverget. Proof: Let i=0 a i be a absolutely coverget series. We will show that i=0 a i coverges by usig Cauchy s Criterio (Propositio 7.2.3). Let ɛ > 0. Let N such that for all > m > N, i=m a i < ɛ. The for all > m > N, i=m a i i=m a i < ɛ. But the coverse of this theorem is false. Ideed, we will see that the series i=1 ( 1)i /i coverges (Corollary 7.4.2), but we kow that it does ot coverge absolutely (Theorem 7.1.3). We will later show that permutig the terms of a series may chage its value. However, this is ot the case with the absolutely coverget series. Theorem 7.3.2 (Rearragemet of the Terms) Let i=0 a i be a absolutely coverget series. Let f : N N be a bijectio. Let b i = a f(i). The i=0 b i is also absolutely coverget ad its sum is equal to i=0 a i Proof: Let s ad t be the partial sums of the series i=0 a i ad i=0 b i respectively. Let a = i=0 a i. We will show that lim t = a as well. Let ɛ > 0. Sice lim m s m = a, there is a N 1 such that s m a < ɛ/2 for all m > N 1. Sice i=0 a i coverges absolutely, there is a N 2 such that for all m > N 2, k=i a i < ɛ/2. Choose a m > max(n 1, N 2 ). Let N be such that {0,..., m} {f(0),..., f(n)}. Now for > N, we have, t a t s m + s m a < t s m + ɛ/2 = (b 0 +... + b ) (a 0 +... + a m ) +ɛ/2 = (a f(0) +...+a f() ) (a 0 +...+a m ) +ɛ/2 k=m+1 a k +ɛ/2 < ɛ/2 + ɛ/2 = ɛ. We ca also partitio a absolutely coverget series i two (or more) pieces:

88 CHAPTER 7. SERIES Theorem 7.3.3 Let us partitio the terms (a i ) i of a absolutely coverget series i=0 a i i two disjoit ad ifiite subsets (b i ) i ad (c i ) i. The i=0 b i ad i=0 c i are absolutely coverget series ad i=0 a i = i=0 b i + i=0 c i. Proof: Let α, β ad γ be the partial sums of the three series i this order. Give, we ca fid m ad p such that every term of the partial sum α appears i oe of the partial sums β m ad γ p. Thus 0 β m + γ p α i=+1 a i, so that lim (β m + γ p α ) = 0. Corollary 7.3.4 Let i a i be absolutely coverget. Let (b i ) i be the positive terms ad (c i ) i the egative terms of (a i ) i. The i a i = i b i + i c i. Exercises. i. Suppose i a i coverge absolutely. Show that i a i i a i. (First Solutio: Let R = i a i. The i a i B(0, R). Sice B(0, R) is closed (why?), the limit i a i is still i B(0, R). Secod Solutio: Let x = i a i. By Exercise v, page 63, i a i = lim x = lim x lim i a i = i a i.) ii. Show that i=0 a i coverges absolutely, the i=0 a2 i coverges (absolutely) as well. (See also Exercise vi, page 89). 7.4 Alteratig Series A series of the form i=0 ( 1)i a i where a i 0 is called alteratig series. Theorem 7.4.1 (Alteratig Series Test) Let (a i ) i be a decreasig sequece such that lim i a i = 0. The i=0 ( 1)i a i coverges. Furthermore, 2+1 2 ( 1) i a i ( 1) i a i ( 1) i a i. for all. i=0 i=0 Proof: Let s = a 0 a 1 +... + ( 1) a. We cosider the sequeces (s 2 ) ad (s 2+1 ) of eve ad odd partial sums. Sice s 2+1 = (a 0 a 1 ) +... + (a 2 a 2+1 ), the sequece (s 2+1 ) is icreasig. Sice s 2 = a 0 (a 1 a 2 )... (a 2 1 a 2 ), the sequece (s 2 ) is decreasig. Sice s 2+1 s 2 = a 2+1 0, s 2+1 s 2. By the Mootoe Covergece Theorem (6.7.2), both the series (s 2 ) ad (s 2+1 ) coverge, say to a ad b respectively. The a b = lim s 2 lim s 2+1 = lim (s 2 s 2+1 ) = lim a 2+1 = 0, so that a = b. Hece the sequece (s ) coverges to a as well (see Exercise iii, page 67). Corollary 7.4.2 i=1 ( 1)i /i coverges to a umber i (0, 1). i=0

7.5. CRITERIA FOR CONVERGENCE 89 Example. [A, page 36] We ow give a example (ecessarily o-absolutely coverget, see Theorem 7.3.2) of a coverget series that, whe the terms are permuted, coverges to a differet umber. We cosider series i=1 ( 1)i /i. Let S be its sum. Multiply the series by 1/2 ad compute S + S/2: 3S/2 = S + S/2 = ( 1) i /i + ( 1) i /2i i=1 ad it is easy to see that the partial sums of the right had side are exactly the partial sums of the sequece 1 + 1/3 1/2 + 1/5 + 1/7 1/4 + 1/9 +... = 1 + i=1 i=1 ( 1 4i 1 1 2i + 1 ), 4i + 1 which is a rearragemet of the alteratig series i=1 ( 1)i /i. Sice this sum is ozero, we see that a differet rearragemet of the terms of a series may chage the sum. Exercises. i. Show that i=1 ( 1)i / i coverges. ii. Prove the Alteratig Series Test by showig that the sequece of partial sums is a Cauchy sequece. iii. Show that 1 (2+1)(2+3) 1 (2+1)(2+3) coverges. (Hit: Note that 1 2+1 1 2+2 = iv. Prove the Alteratig Series Test by usig the Nested Itervals Property (Theorem 3.1.12). v. Fid divergig series i=0 x i ad i=0 y i such that i=0 x iy i coverges. vi. Fid a covergig series i=0 x such that i=0 x2 i 7.5 Criteria for Covergece We first geeralize Theorem 7.1.2: is diverget. Propositio 7.5.1 Let (x i ) i be a sequece such that for some r (0, 1), x i+1 r x i evetually. The i=0 x i coverges absolutely. Proof: We may assume that x i R 0 ad that the coditio x i+1 r x i holds for all i. By iductio x i r i x 0. Thus 0 i=0 x i x 0 i=0 ri. Sice the latter sequece coverges, i=0 x i coverges as well (Corollary 7.2.5).

90 CHAPTER 7. SERIES Corollary 7.5.2 (Ratio Test, d Alembert) Let (x i ) i be a ozero sequece i a Baach space such that lim x i+1/x i exists ad is < 1. The i=0 x i coverges absolutely. Proof: We may assume that x i > 0 for all i. Let s be the limit of the sequece (x i+1 /x i ) i. By hypothesis 0 s < 1. Let ɛ = 1 s 2. The ɛ > 0 ad there is a N such that for all i > N, s x i+1 x i < ɛ. Let r = s + ɛ. The x i+1 x i < r ad the corollary follows from Propositio 7.5.1. Examples. i. The series i=0 z /! coverges for all z i all Baach spaces. ii. Cosider the series 1 + αz + α(α 1) z 2 +... 2! α(α 1)... (α + 1) z +...! where α C. The series coverges for z < 1 accordig to d Alembert (Corollary 7.5.2). Corollary 7.5.3 (d Alembert) If lim i a i+1 /a i < 1, the i=0 a i coverges absolutely. If lim i a i+1 /a i > 1, the i=0 a i diverges. Proof: As the proof of Corollary 7.5.2, usig Lemma 6.10.1. Example. Let { 1/i 2 i is eve a i = 2/(i + 1) 2 i is odd The lim i a i+1 /a i = 2 > 1 ad lim i a i+1 /a i = 1/2 < 1. But i=0 a i coverges. Corollary 7.5.4 For ay z C, the series coverge absolutely. exp(z) = i=0 zi /i! si(z) = i=0 ( 1)i z 2i+1 /(2i + 1)! cos(z) = i=0 ( 1)i z 2i /(2i)! These series the give rise to fuctios from C ito C. expoetiatio, sie ad cosie respectively. They are called Theorem 7.5.5 (Root Test, Cauchy) If lim i a i 1/i < 1, the i=0 a i coverges absolutely. If lim i a i 1/i > 1, the i=0 a i diverges.

7.5. CRITERIA FOR CONVERGENCE 91 Proof: Let r = lim i a i 1/i < 1. Let ɛ = 1 r 2. Note that r + ɛ < 1. By Lemma 6.10.1 there is a N such that for all i > N, a i 1/i < r + ɛ, i.e. a i < (r + ɛ) i. Now our theorem follows from theorems 7.2.5 ad 7.1.2. Corollary 7.5.6 If R = 1/lim i a i 1/i, the i=0 a iz i coverges absolutely if z < R ad diverges if z > R. R = 1/lim i a i 1/i is called the radius of covergece of the series i=0 a iz i. A series of the form i=0 a iz i is called a power series. If R is its radius of covergece, such a series gives rise to a fuctio from the ball of ceter 0 ad radius R ito C. Theorem 7.5.7 Let f(z) = i=0 a iz i be a power series with radius of covergece R. The i=1 ia iz i 1 has the same radius of covergece. Theorem 7.5.8 (Abel, form due to Lejeue-Dirichlet) The (complex or real) series i=0 v ɛ coverges if i. The sum v m +... + v is bouded for all m. ii. The series =0 ɛ ɛ +1 coverges. iii. lim ɛ = 0. Proof: We will use Cauchy criterio of covergece (7.2.3). Let ɛ > 0. For m, let V m, = v m +... + v. By hypothesis, there is a A such that V m, < A for all m. Let N 1 be such that for all m > N 1, ɛ m ɛ m+1 +... + ɛ 1 ɛ < ɛ/2a ad let N 2 be such that for all > N 2, ɛ > ɛ/2a. Let N = max{n 1, N 2 }. For m > N we compute as follows: Sice V k,k+1 V k,k = v k+1, we have, v m ɛ m +...+v ɛ = V m,m ɛ m +(V m,m+1 V m,m )ɛ m+1 +... + (V m, V m, 1 )ɛ = V m,m (ɛ m ɛ m+1 ) + V m,m+1 (ɛ m+1 ɛ m+2 ) +... + V m, 1 (ɛ 1 ɛ ) + V m, ɛ. Thus v m ɛ m +... + v ɛ A( ɛ m ɛ m+1 + ɛ m+1 ɛ m+2 +... + ɛ 1 ɛ + ɛ ) < A(ɛ/2A) + A(ɛ/2A) = ɛ. Corollary 7.5.9 The (complex or real) series i=0 v ɛ coverges if i. The sum v m +... + v is bouded for all m. ii. (ɛ ) is positive, odecreasig ad lim ɛ = 0. Proof: We check that the above coditios are met. We oly have to be cocered with coditio (ii): =0 ɛ ɛ +1 = =0 (ɛ ɛ +1 ) = ɛ 0. We ca obtai the alteratig series test (Theorem 7.4.1) as a cosequece: Corollary 7.5.10 (Alteratig Series Test) Let (v i ) i be a decreasig sequece such that lim i v i = 0. The i=0 ( 1)i v i coverges. Proof: Take v = ( 1) i the corollary above.

92 CHAPTER 7. SERIES Exercises. i. Discuss the covergece ad absolute covergece of the alteratig series 1 1/2 α + 1/3 α 1/4 α +... for various values of α Q. ii. Decimal Expasio. Let r R 0. The there are k Z ad a i {0, 1,..., 9} such that r = i=k a i10 i. iii. Let > 0 be a atural umber. Let r R 0. The there are k Z ad a i {0, 1,..., 1} such that r = i=k a i i. iv. Show that 1/3 = i=1 2 2i. v. Let (a i ) i be a positive decreasig sequece, k a iteger greater tha 1. Show that the series i=1 a i ad i=1 ki a k i either both coverge or both diverge. vi. Show that if the ratio test says that a sequece coverges, so does the root test. vii. Show that if a i > 0 ad lim i ia i exists ad is ozero, the i=0 a i diverges. viii. Assume a i > 0 ad lim i i 2 a i exists. Show that i=0 a i coverges. 7.6 Supplemetary Problems i. Show that the sum of the reciprocals of atural umbers whose decimal expasio does ot cotai a 0, i.e. the series 1/1 +... + 1/9 + 1/11 +... + 1/19 + 1/21 +... + 1/29 + 1/31 +... + 1/99 + 1/111 +... + 1/119 + 1/121 +... + 1/129 +... is coverget. ii. Let i=0 a i is give. For each i N, let p i = a i if a i 0 ad assig p i = 0 if a i < 0. I a similar maer, let q i = a i if a i 0 ad q i = 0 otherwise. a. Show that if i=0 a i diverges, the at least oe of i=0 p i or i=0 q i diverges (to ad resp.) b. Show that if i=0 a i coverges coditioally (i.e. ot absolutely), the i=0 p i = ad i=0 q i =. [A, Ex. 2.7.3, page 68] c. Show that if i=0 a i coverges coditioally ad r R, the there is a bijectio f : N N such that i=0 a f(i) = r. [A, Sectio 2.9] iii. Let x i = 1/1 2 + 1/2 2 + 1/3 2 +... + 1/i 2. a. Show that for all itegers i 1, x i 2 1/i < 2. b. Coclude that the series i=1 1/i2 coverges. c. Show that for a iteger i large eough, i 2 2 i. d. Coclude that i=1 1/i2 1 + 1/4 + 1/9 + 1/8 = 107/72.

7.6. SUPPLEMENTARY PROBLEMS 93 iv. Show that the series i=1 v. Show that the series i=1 1 i 2 +i 1 i 2 +2i coverges. Fid its value. coverges. Fid its value. vi. Which of the followig series coverge? Ca you estimate their sum? 7.6.1 Midterm of Math 152 1 i=1 i 2 i 1 i=3 i 2 3i+2 1 i=0 3i 2 i i=0 3i 2 2 i i=0 3i 3 2 i=0 3i +4 i i=0 2i5i i! i=0 ( 1)i i i+1 i=0 ( 1)i ( 1) i i=1 i i=0 ( 1)i 2i i 3 i i 2 i+1 i. Decide the covergece of the series 1 2 2. ii. Decide the covergece of the series 1 2 + 1. iii. Decide the covergece of the series 1 4 6. iv. Suppose that the series a is coverget. Show that lim a = 0. v. Suppose that (a ) is a positive ad decreasig sequece ad that the series a is coverget. Show that lim a = 0. vi. Fid a positive sequece (a ) such that the series a is coverget but that lim a 0. vii. Suppose that series a is absolutely coverget ad that the sequece (b ) is Cauchy. Show that the series a b is absolutely coverget. viii. Let (a ) be a sequece. Suppose that =1 a a +1 coverges. Such a sequece is called of bouded variatio. Show that a sequece of bouded variatio coverges. i!

94 CHAPTER 7. SERIES 7.6.2 Solutios of the Midterm of Math 152 i. Decide the covergece of the series Aswer: Sice for all > 1, 1 2 2. 1 2 2 = 1 2 2 1 = 1 2 ad sice 1 diverges, the series ii. Decide the covergece of the series Aswer: Sice for > 0, 1 2 + 1. 1 2 2 1 2 + 1 1 2 + = 1 2 2 ad sice 1 diverges, the series iii. Decide the covergece of the series Aswer: Sice for > 1, 1 2 +1 1 4 6. diverges as well. diverges as well. 1 4 6 = 1 4 6 1 2 4 4 /2 = 2 ad sice 1 2 coverges, the series 1 4 6 coverges as well. iv. Suppose that the series a is coverget. Show that lim a = 0. Proof: Let s = a 0 +... + a ad s = a. Thus lim s = s. We have lim a = lim (s s 1 ) = s s = 0. v. Suppose that (a ) is a positive ad decreasig sequece ad that the series a is coverget. Show that lim a = 0. Proof: We kow that lim a = 0. Let s = a 0 +... + a ad s = a. Thus lim s = s.

7.6. SUPPLEMENTARY PROBLEMS 95 Sice (a ) is decreasig, a 2 a +1 +... + a 2 = s 2 s. Thus lim a 2 = lim (s 2 s ) = s s = 0. Hece lim 2a 2 = 0. Also 0 < (2 + 1)a 2+1 (2 + 1)a 2 = 2a 2 + a 2. By the above ad the fact that lim a 2 = 0, the right had side coverges to 0. Hece by the squeezig lemma lim (2 + 1)a 2+1 = 0. From the above two paragraphs it follows that lim a = 0. (Remark: As we will see later, the series 1 =2 l diverges. This example shows that the coditios that (a ) is positive ad decreasig ad that lim a = 0 are ot eough for the series a to be coverget.) vi. Fid a positive sequece (a ) such that the series a is coverget but that lim a 0. Solutio: Take a = 1/ 2 if is ot a square ad a = 1/ if is a square. The (a ) does ot coverge as lim ( 2 + 1)a 2 +1 = 0 ad lim 2 a 2 = 1. O the other had a = o square a + a square a = o square 1/2 + 1/2 < 2 1/2 < 4. vii. Suppose that series a is absolutely coverget ad that the sequece (b ) is Cauchy. Show that the series a b is absolutely coverget. Proof: Sice (b ) is Cauchy the sequece ( b ) is bouded. I fact that is all we eed to coclude. Ideed, let B be a upper boud of the sequece ( b ). The a b B a ad sice a coverges, a b coverges as well. viii. Let (a ) be a sequece. Suppose that =1 a a +1 coverges. Such a sequece is called of bouded variatio. Show that a sequece of bouded variatio coverges. Proof: Sice =1 a a +1 coverges, =1 (a a +1 ) coverges as well. Thus the sequece of partial sums whose terms are 1 (a i a i+1 ) = a 1 a i=1 coverges, say to a. Thus (a ) coverges to a 1 a.

96 CHAPTER 7. SERIES

Chapter 8 Supplemetary Topics 8.1 Liouville Numbers A real umber r is called algebraic (over Q) if there is a ozero polyomial p(x) Q[x] (equivaletly p(x) Z[x]) such that p(r) = 0. Otherwise r is called oalgebraic or trascedetal over Q. Cosider the root α R of a polyomial f(x) Z[x]. Set f(x) = i=0 a ix i where a 0. Lemma 8.1.1 If M = max{ a i /a : i = 0,..., 1} the α < 1 + M. Proof: We may assume α 0. Sice 0 = f(α) = a 0 + a 1 α i +... + a α ad a α 0, we have 1 = a 1α 1 +... + a 1 α + a 0 a α = a 1 a α +... + a 0 a α. Now if α 1 + M, the the absolute value of the right had side is less tha 1 M( 1 + M + 1 (1 + M) 2 +... + 1 (1 + M) ) < 1 M( 1 + M + 1 (1 + M) 2 +... + 1 (1 + M) +...) = M 1 1 + M 1 1 1+M a cotradictio. = 1, Assume ow 0 < α Q is a root of f(x) = i=0 a ix i Q[x] with a 0. Let M be as above. Let q N be ay atural umber large eough so that, 1) 1/q < α, 2) f has o root i (α 1/q, α + 1/q), 3) α + 1/q < 1 + M, ad 4) q > i=1 a i (1 + M) i 1. 97

98 CHAPTER 8. SUPPLEMENTARY TOPICS Note that if q 0 satisfies these coditios the ay q > q 0 also satisfies these coditios. Note also that there are exactly two atural umbers such that α p/q < 1/q: if p is the smallest such atural umber the α 1/q < p/q < α < (p + 1)/q < α + 1/q < 1 + M. Let p N be such that α p/q < 1/q. We will show that α p/q > 1/q +1. Clearly f(p/q) = A/q for some A Z. By (2), A 0. We also have, A/q = f(p/q) = f(α) f(p/q) = (α p/q)f (β) for some β betwee p/q ad α. Note that β < max{p/q, α} < 1 + M. Sice f (x) = i=1 ia ix i 1, f (β) = ia i β i 1 a i β i 1 a i (1 + M) i 1 < q. i=1 Thus, sice A 0, i=1 i=1 1/q A/q = α p/q f (β) < α p/q q. Hece α p/q > 1/q +1. Thus if α is a irratioal root of a polyomial of degree, for large eough q, if α p/q < 1/q the α p/q > 1/q +1. It follows that if α is irratioal ad if there are ifiitely may atural umbers q for which α p/q < 1/q +1 (< 1/q) for some atural umber p (that may deped o q), the α is ot the root of a polyomial of degree. If this happes for ifiitely may itegers, the α is ot algebraic. This is what happes for umbers of the form α = i=1 a i 10 i! where a i = 0, 1,..., 9, but the values of a i are ot all 0 after a while. Because the decimal expasio is ot periodic, a Q. We check the secod coditio. Let be ay atural umber ad m > + 1. p/q = m a i i=1. The 10 i! 0 < α p/q = i=m+1 = a i 10 i! i=m+1 9 10 i! = 9 10 (m+1)! 10 10 (m+1)! < 1 (10 m! ) m = 1/qm. Let q = q m = 10 m! ad i=0 1 10 i = Thus α caot be a algebraic umber. Such a umber is called a Liouville umber.

8.1. LIOUVILLE NUMBERS 99 Theorem 8.1.2 (Liouville) If α = i=1 a i 10 i! where a i = 0, 1,..., 9, but the values of a i are ot all 0 after a while, the α is ot a algebraic umber.

100 CHAPTER 8. SUPPLEMENTARY TOPICS

Chapter 9 Covergece of Fuctios 9.1 Poitwise Covergece of a Sequece of Fuctios Let X be a set ad (Y, d) be a metric space. Let (f ) be a sequece of fuctios from X ito Y. Assume that for all x X, the sequece (f (x)) of the metric space Y coverges. Let f(x) be this limit: f(x) := lim f (x). The the rule x f(x) defies a fuctio f from X to Y. we say that the fuctio f is the poitwise limit of the sequece (f ). I this case we write or sometimes f = lim f f p = lim f to precise that the covergece is poitwise (soo we will defie other kids of covergece). Clearly, if f = g for all, the lim f p = g. Exercises ad Examples. i. Let X = R, Y = R with the usual metric ad f : R R be defied by f (x) = x/. The the zero fuctio is the poitwise limit of the sequece (f ). ii. Let X = Y = (0, 1) ad let f (x) = 1/ for all N\{0}. Ufortuately, the poitwise limit of the sequece (f ) does ot exist because 0 Y. This is of course oly a mior problem. It is eough to chage Y with [0, 1), or eve with R, i which case the poitwise limit is the costat zero fuctio. 101

102 CHAPTER 9. CONVERGENCE OF FUNCTIONS iii. Let X = Y = [0, 1] ad f (x) = x. The the fuctio f defied by f(x) = { 0 if x 1 1 if x = 1 is the limit of the sequece (f ). iv. Let X = Y = R ad let f : R R be defied by f (x) = x /!. The the zero fuctio is the poitwise limit of the sequece (f ). v. Let X = Y = [0, 1]. Let { 2(x k f (x) = k 2 ) if 2 x k+1 2( k+1 k 2 x) if 2 x k+1 2 ad k 2N 2 ad k N \ 2N Draw the graph of f. What is its legth? Show that the zero fuctio is the poitwise limit of the sequece (f ). What is the legth of the limit fuctio? vi. Let X = Y = C (or R). defie f (z) = i=0 zi /i! g (z) = i=0 ( 1)i z 2i+1 /(2i + 1)! h (z) = i=0 ( 1)i z 2i /(2i)! The lim f = exp lim g = si lim f = cos. Theorem 9.1.1 Let Y = R k p with the usual metric. Let lim f = f ad p lim g = g. The i. lim (f + g ) = p f + g ii. For all r R, lim rf p = rf. iii. If Y = R, the lim (f g ) p = fg. iv. If Y = R ad f(x) 0 for all x R, the lim 1/f p = 1/f. Proof: Trivial Note that the first two parts of the theorem above says that the set of poitwise coverget sequeces of fuctios from X ito the Euclidea space R is a real vector space.

