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

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1 Aalysis Notes (oly a draft, ad the first oe!) Ali Nesi Mathematics Departmet Istabul Bilgi Uiversity Kuştepe Şişli Istabul Turkey Jue 22, 2004

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3 Cotets 1 Prelimiaries Biary Operatio Biary Relatios Real Numbers Axioms for Additio Axioms for Multiplicatio Distributivity Axioms for the Order Relatio Totally Ordered Sets Completeess Axiom Other Number Sets Natural Numbers ad Iductio Expoetiatio Factorial Sequeces Itegers ad Ratioal Numbers Expoetiatio Uiqueess of the Real Number System Complex Numbers Real Vector Spaces 43 5 Metric Spaces Examples Defiitio ad Further Examples Normed Real Vector Spaces ad Baach Spaces ad Algebras Ope Subsets of a Metric Space Sequeces ad Limits Defiitio Examples of Covergece i R ad C The Sequece (1/)

4 4 CONTENTS The Sequece (α ) The Sequece (α /!) Covergece ad the Order Covergece ad the Four Operatios More O Sequeces Cauchy Sequeces Covergece of Real Cauchy Sequeces Covergece of Some Sequeces The Sequece ((1 + 1/) ) The Sequeces (2 1/ ) ad ( 1/ ) The Sequece (x ) Divergece to Ifiity Limit Superior ad Iferior Complete Metric Spaces Completio of a Metric Space Supplemetary Problems Series Defiitio ad Examples Easy Cosequeces of the Defiitio Absolute Covergece Alteratig Series Criteria for Covergece Supplemetary Problems Midterm of Math Solutios of the Midterm of Math Supplemetary Topics Liouville Numbers Covergece of Fuctios Poitwise Covergece of a Sequece of Fuctios Uiform Covergece of a Sequece of Fuctios Uiform Covergece of a Series of Fuctios Uiform Covergece ad Metric Limits of Fuctios Covergece of a Family of Fuctios Supplemetary Topics Topological Spaces Defiitio ad Examples Closed Subsets Iterior Closure Base of a Topology Compact Subsets

5 CONTENTS Covergece ad Limit Poits Coected Sets Cotiuity Cotiuity o Metric Spaces Cotiuity o Topological Spaces Cotiuous Fuctios ad R Uiform Cotiuity Uiform Covergece ad Cotiuity Supplemetary Topics A Cotiuous Curve Coverig [0, 1] Differetiable Fuctios Defiitio ad Examples Differetiatio of Complex Fuctios Basic Properties of Differetiable Fuctios Rules of Differetiatio Relatioship Betwee a Fuctio ad Its Derivative Uiform Covergece ad Differetiatio Secod ad Further Derivatives Aalytic Fuctios Power Series Taylor Series Calculatig Taylor Polyomials Aalytic Fuctios Trascedetal Fuctios Expoetiatio ad Trigoometric Fuctios Iverse Trigoometric Fuctios Logarithm Hyperbolic Fuctios Supplemet Trigoometric Fuctios Series Notes Graph Drawig Drawig i Cartesia Coordiates Asymptotes Parametric Equatios Polar Coordiates Geometric Loci Supplemet Lipschitz Coditio A Metric O R

6 6 CONTENTS 15 Riema Itegral Defiitio ad Examples Fudametal Theorem of Calculus How To Itegrate? Power Series Trigoometric Fuctios Itegratio of Complex Fuctio Fuctios Defied by Itegratio Applicatios Applicatio to Series Stirlig Formula Euler s Γ Fuctio Supplemets Stoe-Weierstrass Theorem Fourier Series Hilbert Spaces Fourier Series Topological Spaces (cotiued) Product Topology Homeomorphisms Sequeces i Topological Spaces Sequetial Compactess Supplemets T 0 -Idetificatio Exams Midterm Math 121 (November 2002) Fial Math 121 (Jauary 2003) Resit of Math 121, February Correctio of the Resit of Math 121, February Secod Resit of Math 121, March Midterm of Math 152, April Fial of Math 152, Jue

7 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 Chapters 1 through 19 form the first semester of a four semester course.

8 8 CONTENTS

9 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 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.

11 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 12 CHAPTER 1. PRELIMINARIES

15 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 proves more, amely the followig: Lemma If x, y R satisfy x + y = 0, the y = x. We start to prove some umber of well-kow results: Lemma 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 If x, y R satisfy x + y = 0, the x = y. Proof: Follows directly from lemmas ad Lemma For all x, y R, (x + y) = ( y) + ( x). Proof: We compute directly: (x + y) + (( y) + ( x)) = x + y + ( y) + ( x) = x ( 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 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 (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 (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 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 Give x R, the elemet y as i M3 is uique. The proof of this lemma ca be traslated from the proof of Lemma 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.

17 2.3. DISTRIBUTIVITY 17 As we have oticed, the proof of Lemma proves more, amely the followig: Lemma If x, y R satisfy xy = 1, the y = x 1. Lemma For all x R, (x 1 ) 1 = x. Proof: As i Lemma Lemma If x, y R satisfy xy = 1, the x = y 1. Proof: As i Lemma Lemma For all x, y R, (xy) 1 = y 1 x 1. Proof: As i Lemma 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 (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 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.

21 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 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 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.

24 24 CHAPTER 2. REAL NUMBERS Lemma 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 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 ).

25 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 26 CHAPTER 2. REAL NUMBERS

27 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 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 (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 i. lub(n) = 0, i.e. 0 is the least elemet of N. ii. If x N \ {0} the x 1 N. 27

29 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 (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 (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 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 vii a is the largest elemet of A. Theorem (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 vii, x < + 1 m x, a cotradictio. Similarly m. Thus = m. Theorem (Divisio) For ay, m N, m 0 there are uique q, r N such that = mq + r ad 0 r < m.

30 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 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}, (2 1) = 2. ii. Show that for ay N\{0}, = (+1)(2+1)/6. iii. Show that for ay N, = iv. Show that for ay N \ {0}, (2 1) = 2. v. Show that for ay N\{0}, = (+1)(2+1)/6. vi. Show that for ay N, = vii. Show that for ay N \ {0}, = (2+1)(+2) 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 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.

31 3.1. NATURAL NUMBERS AND INDUCTION 31 Proof: Left as a exercise. Theorem 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 Left as a exercise. The umber s is called the m th -root of r Factorial For N, we defie! by iductio o : Set 0! = 1, 1! = 1 ad ( + 1)! =!( + 1). This just meas that! = 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}, = ( ) 2. viii. Show that for ay N \ {0}, = (+1)( ) 30. Choose k. For, k N ad k, defie ( k ) =! k!( k)!.

32 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. (See Theorem 6.8.1).

33 3.1. NATURAL NUMBERS AND INDUCTION 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 = Thus the sequece (y ) is (1, 1/3, 1/5, 1/7,..., 1/(2 + 1),...). If we take k 0 = 1 ad k = 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 (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.

36 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

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I II III The Study of Factors to the Failure or Success of Applying to Holding International Sport Games Abstract For years, holding international sport games has been Taiwan s goal and we are on the way

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### 绝妙故事

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