9.2. UNIFORM CONVERGENCE OF A SEQUENCE OF FUNCTIONS 103 9.2 Uiform Covergece of a Sequece of Fuctios As we defied the covergece of a series of umbers i=0 a i, we ca also defie the covergece of a series of fuctios i=0 f i as the poitwise limit of the sequece of fuctios i=0 f i. Here f ad each f i are fuctios from a set X ito R k (fixed k) or eve ito ay Baach space (a complete ormed vector space). Thus, we say that the series i=0 f i coverges (poitwise) to f if for all x X, the series i=0 f i(x) coverges to f(x). Let us rewrite the defiitio of poitwise covergece: Let X be a set ad (Y, d) be a metric space. Let (f ) be a sequece of fuctios from X ito Y ad f a fuctio from X ito Y. Assume that f = p lim f. This meas that for all x X, f(x) = lim f (x); i other words, for all x X ad all ɛ > 0, there is a N such that d(f (x), f(x)) < ɛ wheever > N. Here, N depeds o ɛ, but also o x. That is why, oe sometimes writes N ɛ,x istead of N. Whe N is idepedet of x, we say that the covergece is uiform. More precisely, the sequece (f ) coverges uiformly to f if for all ɛ > 0 there is a N (that depeds oly o ɛ) such that for all x X ad > N, d(f (x), f(x)) < ɛ. Whe that is the case, oe writes f = u lim f. u Clearly, if f = g for all, the lim f = g. It is also clear that the uiform covergece is stroger tha the poitwise covergece, i other words, if f = u lim f the f = p lim f. But the coverse fails as the followig example shows. Examples. i. Let X = ( 1, 1) ad Y = R with the usual metric. Cosider the sequece (f ) of fuctios from X ito Y defied by f (x) = 1 x 1 x. It is clear that the poitwise limit of the sequece (f ) is the fuctio f defied by f(x) = 1 1 x. Therefore the uiform limit of this sequece - if it exists at all - should also be the same fuctio. Assume lim f u = f. The for all ɛ > 0 there is a N such that x 1 x = f (x) f(x) < ɛ for all > N ad x X. But whatever is, the fuctio x 1 x is ubouded o X, so that there caot be such a N (idepedet of all x).

104 CHAPTER 9. CONVERGENCE OF FUNCTIONS ii. Let α (0, 1). Let X = ( 1, α) ad Y = R with the usual metric. Cosider the sequece (f ) of fuctios from X ito Y defied by f (x) = 1 x 1 x (as above). It is clear that the poitwise limit of the sequece (f ) is the fuctio f defied by f(x) = 1 1 x. Therefore the uiform limit of this sequece if it exists at all should also be the same fuctio. Ideed it is. Let us prove that. Let ɛ > 0. Sice 0 < α < 1, lim α = 0, so that there is a N such that α < ɛ(1 α) for all > N. (Note that N depeds oly o ɛ). Now, for all > N ad for all x X, we have, Thus lim f u = f. f (x) f(x) = x 1 x < x 1 α < α 1 α < ɛ. iii. Let f : [ 1, 1] R be defied by f (x) = x 1+ 1 2+1. Show that lim f (x) u = x. The secod example above shows that the domai X is importat for the uiform covergece. For this reaso, oe says that the sequece (f ) coverges uiformly o X. Theorem 9.2.1 Let X be ay set ad Y = R with the usual metric. Let f, g, f, g ( N) be fuctios from X ito Y. Let lim f u = f ad lim g u = g. The i. lim (f + g ) u = f + g ii. For all r R, lim rf u = rf. Proof: Trivial. It is false that if (f ) ad (g ) are sequeces fuctios that coverge uiformly, the the sequece (f g ) coverges uiformly. Suppose the sequece (f ) of fuctios from a set ito a metric space coverges uiformly (hece poitwise) to f. Let T = sup x {d(f (x), f(x))} R 0 {ifty}. Take ɛ = 1 i the defiitio to fid a N such that for all > N ad x, d(f (x), f(x)) < 1, i.e. for all > N, T < 1. It follows that if the sequece (f ) of fuctios from a set ito a metric space coverges uiformly (hece poitwise) to f, the the sequece T = sup x {d(f (x), f(x))} R 0 { } is evetually bouded. Lemma 9.2.2 Suppose the sequece (f ) of fuctios from a set ito a metric space coverges poitwise to f. Let T = sup x {d(f (x), f(x))} R 0 {ifty}. The lim f = f if ad oly if lim T = 0.

9.2. UNIFORM CONVERGENCE OF A SEQUENCE OF FUNCTIONS 105 Proof: Trivial. Cosider the set of fuctios F from a set X ito a metric space (Y, d). For f, g F, let d (f, g) = mi{sup x X {d(f(x), g(x))}, 1}. Lemma 9.2.3 (F, d ) is a metric space ad a sequece (f ) from F coverges to a fuctio i f F if ad oly if lim d (f, g) = 0. Proof: Easy. If we cosider the set B of bouded fuctios from X ito the metric space (Y, d), the i the lemma above we ca take d (f, g) = sup x X {d(f(x), g(x))} as the ext lemma will show. If (X, d) is a ormed vector space (V, ), the f = sup{ f(x) : x X} is a orm o the vector space of bouded fuctios from the set X ito V, as it ca easily be checked. Lemma 9.2.4 Suppose the sequece (f ) of fuctios from a set X ito a metric space (Y, d) coverges uiformly to f. Assume that the fuctios f are evetually bouded. The f is bouded. Proof: Take ɛ = 1 i the defiitio of the uiform covergece. Let N be such that for all > N ad all x, d(f (x), f(x)) < 1. Let > N be such that f is bouded. Let b Y ad R R be such that f (x) B(b, R) for all x R. The for all x R, d(f(x), b) d(f(x), f (x)) + d(f (x), b) < 1 + R. Let F be the set fuctios from X ito the metric space (Y, d). Set d (f, g) = sup x X {d(f(x), g(x)), 1}. The d is a distace o F ad a sequece (f ) of fuctios coverge uiformly if ad oly if the sequece coverges i this metric space. The followig criterio à la Cauchy shows the uiform covergece of a sequece without explicitly calculatig the limit: Theorem 9.2.5 (Cauchy Criterio for Uiform Covergece) A sequece (f ) of fuctios from a set X ito a complete metric space (Y, d) coverges uiformly if ad oly if for ay ɛ > 0 there is a N ɛ such that for all, m > N ad for all x X, d(f (x), f m (x)) < ɛ. Proof: By hypothesis, for all x X, (f (x)) is a Cauchy sequece, hece it has a limit. Let f(x) deote this limit. We will prove that lim f u = f. Let ɛ > 0. By hypothesis there is a N, idepedet of x, such that for all, m > N, d(f (x), f m (x)) < ɛ/2. Let x X be give. Sice the sequece (f (x)) coverges to f(x), there is a N x (that may deped o x) such that for all m > N x, d(f m (x), f(x)) < ɛ/2. Let > N be ay atural umber. Let x X be ay. We will show that d(f (x), f(x)) < ɛ, provig that (f ) coverges to f uiformly. Let m x > max{n, N x } be fixed. The d(f (x), f(x)) d(f (x), f mx (x))+d(f mx (x), f(x)) < ɛ/2 + ɛ/2 = ɛ.

106 CHAPTER 9. CONVERGENCE OF FUNCTIONS Theorem 9.2.6 Let (f ) be a sequece of fuctios defied o a subset A of R that coverges uiformly to a fuctio f. If each f is cotiuous at a poit a A, the f is cotiuous at a. Proof: Let ɛ > 0. We wat to fid a δ > 0 such that for x A ad x a < δ the f(x) f(a) < ɛ. For all ad x A we have f(x) f(a) f(x) f (x) + f (x) f (a) + f (a) f(a). We will make each term of the right had side less tha ɛ/3, for large eough ad x close eough to a. = Sice lim f u f, there is a N such that for > N ad x A, f(x) f (x) < ɛ/3. Thus we also have f (a) f(a) < ɛ/3. (Here we use the uiform cotiuity). Pick such a > N. Sice f is cotiuous at a, there is a δ > 0 such that for x A ad x a < δ, f (x) f (a) < ɛ/3. Example. The fuctio f (x) = (1 x ) defied o [ 1, 1] coverges poitwise (but ot uiformly) ad its limit is ot cotiuous at a = 0. This example shows also that poitwise covergece o a compact set does ot imply the uiform covergece. Exercises. i. Show that the fuctios f (x) = x x+1 coverge uiformly o [0, 1]. ii. Determie the poitwise or the uiform covergece of the followig fuctios: i. f (x) = k=0 xk o X = ( 1, 1). ii. f (x) = k=0 xk o X = ( 1/2, 1/2). iii. f (x) = x o X = (0, 1). iv. f (x) = x o X = [ α, α] where α (0, 1). v. f (x) = x 1 x vi. f (x) = x 1 x o X = (0, 1). o X = (0, 1 α) for ay α (0, 1). iii. Let f (x) = 1 1+x a. Fid the set A = {x R : (f (x)) coverges}. For x A, let f(x) = lim f (x). b. What is f? c. Is the covergece uiform? Justify your aswer. d. Discuss the uiform covergece of (f ) i the (ope or closed) itervals cotaied i A.

9.3. UNIFORM CONVERGENCE OF A SERIES OF FUNCTIONS 107 iv. Let f : R R be defied as follows: { x 2 if x f (x) = 2 if x Does (f ) coverge uiformly? v. Let f : R R be defied as follows: { x 2 if x f (x) = 2 if x Does (f ) coverge uiformly? vi. Is part iv of Theorem 9.1.1 true for uiform covergece? vii. Let (f ) be a sequece of fuctios from X ito a metric space Y. Let A, B X. Assume that the sequece (f ) coverges uiformly o A ad also o B. Show that (f ) coverges uiformly o A B. viii. Study the covergece of f (x) = ix. Study the covergece of f (x) = si(x)/. x. Study the covergece of f (x) = x 1+x. x 1+x 2 o itervals of R. xi. Assume (f ) coverges uiformly to a fuctio f o a compact set K R. Let g : K R be a cotiuous fuctio satisfyig g(x) 0. Show that (f /g) coverges uiformly to f/g o K. xii. Give a example of fuctios (f ) ad (g ) that coverge uiformly, but whose product sequece (f g ) does ot coverge uiformly. 9.3 Uiform Covergece of a Series of Fuctios The poitwise or the uiform covergece of a series N f (x) of fuctios is defied i the expected way as the poitwise or the uiform covergece of the partial sums i=0 f i. We defie it more precisely Let f ( N) ad f be fuctios defied o a subset A of R. The series of i=0 f is said to coverge uiformly to f if the sequece of partial sums (s ) where s = f 1 +... + f coverges uiformly to f. Theorem 9.3.1 Assume each f i is cotiuous ad i=0 f i coverges uiformly to f o A. The f is cotiuous o A. Proof: Follows directly from Theorem 9.2.6.

108 CHAPTER 9. CONVERGENCE OF FUNCTIONS Theorem 9.3.2 (Cauchy s Criterio for Uiform Covergece of Series) The series i=0 f coverges uiformly o A R if ad oly if for every ɛ > 0 there exists a N N such that for all N < m <, i=m+1 f < ɛ. Proof: Exercise. Theorem 9.3.3 (Weierstrass M-Test) Let (f i ) i be a sequece of fuctios from a set X ito C. Suppose that f i (x) M i for all x X ad for all i N. If the series i=0 M i coverges, the the sequece i=0 f i coverges uiformly o X. Proof: We will use Theorem 9.3.2. Let ɛ > 0. Sice the series i=0 M i coverges, by Cauchy Criterio for series, there is a N N such that for all N < m <, i=m+1 M i < ɛ. Thus for all N < m < ad all x X, i=m+1 f i(x) i=m+1 M i < ɛ. Theorem 9.3.4 If a power series i=0 a x coverges absolutely at a poit x 0 the it coverges uiformly o the closed ball B(0, x 0 ). Proof: Follows directly from Weierstrass M-Test (Theorem 9.3.3). What happes at the ed poits? Here is the aswer: Theorem 9.3.5 (Abel s Theorem) Let the power series =0 a x coverge at the poit R > 0 (resp. at the poit R < 0). The the series coverges uiformly o the iterval [0, R] (resp. o the iterval [ R, 0]). To prove this theorem, we eed the followig lemma: Lemma 9.3.6 (Abel s Lemma) Let (b ) satisfy b 1 b 2... 0. Let i=0 a i be a series whose partial sums are bouded, by A say. The for all N, a 1 b 1 +... a b Ab 1. Proof: To be prove. Proof of Abel s Theorem. To be prove. Corollary 9.3.7 If a power series coverges poitwise o the set A R, the it coverges uiformly o ay compact set K A. Proof: By Theorem 9.3.4 ad Abel s Theorem (9.3.5). Corollary 9.3.8 A power series is cotiuous at every poit at which it is coverget. Exercises. i. Show that if i=0 f i coverges uiformly, the (f ) coverges uiformly to the zero fuctio. ii. Show that the series i=1 cos(2i x)/2 i coverges uiformly. Therefore the limit fuctio is cotiuous. iii. Show that =1 x / 2 is cotiuous o [ 1, 1].

9.4. UNIFORM CONVERGENCE AND METRIC 109 9.4 Uiform Covergece ad Metric Let X be a set ad (V, ) a ormed vector space. Cosider the ormed vector space B(X, V ) of fuctio from X ito Y which are bouded. (See Exercise iii, page 53). As ay ormed vector space, B(X, V ) is a metric space (Lemma 5.3.1) ad so we may speak about the covergece of a sequece of B(X, V ). Lemma 9.4.1 Let (f ) be a sequece from B(X, V ) ad f B(X, V ). If u lim f = f i the metric of B(X, V ), the lim f = f. Coversely if (f ) is a sequece from B(X, V ) that coverges uiformly to a fuctio f : X V, the f B(X, V ) ad lim f = f i the metric of B(X, V ). Proof: We first show that if (f ) is a sequece from B(X, V ) that coverges uiformly to a fuctio f : X V, the f B(X, V ) Let ɛ > 0. Sice lim f = f i the metric of B(X, V ), there is a N such that for all > N, f (x) f(x) < ɛ, i.e., by the very defiitio of the metric o B(X, V ), sup{ f (x) f(x) : x X} < ɛ. This meas that f (x) f(x) < ɛ for all x X, provig the uiform covergece. Let us ow show the coverse. TO BE COMPLETED 9.5 Limits of Fuctios Let (X, d) be a metric space. Let A X be a subset. A poit x X is called a limit poit of A if ay ope subset of X cotaiig x cotais a elemet of A differet from x. Let (X, d) ad (Y, d ) be two metric spaces, A a subset of X ad f : A Y a fuctio. Let a X be a limit poit of A ad b Y. We say that the limit of f(x) whe x goes to a is b if for all ɛ > 0 there is a δ > 0 such that for all x X, if x a ad d(x, a) < δ, the d (f(x), b) < ɛ. We the write lim f(x) = b. x a If X = R or C ad A is a ubouded subset of X, we would like to be able to speak about the limit at ifiity. The defiitios i R or i C are differet. Let A be a subset of R without upper boud. Let (Y, d ) be a metric space. Let f : A Y be a fuctio. Let b Y. We say that the limit of f(x) whe x goes to is b if for all ɛ > 0 there is a N > 0 such that for all x A, if x > N the d (f(x), b) < ɛ. We the write lim f(x) = b. x The limit whe x goes to is defied similarly. Note that whe A = N, we obtai the defiitio of the limit of a sequece i the metric space Y.

110 CHAPTER 9. CONVERGENCE OF FUNCTIONS Now let A be a ubouded subset of C. Let (Y, d ) be a metric space. Let f : A Y be a fuctio. Let b Y. We say that the limit of f(z) whe z goes to is b if for all ɛ > 0 there is a N > 0 such that for all z A, if z > N the d (f(z), b) < ɛ. We the write Exercises. lim f(z) = b. z i. Show that lim x 5 (x 2 3x + 5) = 15 by usig the defiitio of limits. ii. Let X = R, A = R >0, Y = R ad f : A Y be defied by f(x) = 1/x. Show that lim x 0 f(x) does ot exist. iii. Let X = A = Y = R with the usual metric. Let f : A Y be defied as follows: { 1 if x < 0 f(x) = 1 if x 0 Show that lim x 0 f(x) does ot exist. iv. Let A = (0, 1), X = [0, 1], Y = R with the usual metric. Let f : A Y be defied as follows: { 0 if x Q f(x) = 1 if x Q Show that lim x 1 f(x) does ot exist. Theorem 9.5.1 The limit lim x a f(x) of a fuctio f is uique wheever it exists. Theorem 9.5.2 The relatioship of the limits with the fuctioal operatios. 9.6 Covergece of a Family of Fuctios If f(x, y) is a fuctio that depeds o two parameters, sometimes we may wat to view the parameter x as a idex set. I this case we may write f x (y) istead of f(x, y). Suppose f is such a fuctio goig from A Y ito Z where A X ad X ad Z are metric spaces. Suppose that x 0 X is a limit poit of A. Let f : Y Z be a fuctio. The we say that f is the poitwise limit p of the family (f x (y)) x A whe x approaches x 0 ad we write lim x x0 f x = f if for all y Y, lim x x0 f x (y) = f(y). This makes sese because for fixed y Y, x f x (y) is map from the metric space X ito the metric space Z. More formally this meas the followig: For all y Y ad ɛ > 0, there is a δ = δ ɛ,y > 0 such that if x B(x 0, δ) \ {x 0 } the f x (y) B(f(y), ɛ). We emphasize the fact that δ depeds o y. If we ca choose δ idepedet of y, the we say that the covergece is uiform.

9.7. SUPPLEMENTARY TOPICS 111 If x varies over R or C, we would like to be able to make x to. This is p easy to do. Assume first A R has o upper boud. Defie lim x f x = f if lim x f x (y) = f(y) for all y Y, i.e. if for all y Y ad ɛ > 0 there is a N such that for all x A, if x > N = N ɛ,y the d (f x (y), f(y)) < ɛ. Limit whe x goes to is defied similarly. If A C is a ubouded subset, the p we defie lim x f x = f if limx f x (y) = f(y) for all y Y, i.e. if for all y Y ad ɛ > 0 there is a N = N ɛ,y such that for all x A, if x > N the d (f x (y), f(y)) < ɛ. The covergece is said to be uiform if N ca be chose idepedetly of y. Theorem 9.6.1 The uiform limit of cotiuous fuctios is cotiuous. NOTE: I should thik about defiig these covergece defiitios oce for all, without specific cases. 9.7 Supplemetary Topics Theorem 9.7.1 (Weierstrass) Let (f i ) i be a sequece of fuctios from a ubouded subset X of R ito C. Suppose that f i (x) M i for all x X ad for all i N. Assume that the series i=0 M i coverges. The the sequece i=0 f i(x) coverges uiformly o X. Assume also that lim x f i (x) = l i for all i. The i=0 l i coverges absolutely ad lim x i=0 f i(x) = i=0 l i. Proof: The first part is just Weierstrass M-Test. We prove the last part. Clearly l i M i, so that the absolute covergece occurs. Let ɛ > 0. Let be such that i=+1 M i < ɛ/4. Choose x 0 such that for x > x 0 ad i = 0, 1,...,, f i (x) l i < ɛ/2( + 1). We have i=0 f i(x) i=0 l i = i=0 (f i(x) l i ) i=0 f i(x) l i = i=0 f i(x) l i + i=+1 f i(x) l i i=0 f i(x) l i + 2 i=+1 M i ɛ/2( + 1) + 2ɛ/4 = ɛ. < i=0 This proves the theorem. Example. i. Let N >0 ad z C. We have (1 + z/) = i=0 ( i ) (z/) i. Set f 1 () = 1 + z,

112 CHAPTER 9. CONVERGENCE OF FUNCTIONS ad for i > 1, ( ) (z/) f i () = i i = ( 1)...( i+1) i! (z/) i if i 0 if i > Sice, for i, ( f i () = zi 1 1 ) (... 1 i 1 ), i! we have lim f i() = z i /i! =: M i ad f i () M i ad i M i coverges by Example i, page 90. Thus the coditios are realized ad so lim (1 + z/) exists ad is equal to i=0 zi /i!.

Chapter 10 Topological Spaces 10.1 Defiitio ad Examples A set X together with a set τ of subsets of X is called a topological space if τ satisfies the three coditios of Propositio 5.4.3: i., X τ. ii. If U 1,..., U τ the U 1... U τ. iii. If U i τ for all i the i U i τ. The elemets of τ are called the ope subsets of the topological space X. Every metric space iduces a topological space as stated i Propositio 5.4.3. But the coverse is false as the followig example shows: Let X be ay set. Let ad X be the oly ope subsets, i.e. let τ = {, X}. The X becomes a topological space. If X 2 the there is o metric o X that gives this topology. The topology o X is called the coarsest topology or the weakest topology o X. A topological space whose topology is iduced by a metric is called a metrisable topology. Sometimes two differet metrics o the same set give the same topologies, i.e. the same ope sets. I this case we say that the two metrics are equivalet. Exercises. i. Show that the followig metrics o R 2 are equivalet. ii. Show that every metric d o a set X is equivalet to the metric d defied by d (x, y) = max{d(x, y), 1). iii. Show that for equivalet metrics, Cauchy sequeces, coverget sequeces ad limits do ot chage. Discrete Topology. I the topology iduced by the discrete metric, every subset is ope. This is called the discrete or the fiest topology o X. 113

114 CHAPTER 10. TOPOLOGICAL SPACES Iduced Topology. Let X be a topological space ad Y X. By lettig the ope subsets of Y to be the traces of ope subsets of X o Y, we defie a topology o Y. Thus a ope subset of Y is of the form U Y where U is a ope subset of X. This topology o Y is called the restricted or the iduced topology. Note that the ope subsets of Y ad X are differet. So, we must be careful whe speakig about the ope subsets of Y. For this reaso, we may say that a subset of Y is X-ope or Y -ope. But if Y is ope i X, the subsets of Y which Y -ope are exactly X-ope subsets of Y. If the topology o X is iduced by a metric ad Y X, the the restricted topology o Y is iduced by the restricted metric o Y. Order Topology. TO BE DEFINED. Topology Geerated. Let X be a set. Let (τ i ) i be a family of topologies o X, i.e. each τ i is the set of ope subsets of a topology o X. The i τ i defies a topology o X. A ope subset of i τ i is a subset of X that is ope i all these topologies. Let be a set of subsets of a set X. Cosider the itersectio of all the topologies that cotai. By the previous item this is a topology. It is the smallest topology that cotais the subsets of as ope sets. We will deote this topology by ad call it the topology geerated by. We ca describe the topology above more precisely. Cosider the subsets which are arbitrary uios of fiite itersectios of elemets from. These subsets form the ope subsets of a topology ad this topology is obviously the smallest topology that cotais all the elemets of. Examples. Let X be a set. The topology geerated by cosists of ad X. If A X, the topology geerated by {A} is {, A, X}. If A, B X, the topology geerated by {A, B} is {, A B, A, B, A B, X}. The topology geerated by the sigleto sets is the set of all subsets of X. Exteded Real Lie Exercises. i. Show that i a metric space a fiite subset is closed. ii. Let X be ay set. Show that the cofiite subsets of X defie a topology o X. Show that this topology is ot metrisable if X is ifiite. iii. Show that a pseudometric o a set defies a topology i a atural way. (d : X X R 0 is a pseudometric if d(x, x) = 0, d(x, y) = d(y, x) ad d(x, y) d(x, z) + d(z, y).

10.2. CLOSED SUBSETS 115 iv. The topologies defied o R by the metrics d p ad d are all equal, i.e. ay ope subset with respect to ay of them is also ope i the other oe. v. Let X be a set ad A, B, C be three subsets of X. Show that the topology geerated by {A, B, C} has at most??? ope subsets. vi. Let X = R ad the set of ope ad bouded itervals of R. Show that the topology geerated by is the topology iduced by the usual metric. vii. Let X = R ad the set of itervals of the form [a, b) where a, b R. Is the topology geerated by metrisable? viii. Let X = R ad the set of itervals of the form [a, b) where a, b Q. Is the topology geerated by equal to the topology above? ix. Let X be a set. Let d ad d be two distace fuctios o X. Show that the two metrics geerate the same topology if ad oly if ay sequece of X that coverges for oe of the metrics, coverges to the same elemet for the other metric. 10.2 Closed Subsets The complemet of a ope subset of a topological space is called a closed subset. The properties of ope subsets reflect upo closed subsets aturally: Propositio 10.2.1 Let X be a topological space. The i. ad X are closed subsets. ii. A fiite uio of closed subsets is a closed subset. iii. A arbitrary itersectio of closed subsets is closed. 10.3 Iterior Let X be a topological space. Let A X. The the uio A of all the ope subsets of X which are i A is also a ope subset of X which is i A. Thus A is the largest ope subset of X which is i A, i other words it cotais all the ope subsets (i X) of A. The set A is called the iterior of A. Lemma 10.3.1 Let A, B X. The the followig hold: i. A is ope if ad oly if A = A. I particular A = A. ii. If A B the A B. iii. A B = (A B). iv. A B (A B). Proof: i. Clearly, if A is ope the A is the largest ope subset of itself, so A = A. ii. Sice A A B, A is a ope subset of B. Thus A B.

116 CHAPTER 10. TOPOLOGICAL SPACES 10.4 Closure Let X be a topological space. Let A X. The the itersectio A of all the closed subsets of X that cotai A is also a closed subset of X that cotais A. Thus A is the smallest closed subset of X which cotais A. The set A is called the closure of A. Lemma 10.4.1 Let A, B X. The the followig hold: i. A is closed if ad oly if A = A. I particular A = A. ii. If A B the A B. iii. A B A B. iv. A B = A B. Lemma 10.4.2 x A if ad oly if every ope subset cotaiig x itersects A otrivially. Exercises. i. Fid a example where the iclusio of Lemma 10.3.1.iii is strict. ii. Fid a example where the iclusio of Lemma 10.4.1.iii is strict. iii. Show that (A) c = (A c ) ad that (A ) c = A c. iv. Fid the iterior ad the closure of a arbitrary subset i the fiest topology. v. Fid the iterior ad the closure of a arbitrary subset i the coarsest topology. vi. Fid the closure ad the iterior of a arbitrary iterval i the usual topology of R. 10.5 Base of a Topology Let X be a topological space. Let B be a set of ope subsets of X such that every ope subset of X is a uio of ope sets from B. The B is called a base of the topology. Obviously, the set of ope subsets of X form a base of the topological space X. Examples. 1. Assume X is a metric space. The the ope balls of X form a base of the topology. 2. Assume X = R with the Euclidea topology. The the balls of the form B(a, q) such that a Q ad q Q form a base of the topology. Note that this base is coutable. 3. Let be a set of subsets of a set X. The topology geerated by has a base cosistig of sets of the form U 1... U where N ad U i.

10.6. COMPACT SUBSETS 117 Let X be a topological space ad x X. Let B x be a set of ope subsets of X that cotai x such that every ope subset of X that cotais x cotais oe of the of ope sets from B x. The B x is called a base of the topology at the poit x. Assume X is a metric space ad x X. The the ope balls of the form B(x, q) where q Q form a base of the topology at x. Note that this base at x is coutable. 10.6 Compact Subsets A topological space X is called compact if ay ope coverig of X has a fiite subcoverig. A subspace A of X is compact if A is compact with the iduced topology. If X has fiitely may ope sets the ay subset of X is compact. A subset A of a topological space is called compact if A is compact as a topological space, i.e. if ay family of ope subsets of X that covers A has a fiite subcover that covers A. Sigleto subsets of a topological set are certaily compact. Clearly the uio of fiitely may compact subsets of a topological space is compact. Lemma 10.6.1 Let K be a closed subset of a compact topological space X. The K is compact. Proof: Let (U i ) i I be a ope coverig of K. The (U i ) i I {K c } is a ope coverig of X. Hece a fiite umber of (U i ) i I {K c } cover X. Hece a fiite umber of (U i ) i I cover K. Propositio 10.6.2 A compact subset of a metric space is closed ad bouded. Proof: Let K be a compact subset of the metric space (X, d). Let a X be ay subset. The the cocetric ope balls (B(a, )) N cover K. Therefore a fiite umber of them, hece oly oe of them, covers K. This shows that K is bouded. We ow show that K is closed, by showig that its complemet K c is ope. Let a K c. The the cocetric ope subsets (B(a, 1/) c ) N cover K. Hece oly oe of them cover K. If K B(a, 1/) c, the B(a, 1/) K c. Thus K c is ope ad K is closed. Exercises. i. The coverse of the above propositio is false. Fid a couterexample. Theorem 10.6.3 (Heie-Borel Theorem) A closed ad bouded subset of R (with the usual metric) is compact.

118 CHAPTER 10. TOPOLOGICAL SPACES Proof: Assume first that = 1. Let K be a closed ad bouded subset of R. By Lemma 10.6.1, we may assume that K = [a, b] is a closed iterval. Assume [a, b] is ot compact. Let (U i ) i I be a ope coverig of [a, b] without a fiite subcover. We wat to choose sequeces (x ) N ad (y ) N so that for all, i) x x +1 < y +1 y, ii) 0 < y x < (b a)/2, iii) No fiite subcover of (U i ) i I covers [x, y ]. Let x 0 = a ad y 0 = b. Assume x 0, x 1,..., x ad y 0, y 1,..., y are chose so that (i, ii, iii) hold, except of course (i) for (sice x +1 is ot defied yet). We choose x +1 ad y +1 as follows. Oe of the two itervals [x, x +y 2 ] ad [ x +y 2, y ] is ot covered by a fiite subcover of (U i ) i I. If [x, x +y 2 ] is ot covered by a fiite subcover of (U i ) i I, let x +1 = x ad y +1 = x +y 2. Otherwise, [ x +y 2, y ] is ot covered by a fiite subcover of (U i ) i I ad i this case we let x +1 = x +y 2 ad y +1 = y. Clearly (i, ii, iii) hold. By Theorem 3.1.12 (see also Exercise x, page 58), N [x, y ] is a sigleto set, say {c}. Sice c [a, b], there is a i I such that c U i. Let ɛ > 0 be such that [c ɛ, c + ɛ] U i. By (ii) there is a N such that y x < ɛ. The [x, y ] [c ɛ, c + ɛ] U i as oe ca show easily. But this cotradicts (iii), because oly oe of the U i s suffices to cover [x, y ]. The proof for > 1 is similar, istead of the itervals [x, y ] be fid closed cubes (C ) by dividig the previous cube C to 2 equal parts ad choosig C +1 the oe which is ot covered by a fiite subcover of (U i ) i I. Theorem 10.6.4 If (K ) N is a descedig sequece of oempty compact subsets of R, the K. Proof: Assume otherwise. The K c = R, so that (K) c N is a ope coverig of K 0. Thus a fiite subsequece of (K) c N covers K 0, hece K 0 K. c Sice K K 0, this implies that K =, a cotradictio. Exercises. i. Let (K i ) i be a family of oempty compact subsets of a topological space. Show that if a fiite itersectio of the K i s is oempty, the i K i. ii. X be a complete metric space. Let p be a elemet ot i X. Show that the metric of X caot be exteded to a metric o X {p} without p beig isolated. Coclude that if f : R R >0 is a fuctio satisfyig f(x) f(y) x y f(x) + f(y) for all x, y R the f is bouded away from 0, i.e. there is a ɛ > 0 such that f(x) > ɛ for all x R. (Hit: Set d(x, p) = f(x) for x R). iii. Let F R be a oempty closed subset. Let A R. Show that there is a B F such that d(a, B) = if{d(a, P ) : P F }. (Proof: Let d = if{d(a, P ) : P F }. Let ɛ > 0. The B(A, d + ɛ) F is a closed ad

10.7. CONVERGENCE AND LIMIT POINTS 119 bouded subset of R, hece it is compact. The distace fuctio f from the compact subset B(A, d + ɛ) F ito R defied by f(p ) = d(a, P ) is cotiuous, so it attais its ifimum d.) 10.7 Covergece ad Limit Poits Exercises. For each of the topological spaces (X, τ), describe the coverget sequeces ad discuss the uiqueess of their limits. i. τ = (X). ( (X) is the set of all subsets of X, 2 pts.). Aswer: Oly the evetually costat sequeces covergig to that costat. ii. τ = {, X} (2 pts.). Aswer: All sequeces coverge to all elemets. iii. a X is a fixed elemet ad τ is the set of all subsets of X that do ot cotai a, together with X of course. (5 pts.) Aswer: First of all, all sequeces coverge to a. Secod: If a sequece coverges to b a, the the sequece must be evetually the costat b. iv. a X is a fixed elemet ad τ is the set of all subsets of X that cotai a, together with of course. (5 pts.) Aswer: Oly the evetually costat sequeces coverge to a. A sequece coverge to b a if ad oly if the sequece evetually takes oly the two values a ad b. v. τ is the set of all cofiite subsets of X, together with the of course. (6 pts.) Aswer: All the sequeces without more tha oe ifiitely repeatig terms coverge to all elemets. Evetually costat sequeces coverge to the costat. There are o others. vi. Let τ be the topology o R geerated by {[a, b) : a, b R}. Compare this topology with the Euclidea topology. (3 pts.) Is this topology geerated by a metric? (20 pts.) Aswer: Ay ope subset of the Euclidea topology is ope i this topology because (a, b) = =1[a + 1/, b). But of course [0, 1) is ot ope i the usual topology. Assume a metric geerates the topology. Note that [0, ) is ope as it is the uio of ope sets of the form [0, ) for N. Thus the sequece ( 1/) caot coverge to 0. I fact for ay b R, o sequece ca coverge to b from the left. Thus for ay b R there is a ɛ b > 0 such that B(b, ɛ b ) [b, ). Let b 0 be ay poit of R. Let ɛ 0 > 0 be such that B(b 0, ɛ 0 ) [b 0, ). Sice {b 0 } is ot ope, there is b 1 B(b 0, ɛ 0 ) \ {b 0 }.

120 CHAPTER 10. TOPOLOGICAL SPACES Let 0 < ɛ 1 < ɛ 0 /2 be such that B(b 1, ɛ 1 ) [b 1, ) B(b 0, ɛ 0 ). Iductively we ca fid (b ) ad (ɛ ) such that B(b, ɛ ) [b, ) B(b 1, ɛ 1 )\ {b 1 } ad ɛ < ɛ 0 /2. The (b ) is a strictly icreasig coverget sequece, a cotradictio. 10.8 Coected Sets A topological space is called coected if it is ot the uio of two oempty disjoit subsets. TCoclude hus a topological space X is coected if wheever U ad V are disjoit ope subsets of X whose uio is X, oe of the ope subsets must be the emptyset. Otherwise X is called discoected. Note that if X = U V with U V =, the U ad V are also closed. Note also that a sigleto set is coected. Lemma 10.8.1 The uio of two coected subspace with a oempty itersectio is coected. The uio of a chai of coected subsets is coected. Lemma 10.8.2 A ope ad coected subset of a topological space is cotaied i a maximal ope ad coected subset, which is ecessarily closed. Theorem 10.8.3 A subset X of R is coected if ad oly if X is a iterval (all possible kids). Defie totaly discoected. Show Q is totally discoected. Show Z ad Z p are totally discoected with the p-adic metric.

Chapter 11 Cotiuity 11.1 Cotiuity o Metric Spaces Let (X, d) ad (Y, d ) be two metric spaces, f : X Y a fuctio ad a X. We say that f is cotiuous at a if for ay ɛ > 0 there is a δ > 0 such that for all x X if d(x, a) < δ the d (f(x), f(a)) < ɛ. The last coditio says that if x B(a, δ) the f(x) B(f(a), ɛ), i.e. that f(b(a, δ)) B(f(a), ɛ), i.e. that B(a, δ) f 1 (B(f(a), ɛ)). Thus f : X Y is cotiuous at a X if ad oly if a is i the iterior of f 1 (B(f(a), ɛ)) for ay ɛ > 0. It follows that f : X Y is cotiuous at a X if ad oly if a is i the iterior of f 1 (V ) for ay ope subset V of Y cotaiig f(a). We ote this for future referece: Lemma 11.1.1 Let (X, d) ad (Y, d ) be two metric spaces, f : X Y a fuctio ad a X. The f : X Y is cotiuous at a X if ad oly if a is i the iterior of f 1 (V ) for ay ope subset V of Y cotaiig f(a). Note that if a X is a isolated poit of X, i.e. if B(a, δ) = {a} for some δ > 0 the f is always cotiuous at a. A poit of X which is ot a isolated poit is called a limit poit of X. Lemma 11.1.2 Let (X, d) ad (Y, d ) be two metric spaces, f : X Y a fuctio ad a X a limit poit of X. The f is cotiuous at a if ad oly if lim x a f(x) = f(a). Proof: This is immediate. Lemma 11.1.3 Let (X, d) ad (Y, d ) be two metric spaces, f : X Y a fuctio ad a X. The f is cotiuous at a if ad oly if for ay sequece (x ) of X covergig to a, (f(x )) is a sequece of Y covergig to f(y). Proof: Suppose f is cotiuous at a. Let (x ) be a sequece of X covergig to a. Let ɛ > 0. Let δ > 0 be such that f(b(a, δ)) B(f(a), ɛ). Let N be such that for all > N, d(x, a) < δ, i.e. x B(a, δ). Hece f(x ) 121

122 CHAPTER 11. CONTINUITY f(b(a, δ)) B(f(a), ɛ), i.e. d (f(x ), f(a)) < ɛ. This shows that the series (f(x )) coverges to f(a). Coversely, suppose that for ay sequece (x ) of X covergig to a, (f(x ) is a sequece of Y covergig to f(y). Assume that f is ot cotiuous at ɛ. Let ɛ > 0 be such that for every δ > 0, there is a x δ B(a, δ) for which f(x δ ) B(f(a), ɛ). Choose δ = 1/ for N >0 ad let y = x 1/. The (y ) coverges to a because d(y, a) = 1/. But d(f(y ), f(a)) > ɛ ad so the sequece (f(y )) does ot coverge to f(a). This is a cotradictio. Let (X, d) ad (Y, d ) be two metric spaces ad f : X Y a fuctio. We say that f is cotiuous if f is cotiuous at every poit of X. Lemma 11.1.4 Let (X, d) ad (Y, d ) be two metric spaces ad f : X Y a fuctio. The followig coditios are equivalet: i. f is cotiuous ii. The preimage of every ope subset of Y is a ope subset of X. iii. The preimage of ay ope ball i Y is a ope subset of X. iv. The preimage of every closed subset of Y is a closed subset of X. v. For ay coverget sequece (x ) of X, f(lim x ) = lim f(x ). Proof: (i ii) Suppose f is cotiuous. It is eough to show that the preimage of a ope ball of Y is a ope subset of X. Let b Y ad r > 0. Let a f 1 (B(b, r)). The f(a) B(b, r). Let ɛ = r d(f(a), b) > 0. The B(f(a), ɛ) B(b, r). Sice f is cotiuous at a, there is a δ > 0 such that B(a, δ) f 1 (B(f(a), ɛ)) f 1 (B(b, r)). It follows that f 1 (B(b, r)) is ope. (ii i) Suppose ow that the preimage of every ope subset of Y is a ope subset of X. Let a X. Let ɛ > 0. Sice f 1 (B(f(a), ɛ)) is ope ad a is i this set, there is a δ > 0 such that B(a, δ) f 1 (B(f(a), ɛ)). It follows that f is cotiuous at a. The rest of the equivaleces are ow immediate. Note that i the lemmas 11.1.1 ad 11.1.4 above the existece of a metric o X ad Y disappeared, oly topological cocepts are left. This will be the basis for extedig the cocept of cotiuity of a fuctio betwee metric spaces to the cocept of cotiuity of a fuctio betwee topological spaces. We will do this i the ext subsectio. Corollary 11.1.5 Cotiuity of fuctios betwee metric spaces depeds o the topologies geerated by the metrics rather tha o the metrics themselves. I other words cotiuity of fuctios is preserved uder equivalet metrics. Lemma 11.1.6 Let (X, d) be a metric space. The the map d : X X R is cotiuous. Proof: It is eough to show that the iverse image of ay ope bouded iterval (r, s) is ope. Let us take the sup distace o X X. Let (a, b) X X be such that d(a, b) (r, s). Let x B(a, s) \ B(a, r) ad y B(b, s) \ B(b, r) The r < sup(d(a, x), d(b, y)) = d((a, b), (x, y)) = sup(d(a, x), d(b, y)) < s. Sice B(a, s) \ B(a, r) ad B(b, s) \ B(b, r) are ope i X, this proves that d is cotiuous.

11.2. CONTINUITY ON TOPOLOGICAL SPACES 123 Exercises. i. Let f : [a, b] [a, b] be a fuctio with the followig property: There exists a real umber c (0, 1) such that for ay x, y [a, b] oe has a) Show that f is cotiuous, ad f(x) f(y) < c x y. b) Show that there exists a x [a, b] such that f(x) = x. ii. Assume that X is a complete metric space which is also coected. Show that for ay a X ad ay r 0 there is a b X such that d(a, b) = r. Coclude that if x has more tha oe poit, show that X must be ucoutable. 11.2 Cotiuity o Topological Spaces Let X ad Y be two topological spaces. A fuctio f : X Y is called cotiuous if f 1 (V ) is ope for all ope subsets V of Y. The fuctio f is said to be cotiuous at a if a is i the iterior of f 1 (V ) for ay ope subset V of Y cotaiig f(a). By lemmas 11.1.1 ad 11.1.4, this geeralizes cotiuity i metric spaces. Lemma 11.2.1 Let X ad Y be two topological spaces. A fuctio f : X Y is cotiuous if ad oly if f is cotiuous at a for all a X. Proof: Left as a exercise. Examples. i. Let X be the discrete topological space (i.e. every subset is ope). The ay fuctio from X ito ay topological space is cotiuous. ii. Let X be ay topological space. The idetity fuctio Id X from X ito X is cotiuous. iii. Oe should ote that the idetity map Id X : X X from ay topological space ito itself is cotiuous if oe cosiders the domai X ad the arrival set X with the same topologies (i.e. the same ope subsets). Otherwise this may be false. For example if X 1 deotes the topological space o the set X where oly X ad are ope ad X 1 deotes the discrete topological space o X, the the idetity map Id X : X 1 X 2 is ot cotiuous uless X = 1. iv. Let X be a set. Let X 1 ad X 2 be two topologies o the set X. The the map Id X : X 1 : X 2 is cotiuous if ad oly if ay subset of X which is ope for the topological space X 2 is ope i the topological space X 1. I this case we say that X 1 is a refiemet of X 2.

124 CHAPTER 11. CONTINUITY v. Let X ad Y be ay two topological spaces. The ay costat map from X ito Y is cotiuous. Propositio 11.2.2 i. Compositio of cotiuous fuctios is cotiuous. ii. A fuctio is cotiuous if ad oly if the iverse image of a closed set is closed. iii. Let X ad Y be topological spaces. Let B be a base of Y. The a fuctio f : X Y is cotiuous if ad oly if the iverse image of a set i B is ope i X. iv. A fuctio f : X Y is cotiuous if ad oly if the iverse image of a closed subset of Y is closed i X. Let X be a set, X i be topological spaces ad f i : X X i be fuctios. The there is a smallest/weakest topology o X that makes all the fuctios f i cotiuous. This topology is geerated by the set Exercises. {f 1 i (U) : i I ad U X i is ope}. i. By usig the defiitio of cotiuity, show that the fuctio f(x) = x x 1 is cotiuous i its domai of defiitio. ii. Let X be a topological space. Give two cotiuous umerical fuctios f ad g o X, show that max{f(x), g(x)} is also cotiuous. iii. Show that if + : R R R ad : R R R are cotiuous. iv. Show that if f, g : X R are cotiuous, the so is f + g. v. Show that if f, g : X C are cotiuous, the so are f + g ad fg. vi. Show that ay polyomial map from C ito C is cotiuous. vii. Show that the map x 1/x from (0, 1) ito R is cotiuous. viii. Show that the map x (x 1)2 x(x 2) is cotiuous from (0, 2) ito R. ix. Show that the map x 1 1+x 2 is a homeomorphism from (0, ) oto (0, 1). x. Show that there is o homeomorphism from (0, 1) oto [0, 1). xi. Show that the map that seds ratioal umbers to 0 ad irratioal umbers to 1 is ot cotiuous. xii. Let f : R R be a cotiuous fuctio. Let Z = {r R : f(r) = 0}. Show that 1/f : R \ Z R is cotiuous. xiii. Let X be a topological space ad Y X. Show that the smallest topology o Y that makes the iclusio i : Y X is the iduced topology.

11.2. CONTINUITY ON TOPOLOGICAL SPACES 125 xiv. Let X, Y be topological spaces, f : X Y a cotiuous fuctio, A X ad f(a) B Y. Show that f A : A B is cotiuous. xv. Let X, Y be topological spaces, f : X Y a fuctio ad f(x) B Y. Show that f : X Y is cotiuous if ad oly if f : X B is cotiuous. xvi. Cosider the spaces R ad R m with their usual topology. Let X R ad Y R m. Let f : X Y be a fuctio. Show that f is cotiuous if ad oly if each f i = π i f : X R is cotiuous. xvii. Let f : X Y be cotiuous ad B Y. Show that f 1 (B ) = f 1 (B) ad f 1 (B) = f 1 (B). xviii. II. Subgroup Topology o Z. Let τ = {Z + m :, m Z, 0} { }. We kow that (Z, τ) is a topological space. a. Let a Z. Is Z \ {a} ope i τ? Aswer: Yes. 0,±1 Z (3Z + 2) = Z \ {1}.Traslatig this set by a 1, we see that Z \ {a} is ope. b. Fid a ifiite o ope subset of Z. Aswer: The set of primes is ot a ope subset. Because otherwise, for some a 0 ad b Z, the elemets of az + b would all be primes. So b, ab + b ad 2ab + b would be primes, a cotradictio. c. Let a, b Z. Is the map f a,b : Z Z defied by f a,b (z) = az + b cotiuous? (Prove or disprove). Aswer: Traslatio by b is easily show to be cotiuous. Let us cosider the map f(z) = az. If a = 0, 1, 1 the clearly f is cotiuous. Assume a 0, ±1 ad that f is cotiuous. We may assume that a > 1 (why?) Choose a b which is ot divisible by a. The f 1 (bz) is ope, hece cotais a subset of the form cz + d. Therefore a(cz + d) bz. Therefore ac = ±b ad so a divides b, a cotradictio. Hece f is ot cotiuous uless a = 0, ±1. d. Is the map f a,b : Z Z defied by f(z) = z 2 cotiuous? (Prove or disprove). e. Is the topological space (Z, τ) compact? (Prove or disprove). Aswer: First Proof: Note first the complemet of ope subsets of the form az + b are also ope as they are uios of the form az + c for c = 0, 1,..., a 1 ad c b mod a. Now cosider sets of the form U p = pz + (p 1)/2 for p a odd prime. The p U p = because if a p U p the for some x Z \ { 1}, 2a + 1 = px + p, so that a is divisible by all primes p ad a = 0. But if a = 0 the (p 1)/2 is divisible by p, a cotradictio. O the other had o fiite itersectio of the U p s ca be emptyset as (az + b) (cz + b) if a ad b are prime to each other (why?) Hece (U c p) p is a ope cover of Z that does ot have a fiite cover. Therefore Z is ot compact.

126 CHAPTER 11. CONTINUITY First Proof: Let p be a prime ad a = a 0 + a 1 p + a 2 p 2 +... be a p-adic iteger which is ot i Z. Let b = a 0 + a 1 p + a 2 p 2 +... + a 1 p 1. The p Z + b = but o fiite itersectio is empty. We coclude as above. xix. Let f : R R be the squarig map. Suppose that the arrival set is edowed with the usual Euclidea topology. Fid the smallest topology o the domai that makes f cotiuous. (5 pts.) Aswer: The smallest such topology is the set {U U : U ope i the usual topology of R}. xx. Let (V, ) be a ormed real metric space. Show that the maps V V V ad R V V defied by (v, w) v + w ad (r, v) rv respectively are cotiuous. xxi. Show that a isometry betwee metric spaces is a homeomorphism betwee the topological spaces that they iduce. We ed this subsectio with easy but importat results. Theorem 11.2.3 i. The image of a compact set uder a cotiuous map is compact. ii. The image of a coected set uder a cotiuous map is compact. Proof: Let f : X Y be a cotiuous map. i. Let K X be a compact subset of X. Let (V i ) i I be a ope coverig of f(k). The (f 1 (V i )) i I is a ope coverig of f 1 (f(k)). Sice K f 1 (f(k)), (f 1 (V i )) i I is also a ope coverig of K. Hece K is covered by f 1 (V i1 ),..., f 1 (V ik ) for some i 1,..., i k I, i.e. K f 1 (V i1 )... f 1 (V ik ). By applyig f to both sides, we get f(k) f(f 1 (V i1 )... f 1 (V ik )) = f(f 1 (V i1 ))... f(f 1 (V ik )) V i1... V ik. This shows that f(k) is compact. ii. Let f(x) U V where U ad V are ope i Y ad U V f(x) =. The X = f 1 (f(x)) f 1 (U) f 1 (V ) ad = f 1 (U V f(x)) = f 1 (U) f 1 (V ) f 1 (f(x)) = f 1 (U) f 1 (V ) X = f 1 (U) f 1 (V ), so either f 1 (U) = of f 1 (V ) =, i.e. either U f(x) = or V f(x) =. Corollary 11.2.4 If K is a compact set ad f : K Y a cotiuous bijectio ito a topological space Y the f is a homeomorphism from K oto f(k). Proof: Let F K be closed. The F is compact. So f(f ) is compact i the compact subset f(k). So f(f ) is closed i f(k). Thus f 1 is cotiuous.

11.3. CONTINUOUS FUNCTIONS AND R 127 11.3 Cotiuous Fuctios ad R Cotiuous fuctios assume their extreme values o compact subsets: Theorem 11.3.1 Let X be a compact topological space ad f : X R be cotiuous. The there are a, b X such that f(a) f(x) f(b) for all x X. Proof: By Theorem 11.2.3, f(x) is a compact subset of R. By Propositio 10.6.2, f(x) is closed ad bouded. Clearly, a closed ad bouded subset of R cotais its upper boud. Thus f(x) has a maximal elemet, say f(b). Similarly for the existece of a. A cotiuous real valued fuctio which is positive oce, remais positive for a while, i.e. if a cotiuous fuctio is positive at a poit the it is positive i a eighborhood of this poit: Lemma 11.3.2 Let f : (a, b) R ad c (a, b). Assume that f is cotiuous at c. If f(c) > 0 the f > 0 i a ope eighborhood of c. If f(c) < 0 the f < 0 i a ope eighborhood of c. Proof: Let ɛ = f(c). Sice f is cotiuous at c, there is a δ > 0 such that if x (c δ, c + δ) the f(x) (f(c) ɛ, f(c) + ɛ) = (0, 2ɛ). A cotiuous real valued fuctio which assumes two values assumes all the values betwee them: Theorem 11.3.3 (Itermediate Value Theorem) If f : [a, b] R is cotiuous ad d is betwee f(a) ad f(b), the d = f(c) for some c [a, b]. Proof: Replacig f by f d, we may assume that d = 0. Replacig f by f if ecessary, we may assume that f(a) 0 f(b). We may further assume that f(a) < 0 < f(b). Cosider the set A := {x [a, b] : f(x) < 0}. Sice a A, A. Let c = sup(a) [a, b]. By Lemma 11.3.2 a < c < b. We will show that f(c) = 0. If f(c) < 0 the we get a cotradictio by Lemma 11.3.2. If f(c) > 0 the we agai get a cotradictio by Lemma 11.3.2. Thus f(c) = 0. The image of a compact ad coected subset uder a cotiuous real valued fuctio is a closed iterval. Prove this uder this geerality. Corollary 11.3.4 If f : [a, b] R is cotiuous, the f([a, b]) = [m, M] where m = if(f([a, b])) ad M = sup(f([a, b])). Theorem 11.3.5 Let f : [a, b] [m, M] be oe to oe, oto ad cotiuous. The f 1 : [m, M] [a, b] is cotiuous also.

128 CHAPTER 11. CONTINUITY Exercises. i. [Cauchy] Show that for all real umbers r 1,..., r, ( i r i) 2 i r2 i. (Solutio: Replacig r i by r i, we may assume that all r i 0. If i r i = 0, the result is obvious. Replacig r i by r i / i r i, we may assume that i r i = 1. So we have to show that if i r i = 1, the i r2 i 1/. Let A = {(r 1,..., r ) R 0 : i r i = 1} ad let f : A R be defied by f(r 1,..., r ) = i r2 i. The A is a compact subset of R because A is a closed subset of the compact set [0, 1] (Heie-Borel Theorem, Theorem 10.6.3), beig the iverse image of 1 uder the cotiuous map (r 1,..., r ) i r i. Thus f assumes its miimal value. We will show that f(r 1,..., r ) f(1/,..., 1/) = 1/. Let R = (r 1,..., r ) be the poit where the miimal value is assumed. Assume r i r j for some i ad j. Assume r i < r j. Let 0 < ɛ < r j r i. Now look at the poit S where all the coordiates are the same as the coordiates of R except that the i-th coordiate is r i + ɛ ad the j-th coordiate is r j ɛ. The S A ad a easy calculatio shows that f(s) < f(r), a cotradictio. Thus r i = r j for all i, j. Sice i r i = 1, this implies that r i = 1/ for all.) ii. Let f : [a, b] [a, b] be a cotiuous fuctio. Show that f has a fixed poit. iii. Show that if f is oe to oe ad cotiuous o [a, b], the f is strictly mootoe o [a, b]. iv. Let f be a cotiuous umerical fuctio o the closed iterval [a, b]. Let x 1,..., x be arbitrary poits i [a, b]. Show that f(x 0 ) = (f(x 1 ) +... + f(x )) for some x 0 [a, b]. v. Show that there does ot exist ay cotiuous fuctio f : R R which assumes every x R twice. vi. Does there exist a cotiuous fuctio f : R R which assumes every real umber three times? (Yes!) vii. Assume f, g : [0, 1] [0, 1] be cotiuous fuctios ad assume that f(0) g(0) ad f(1) g(1). Show that there exists a x [0, 1] such that f(x) = g(x). viii. Assume that f is a cotiuous real-valued fuctio o [0, 1] ad that f(0) = f(1). If N is positive show that there is a poit x i [0, 1] such that f(x) = f(x + 1 ). 11.4 Uiform Cotiuity Exercises. i. Assume (f ) coverges uiformly to f o a subset A of R ad that each f is uiformly cotiuous o A. Show that f is uiformly cotiuous o A.

11.5. UNIFORM CONVERGENCE AND CONTINUITY 129 11.5 Uiform Covergece ad Cotiuity Theorem 11.5.1 Uiform limit of a family of cotiuous fuctios is cotiuous. 11.6 Supplemetary Topics 11.6.1 A Cotiuous Curve Coverig [0, 1] 2 Propositio 11.6.1 [0, 1] ad [0, 1] 2 are ot homeomorphic. Proof: Take away oe poit from [0, 1]. Corollary 11.6.2 There is o cotiuous bijectio from [0, 1] oto [0, 1] 2. Proof: From Corollary 11.2.4 ad Propositio 11.6.1. Theorem 11.6.3 (Peao, 1890) There is a cotiuous map (curve) from [0, 1] oto [0, 1] 2. Proof: Let I = [0, 1]. Subdivide I ito four closed itervals of equal legth [0, 1/4], [1/4, 1/2], [1/2, 3/4] ad [3/4, 1]. Subdivide I 2 ito four smaller squares of legth 1/2 1/2: [0, 1/2] [0, 1/2], [0, 1/2] [1/2, 1], [1/2, 1] [1/2, 1] ad [1/2, 1] [0, 1/2]. Make a correspodece betwee the four smaller itervals ad the four smaller squares i the order they appear above. Call this stage 1. At stage 2, subdivide each segmet ad each square ito four equal parts agai. Make the correspodece i such a way that the itersectig itervals correspod to eighborig squares. Cotiue i this fashio. Now for ay poit x of [0, 1], cosider the set of itervals that x belogs to. The poit x will belog to either oe or two of the four itervals of stage, of total legth 1/2 1. To these itervals correspod closed squares of stage, oe iside the other ad of total area /2 2 1. Because of the size of the squares, these correspodig squares itersect at a uique poit, say f(x). The map f so costructed is a map from I oto I 2, but is ot a bijectio. This map is cotiuous, because for x, y I close to each other, the images f(x) ad f(y) are also close to each other, as ca be easily checked. I fact if x y < 1/4 the d(f(x), f(y)) 2 2/2. Note that I ad I 2 are ot homeomorphic, because takig oe poit away from I discoects I, a pheomeo that does ot occur i I 2. Ideed the map f 1 defied above is ot cotiuous, as the images uder f 1/2 of two poits close to the ceter of I 2 which belog to the ceters of opposite squares [0, 1/2] [0, 1/2] ad [1/2, 1] [1/2, 1] are ever at a distace < 1/4. Questio. Is there a cotiuous bijectio from I oto I 2?

130 CHAPTER 11. CONTINUITY Exercises. i. Show that a subgroup of R is either discrete, i which case it is oegeerated, or is dese. ii. Let A R. Show that Z + αz is discrete if ad oly if α Q.

Chapter 12 Differetiable Fuctios 12.1 Defiitio ad Examples DO ALSO DIFFERENTIATION IN C, R ad i ormed vector spaces more geerally, all at oce Let X R, f : X R a fuctio ad a X. We say that f is differetiable at a if f(a + h) f(a) lim h 0 h exists. We the write f f(a + h) f(a) (a) = lim. h 0 h If X is ope, the fuctio f is called differetiable o X if f is differetiable at all the poits of X. Such a fuctio defies a fuctio f : X R called the derivative of f. Examples. i. Let N ad f : R R be defied by f(x) = x. The lim h 0 f(x+h) f(x) (x+h) h = lim x h 0 h ) i=0 x i i h i x = lim h 0 ) h 1 i=0 x i i h i = lim h 0 ( h ) 1 = lim h 0 i=0 x i i h i 1 ( ) = lim h 0 x 1 1 h 0 = x 1. 131

132 CHAPTER 12. DIFFERENTIABLE FUNCTIONS ii. Let f : [0, 1] [0, 1] be defied by f (x) = x. The the sequece of differetiable fuctios (f ) coverges to the fuctio f defied by f(x) = ad f is ot differetiable. { 0 for 0 x < 1 1 for x = 1 Lemma 12.1.1 If f is differetiable at a the f is cotiuous at a. Thus a differetiable fuctio is cotiuous. Exercises. i. Let f : R R be the squarig fuctio f(x) = x 2. Show that f is differetiable ad f (x) = 2x. ii. Let N \ {0} ad f : [ 1, 1] R be defied by f (x) = x 1+1/. Show that f is differetiable o [ 1, 1]. iii. Show that the fuctio f(x) = { x 2 si(1/x) if x 0 0 if x = 0 is differetiable but f is ot cotiuous. 12.2 Differetiatio of Complex Fuctios 12.3 Basic Properties of Differetiable Fuctios From ow o we let f ad a be as above. Lemma 12.3.1 If f is differetiable at a ad if f(a) is maximal the f (a) = 0. Exercises. i. Let f : U R (or C) be differetiable at a U, where U is ope. Assume that there is a sequece (x ) of U covergig to a such that f(x ) = f(x m ) all, m. The f(a) = f(x ) ad f (a) = 0. (Hit: Follows almost directly from the defiitio of derivatives). ii. Let f : [a, b] R be differetiable. Assume that f (x) 0 for x [a, b]. The f has fiitely may zeroes i [a, b]. (Hit: See Exercise i, page 132).

12.4. RULES OF DIFFERENTIATION 133 12.4 Rules of Differetiatio 12.5 Relatioship Betwee a Fuctio ad Its Derivative Letf : X R be a fuctio from a topological space ito R. We say that a poit x is a local maximum (resp. local miimum) of f if there is a ope subset U of X cotaiig x such that for all x U, f(x) f(x ) (resp. f(x) f(x )). A absolute maximum (resp. absolute miimum) of f is a poit x such that x U, f(x) f(x ) (resp. f(x) f(x )). A extremum (local or absolute) poit is a maximum or a miimum poit. Theorem 12.5.1 Let x X R ad f : X R be a fuctio differetiable at x. If x is a local extremum the f (x ) = 0. Proof: Suppose x is a local maximum. The there is a δ > 0 such that (x δ, x + δ) X ad Now f(x) f(x ) if x (x δ, x ) f(x) f(x ) if x (x, x + δ). f(x +h) f(x ) h 0 for h (0, δ) f(x +h) f(x ) h 0 for h ( δ, 0). Thus f (x ) = lim = f(x +h) f(x ) h 0. If x is a local miimum the proof is similar. Theorem 12.5.2 (Rolle s Theorem) For a < b, let f : [a, b] R be differetiable o (a, b). If f(a) = f(b) the f (c) = 0 for some c (a, b). Proof: If f is costat the f = 0 ad there is othig to prove. Assume f is ot a costat. By Theorem 11.3.1, f has a absolute maximum ad a absolute miimum. Sice f is ot costat, these are two distict poits. Sice f(a) = f(b), oe of these poits, say c, must be i the ope iterval (a, b). By Theorem 12.5.1, f (c) = 0. The ext theorem is oe of the most importat result i this series. We will have several opportuities to use it. Theorem 12.5.3 (Mea Value Theorem of Differetial Calculus) For a < b, let f : [a, b] R be differetiable o (a, b). The for some c (a, b). f(a) f(b) a b = f (c)

134 CHAPTER 12. DIFFERENTIABLE FUNCTIONS Proof: For t [0, 1], let g(t) = f(ta + (1 t)b) tf(a) + (1 t)f(b). The g(0) = g(1) = 0. By Rolle s Theorem (12.5.2), g (t ) = 0 for some t (0, 1). But g (t) = (a b)f (ta + (1 t)b) f(a) + f(b). Thus 0 = g (t ) = (a b)f (t a + (1 t )b) f(a) + f(b). If c = t a + (1 t )b, the f (c) = f(a) f(b) a b. Theorem 12.5.4 If f is differetiable o (a, b) ad f (x) 0 for every x (a, b) the f is mootoe decreasig o (a, b). Proof: Let a < x 1 < x 2 < b. By the Mea Value Theorem of Differetial Calculus (12.5.3), there is a x (x 1, x 2 ) such that f (x ) = f(x 2) f(x 1 ) x 2 x 1. Sice f (x ) 0 ad x 2 x 1 > 0, f(x 2 ) f(x 1 ) 0. The fact that f (c) > 0 at some c does ot imply that f is icreasig aroud c. It is easy to fid a couterexample, fid oe. 12.6 Uiform Covergece ad Differetiatio The fact that a sequece of differetiable fuctio (f ) coverges uiformly to a fuctio f, does ot imply that f is differetiable. Example. Let f : [ 1, 1] R be defied by f (x) = x 1+ 1 2+1. The each f is differetiable o [0, 1] (see Exercise ii, page 132). Also lim f (x) u = x. (See Exercise iii, page 132). But the fuctio f(x) = x is ot differetiable at x = 0. Questio. Suppose (f ) is a sequece of differetiable fuctios that coverge uiformly to a differetiable fuctio f. Is it true that f p = lim f? Theorem 12.6.1 Let the sequece of differetiable fuctios (f ) coverge poitwise to f o the closed iterval [a, b]. Assume that the sequece (f ) coverges uiformly. The f is differetiable ad f = lim f. Proof: Let lim f u = g. Let c [a, b]. We wat to show that f(x) f(c) lim = g(c). x c x c Thus give a ɛ > 0, we wat to fid a δ > 0 such that f(x) f(c) g(c) x c < ɛ for all x [a, b] that satisfies 0 < x c < δ. Let ɛ > 0 be give. By the triagular iequality, for ay, we have, f(x) f(c) x c g(c) f(x) f(c) x c f(x) f(c) x c + f(x) f(c) x c f (c) + f (c) g(c).

12.6. UNIFORM CONVERGENCE AND DIFFERENTIATION 135 We will make each term of the right had side less tha ɛ/3. It is easy to make the secod term small by choosig δ small eough. It is also easy to make the last term small by choosig large eough. The mai problem is with the first term, that we deal first. Assume c < x. The proof is the same if x < c. Let ad m be ay atural umbers. Applyig the Mea Value Theorem to the fuctio f f m o the iterval [c, x] we ca fid a α = α m, (c, x) such that f m(α) f (α) = f m(x) f (x) x c f m(c) f (c). x c O the other had, by the Cauchy Criterio for Uiform Covergece (Theorem 9.2.5), there exists a N 1 such that for all, m > N 1 ad x [a, b], I particular, f m(x) f (x) < ɛ/3. f m(α) f (α) < ɛ/3. (It is worth while to otice that, sice α depeds o ad m, we really eed the uiform covergece of (f ) for this part of the argumet). Thus f m (x) f (x) f m(c) f (c) x c x c < ɛ/3 for all, m > N 1 ad all x [a, b]. Now let m go to ifiity, to get (1) f(x) f(c) f (x) f (c) x c x c < ɛ/3 for all > N 1 ad ay x [a, b]. Sice lim f = g, there is a N 2 large eough so that, (2) f (c) g(c) < ɛ/3 for all > N 2. Let = max{n 1, N 2 } + 1. Note that (1) ad (2) hold for this. Choose δ > 0 such that if x [a, b] ad x c < δ, the (3) f (x) f (c) x c Now (1), (2) ad (3) give us the result. f (c) < ɛ/3. I the theorem above we do ot really eed the poitwise covergece of (f ) o [a, b], it is eough to kow that (f (x 0 )) coverges for some to get the poitwise, i fact the uiform covergece o [a, b]. Theorem 12.6.2 Let (f ) be a sequece of differetiable fuctios defied ad assume that the sequece (f ) coverges uiformly o [a, b]. If there exists a poit x 0 [a, b] such that (f (x 0 )) is coverget. The (f ) coverges uiformly o [a, b] ad lim f is differetiable ad (lim f ) = lim f.

136 CHAPTER 12. DIFFERENTIABLE FUNCTIONS Proof: I view of the previous theorem, it is eough to prove the uiform covergece of (f ) o [a, b]. COMPLETE THE PROOF. Exercises. 12.7 Secod ad Further Derivatives

Chapter 13 Aalytic Fuctios 13.1 Power Series We collect the mai results ad a few more about power series. This will show that the power series wheever ad wherever they coverge behave like polyomials, as much as they ca. Theorem 13.1.1 Let i=0 a iz i be a power series with R = lim{1/ a i 1/i } R 0 { } its radius of covergece. Let 0 < r < R. The the followig hold. i. If z < R, the i=0 a iz i coverges absolutely. If z > R, the i=0 a iz i diverges. ii. i=0 a iz i coverges uiformly o B(0, r). iii. The radius of covergece of i=1 ia iz i 1 coverges uiformly o B(0, r). iv. i=0 a iz i is ifiitely differetiable i B(0, R) ad ( a i z i ) = ia i z i 1. i=0 v. If f(z) = i=0 a iz i i B(0, R), the f is ifiitely differetiable i B(0, R) ad a i = f (i) (0)/i!. I other words, the coefficiets of a power series are uique ad if the fuctio f(z) is a power series, the f(z) = i=0 i=1 f (i) i i!. z Proof: Part (i) is Corollary 7.5.6. Part (ii) is by Propositio 6.8.3 ad part (i). iii. Sice i=0 a iz i coverges absolutely for z = r, the series i=0 a ir i coverges. By Weierstrass M-Test (Theorem 9.3.3), the series f = i=0 a iz i coverges uiformly o B(0, r). 137

138 CHAPTER 13. ANALYTIC FUNCTIONS iv. By parts (i, iii), the series i=1 ia iz i 1 coverges uiformly o B(0, r). Let f = i=0 a iz i. By Theorem 12.6.1, f = i=0 a iz i is differetiable ad f = lim f = lim i=1 ia iz i 1 = i=1 ia iz i 1 for z B(0, r), so also for all z B(0, R). v. Direct applicatio of the previous facts. Corollary 13.1.2 Cosider the ifiite series i=0 a i(z a) i with 0 < R = lim{1/ a i 1/i } R 0 { } its radius of covergece. Let 0 < r < R. The the followig hold. i. If z B(a, R), the i=0 a i(z a) i coverges absolutely. If z B(a, R), the i=0 a i(z a) i diverges. ii. The radius of covergece of i=0 a i(z a) i is also R. iii. i=0 a i(z a) i coverges uiformly o B(a, r). iv. i=0 a i(z a) i is ifiitely differetiable i B(a, R) ad ( a i (z a) i ) = ia i (z a) i 1, i=0 the covergece beig uiform i B(a, r). v. If f(z) = i=0 a i(z a) i for z B(a, R), the f is ifiitely differetiable i B(a, R) ad a i = f (i) (a)/i!. Exercises. i=1 i. Let i=1 a i/ i be coverget for all > 1. Show that lim i=0 a i/ i = a 0. (Solutio: Cosider the power series f(x) := i=1 a ix i which is coverget at x = 1/2, thus its radius of covergece is at least 1/2. Thus f(x) is cotiuous at 0. It follows that for all ɛ > 0 there is a δ > 0 such that if x < δ ad x 0 the f(x) f(0) < ɛ. Let 1/δ 1 < N = [1/δ] 1/δ. The for > N, 1/ 1/(N + 1) < δ ad so f(1/) f(0) < ɛ, i.e. i=1 a i/ i a 0 < ɛ. Thus lim i=0 a i/ i = a 0.) 13.2 Taylor Series As see above, the power series i=0 a iz i ad their cousi i=0 a i(z a) i are easy to hadle, they behave almost like polyomials. Accordig to these results, it would be helpful if every fuctio could be expressed as such a ifiite series aroud ay a. But sice such a series is ifiitely differetiable if it coverges aroud a, a fuctio which is ot ifiitely differetiable aroud a caot be expressed as such a series. As it happes, eve this coditio is ot sufficiet for a fuctio to be expressed as such a series. We will see a example of such a fuctio later. A fuctio f defied o a ope subset U (of R or of C) is said to be aalytic at a poit a U if there is a sequece (a i ) i such that f(z) = i=0 a i(z a) i for z B(a, R) for some R > 0. The coefficiets a i are uique ad equal to

13.2. TAYLOR SERIES 139 φ (i) (a)/i! as Theorem 13.1.1 shows. Also we ca fid a maximal R where the equality f(z) = i=0 a i(z a) i holds. If f is a complex fuctio, the the fact that f is differetiable o U oce is eough to isure that for ay a U, f = i=0 a i(z a) i for z B(a, R) for some R > 0. This result is part of a subject called complex aalysis ad will be see somewhere else. The situatio i R is much more mysterious ad is the subject of this sectio. Lemma 13.2.1 Let U be a ope subset of R ad f : U R a (+1)-times differetiable fuctio. Let [a, b] U. The f(b) = f(a)+ f (a) 1! for some c (a, b). Proof: Let (b a)+ f (a) 2! g(x) = f(b) f(x) (b x)f (x) (b x)2 f (x) 2! (b a) 2 +...+ f () (a) (b a) + f (+1) (c)! ( + 1)! (b a)+1... (b x) f () (x) (b x)+1 γ! ( + 1)! where γ is chose so that g(a) = 0. Sice g(b) = 0 also, by Rolle s Theorem (12.5.2), there is a c (a, b) such that g (c) = 0. Note that, after simplificatio, g (x) = [f(b) = i=1 (b x) i f (i) (x) i=0 i! (b x)+1 γ (+1)! ] (b x) i 1 f (i) (x) (i 1)! i=0 = (b x) f (+1) (x)! + (b x) γ!. (b x) i f (i+1) (x) i! + (b x) γ! Thus γ = f (+1) (c). Now the equatio g(a) = 0 gives the desired result. Corollary 13.2.2 Let U be a ope subset of R ad f : U R a ( + 1)- times differetiable fuctio. Let [a, b] U. The for all x [a, b) there is a ξ (x, b) (that depeds o a ad x) such that f(x) = f(a)+ f (a) 1! (x a)+ f (a) 2! (x a) 2 +...+ f () (a) (x a) + f (+1) (ξ)! ( + 1)! (x a)+1. It follows that the ifiitely differetiable real fuctio f is equal to its Taylor series f (i) (z a) i i! i=0 aroud a (so is aalytic) if ad oly if f (+1) (ξ) (+1)! (z a) +1 coverges to 0 as goes to ifiity. The term ( + 1)! is helpful to make the quatity small. If z is close to a, (z a) +1 ca also be made small. Oly the term f (+1) (ξ) is bothersome. Note that here ξ depeds o, a ad x.

140 CHAPTER 13. ANALYTIC FUNCTIONS 13.2.1 Calculatig Taylor Polyomials Theorem 13.2.3 Suppose that o a ope subset we have f(x) = a i x i + x ɛ(x) i=0 where lim x ɛ(x) = 0 ad f is ( + 1)-times differetiable. The T f (x) = i=0 a ix i.???? Taylor Polyomials of 1/ cos x We start otig that sice 1/ cos x is a eve fuctio, its derivatives of odd degree must be 0 at x = 0. First Method. Let f(x) = 1/ cos x ad start computig its derived series. This may take a log time ad is ot advised. Secod Method. Let us compute the first 6 (or 7) terms of the Taylor series of 1/ cos x aroud 0. For this we take the iverse of the Taylor series for cos x aroud 0. 1 1 cos x = 1 x 2 /2!+x 4 /4! x 6 /6!+x 7 f(x) = 1 1 (x 2 /2! x 4 /4!+x 6 /6! x 7 f(x)) = 1 + (x 2 /2! x 4 /4! + x 6 /6! x 7 f(x))+ +(x 2 /2! x 4 /4! + x 6 /6! x 7 f(x)) 2 + +(x 2 /2! x 4 /4! + x 6 /6! x 7 f(x)) 3 = 1 + (x 2 /2! x 4 /4! + x 6 /6!) + (x 4 /4 x 6 /4!) + x 6 /8 + x 7 g(x) = 1 + x 2 /2 x 4 (1/4! 1/4) + x 6 (1/6! 1/4! + 1/8) + x 7 g(x) = 1 + x 2 /2 + 5x 4 /24 + 61x 6 /720 + x 7 g(x), where f ad g are fuctios that coverge to 0 whe x goes to 0. Taylor Polyomial of ta x Note first that sice ta x is a odd fuctio, its derivatives of eve degree must be 0. First Method. Let f(x) = ta x ad start computig its derived series. This is log ad tirig ad is ot advised. Secod Method. We use the Taylor polyomial of 1/ cos x that has bee computed above: ta x = si ( x 1 cos x ) ( ) = x x3 3! + x5 5! + x 6 f(x) 1 + x2 2 + 5x4 24 + 61x6 720 + x6 g(x) = x + ( ) 1 3! + 1 x 3 + ( 1 5! 1 12 + 24) 5 x 5 + x 6 h(x) = x + 1 3 x3 + 2 15 x5, where f, g ad h are fuctios that coverge to 0 whe x goes to 0. Third Method for those who kow how to itegrate. Sice ta x = 1/ cos 2 x, if we kow the Taylor polyomial of 1/ cos 2 x, we ca itegrate to

13.2. TAYLOR SERIES 141 fid the Taylor polyomial of ta x, except may be the first term. But sice ta 0 = 0, the first term must be 0. Fourth Method. Sice ta x is ( l(cos x)) = ta x, if we kow the Taylor polyomial of l(cos x), the by differetiatig, we ca fid the Taylor polyomial of ta x: ( l(cos x) = l ( Hece = l = 1 x2 1 ( x 2 ta x = ( l(cos x)) = ) 6! x 7 f(x) )) 6! + x 7 f(x) ) 6! + x 7 f(x) 1 2 ( x 2 2! x4 4! + x6 2! + x4 4! x6 2! x4 4! + x6 ( ) 3 1 x 2 3 2! x4 4! + x6 6! + x 7 f(x) = 1 2 x2 1 12 x4 2 90 x6 + x 7 g(x) ( ) 2 x 2 2! x4 4! + x6 6! + x 7 f(x) ( 1 2 x2 + 1 12 x4 + 2 ) 90 x6 + x 7 g(x) = x+ 1 3 x3 + 2 15 x5 +x 6 h(x) where the limits of g ad h whe x goes to 0 are 0. Taylor Polyomial of si 3 x We compute up to ith degree: ( ) 3 ( si 3 x (x) = 1! x3 3! + x5 5! x7 7! + x 8 f(x) = x 3 1 1! x2 [ ( = x 3 1 + 3 1 1 ( 1) 1!1!3! x 2 + 3 1 1 1 1!1!5! + 3 1 ( 1) ( 1) ( 1!3!3! + ( 1) ( 1) ( 1) 3!3!3! + 3! 1 ( 1) 1 1!3!5! + 3 1 1 ( 1) 1!1!7! = x ( 3 1 1 2 x2 + 13 120 x4 41 3024 x6 + x 7 ɛ (x) ) = x 3 1 2 x5 + 13 120 x7 41 3024 x9 + x 10 ɛ (x) where the limits of f, ɛ ad ɛ whe x goes to 0 are 0. Exercises. ) x 4 + 3! + x4 5! x6 ) x 6 + x 7 ɛ (x) 7! + x 7 f(x) i. Show that the derivatives of eve (resp. odd) degree of a aalytic fuctio which is odd (resp. eve) must be 0. ii. Let f : R >0 R be twice differetiable. Assume that f > 0 ad f < 0 the f caot be always egative. Proof: Assume f (x) < 0 for all x. Let a R >0 be fixed. The for all x > a, f(x) = f(a) + (x a)f (a) + (b x) 2 f (c)/2! for some c. We ca choose x large eough so that f(a) + (x a)f (a) < 0. The f(x) < 0. iii. If f(x) = 1 1+x ad U = ( 1, ), solve the first two questios. ] ) 3

142 CHAPTER 13. ANALYTIC FUNCTIONS iv. a. Estimate the error made i replacig the fuctio exp o the iterval [0, 1] by its Taylor polyomial of degree 10. b. O what iterval [0, h] does the fuctio exp differ from its Taylor polyomial of degree 10 by o more tha 10 7? c. For what value of does the fuctio exp differ from its Taylor polyomial of degree by o more tha 10 7 o the iterval [0, 1]? 13.3 Aalytic Fuctios If f is ifiitely differetiable ad a, x U, we set f () (a) (T f)(x) = (x a).! =0 Note that (T f)(x) may or may ot be coverget for a give x. But if T f is coverget, the T f ad f have the same th -derivatives at 0 for all. Does this coditio implies that T f = f. Thus we have two questios: 1. For what values of x is (T f)(x) coverget? Is T f coverget for all x U. 2. I case (T f)(x) is coverget o U, do we have (T f)(x) = f(x) for all x? The aswer to both questios is egative as the ext examples show. Examples. i. Let f be the real fuctio f(x) = 1 1+x. The T f(x) = 2 i=0 ( 1)i x 2i. It ca be easily checked that T f coverges for x ( 1, 1) ad diverges elsewhere, although f is defied everywhere. Thus the aswer to the first questio is egative. ii. Let f(x) = { e 1/x 2 if x 0 0 if x = 0 The f is ifiitely differetiable ad f () (0) = 0 for all N. Therefore T f = 0 ad T f(x) = f(x) if ad oly if x = 0. Thus the aswer to the secod questio is also egative. A fuctio f defied o a ope subset U (of R or of C) is said to be aalytic at a poit a U if there is a sequece (a i ) i such that f(z) = i=0 a i(z a) i for z B(a, R) for some R > 0. The coefficiets a i are uique ad equal to φ (i) (a)/i! as Theorem 13.1.1 shows. Also we ca fid a maximal R where the equality f(z) = i=0 a i(z a) i holds. Propositio 13.3.1 If f : B(0, R) C is aalytic ad f = 0 o R B(0, R) the f = 0. Proof:. By Exercise i, page 132, f = 0 o R. Thus f () (0) = 0 for all ad so f = T f = 0.

13.4. TRANSCENDENTAL FUNCTIONS 143 13.4 Trascedetal Fuctios We defie exp(z) = =0 z /! si(z) = =0 ( 1) z 2+1 /(2 + 1)! cos(z) = =0 ( 1) z 2 /(2)! cosh(z) = =0 z2+1 /(2 + 1)! sih(z) = =0 z2 /(2)! Each of these fuctios are aalytic o C as it ca be checked by usig Theorem 13.1.1. Also they all take real values at real umbers, i.e. they ca also be cosidered as real fuctios. Exercises. i. Show that lim (1 + z/) = exp(z). (See Example i, page 111). ii. Show that si z = exp(iz) exp( iz) 2i. 13.4.1 Expoetiatio ad Trigoometric Fuctios We have exp(0) = 1, cos(0) = 1, si(0) = 0. By Theorem 7.5.7, exp = exp, si = cos, cos = si, cosh = sih, sih = cosh, cos 2 (x) + si 2 (x) = 1. Theorem 13.4.1 exp(x + y) = exp(x) exp(y). Proof: Sice exp(z) = =0 z /!, we have to prove that the limit of ( x i /i!)( y i /i!) i=0 i=0 (x + y) i /i! i=0 is 0 as. We first compute the left had side: ( x i /i!)( y i /i!) = i=0 i=0 Now we compute the left had side: (x + y) i /i! = i=0 i i=0 j=0 ( i j ) x j y i j /i! = i=0 j=0 i i=0 j=0 x i y j i!j!. x j y i j (i j)!j! = i i=0 k=0 x j y k k!j!. Aother proof of Theorem 13.4.1: Let a R. Let f(x) = exp(x + a) ad g(x) = exp(x) exp(a). The f(0) = g(0) ad f = f ad g = g. Thus f () (0) = g () (0) all. So f = g by Theorem 13.1.1. Theorem 13.4.2 exp(iz) = cos z + i si z for all z C.

144 CHAPTER 13. ANALYTIC FUNCTIONS Proof: Easy. Corollary 13.4.3 si(x + y) = si(x) cos(y) + cos(x) si(y) ad cos(x + y) = cos(x) cos(y) si(x) si(y) for all x, y C. Proof: If x, y R, this follows from Theorems 13.4.2 ad 13.4.1. Now apply Propositio 13.3.1 to the aalytic fuctios si(x + a) si(x) cos(a) + cos(x) si(a) ad cos(x + a) cos(x) cos(a) si(x) si(a). Theorem 13.4.4 The fuctios si ad cos restricted to R have a commo period. Proof: Assume cos(x) > 0 for all x. The si is icreasig. Sice si(0) = 0, si(x) > 0 for all x > 0. Thus cos = si < 0 o R >0. Also, cos (x) = cos(x) < 0 ad this cotradicts Exercise ii, page 141. Thus cos(x) = 0 for some x > 0. The si(x) = ±1. So cos(2x) = cos 2 (x) si 2 (x) = 1 ad si(2x) = 0. Thus A := {α R >0 : si(α) = 0} is oempty. By the formulas si(x + y) = si x cos y + si y cos x ad si( x) = si x, we see that the set A of zeroes of si x = 0 is a additive subgroup of R. Sice si is cotiuous, this subgroup is closed. If it had a accumulatio poit, the accumulatio poit itself would be i A, but the 0 would be a accumulatio poit. Let (a, b) be a ope iterval. Assume 0 < a < b. Let α A be such that 0 < α < b a. The α (a, b) A for some N. So A is dese i R. Sice A is also closed, this implies that A = R ad si = 0, a cotradictio. Hece A is discrete. Let π > 0 be the least positive elemet of A. It is easy to show that the period of si is 2π. Also follows from Lemma i. DETAILS It follows from the above proof that a subgroup of R is either dese i R or is geerated by oe elemet. Fid a upper boud for π. The umber e. We let e = exp(1) = i=0 1/i!. Propositio 13.4.5 e is irratioal. Proof: Assume e = m/ for m ad > 0 itegers. The the umber ( ) N :=! e 1/i! =!/i! > 0 i=0 is a positive atural umber ad we have i=+1 N = i=+1!/i! = i=1 1/( + 1)... ( + i) < i=1 1/( + 1)i = 1 +1 i=0 1/( + 1)i 1 1 = = 1/, +1 1 1 +1

13.4. TRANSCENDENTAL FUNCTIONS 145 implyig N = 0, a cotradictio. See [AZ, page 28] for the fact that e q is irratioal for all q Q \ {0}. The propositio above is a cosequece of a more geeral result: Theorem 13.4.6 If (a i ) i is a sequece of zeroes ad oes which is ot evetually zero, the i=0 a i/i! is irratioal. Proof: The same as above. Take this proof to its appropriate place. Exercises. i. Fid cos(15 ). ii. Express si(4x) ad cos(4x) i terms of si x ad cos x. formula). (Prove your iii. Let f(x) = x 3 3x + 5. Show that f(l a) = 6 for some a > 1. iv. Show that exp(2iz) 1 exp(2iz)+1 = i ta z for all z C. v. Describe the set α R such that the subgroup 1, α = Z + αz is discrete. (Solutio: If α = p/q Q, the it is easy to show that 1, α = 1/q is discrete. Otherwise, the elemets of the sequece (α [α]) N are distict ad are all i [0, 1) 1, α. Thus 1, α has a accumulatio poit i [0, 1). Thus 0 is a accumulatio poit of 1, α. Now for 0 β < γ i R, choose α 1, α (0, γ β). For some k N, β < kα γ ad kα (β, γ)). vi. Assumig π is irratioal, show that si(z) is dese i [ 1, 1]. (Solutio: Let y [ 1, 1]. Let x R be such that si x = y. Sice Z + 2πZ is dese i R, there is a sequece a +2πb with a, b Z such that lim (a + 2πb ) = x. Sice si is cotiuous, y = si x = si(lim a + 2πb ) = lim si(a + 2πb ) = lim si a.) vii. Show that =1 si is coverget. (Solutio: We will apply Abel s Theorem (Theorem 7.5.8). We oly eed to show that the partial sums of m j= si are bouded. Sice si = Im(ɛi ), it is eough to show that the partial sums m j= eij are bouded. We will prove that the sums m j=0 eij are bouded. This will prove the result sice m j= eij = m j=0 eij 1 j=0 eij m j=0 eij + 1 j=0 eij. Now we compute: m j=0 eij = ei(m+1) 1 e i 1 2 e i 1.) viii. Fid the radius of covergece of =1 si z. (Solutio: Sice si z z, the radius of covergece 1. If the radius of covergece is > 1, the the radius of covergece of the derived series =0 si z 1 is also > 1. But by Exercise vi, page 145, the geeral term si z does ot coverge to 0 if z < 1).

146 CHAPTER 13. ANALYTIC FUNCTIONS ix. Show that lim (1 + 1/) = e. x. Fid the followig limits: lim x (1 + 1/x) x lim x (1 1/x) x lim (1 1/) lim (1 1/ 2 ) 2 lim (1 1/ 2 ) lim (1 1/) 2 lim x (1 + c/x) x 13.4.2 Iverse Trigoometric Fuctios 13.4.3 Logarithm 13.4.4 Hyperbolic Fuctios 13.5 Supplemet 13.5.1 Trigoometric Fuctios By Example i, page 111, lim (1 + z/) = exp(z). Thus si z = exp(iz) exp( iz) 2i 1 = lim 2i [(1 + iz/) (1 iz/) ]. ( O the ) other had, we ote that (1 + iz/) (1 iz/) = 0 if ad oly if ( ) 1+iz/ 1 iz/ = 1 if ad oly if +iz iz = 1 if ad oly if +iz iz = exp(2ikπ/) for some k = 0, 1,..., 1 if ad oly if (as a easy calculatio shows) z = i exp(2πik/) 1 exp(2πik/)+1 = ta kπ by Exercise iv, page 145. ppppp DEVAM EDECEK, VALIRON, SAYFA 41 13.5.2 Series Propositio 13.5.1 For x 2kπ (k Z) ad (ɛ ) a decreasig sequece whose limit is 0, the series =0 ɛ e ix is coverget. Proof: We will apply Theorem 7.5.8. Let x 2kπ for ay k Z ad v = v (x) = e ix. For m <, v m +... + v = e imx +... + e ix = e imx (1 + e ix +... + e i( m)x imx 1 ei( m+1)x = e 1 e i x, so that v m +... + v 2 1 e ix. Hece the coditios of Theorem 7.5.8 are met. Corollary 13.5.2 For x 2kπ (k Z) ad (ɛ ) a decreasig sequece whose limit is 0, the series =0 ɛ si(x) ad i=0 ɛ cos(x) are coverget.

13.6. NOTES 147 Exercises. i. Show that k=1 si(kx)/k2 is uiformly coverget o (, ). ii. Show that k=0 (xe x ) k coverges uiformly o [0, 2]. iii. Show the followig: si 2 (z) + cos 2 (z) = 1 exp(iz) = cos(z) + i si(z) cos is eve, si is odd sih 2 (z) cosh 2 (z) = 1 exp(z) = cosh(z) + i sih(z). iv. Let X be a set with elemets. Let 0 p := p() 1. We select a radom subset A := A(p) of Xi such a way that the probability that a elemet x X is i A is p. i. Calculate the probability that A has ay elemet at all. ii. What is the expected umber of elemets of A? iii. Let p() = 1/. What is the probability that A(p) whe teds to ifiity? iv. Assume that p = p() 1/. Show that the probability that A(p) = whe teds to ifiity is 1. v. Assume that p = p() 1/. Show that the probability that A(p) = whe teds to ifiity is 0. 13.6 Notes Aalytic fuctios are ot the oly fuctios whose sequece (f () (x 0 )) of derivatives determie the fuctio i a eighborhood of x 0. The fuctios defied ad ifiitely differetiable o [a, b] such that if M = max{ f () (x) : x [a, b]}, the is diverget (Dejoy 1921 ad Carlema). Such fuctios are called quasi aalytic. 1/M 1/

148 CHAPTER 13. ANALYTIC FUNCTIONS

Chapter 14 Graph Drawig 14.1 Drawig i Cartesia Coordiates 14.1.1 Asymptotes Exercises. i. Draw with as much care as possible the graph of f(x) = x 2 (x 1)(x+2). ii. Let f(x) = x + 2x 2 si(1/x) if x = 0 ad f(0) = 0. i. Show that f (0) = 1. ii. Show that ay iterval cotaiig 0 also cotais poits where f (x) < 0, so that f caot be icreasig o ay iterval aroud 0 although f (0) > 0. 14.2 Parametric Equatios 14.3 Polar Coordiates 14.4 Geometric Loci Exercise. [BR] Let A = (1, 0). Fid the set of poits M of the plae R 2 such that, if P ad Q desig the projectios of M oto the axes x ad y respectively, the poit S of itersectio of AQ ad P M is at distace 1 from A. First Solutio. Let M = (x 0, y 0 ). The P = (x 0, 0) ad Q = (0, y 0 ). The lie AQ has equatio y = 0 y0 1 0 x+y 0. Thus the poit S has coordiates (x 0, y 0 x 0 + y 0 ). The square of its distace from A(1, 0) is (x 0 1) 2 + ( y 0 x 0 + y 0 ) 2, which we wat to be 1. Thus the equatio of the curve i Cartesia coordiates is (x 1) 2 + ( yx + y) 2 = 1, or (x 1) 2 (1 + y 2 ) = 1, or y 2 1 = (x 1) 1, i.e. ( ) 2 1/2. 1 y = ± (x 1) 1 Note that x 1 ad 1 2 (x 1) 1 0, i.e. x [0, 2] \ {1}. 2 149

150 CHAPTER 14. GRAPH DRAWING Note also that the curve is symmetric with respect to the x axis. The lie x = 1 is a asymptote ad the curve is also symmetrical with respect to the lie x = 1. It is possible to draw this curve, eve if some paiful work is ecessary. Secod Solutio. Let θ be the agle P AQ. The oe ca see immediately that x(θ) = 1 cos θ y(θ) = ta(θ) where θ [0, 2π]\{π/2, 3π/2}. It is eough to draw the curve whe θ [0, π/2) because of the obvious symmetries. It is easy to see that i) x(0) = y(0) = 0. ii) x ad y both icrease whe θ icreases from 0 to π/2. iii) Whe θ goes to π/2, x goes to 1 ad y goes to. Thus there is a asymptote at θ = π/2, which is the lie x = 1. Differetiatig with respect to θ, we fid Thus dx/dθ = si θ dy/dθ = 1/ cos 2 θ. dy/dx = (dy/dθ)/(dx/dθ) = 1/ cos 2 θ si θ. It follows that whe θ [0, π/2), y is a icreasig fuctio of x. It also follows that the y axis is taget to the curve (case θ = 0). Now we fid d 2 y/dx 2 : d 2 y/dx 2 = d(dy/dx)/dx = (d(dy/dx)/dθ)/(dx/dθ) = (1/ cos 2 θ si θ) / si θ = 2 si2 θ cos 2 θ cos 3 θ si 3. θ There is a iflectio poit whe 2 si 2 θ cos 2 θ = 0, i.e. whe y = ta(θ) = 1/ 2. We ca compute the first coordiate x of the iflectio poit: 1/ cos 2θ 1 = ta 2 θ = 1/2, cos θ = 2/3 ad x = 1 cos θ = 1 2/3. The graph is cocave dow before (1 2/3, 1/ 2) ad cocave up afterwards. Exercise. [BR] To a poit M = (X, Y ), Y 0, we associate a poit m as follows: The lie OM itersects the lie y = a i H. The lies passig from M ad H ad parallel to the x ad y axes itersect at m. Give m, we ca also fid M. 1. Calculate the coordiates of m i terms of the coordiates of M. 2. Fid the curve that the poit m traces whe M moves o the circle X 2 + Y 2 2RX = 0. 3. Fid the curve that the poit M traces whe m moves o the circle x 2 + y 2 2Ry = 0. Solutio.

14.5. SUPPLEMENT 151 14.5 Supplemet Theorem 14.5.1 (Cauchy s Mea-Value Theorem) If f ad g are both cotiuous o [a, b] ad differetiable o (a, b), the there exists a c (a, b) such that f (c)[g(b) g(a)] = g (c)[f(b) f(a)]. Proof: Let F : [a, b] R be defied by F (x) = f(x)[g(b) g(a)] g(x)[f(b) f(a)]. The F is cotiuous o [a, b] ad differetiable o (a, b). It is easy to check that F (a) = F (b). By Rolle s Theorem (Theorem 12.5.2), there is c (a, b) such that F (c) = 0. Note that Theorem 12.5.3 follows from the above oe by takig g(x) = x. Lemma 14.5.2 If f is differetiable the f satisfies the Mea Value Theorem, i.e. if a, b U, f (a) < γ < f (b), the f (c) = γ for some c (a, b). Proof: Sice f (a) < γ < f (b), there is a h > 0 such that f(a + h) f(a) h < γ < f(b + h) f(b). h Fix such a h > 0. The fuctio x f(x+h) f(x) h is cotiuous o (a, b). By the Itermediate Value Theorem (Theorem 11.3.3) there is a x = x(h) such that f(x+h) f(x) h = γ. Also, by Lemma 12.5.3, there is a c (x, x + h) such that f(x+h) f(x) h = f (c). Thus γ = f (c). Theorem 14.5.3 If f exists ad is bouded o some iterval I, the f is uiformly cotiuous o I. 14.5.1 Lipschitz Coditio 14.5.2 A Metric O R Lemma 14.5.4 i. For i = 1,...,, let (X i, δ i ) be a metric space. Let X = X 1... X. Let p > 1 be ay real umber. For x, y X let d p (x, y) = ( i=1 δ i(x i, y i ) p ) 1/p. The (X, d p ) is a metric space. ii. Let everythig be as above. Set d (x, y) = max{δ i (x i, y i ) : i = 1,..., }. The (X, d ) is a metric space. Proof: i. Settig a i = δ i (x, y), b i = δ i (x, z), c i = δ i (y, z), it is eough to show that if b i, c i 0, the ( (b i + c i ) p ) 1/p ( b p i )1/p + ( c p i )1/p. i=1 i=1 i=1

152 CHAPTER 14. GRAPH DRAWING Takig p-th powers, we eed to show that (b i + c i ) p (( b p i )1/p + ( c p i )1/p ) p. i=1 i=1 We first compute the left had side: (b i + c i ) p = i=1 p i=1 j=0 ( ) p b j j i cp j i = We ow compute the right had side: (( b p i )1/p + ( c p i )1/p ) p = i=1 = i=1 i=1 i=1 j=1 p j=0 b p i + p 1 c p i + ( ) p ( j i=1 b p i + c p i + p 1 ( ) p b j j i cp j i. i=1 i=1 j=1 i=1 ( ) p ( j i=1 b p i )j/p ( c p i )(p j)/p = i=1 Simplifyig, we see that we have to show that i=1 b j i cp j i ( b p i )j/p ( i=1 i=1 b p i )j/p ( c p i )(p j)/p. i=1 i=1 c p i )(p j)/p for all j = 1,..., p 1. Settig r = b p i, s = cp i, α = j/p ad β = (p j)/p, we see that it is eough to show that ri α s β i i=1 ( r) α ( s) β i=1 i=1 if 0 r i, 0 s i, 0 < α, β < 1 ad α + β = 1. We may assume that some r i 0 ad some s j 0, thus i=1 r i ad i=1 r i are ozero. Dividig both sides by ( i=1 r i) α ( i=1 r i) β, we see that we have to show that r i ( i=1 r ) α s ( i i i=1 s ) β 1. i By settig i=1 u i = r i i=1 r i we see that we have to prove that ad v i = u α i v β i = 1 i=1 r i i=1 s, i if 0 u i, 0 v i, i=1 u i = i=1 v i = 1, 0 < α, β < 1 ad α + β = 1.

14.5. SUPPLEMENT 153 Set f(α) = i=1 uα i vβ i = 1. Note that f(0) = f(1) = 1. The a easy calculatio shows that f (α) = i = 1 (l(u i )u α i v 1 β i l(v i )u α i v 1 β i ) = i = 1 u α i v 1 β i (l(u i ) l(v i )), ad that f (α) = i=1 (l(u i ) l(v i )) 2 u α i v 1 β i ). Thus f is cocave up, hece f(α) f(0) = f(1) = 1 for all α (0, 1).

154 CHAPTER 14. GRAPH DRAWING

Chapter 15 Riema Itegral 15.1 Defiitio ad Examples 15.2 Fudametal Theorem of Calculus 15.3 How To Itegrate? 15.3.1 Power Series How to itegrate polyomials ad power series... 15.3.2 Trigoometric Fuctios Exercises. i. Let I = π/2 0 si x dx. Show that I 2 = 1 3 5... (2 1) 2 4 6... (2) π 2 ad that I 2+1 = 2 4 6... (2) 1 3 5... (2 + 1). ii. [Wallis Formula] Let I be as above. Show that (I ) is a decreasig sequece. (Hit: (si x) is a decreasig sequece of fuctios). Coclude that (2 4 6... (2)) 2 (3 5... (2 1)) 2 (2 + 1) < π 2 < (2 4 6... (2 2))2 (2) (3 5... (2 1)) 2. Coclude that π 2 = lim (2 4 6... (2)) 2 (3 5... (2 1)) 2 2 155

156 CHAPTER 15. RIEMANN INTEGRAL ad that (2 4 6... (2)) 4 2 4 (!) 4 π = lim ((2)!) 2 = lim ((2)!) 2. The last equality is kow as Wallis formula. 15.4 Itegratio of Complex Fuctio Defie b f(t, z)dt. a 15.4.1 Fuctios Defied by Itegratio Theorem 15.4.1. Let f(z, t) be a fuctio where z A C ad t I, a iterval of R. Suppose I f(z, t)dt exists for all z A. The z f(z, t)dt is I a cotiuous fuctio from A ito C. 15.5 Applicatios 15.5.1 Applicatio to Series Theorem 15.5.1 (Itegral Test, Cauchy) Let f be a oicreasig positive real valued fuctio which is defied for x 1. The the series =1 f() coverges if ad oly if f(x)dx exists. Also, 1 ad 1 f(x)dx f() f(1) + =1 f() = =1 where +1 f(x)dx R f(x)dx. Proof: For < m we have, m f(i) + R i=1 f(x)dx f() + f( + 1) +... + f(m 1) 1 f(x)dx m 1 1 f(x)dx. This proves the first part. The secod part follows by takig = 1 ad 2 ad sedig m to ifiity i the above iequalities. The third part is easy as well. Assume ow the series diverges but that f(x) coverges to 0 whe x goes to ifiity. Let m = m() be ay iteger valued fuctio. The m m 0 f()+f(+1)+...+f(m 1) f(x)dx f(x)dx f(x)dx, 1 m 1

15.5. APPLICATIONS 157 so that whe teds to ifiity, we get, lim ( f() + f( + 1) +... + f(m 1) For example if we take m = 2 ad f(x) = 1/x, ( 1 lim + 1 1 + +... + 1 2 1 Sice 2 2 m ) f(x)dx = 0. ) dx/x = 0. dx/x = l(2) l() = l(2), we have, ( 1 lim + 1 1 + +... + 1 ) = l(2). 2 More geerally, let For m > we have σ = f(1) + f(2) +... + f() σ σ m = m 1 f(x)dx. f(x)dx (f( + 1) +... + f(m)). By the theorem this umber is oegative (replace ad m by +1 ad m+1 respectively) ad σ σ m = σ σ m = m f(x)dx (f() +... + f(m 1)) + f() f(). Sice lim f() = 0, the double sequece (σ m σ ) m, coverges to 0. Therefore the sequece (σ ) has a limit, say c f. Thus ( ) f(i) f(x)dx = c f. lim i=1 I particular if f(x) = 1/x, we see that (1 + 12 +... +... + 1 ) l() lim exists. This costat is called Euler s costat or Euler-Mascheroi costat. Sice the sequece is icreasig, the Euler costat is a positive umber. We ca ow state the followig result: Theorem 15.5.2 Let f(x) be a positive real valued oicreasig fuctio defied for x 1 ad which coverges to 0 as x goes to ifiity. The ( ) f(i) f(x)dx lim i=1 1 1

158 CHAPTER 15. RIEMANN INTEGRAL exists. I particular (1 + 12 +... +... + 1 ) l() = 0.57721... exists. lim Corollary 15.5.3 The series ɛ 0. =1 1 1+ɛ coverges if ɛ > 0 ad diverges if Proof: Follows from Theorem 15.5.1. ( ) Corollary 15.5.4 If lim 1 + l u l < 0 the the series u coverges ( ) l u absolutely. If lim 1 + l < 0 the the series u coverges absolutely. Proof: Follows from Corollary 15.5.3. Theorem 15.5.5 (Raabe ad Duhamel) If lim ( u u +1 1 ) > 1, the the series ( ) u u coverges absolutely. If u > 0 ad lim u +1 1 < 1 the the series u diverges. ( ) u Proof: I the first case, there is a ɛ > 0 such that u +1 1 > 1 + ɛ evetually. Thus u u +1 > 1 + 1+ɛ (1 + 1/) δ+1 evetually. Thus ( ) u +1 δ+1 u < = + 1 evetually. Let 0 < δ < ɛ. The 1 + 1+ɛ > 1/( + 1)δ+1 1/ δ+1 evetually. Sice 1/1+δ coverges, the result follows from Corollary 7.2.6. Assume ow we are i the secod case. The ( v v 1 1 ) < 1 evetually. Thus v 1 v > 1/(+1) 1/ evetually. Sice 1/ diverges, the secod part follows from Theorem 7.2.6 as well. Example. Cosider the series 1 + αz + α(α 1) z 2 +... 2! α(α 1)... (α + 1) z +...! where α C. The series coverges for z < 1 accordig to d Alembert (Corollary 7.5.2). Suppose ow z = 1. If u is the term of z, the u = + 1 α. u +1 Thus whe z = 1, by Raabe ad Duhamel s Covergece Rule, the series coverges ( absolutely ) for R(α) > 0, because a simple computatio shows that u u 1 1 coverges to 1 + R(α). If α R <0 ad x = 1 the series diverges as it ca be checked as above.

15.5. APPLICATIONS 159 Exercises. i. Show that the Euler costat is < 1. 15.5.2 Stirlig Formula We will try to fid a approximatio for!. Our first try: Lemma 15.5.6 lim l(!) l = lim l(!) l = 1. e Proof: Note that l(!) = i=1 l i. By the itegratio method, we have l < l + 1 = l x dx 1 i=1 l i +1 l x dx = ( + 1) l( + 1) ; 1 furthermore the quotiet of the first term to the last term coverges to 1: l l 1 (+1] l(+1) = (1+1/) l(+1) 1 This gives the result. = 1 1 l l(+1)/ l 1/ l l 1 l(+1) 1 l l(+1) 1/ 1/+1 1. Theorem 15.5.7 lim! (/e) 2π = 1. We ca obtai a better approximatio: Theorem 15.5.8! = (/e) 2π(1+1/12+f()/ 2 ) where f is a fuctio that is bouded. The proof will show that we ca exted the approximatio. But this will be doe by aother method i the ext subsectio. Questio: Does lim f() exist? Proof: Cosider the sequece For > 1, we have, s = l(!) (l 1) = l(!) l e. u := s s 1 = 1 + ( 1) l(1 1/) = 1 ( 1)(1/ + 1/2 2 + 1/3 3 +...) = 1 i=1 i(i+1) = 1/2 + 1/6 2 + g()/ 3 i where g() = i=3 1/i(i+1)i 3 is a positive valued fuctio whose limit whe goes to ifiity is fiite (see Exercise i, page 138). We ow itroduce (1) S := s 1 2 l = l(!) (l 1) 1 2 l

160 CHAPTER 15. RIEMANN INTEGRAL (See Exercise i, page 161). The v := S S 1 = u + 1 2 = = 1/2 + 1/6 2 + g()/ 3 + 1 2 l(1 1/) 1/2 + 1/6 2 + g()/ 3 1 2 (1/ + 1/22 + 1/3 3 +...) = (1/6 2 1/4 2 ) + g() (1/3+1/4+1/52 +...)/2 3 = 1/12 2 + h()/ 3 l 1 where lim h() is a costat. Settig v 1 = 1, we have S = v i = 1 12 i=1 1/i 2 + i=1 i=2 h(i) i 3. Thus the sequece (S ), i.e. the series i=1 v i has a limit say S. The S = S R where R = i=+1 v i = 1 12 i=+1 1 i 2 + i=+1 h(i) i 3. Now we have two remarks: First we show that 2 h(i) i=+1 i coverges to a 3 fiite umber whe goes to ifiity. Sice (h(i)) i is bouded, we may suppose h(i) = 1. By the itegral test, we have, 2 i=+1 1 i 3 = 2 i=1 1 ( + i) 3 < 1 2 2 ( + x) 3 dx = 2 2( + 2) 2, which shows what we wat. Thus we may replace i=+1 above by t()/ 2 where t() is a fuctio with a fiite limit: h(i) i 3 i the formula (2) R = i=+1 Here is our secod remark: so that (3) 1 + 1 = v i = 1 12 i=+1 1 i 2 + t() 2. 1 i 1 i + 1 = 1 i(i + 1) < 1 i 2 < 1 i(i 1) = 1 i 1 1 i, i=+1 ( 1 i 1 ) < i + 1 i=+1 1 i 2 < i=+1 ( 1 i 1 1 ) = 1 i. Let ( k() = 2 1 12 i=+1 ) 1 i 2.

15.5. APPLICATIONS 161 By (3), 0 < k() < 2 12 ( 1 1 +1 ) = 1 12(1+1/) < 1. Thus k() remais bouded. (Questio: Does it have a limit?) Note that (4) i=+1 1 i 2 = 1 12k() 2. Now by (1), (2) ad (4), l(!) (1) = S + l ( ) e + 1 2 l = S R + l ( ) e + 1 2 l (2) = S + 1 12 i=+1 1/i2 t() + l ( ) 2 e + 1 2 l (4) = S + 1 12 k() t() 2 + l ( ) 2 e + 1 2 l. By expoetiatig ad usig e x = 1 + x + x 2 /2! +..., we get, (! = e S ) ( 1 + 1 e 12 + f() ) 2 for some f() that remais bouded whe goes to ifiity. It remais to evaluate A := e S ad we use Wallis formula for this. We replace! by the formula foud above i Wallis formula, which we recall from page 155: 2 4 (!) 4 π = lim ((2)!) 2. We easily fid A = 2π. Thus! = ( ) ( 2π 1 + 1 e 12 + f() ) 2. Exercises. i. Show that s 1 2 l ca be made as close to a costat. (Solutio: We cotiue with the otatio of the proof of the theorem above. Note that s 1 = s s 1 = i=2 u i = i=2 (1/2i+1/6i2 +g(i)/i 3 ) = 1 2 i=2 1/i+ 1 6 i=2 1/i2 + i=2 g(i)/i3. Sice g(i) is bouded ad sice i=2 1/i2 ad i=2 1/i3 coverge, for large eough 1 6 i=2 1/i2 + i=2 g(i)/i3 ca be made as close to a costat as we wish to. O the other had i=2 1/i = i=1 1/i 1, ca be made as close to l 1 as we wish to (Theorem 15.5.2). Thus s ca be made close to A + 1 2 l for some costat A.) 15.5.3 Euler s Γ Fuctio Cosider the itegral Γ(z) = 0 e t t z 1 dt

162 CHAPTER 15. RIEMANN INTEGRAL where z C (but the reader may opt to take z R) ad, for 0 < a < b, the fuctio F (z) = F a,b (z) = b a e t t z 1 dt, which is a cotiuous fuctio of z by Theorem 15.4.1. Fixig a > 0 ad regardig this as a family of fuctios parametrized by b > a, we claim that for z varyig over a set whose real part is bouded above, F (z) coverges uiformly whe b. We have to show that for all ɛ > 0 there is b 0 such that for all b, b > b 0 ad all z i the right domai, b b e t t z 1 dt < ɛ. Ideed, let ɛ > 0 ad A be the upper boud for the real part of z. Let b 0 be such that t A 1 < e t/2 for t > b 0 ad e b0/2 < ɛ/4. The we have e t t z 1 = e t t x 1 t iy = e t t x 1 e iy l t = e t t x 1 e t t A 1 < e t/2, so that for b, b > b 0, b b e t t z 1 dt b e t t z 1 dt < b b b e t/2 dt = 2e b /2 + 2e b/2 < 4e b0/2 < ɛ. It follows that a e t t z 1 dt is a cotiuous fuctio of z for a > 0. Now we fix b ad view F (z) as a family of fuctios parametrized by 0 < a < b. We claim that the family F (z) coverges uiformly whe a 0 o ay set i which the real part of z is bouded below. For this we have to show that for ay ɛ > 0 there is a δ > 0 such that for all 0 < a < a < δ ad for all z C i the right domai (the real part above some fixed A > 0), a a e t t z 1 dt < ɛ. Ideed let ɛ > 0. Let A be the lower boud for the real part of z. For t [0, 1), we have e t t z 1 = e t t x 1 < t x 1 t A 1. Let δ < 1 be such that δ A < ɛa/2. The for all 0 < a < a < δ < 1, a a e t t z 1 dt a e t t z 1 dt a a a ta 1 dt = a A /A a A /A < 2a A /A < 2δ A /A < ɛ. Thus b 0 e t t z 1 dt is a cotiuous fuctio of z for a > 0. It follows from above that Γ(z) is well-defied ad is a cotiuous fuctio of z o the domai {z C : R(z) > 0}. By itegratig by parts, we have (du = e t dt, v = t z ), Γ(z + 1) = e t t z dt = [ e t ] 0 0 0 ( e t )zt z 1 dt = [ 1 + 1] + z e t t z 1 dt = zγ(z). 0

15.5. APPLICATIONS 163 We deduce that for N \ {0}, Γ(z + ) = (z + 1)(z + 2) zγ(z). O the other had Γ(1) = 0 e t dt = [ e t ] 0 = 1. Thus, settig z = 1 above we get, Γ( + 1) =! We proved the followig: Theorem 15.5.9 The fuctio Γ(z) = 0 e t t z 1 dt where z C is well-defied ad cotiuous for R(z) > 0. Furthermore ad Γ( + 1) =! Γ(z + ) = (z + 1)(z + 2) zγ(z) The fuctio Γ is called Euler s Gamma fuctio. We will soo exted its domai of defiitio from R(z) R >0 to R(z) R \ ( N). For this purpose we write Γ(z) = 1 0 e t t z 1 dt + 1 e t t z 1 dt. As we have see above the secod part is a cotiuous fuctio of z for all z C. We deal with the first oe. Let us replace e t by its Taylor series =0 ( 1) t /!. This series coverges uiformly for t [0, 1], thus also whe we multiply it with t z 1, hece we ca itegrate it term per term to get: 1 0 e t t z 1 dt = 1 0 ( =0 ( 1) t +z 1 /!)dt = = =0 [( 1) t +z /!( + z)] 1 0 = =0 ( 1) /!( + z) =0 [ 1 0 ( 1) t +z 1 /!]dt Such a series coverges uiformly o every compact subset ( closed bouded subset) ot cotaiig the opositive itegers accordig to Weierstrass M-test (Theorem 9.3.3), because the + z > δ > 0 for some δ ad so ( 1) /!( + z) = 1/! + z < 1/δ! ad =0 1/δ! coverges. Thus Γ(z) = ( 1) /!( + z) + =0 1 e t t z 1 dt. But this ew expressio is well-defied ad cotiuous for all z C \ ( N). We still call it Γ(z). Now we check that the fuctioal equality Γ(z + ) = (z + 1)(z + 2) zγ(z)

164 CHAPTER 15. RIEMANN INTEGRAL holds for the ew defiitio for all z C \ ( N). It is eough to show it for = 1. Ideed, Γ(z + 1) = =0 = =0 = =0 = =0 = z =1 ( 1)!(+z+1) + ( 1) ((+z+1) z) (+1)!(+z+1) ( 1) ((+z+1) z) 1 e t t z dt + 1 e t t z dt (+1)!(+z+1) + [ e t t z ] 1 + z e t t z 1 dt 1 ( 1) (+1)! + z ( 1) =1!(+z) + e 1 + z e t t z 1 dt 1 ( 1)!(+z) + z e t t z 1 dt = zγ(z). 1

Chapter 16 Supplemets 16.1 Stoe-Weierstrass Theorem Theorem 16.1.1 Let K be a compact metric space. Let A C(K) be a subalgebra. Suppose 1) A separates poits. 2) A cotais the costat fuctios. The A is dese i C(K) with respect to the uiform orm. Ex: Fid maximal ideals of C([a, b]). Aswer: Oly the expected oes. 165

166 CHAPTER 16. SUPPLEMENTS

Chapter 17 Fourier Series 17.1 Hilbert Spaces Let H be a real or complex vector space with a scalar product (, ). The x = (x, x) defies a ormed vector space. Suppose H is separable (has a coutable dese subset) ad complete. H is called a separable Hilbert space. Choose a coutable dese subset, the a coutable maximal liearly idepedet subset, the orthogoalize this subset by Gram-Schmidt. Theorem 17.1.1 Let H be a Hilbert space ad M H be a closed subspace. The there is a uique y M such that x y = d(x, M). Theorem 17.1.2 Ay Hilbert space is isomorphic to l 2. 17.2 Fourier Series Theorem 17.2.1 The fuctio 1 d 2 (f, g) = f g 2 = 2π π π f(x) g(x) 2 dx defies a pseudometric o the set of itegrable fuctios o the iterval [ π, π]. Theorem 17.2.2 Let f be a itegrable fuctio o [ π, π]. Let φ (x) = exp(ix) ad c = 1 π f(x) exp( ix)dx = f, φ. 2π π Let s (f; x) = N N c φ (x). The ( π ) f(x) s N (f; x) 2 dx = 0. lim N π 167

168 CHAPTER 17. FOURIER SERIES Furthermore for all, m Z φ m, φ = δ,m The L 2 -orm: f 2 = π π f(x) 2 dx. Lemma 17.2.3 f bouded itegrable. ɛ > 0. The there is a polygoal fuctio p o [ π, π] such f p 2 < ɛ. Proof: Trivial. Theorem 17.2.4 Let f ad g be two itegrable fuctios o [ π, pi]. f(x) c φ ad g d φ be its Fourier series. The Let f, g = c d =ifty ad f 2 = c 2. =ifty Lemma 17.2.5 If f is a bouded fuctio o [ π, π] which is itegrable the for all ɛ > 0 there is a h C([ π, π]) such that f h 2 < ɛ. R = Itegrable fuctios. C([ π, π]) = Cotiuous fuctios. F = Fiite liear combiatios of φ = φ : Z. s N (f; x) F N = φ : N N. L 2 = {f : π π f 2 dx < }. It t N F N the f s N f t N.

Chapter 18 Topological Spaces (cotiued) 18.1 Product Topology For each i i a idex set I, let X i be a topological space. Let X = i I X i. Cosider the smallest topology o X that makes all the projectio maps π i : X X i cotiuous. This is called the product topology. Exercises. i. The ope sets {x i I X i : x i1 U i1,..., x i U i } ( N ad U ij X ij ope) form a base of the topology. ii. If the topological spaces X i are discrete, the the ope sets {x i I X i : x i1 = a i1,..., x i = a i } ( N ad a ij X ij ) form a base of the product topology. iii. If I = N ad the topological spaces X i are discrete, the the ope sets {x i I X i : x 1 = a 1,..., x = a } ( N ad a i X i ) form a base of the product topology. iv. If X i = R (with the usual topology), the the ope sets {x i I X i : x i1 (a i1, b i1 ),..., x i (a i, b i1 )} 169

170 CHAPTER 18. TOPOLOGICAL SPACES (CONTINUED) ( N ad a ij, b ij X ij ) form a base of the product topology. v. Assume I is fiite ad each X i is a metric space. Show that the product topology o I X i is give by ay of the product metrics. vi. Suppose I = ω = N ad X i = N has the discrete topology. The, viewig ω X i as the set of fuctio from N ito N, we have: a. ω X i is metrisable by the followig metric d(x, y) = 1/( + 1) if is the smallest iteger for which x y. b. The set of ijective fuctios from ω ito ω is a closed subset of ω X i. c. Neither the set of surjective fuctios or Sym(ω) is a closed subset of ω X i. Problem 18.1.1 What is the closure of Sym(ω) i ω ω? Problem 18.1.2 Let Sym f (ω) be the set of permutatios of Sym(ω) that move oly fiitely may elemets of ω. Clearly Sym f (ω) is a subgroup of Sym(ω). What is its closure i Sym(ω)? Hausdorff Spaces. Let X be a topological space. If for ay two distict poits x ad y of X there are disjoit ope sets U x ad U y cotaiig x ad y, the we say that X is Hausdorff. The coarsest topology o a set with at least two poits is ot Hausdorff. Propositio 18.1.3 A metric space is Hausdorff. Exercises. i. Assume that each X i is a Hausdorff space. Show that the product topology i I X i is Hausdorff. 18.2 Homeomorphisms The otio of isomorphism betwee topological spaces is defied as follows: Two topological spaces X ad Y are called homeomorphic (i.e. isomorphic) if there is cotiuous bijectio f : X Y whose iverse is also cotiuous. Such a map is called a homeomorphism. 18.3 Sequeces i Topological Spaces Let X be a set, (x ) a sequece i X ad x X. We say that x is a limit of the sequece (x ) if for ay ope subset U cotaiig x, there is a atural umber N such that x U wheever > N. Lemma 18.3.1 Let A X be closed, (a ) a sequece from A ad x a limit of (a ). The x A.

18.4. SEQUENTIAL COMPACTNESS 171 Examples. 1. If the sequece is evetually costat, i.e. if there is a x ad a atural umber N such that x = x for > N, the x is a limit of the sequece (x ). 2. Let X have the coarsest topology. The ay poit of X is a limit of ay sequece of X. 3. Let X have the discrete topology. The a sequece has a limit if ad oly if the sequece is evetually costat, i.e. if there is a x ad a atural umber N such that x = x for > N. Propositio 18.3.2 I a Hausdorff topological space a sequece has at most oe limit. x. I this case we say that (x ) is a covergig sequece ad we write lim x = Propositio 18.3.3 Let X ad Y be two topological spaces. Let f : X Y. Show that f is cotiuous if ad oly if for ay coverget sequece (x ) of X, (f(x )) is coverget ad f(lim x ) = f(lim f(x )). 18.4 Sequetial Compactess Theorem 18.4.1 A metric space is compact if ad oly if it is sequetially compact. The coverse fails: Example. 18.5 Supplemets 18.5.1 T 0 -Idetificatio A topological space X is called T 0 if for ay distict x, y X there is a ope subset that cotais oly oe of the two poits. Lemma 18.5.1 A topological space X id T 0 if ad oly if for ay two distict x ad y i X, {x} = {x}. Proof: Left as a exercise. I a topological space defie x y if ad oly if {x} = {x}. This is a equivalece relatio. O X/ put the largest topology that makes the projectio map X X/ cotiuous. Propositio 18.5.2 X/ is a T 0 -topological space. Proof: HW. Problem 1. Fid the uiversal property of X/ that characterizes it. Problem 2. Try to do the same with T 1 -spaces.

172 CHAPTER 18. TOPOLOGICAL SPACES (CONTINUED)

Chapter 19 Exams 19.1 Midterm Math 121 (November 2002) i. Which of the followig are ot vector spaces over R (with the compoetwise additio ad scalar multiplicatio) ad why? (2+2+2+2+2+5 pts.) V 1 = {(x, y, z) R 3 : xy 0} V 2 = {(x, y, z) R 3 : 3x 2y + z = 0} V 3 = {(x, y, z) R 3 : xyz Q} V 4 = {(x, y) R 3 : x + y 0} V 5 = {(x, y) R 2 : x 2 + y 2 = 0} V 6 = {(x, y) C 2 : x 2 + y 2 = 0} Aswers. V 1 is ot a vector space because e.g. ( 1, 0, 0) V 1, (0, 1, 0) V 2 but their sum ( 1, 1, 0) V 1. V 2 is a vector space. V 3 is ot a vector space because e.g. (1, 1, 1) V 3, but 2(1, 1, 1) V 3. V 4 is ot a vector space because e.g. (1, 2, 1) V 4, but (1, 2, 1) V 4. V 5 is a vector space because V 5 = {(0, 0, 0)}. V 6 is ot a vector space because e.g. (1, i) V 6, (1, i) V 6, but their sum (1, 0) V 6. ii. O the set X = {2, 3..., 100} defie the relatio x y by x y ad x divides y. a) Show that this defies a partial order o X. (3 pts.) b) Is this a liear order? (2 pts.) c) Fid all the maximal ad miimal elemets of this poset. (5 pts.) Aswers. a) Yes, this is a partial order: Clearly x x for ay x. Sice divisio is trasitive, is trasitive as well. (The details are left). 173

174 CHAPTER 19. EXAMS b) No, because 2 ad 3 are ot comparable. c) The prime umbers are miimal elemets. The maximal elemets are the umbers which are greater tha 50. For example 53 is both miimal ad maximal. iii. O R R defie the relatio as follows (x, y) (x 1, y 1 ) by either y < y 1, or y = y 1 ad x < x 1. a) Show that this is a liear order. (5 pts.) b) Does every subset of this liear order which has a upper boud has a least upper boud? (5 pts.) Aswers. a) Yes! b) No. For example the set R {0} is bouded above by (0, 1) but it does ot have a least upper boud. iv. For each N, let a ad b be two real umbers. Assume that for each, a a +1 b +1 b. Show that N [a, b ] = [a, b] for some real umbers a ad b. (10 pts.) Proof: Sice the set {a : N} is bouded above by b 0, it has a least upper boud, say a. Similarly the set {b : N} has a greatest lower boud, say b. I claim that N [a, b ] = [a, b]. If x a, the x a for all. Likewise, if x b, the x b for all. Hece, if x [a, b], the x [a, b ] for all. Coversely, let x N [a, b ]. The a x b for all. Thus x is a upper boud for {a : N} ad a lower boud for {b : N}. Hece a x b. v. Show that for ay atural umber ad for ay real umber x [0, 1), (10 pts.) (1 x) 1 x + Proof: We proceed by iductio o. If = 0, the both sides are equal to 1. ( 1) x 2. 2 Suppose we kow the result for, i.e. suppose we kow that for ay real umber x [0, 1), (1 x) 1 x + ( 1) x 2. 2 We will prove that for ay real umber x [0, 1), (1 x) +1 1 ( + 1)x + ( + 1) x 2. 2

19.1. MIDTERM MATH 121 (NOVEMBER 2002) 175 Let x [0, 1). Sice (1 x) 1 x + ( 1) 2 x 2 ad sice 1 x > 0, multiplyig by 1 x both sides we get (1 x) +1 = (1 x) (1 x) (1 x + ( 1) 2 x 2 )(1 x) = 1 ( + 1)x + (+1) 2 x 2 ( 1) 2 x 3. Thus (1 x) +1 1 (+1)x+ (+1) 2 x 2 ( 1) 2 x 3. Sice x 0, (1 x) +1 1 ( + 1)x + (+1) 2 x 2 ( 1) 2 x 3 1 ( + 1)x + (+1) 2 x 2. vi. a) Show that for ay complex umber α there is a polyomial of the form p(x) = X 2 + ax + b R[X] such that p(α) = 0. (Note: a ad b should be real umbers). (10 pts.) b) What ca you say about a ad b if α = u + iv for some u, v Z? (5 pts.) Proof: a) Let α be a complex umber. The p(x) := (x α)(x α) = x 2 (α + α)x + αα R[x] ad it is easy to check that p(α) = 0. b) It is clear that if α = u + iv for some u, v Z, the p(x) Z[x]. Secod Proof of part a. Write α = u + iv where u ad v are real umbers. The α 2 = u 2 v 2 + 2uvi. Thus α 2 2uα = (u 2 v 2 + 2uvi) 2u(u + iv) = u 2 v 2, so that α 2 2uα + (u 2 + b 2 ) = 0. Hece α is a root of the polyomial p(x) = x 2 2ux + (u 2 + b 2 ) R[x]. Part b follows from this immediately. vii. a) Show that for ay α C there is a β C such that β 2 = α. (15 pts.) b) Show that for ay α, β C there is a x C such that x 2 +αx+β = 0. (10 pts.) Proof: a. Let α = a + bi. We try to fid β C such that β 2 = α, i.e. we try to fid two real umbers x ad y such that (x + iy) 2 = a + bi. We may assume that α 0 (otherwise take β = 0). Thus a ad b caot be both 0. After multiplyig out, we see that this equatio is equivalet to the system x 2 y 2 = a 2xy = b Sice x = 0 implies a = 0 = y = b, x must be ozero. Thus we have y = b/2x ad so the above system is equivalet to the followig: x 2 (b/2x) 2 = a y = b/2x Equalizig the deomiators i the first oe, we get the followig equivalet system: 4x 4 4ax 2 b 2 = 0 y = b/2x So ow the problem is about the solvability of the first equatio 4x 4 4ax 2 b 2 = 0. (Oce we fid x, which is ecessarily ozero, we set y = b/2x). Settig z = x 2, we see that the solvability of 4x 4 4ax 2 b 2 = 0

176 CHAPTER 19. EXAMS is equivalet to the questio of whether 4z 2 4az b 2 = 0 has a oegative solutio. Sice the last oe is a quadratic equatio over R, it is easy to aswer this questio. There are two possible solutios: z = a ± a 2 + b 2 ad oe of them z = a + a 2 + b 2 is oegative (eve if a is egative). Thus we ca take x = a + a 2 + b 2 ad y = b/2x. b. We first compute as follows: 0 = x 2 + αx + β = x 2 + αx + α 2 /4 + (β α 2 /4) = (x + α/2) 2 + (β α 2 /4). Thus if z C is such that z 2 = α 2 /4 β (by part a there is such a z), the x := z α/2 is a root of x 2 + αx + β. viii. Suppose X ad Y are two subsets of R that have least upper bouds. Show that the set X + Y := {x + y : x X, y Y } has a least upper boud ad that sup(x + Y ) = sup(x) + sup(y ). (15 pts.) Proof: Let a ad b be the least upper bouds of X ad Y respectively. Thus x a for all x X ad y b for all y Y. It follows that x + y a + b for all x X ad y Y, meaig exactly that a + b is a upper boud of X + Y. Now we show that a + b is the least upper boud of X + Y. Let ɛ > 0 be ay. We eed to show that a + b ɛ < x + y for some x X ad y Y. Sice a is the least upper boud of X, there is a x X such that a ɛ/2 < x. Similarly there is a y Y such that b ɛ/2 < y. Summig these two, we get a + b ɛ < x + y. ix. We cosider the subset X = {1/2 : N} {0} of R together with the usual metric, i.e. for x, y X, d(x, y) is defied to be x y. Show that the ope subsets of X are the cofiite subsets 1 of X ad the oes that do ot cotai 0. (20 pts.) Proof: We first show that the sigleto set {1/2 } is a ope subset of X. This is clear because B(1/2, 1/2 +1 ) = {1/2 }. It follows that ay subset of X that does ot cotai 0 is ope. Now let U be ay cofiite subset of X. We proceed to show that U is ope. If 0 U, the we are doe by the precedig. Assume 0 U. Sice U is cofiite, there is a atural umber such that for all, 1/2 U, i.e. B(0, 1/2 ) U. Now U is the uio of B(0, 1/2 ) ad of a fiite subset ot cotaiig 0. Thus U is ope. For the coverse, we first show that a oempty ope ball is of the form described i the statemet of the questio. If the ceter of the ball is 0, the the ball is cofiite by the Archimedea property. If the ceter of the ball is ot 0, the either the ball does ot cotai 0 or else it does cotai 0, i which case the ball must be cofiite. To fiish the proof, we must show that a arbitrary uio of ope balls each of which does ot cotai 0 caot cotai 0. But this is clear! 1 A subset Y of X is called cofiite if X \ Y is fiite.

19.2. FINAL MATH 121 (JANUARY 2003) 177 19.2 Fial Math 121 (Jauary 2003) Justify all your aswers. A ojustified aswer will ot receive ay grade whatsoever, eve if the aswer is correct. DO NOT use symbols such as,,. Make full seteces with correct puctuatio. You may write i Turkish or i Eglish. i. Let (a ) be a coverget sequece of real umbers. a. Does the sequece (a 2 ) coverge ecessarily? (2 pts.) Sice (a 2 ) is a subsequece of the covergig sequece (a ), both sequeces coverge to the same limit. Remarks. We have see i class that a subsequece of a covergig sequece coverges. Cotrary to what some of you thik, a 2 2a! b. Assume a 0 for all. Does the sequece (a /a +1 ) coverge ecessarily? (2 pts.) No, the sequece (a /a +1 ) may ot coverge if lim a = 0. For example, choose { 1/ if is eve a = 1/ 2 if is odd Clearly lim a = 0, but a = a +1 (+1) 2 +1 2 if is eve if is odd Ad the subsequece (+1)2 diverges to, although the subsequece +1 2 coverges to 0. O the other had, if the limit of the sequece (a ) is ozero, say l, the the sequece (a /a +1 ) coverges to 1 because lim a /a +1 = lim a / lim a +1 = l/l = 1. Note that the last part uses the fact that l is ozero. ii. Fid the followig limits ad prove your result usig oly the defiitio. (30 pts.) 2 5 a. lim 5+2. We claim that the limit is 2/5. Let ɛ > 0. Sice R is Archimedea, there is a N such that 2 < ɛn. Now for > N, 2 5 5+2 2 5 = 29 5(5+2) < 29 25 < 2/ < 2/N < ɛ. 2 b. lim 2 5 5+2. We claim that the limit is. lim 2 2 5 5 2 For this it is eough to prove that =. Let A be ay real umber. Let N = max(5a, 2).

178 CHAPTER 19. EXAMS For > N we have 22 5 5 2 proves that lim 2 2 5 5 2 =. > 22 5 5 > 22 2 5 = 2 5 = 5 > N 5 A. This 2 c. lim 2 5 3 +2. We claim that the limit is 0. Let ɛ > 0. Let N 1 be such that 2 > ɛn 1 (Archimedea property of R). Let N = max(1, N). The for all > N, 2 2 5 3 + 2 = 22 5 3 + 2 < 22 3 + 2 22 3 = 2 < 2 N < ɛ. Note that the first equality is valid because 3 5 0, the secod iequality is valid because 3 + 2 > 0. iii. Fid (16 pts. Justify your aswers). a. lim (1/2 + 1/). Note that for 3, 0 < 1/2 + 1/ 1/2 + 1/3 = 5/6. Thus the sequece ((1/2+1/) ) is evetually squeezed betwee the zero costat sequece ad the sequece ((5/6) ). Sice lim (5/6) = 0 (because 5/6 < 1), lim (1/2 + 1/) = 0. b. lim (3/2 7/). Sice 3/2 > 1 ad lim 7/ = 0, there is a N such that 7/N < 1/2 = 3/2 1. I fact, it is eough to take N = 15. The for all N, 3/2 7/ 3/2 7/N > 1 ad so (3/2 7/) (3/2 7/N). Therefore the sequece ((3/2 7/) ) is greater tha the sequece ((3/2 7/N) ). Sice 3/2 7/N > 1, the sequece ((3/2 7/N) ) diverge to. Hece lim (3/2 7/) =. iv. Fid lim ( 2 1 3 5 ) 2 1 2 3. (10 pts. Justify your aswer). Assume > 5. The we have, 3 5 > 3 2 > 3 3 /2 = 3 /2 > 0. Therefore, 2 1 3 5 = 2 1 3 5 < 2 3 5 < 2 3 /2 = 2/. Also 2 1 2 3 > 2 1 2 > 2 2 = 1 2 > 2. Hece It follows that lim ( ( 2 ) 1 2 1 2 3 3 < (2/) 2. 5 ) 2 1 2 2 3 1 3 5 = 0. v. Show that the series =1 (1/) coverges. Fid a upper boud for the sum. (10 pts.) Sice for 2, 1/ 1/2, we have =1 (1/) 1 + =2 (1/) 1 + =1 (1/2) = 1 + 1 2 =0 (1/2) = 1 + 1/2 = 3/2.

19.3. RESIT OF MATH 121, FEBRUARY 2003 179 vi. Let (a ) be a sequece of oegative real umbers. Suppose that the sequece (a 2 ) coverges to a. Show that the sequece (a ) coverges to a. (15 pts.) Note first that, sice a 0, a 0 as well. Let ɛ > 0. Case 1: a > 0. Sice lim a 2 = a, there is a N 2 such that for all > N, a 2 a < ɛa. Now for all > N, a a = a2 a a +a a2 a a < ɛ. Case 2. a = 0. Sice the sequece (a 2 ) coverges to 0, there is a N such that for all > N, a 2 < ɛ 2. So (ɛ + a )(ɛ a ) = ɛ 2 a 2 > 0. Sice ɛ > 0 ad a 0, we ca divide both sides by ɛ + a to get ɛ a > 0, i.e. a < ɛ. Sice a 0, this implies a < ɛ. vii. We have see i class that the sequece give by a = ((1 + 1/) ) coverges to a real umber > 1. Let e be this limit. Do the followig sequeces coverge? If so fid their limit. (15 pts. Justify your aswers). a) lim (1 + 1 +1). ) lim (1 + 1 (1+ +1 = 1 lim +1) +1 = lim (1+ +1) 1 +1 = e 1+ 1 +1 lim 1+ 1 1 = +1 e. The first equality is algebraic. The secod equality holds because the limits of the umerator ad the deomiator ( exist ad they are ozero. ( ) ) +1 The third equality holds because 1 + 1 +1 is a subsequece of (( 1 + 1 ) ). ( b) lim 1 + 1 3. 2) ( ) ( lim 1 + 1 3 (1 ) 2 = lim + 1 2 3/2 2) = e 3/2 by Questio vi. ( c) lim 1 + 1 2 ). ( Sice lim 1 + 1 ) = e > 1, there is a N1 such that for all > N 1, ( 1 + ) 1 e < e 1 2, so 1 < r := e 2 + 1 2 = e e 1 2 < (. 1 + ) 1 Thus, r < (( 1 + ) 1 ) ( = 1 + 1 2 ). Thus = limr r ( ) 1 + 1 2 (because r > 1). It follows that ( 1 + ) 1 2 =. 19.3 Resit of Math 121, February 2003 i. Fid a sequece either decreasig or icreasig that coverges to 1. (2 pts.) ii. Let (a ) be a coverget sequece of real umbers. Suppose that a Z for all. Is it true that lim a Z? (4 pts.)

180 CHAPTER 19. EXAMS iii. Let (a ) be a coverget sequece of real umbers. Suppose that a Q for all. Is it true that lim a Q? (3 pts.) iv. Let (a ) be a coverget sequece of real umbers. Suppose that 5a /2 N for all. What ca you say about lim a? (2 pts.) v. Let (a ) be a sequece of real umbers such that the subsequece (a 2 ) coverges. Does the sequece (a ) coverge ecessarily? (2 pts.) vi. Let (a ) be a sequece of real umbers such that the sequece (a 2 ) coverges to 1. Does the sequece (a ) coverge ecessarily? (2 pts.) vii. Let (a ) be a sequece of real umbers such that the subsequeces (a 2 ) ad (a 2+1 ) both coverge. Does the sequece (a ) coverge ecessarily? (2 pts.) viii. Let (a ) be a sequece of real umbers such that the sequece (a 2 ) coverges to 0. Does the sequece (a ) coverge ecessarily? (8 pts.) ix. Let (a ) be a sequece of real umbers such that lim a =. Is it true that lim a 2 =? (3 pts.) x. Assume lim a exists ad a 0 for all. Does the sequece (a 2 /a 2+1 ) coverge ecessarily? (5 pts.) xi. Fid the followig limits ad prove your result usig oly the defiitio. (30 pts.) a. lim 3+105 5 79 b. lim 2 5+3 100+2 c. lim 8 2 3 89 xii. Fid (16 pts. Justify your aswers). a. lim ( 2 3 + 6 2 +1 b. lim ( 5 4 7 5 ) xiii. Fid lim ( 2 1 3 5 ) 3 ) 2 1 2 3. (10 pts. Justify your aswer). xiv. Show that the series ( /3 =1 +1) coverges. Fid a upper boud 2 for the sum. (10 pts.) xv. Let (a ) be a sequece of real umbers. Assume that there is a r > 1 such that a +1 r a for all. What ca you say about the covergece or the divergece of (a )? (6 pts.)

19.4. CORRECTION OF THE RESIT OF MATH 121, FEBRUARY 2003181 19.4 Correctio of the Resit of Math 121, February 2003 i. Fid a sequece either decreasig or icreasig that coverges to 1. (2 pts.) Aswer: Let a = 1 + ( 1). It is clear that lim a = 1. Sice the subsequece (a 2 ) is decreasig ad coverges to 1 ad the subsequece (a 2 ) is icreasig ad coverges to 1, the sequece (a ) is either icreasig or decreasig. ii. Let (a ) be a coverget sequece of real umbers. Suppose that a Z for all. Is it true that lim a Z? (4 pts.) Aswer: Yes, it is true. I fact this is true eve for Cauchy sequeces: A Cauchy sequece (a ) whose terms are i Z is evetually costat, i.e. there is a N such that a = a N for all N, ad this implies of course that lim a = a N Z. So, let us show that the Cauchy sequece (a ) is evetually costat. I the defiitio of Cauchy sequeces, take ɛ = 1/2. Thus, there is a M such that for all, m > M, a a m < 1/2. But sice a ad a m are i Z, this meas that for all, m > M, a a m = 0, i.e. that a = a m. Now take N = M + 1. iii. Let (q ) be a coverget sequece of real umbers. Suppose that q Q for all. Is it true that lim q Q? (3 pts.) Aswer: Of course ot! I fact every real umber is the limit of a ratioal sequece. Ideed, let r R. Let N \ {0}. Sice Q is dese i R, there is a ratioal umber q (r 1/, r). Sice r 1/ < q < r, by the Sadwich Lemma, lim q = r. iv. Let (a ) be a coverget sequece of real umbers. Suppose that 5a /2 N for all. What ca you say about lim a? (4 pts.) Aswer: Let lim a = r. The lim 5a /2 = 5r/2. By hypothesis ad by part ii, 5r/2 Z. Thus r = 2/5 for some N. v. Let (a ) be a sequece of real umbers such that the subsequece (a 2 ) coverges. Does the sequece (a ) coverge ecessarily? (2 pts.) Aswer: Of course ot! We ca have a 2 = 1/ ad a 2+1 =. vi. Let (a ) be a sequece of real umbers such that the sequece (a 2 ) coverges to 1. Does the sequece (a ) coverge ecessarily? (2 pts.) Aswer: Of course ot! We ca have a = ( 1). The (a ) is a sequece of alteratig oes ad mius oes, so that it diverges. Ad sice a 2 = 1, the sequece (a 2 ) coverges to 1.

182 CHAPTER 19. EXAMS vii. Let (a ) be a sequece of real umbers such that the subsequeces (a 2 ) ad (a 2+1 ) both coverge. Does the sequece (a ) coverge ecessarily? (2 pts.) Aswer: Of course ot! We ca have a = ( 1). The (a ) is a sequece of alteratig oes ad mius oes, so that it diverges. Ad sice a 2 = 1 ad a 2+1 = 1, the sequece (a 2 ) coverges to 1 ad the sequece (a 2+1 ) coverges to 1. viii. Let (a ) be a sequece of real umbers such that the sequece (a 2 ) coverges to 0. Does the sequece (a ) coverge ecessarily? (8 pts.) Aswer: Yes! Let ɛ > 0. Let ν = ɛ. Sice the sequece (a 2 ) coverges to 0, there is a N such that for all > N, a 2 < ν, i.e. a 2 < ɛ 2. Sice a ad ν are positive, this implies that a < ɛ. Thus there is a N such that for all > N, a < ɛ; i.e. the sequece (a ) coverges to 0. ix. Let (a ) be a sequece of real umbers such that lim a =. Is it true that lim a 2 =? (3 pts.) Aswer: Yes! Let A be ay real umber. lim a =, there is a N such that for all > N, a > A. The for 2 > N, a 2 > A. x. Assume lim a exists ad a 0 for all. Does the sequece (a 2 /a 2+1 ) coverge ecessarily? (5 pts.) Aswer: No, the sequece (a 2 /a 2+1 ) may ot coverge if lim a = 0. For example, choose { 1/ if is eve a = 1/ 2 if is odd Clearly lim a = 0, but a = a +1 (+1) 2 +1 2 if is eve if is odd Ad the subsequece (+1)2 diverges to, although the subsequece +1 2 coverges to 0. O the other had, if the limit of the sequece (a ) is ozero, say l, the the sequece (a 2 /a 2+1 ) coverges to 1 because lim a 2 /a 2+1 = lim a 2 / lim a 2+1 = l/l = 1. Note that the last part uses the fact that l is ozero. xi. Fid the followig limits ad prove your result usig oly the defiitio. (30 pts.) a. lim 3+105 5 79 Aswer: lim 3+105 5 79 = 3 5.

19.4. CORRECTION OF THE RESIT OF MATH 121, FEBRUARY 2003183 Proof: Let ɛ > 0. Let N 1 be such that 32 < ɛn 1. Let N = max(n 1, 395). Now for > N, we have, 3 + 105 5 79 3 5 = 762 25 395 = 762 25 395 762 24 < 32 < 32 < ɛ. N 1 The first equality is simple computatio. The secod equality follows from the fact > N 395 > 16 (so that 25 395 > 0). The third iequality follows from the fact that > N 395, so that 25 395 25 = 24.The fourth iequality is also a simple computatio. b. lim 2 5+3 100+2 Aswer: lim 2 5+3 100+2 =. Proof: It is eough to show that lim 2 5+3 100 2 =. We first ote that the two roots of 2 5 + 3 are 5± 25 12 2 = 5± 13 2, so that if 5 > 9 2 = 5+ 16 2 > 5+ 13 2, the 2 5 + 3 > 0. Now let A R be ay real umber. Let N = max(100a + 5, 5). Now, for all > N, 2 5 + 3 100 2 > 2 5 + 3 100 > 2 5 100 = 5 100 > N 5 100 = A. Here, the first iequality follows from the fact that > N 5, so that 2 5 + 3 > 0. c. lim 8 2 3 89. 8 Aswer: lim 2 3 89 = 0. Proof: Let ɛ > 0. Let N = max(1/ɛ, 89). Now for > N, 8 2 3 89 < 2 3 = 1 2 2 1 < 1 < 1 2 < ɛ. xii. Fid (16 pts. Justify your aswers). ( 3. 2 a. lim 3 + 6 +1) 2 Aswer: lim ( 2 3 + 6 2 +1) 3 = 0. 6 2 +1 Proof: We use the fact that 2/3 < 1. Sice lim ( 6 a N such that for all > N, 2 +1 < 1/6. The 0 8 2 3 89 = = 0, there is 3 2 3 + 6 +1) = 2 (2/3+1/6) 3 = (5/6) 3. By Sadwich Lemma lim ( 2 3 + 6 2 +1) 3 = 0. b. lim ( 5 4 7 5 ). Aswer: lim ( 5 4 7 5 ) =.

184 CHAPTER 19. EXAMS 7 Proof: We use the fact that 5/4 > 1. Sice lim = 0, there is a N 5 7 such that for all > N, < 1/8. The ( 5 5 4 ) 7 > (5/4 1/8) = 5 (9/8) (9/8). Sice (9/8) > 1, lim (9/8) =. The result follows. xiii. Fid lim ( 2 1 3 5 Aswer: lim ( Proof: Sice lim ( > N 1, ) 2 1 2 3. (10 pts.). ) 2 1 2 2 3 1 3 5 = 0. ) = 0, there is a N 1 such that for all 2 1 3 5 2 1 3 5 < 1/2. O the other had, for > 3, 2 1 2 3 < 2 1 Let N = max(3, N 1 ). Now for > N, ( ) 2 1 2 2 3 1 3 5 <. < (1/2) 2 1 2 3 < (1/2). Sice the right had side coverges to 0, by Sadwich Lemma, ( ) 2 1 0 lim 2 2 3 1 = 0. (For the first iequality, oe eeds the 3 5 fact that 3 5 > 0 for 2. This follows from the facts that 2 3 2 5 = 1 > 0 ad 3 5 < ( + 1) 3 ( + 1) 5. Ad this last iequality is easy to show). xiv. Show that the series ( /3 =1 +1) coverges. Fid a upper boud 2 for the sum. (10 pts.) Aswer: ( ) /3 =1 2 +1 = =1 (1/)/3 < =1 1/2/3 = =1 1/23/3 + 3+1 =0 1/2 3 + 3+2 =0 1/2 3 = =1 1/2 + 1 2 1/3 =0 1/2 + 1 2 1/3 =0 1/2 = 1/2+2 1/3 +2 2/3 < 5. xv. Let (a ) be a sequece of real umbers. Assume that there is a r > 1 such that a +1 r a for all. What ca you say about the covergece or the divergece of (a )? (6 pts.) Aswer: The sequece diverges. Furthermore the sequece diverges to if it is evetually positive ad to if it is evetually egative. Proof: Oe ca show by iductio o that a > r a 0. Thus lim a = (because r > 1). It should ow be clear that the aswer is valid. 19.5 Secod Resit of Math 121, March 2003 i. Let (a ) ( be a covergig ( sequece( of ozero ) real umbers. Do the sequeces a a +1 ), 1 a 1 a +1 ad a coverge? (6 pts.) ) 1+a 2 +1

19.5. SECOND RESIT OF MATH 121, MARCH 2003 185 ii. Let (a ) be a coverget sequece of atural umbers. Is it true that lim a N? (4 pts.) iii. Let x > 1. Discuss the covergece of (x /!) (4 pts.) iv. Let x R. Discuss the covergece of (x! /!) (8 pts.) v. Let (a ) be a coverget sequece of real umbers. Suppose that lim a Q. Is it true that a Q for ifiitely may? (3 pts.) vi. Let (a ) be a sequece of real umbers such that the subsequeces (a 2 ), (a 2+1 ) ad (a 7+1 ) all coverge. Does the sequece (a ) coverge ecessarily? (5 pts.) vii. Let (a ) be a sequece of real umbers such that the sequece (a 2 ) coverges. Discuss the covergece of the sequece (a ). (5 pts.) viii. Let (a ) be a sequece of real umbers such that lim a =. Is it true that lim a 2 =? (2 pts.) ix. Let (a ) be a sequece of real umbers such that lim a =. Discuss the covergece of lim 1/a 2 = 0. (10 pts.) x. Fid the followig limits ad prove your result usig oly the defiitio. (18 pts.) a. lim 3 2 5 5 2 +3. b. lim 2 5+3 4+2 c. lim 5 2 4 2 3 +2 xi. Fid lim ( 2 1 3 5 ) 2 1 2 3. (10 pts. Justify your aswer). xii. Show that the series =0 ( 1) z 2+1 /(2 + 1)! coverges for all z R. (4 pts.) xiii. Show that the series ( /3 =1 +1) coverges. Fid a upper boud 2 for the sum. (10 pts.) xiv. Let (a ) be a sequece of real umbers. Assume that there is a r > 1 such that a +1 r a for all. What ca you say about the covergece or the divergece of (a )? (6 pts.)

186 CHAPTER 19. EXAMS 19.6 Midterm of Math 152, April 2004 i. Decide the covergece of the series ii. Decide the covergece of the series iii. Decide the covergece of the series 1 2 2. 1 2 + 1. 1 4 6. iv. Suppose that the series a is coverget. Show that lim a = 0. v. Suppose that (a ) is a positive ad decreasig sequece ad that the series a is coverget. Show that lim a = 0. vi. Fid a positive sequece (a ) such that the series a is coverget but that lim a 0. vii. Suppose that series a is absolutely coverget ad that the sequece (b ) is Cauchy. Show that the series a b is absolutely coverget. viii. Let (a ) be a sequece. Suppose that =1 a a +1 coverges. Such a sequece is called of bouded variatio. Show that a sequece of bouded variatio coverges. 19.7 Fial of Math 152, Jue 2004 I. Coverget Sequeces. For each of the topological spaces (X, τ), describe the coverget sequeces ad discuss the uiqueess of their limits. i. τ = (X). ( (X) is the set of all subsets of X, 2 pts.). Aswer: Oly the evetually costat sequeces covergig to that costat. ii. τ = {, X} (2 pts.). Aswer: All sequeces coverge to all elemets. iii. a X is a fixed elemet ad τ is the set of all subsets of X that do ot cotai a, together with X of course. (5 pts.) Aswer: First of all, all sequeces coverge to a. Secod: If a sequece coverges to b a, the the sequece must be evetually the costat b.

19.7. FINAL OF MATH 152, JUNE 2004 187 iv. a X is a fixed elemet ad τ is the set of all subsets of X that cotai a, together with of course. (5 pts.) Aswer: Oly the evetually costat sequeces coverge to a. A sequece coverge to b a if ad oly if the sequece evetually takes oly the two values a ad b. v. τ is the set of all cofiite subsets of X, together with the of course. (6 pts.) Aswer: All the sequeces without ifiitely repeatig terms coverge to all elemets. Evetually costat sequeces coverge to the costat. There are o others. II. Subgroup Topology o Z. Let τ = {Z + m :, m Z, 0} { }. We kow that (Z, τ) is a topological space. i. Let a Z. Is Z \ {a} ope i τ? (5 pts.) Aswer: Yes. 0,±1 Z (3Z + 2) = Z \ {1}.Traslatig this set by a 1, we see that Z \ {a} is ope. ii. Fid a ifiite o ope subset of Z. (5 pts.) Aswer: The set of primes is ot a ope subset. Because otherwise, for some a 0 ad b Z, the elemets of az + b would all be primes. So b, ab + b ad 2ab + b would be primes, a cotradictio. iii. Let a, b Z. Is the map f a,b : Z Z defied by f a,b (z) = az + b cotiuous? (Prove or disprove). (10 pts.) Aswer: Traslatio by b is easily show to be cotiuous. Let us cosider the map f(z) = az. If a = 0, 1, 1 the clearly f is cotiuous. Assume a 0, ±1 ad that f is cotiuous. We may assume that a > 1 (why?) Choose a b which is ot divisible by a. The f 1 (bz) is ope, hece cotais a subset of the form cz + d. Therefore a(cz + d) bz. Therefore ac = ±b ad so a divides b, a cotradictio. Hece f is ot cotiuous uless a = 0, ±1. iv. Is the map f a,b : Z Z defied by f(z) = z 2 cotiuous? (Prove or disprove). (5 pts.) Aswer: No! Left as a exercise. v. Is the topological space (Z, τ) compact? (Prove or disprove).(15 pts.) Aswer: First Proof: Note first the complemet of ope subsets of the form az + b are also ope as they are uios of the form az + c for c = 0, 1,..., a 1 ad c b mod a. Now cosider sets of the form U p = pz + (p 1)/2 for p a odd prime. The p U p = because if a p U p the for some x Z \ { 1}, 2a + 1 = px + p, so that a is divisible by all primes p ad a = 0. But if a = 0 the (p 1)/2 is divisible by p, a cotradictio. O the other had o fiite itersectio of the U p s

188 CHAPTER 19. EXAMS ca be emptyset as (az + b) (cz + b) if a ad b are prime to each other (why?) Hece (U c p) p is a ope cover of Z that does ot have a fiite cover. Therefore Z is ot compact. First Proof: Let p be a prime ad a = a 0 + a 1 p + a 2 p 2 +... be a p-adic iteger which is ot i Z. Let b = a 0 + a 1 p + a 2 p 2 +... + a 1 p 1. The p Z + b = but o fiite itersectio is empty. We coclude as above. III. Miscellaeous. i. Let f : R R be the squarig map. Suppose that the arrival set is edowed with the usual Euclidea topology. Fid the smallest topology o the domai that makes f cotiuous. (5 pts.) Aswer: The smallest such topology is the set {U U : U ope i the usual topology of R}. ii. Let τ be the topology o R geerated by {[a, b) : a, b R}. Compare this topology with the Euclidea topology. (3 pts.) Is this topology geerated by a metric? (20 pts.) Aswer: Ay ope subset of the Euclidea topology is ope i this topology because (a, b) = =1[a + 1/, b). But of course [0, 1) is ot ope i the usual topology. Assume a metric geerates the topology. Note that [0, ) is ope as it is the uio of ope sets of the form [0, ) for N. Thus the sequece ( 1/) caot coverge to 0. I fact for ay b R, o sequece ca coverge to b from the left. Thus for ay b R there is a ɛ b > 0 such that B(b, ɛ b ) [b, ). Let b 0 be ay poit of R. Let ɛ 0 > 0 be such that B(b 0, ɛ 0 ) [b 0, ). Sice {b 0 } is ot ope, there is b 1 B(b 0, ɛ 0 ) \ {b 0 }. Let 0 < ɛ 1 < ɛ 0 /2 be such that B(b 1, ɛ 1 ) [b 1, ) B(b 0, ɛ 0 ). Iductively we ca fid (b ) ad (ɛ ) such that B(b, ɛ ) [b, ) B(b 1, ɛ 1 )\ {b 1 } ad ɛ < ɛ 0 /2. The (b ) is a strictly icreasig coverget sequece, a cotradictio. iii. Show that the series i=0 x /! coverges for ay x R (10 pts.) Show that the map exp : R R defied by exp(x) = i=0 x /! is cotiuous. (10 pts.) Aswer: For the first part show that the sequece of partial sums is Cauchy. The secod part is easy as well, just write dow.

Bibliography [A] Stephe Abbott, Uderstadig Aalysis, Spriger 2001. [AZ] [BR] [PZ] [V] Marti Aiger ad Güter M. Ziegler, Proofs from THE BOOK, secod corrected pritig, Spriger 1999. G. Bouligad ad J. Rivaud, L eseigemet des mathématiques géérales par les problèmes, Librairie Vuibert, 4th editio, 1968. William R. Parzyski ad Philip W. Zipse, Itroductio to Mathematical Aalysis, McGraw-Hill 1987. Georges Valiro, Théorie des Foctios, Masso et C ie, 1941, third editio 1966. 189

Idex <, 13 ω R, 53 +, 13, 14, 13 >, 13, 13 0 1, 16 0, 13, 14 0 0, 30 1, 13 2, 20 Abel s Lemma, 120 Abel s Theorem, 120 abelia group, 16 absolute covergece, 87 absolute extremum, 144 absolute maximum, 144 absolute miimum, 144 absolute value of a real umber, 21 additio, 14 16 additio of real umbers, 13 algebraic umbers, 95 alteratig series, 88 aalytic fuctios, 150, 154 Archimedea Property, 29 associative biary operatio, 9 associativity of series, 85 associativity of the additio, 14 associativity of the multiplicatio, 16 Baach algebra, 53 Baach Space, 53 base of a topology, 128 biary operatio, 9 Biary Relatios, 10 11 Bolzao-Weierstrass Theorem, 71, 73 bouded away from a, 64 bouded sequece, 59 bouded set, 59 C, 39 cartesia product, 43 Cauchy Codesatio Test., 87 Cauchy sequece, 68 Cauchy s Mea-Value Theorem, 163 ceter of a ball, 54 chai, 21 circle, 54 closed subset, 127 closure, 128 coefficiets of a polyomial, 65 commutative biary operatio, 9 commutative group, 16 commutativity, 17 compact, 129 Compariso Test, 86 complete metric space, 79 completeess, 23 complex umbers, 39 42 coditioal covergece, 92 cojugate of a complex umber, 41 coected topological space, 131 costat sequece, 57 cotiuous at a poit, 133 cotiuous fuctio, 134, 135 cotractive sequece, 75 covergece, 57 covergece of series, 83 coverget sequece, 57 cosie, 90 190

INDEX 191 decimal expasio, 92 decreasig sequece, 33 degree, 65 dese, 34 derivative, 143 differetiable at a poit, 143 discoected, 131 discrete metric, 51 discrete topology, 125 distributivity, 17 18 divergece of series, 83 divergece to ifiity, 76 diverget sequece, 57 divisio, 36 divisio of itegers, 29 e, 74, 156 equivalece relatio, 11 Euclidea metric space, 51 Euler s costat, 169 Euler s Gamma fuctio, 175 Euler-Mascheroi costat, 169 evetually costat sequece, 57 expoetiatio, 30, 90 exteded real lie, 126 extremum, 144 factorial, 31 Fiboacci sequece, 82 field, 18 fiest topology, 125 Gamma fuctio, 175 geometric series, 85 glb(a), 22 group, 14 harmoic series, 85 Hausdorff space, 182 Heie-Borel Theorem, 129 Hilbert space, 179 homeomorphism, 182 i, 40 Id A, 10 idetity elemet, 14, 16 idetity elemet of a biary operatio, 9 icreasig sequece, 33 iduced topology, 126 iductive, 27 if(a), 22 itegers, 34 itegral part, 29, 36 Itegral Test, 168 iterior, 127 Itermediate Value Theorem, 138 itervals, 21 iverse, 14, 16 iverse elemet, 14, 16 iverse elemet i a biary operatio, 10 iverse of a fuctio (bijectio), 10 isolated poit, 133 less tha, 13 lexicographique orderig, 23 lim, 78 limit, 57, 182 limit at ifiity, 123 limit iferior, 78 limit of a fuctio, 121, 122 limit poit, 121, 133 limit superior, 78 lim, 78 liearly ordered set, 21 Liouville umber, 96 Lipschitz coditio, 163 local extremum, 144 local maximum, 144 local miimum, 144 lub(a), 22 maximal elemet, 22 Mea Value Theorem of Differetial Calculus, 145 metric space, 50 metrisable topology, 125 mius, 14 Mootoe Covergece Theorem, 70 mootoe sequece, 70 m th -root, 31

192 INDEX multiplicatio, 16 17 multiplicatio of real umbers, 13 Natural Numbers, 27 30 atural umbers, 27 ested itervals property, 33 oalgebraic umbers, 95 odecreasig sequece, 33, 70 oicreasig sequece, 33, 70 oegative, 20 ormed real vector space, 52 th power, 30 oe, 13 ope ball, 54 ope subset, 125 ope subset of a metric space, 54 order, 13, 19 21 ordered field, 19 p-adic metric, 51 partial sums, 83 Peao, 140 poitwise covergece of a family of fuctios, 122 poitwise covergece of a series of fuctios, 115, 119 poitwise limit, 113, 122 polyomial, 44, 65 power series, 91 product, 13 product topology, 181 R, 13 R, 16, 20 R <0, 20 R 0, 20 R >0, 20 R 0, 20 Raabe ad Duhamel s Covergece Rule, 170 radius of a ball, 54 radius of covergece, 91, 149, 150 Ratio Test, 90 ratioal umbers, 34 Real Numbers, 13 24 real sequece, 33 real vector space, 44 real vector spaces, 43 47 rearragig the terms of a series, 84, 87 reflexivity, 11 Rolle s Theorem, 145 Root Test, 90 scalar multiplicatio, 43 scalar product, 49 sequece, 33 sequetial compactess, 183 series, 83 series of fuctios, 115 sie, 90 square root, 34, 73 strictly decreasig sequece, 33 strictly icreasig sequece, 33 strictly egative, 20 strictly positive, 20 subsequece of a sequece, 33 subspace, 46 sum, 13 sum of vectors, 43 sup(a), 22 sup metric, 51 Sym(A), 10 symmetry, 11 T 0 -topological space, 183 Taylor series, 151 (T f)(x), 154 topological space, 125 topology geerated by, 126, 128 totally ordered set, 21 totaly discoected, 131 trascedetal umbers, 95 trasitivity, 11 triagular iequality, 50 ultrametric, 51 uiform covergece, 115 uiform covergece of a family of fuctios, 122 uiform covergece of a series of fuctios, 119

INDEX 193 uiqueess of the real umber system, 37 usual metric, 51 vector, 43 vector space over R, 44 Wallis formula, 167 Weierstrass M-Test, 120, 13 x 1, 16 x 2, 19 Z, 34 zero, 13, 14 zero sequece, 45