Real Analysis I Handouts PDF

Virtual University of Pakistan Real Analysis I (MTH621) Salman Amin Malik Virtual University Learning Management System To my unknown students ii About the instructor Dr. Malik did his MS and PhD (Mathematics) from University of La Rochelle, La Rochelle, France in 2009 and 2012, respectively. Prior to MS and PhD, Dr. Malik completed his MPhil and MSc (Mathematics) from Department of Mathematics, University of the Punjab, Lahore, Pakistan. He has been aliated with several universities in Pakistan and abroad. He has the experience of teaching a wide range of mathematics courses at undergraduate and graduate level. Dr. Malik has published several research articles in international journals and conferences. His area of research includes the study of dierential equations with nonlocal operators and their applications to image processing. He is also interested in inverse problems related to reaction-diusion equations with nonlocal integro- dierential operators and boundary conditions. These models have numerous appli- cations in anomalous diusion/transport, biomedical imaging and non-destructive testing. About the handouts The books followed during this course are: W. Rudin, Principles of Mathemat- ical Analysis, Third Edition, McGraw-Hill, 1976. ISBN: 9780070542358. and W. F. Trench, Introduction to Real Analysis, Pearson Education, 2013. Con- sequently, the most of the examples considered in these notes are from the above mentioned books and their exercises, but not restricted to those books only. If you nd any typing error in the text kindly report to me by writing an email to [email protected]. iii Course Information Title and Course Code: Real Analysis I (MTH621) Number of Credit Hours: 3 credits Course Objective: The Real Analysis I is the rst course towards the rigorous (formal) treatment of the fundamental concepts of mathematical analysis. This course could be considered as the fundamental course in pursue of mathematical study at undergraduate or master level. Although, the topics of the Real Analysis I are self contained but someone having knowledge of Calculus (single and multivari- able) and Dierential Equations will be comfortable with the contents of the course. These subjects could be considered as prerequisite of the Real Analysis I. Learning Outcomes (for the whole course) Upon completion of this course students will be able to Understand the set theoretic statements, the real and complex number sys- tems. Apply principle of mathematical induction, ordered sets. Decide about the convergent or divergent sequences and series. Dene the limit of a function, prove various theorems about limits, sequences and functions. Check the continuity of real valued functions, prove various theorems about continuous functions with emphasize on the proofs. Understand the derivative of a function, proof of various theorems about dif- ferentiability of the function (LO6). Prove and apply Bolzano-Weierstrass theorem, Mean value theorem. Dene Riemann integral, Riemann sums, proof of various results about the Riemann integrals. Prerequisites: Calculus with Analytical Geometry The textbooks for this course: W. Rudin, Principles of Mathematical Analysis, Third Edition, McGraw-Hill, 1976. ISBN: 9780070542358. W. F. Trench, Introduction to Real Analysis, Pearson Education, 2013. Reference books: iv A. N. Kolmogorov and S. V. Fomin, Introductory Real Analysis, Revised English Edition Translated and Edited by R. A. Silverman, Dover Publication, Inc. New York. R. G. Bartle and D. R. Sherbert, Introduction to Real Analysis, Third Edition, 2000, John Wiley & Sons Inc. The real number system Sequences and Series Limits, Continuity and Dierentiability The integration Contents 1 The Real Number System 2 1.1 Basic Set Theory............................. 2 1.1.1 Universal Set........................... 2 1.1.2 Set Builder Notion........................ 2 1.2 Number Theory.............................. 4 1.3 Principle of Mathematical Induction.................. 5 1.3.1 An initial segment........................ 7 1.4 Finite and Innite Set.......................... 8 1.5 Field.................................... 9 1.6 Dedekind Innite Set........................... 9 1.7 The Set of Rational Numbers...................... 10 1.8 Ordered Set................................ 11 1.8.1 Partial Order........................... 12 1.9 Lower Bound............................... 12 1.9.1 Inmum.............................. 12 1.10 Upper Bound............................... 12 1.10.1 Supremum............................. 12 1.11 Least Upper Bound Property...................... 13 1.12 The Completeness Axiom........................ 14 1.13 The Archimedean Property of R.................... 15 1.14 Dense Set in R.............................. 15 1.15 The Set of Rational Numbers is not Complete............. 16 1.16 Applications of Properties of Field................... 17 1.17 The Extended Real Number System.................. 19 1.18 Principle of Mathematical Induction.................. 21 1.18.1 Principle of Mathematical Induction For Z........... 23 1.19 Generalization of Union and Intersection................ 24 1.20 Open Coverings.............................. 27 2 Sequences and Series 29 2.1 Sequences................................. 29 2.1.1 Convergent Sequence....................... 30 2.2 Bounded Sequence............................ 31 2.3 Monotonic Sequences........................... 32 2.4 Some Special Sequences......................... 33 2.5 Subsequence................................ 36 2.6 Limit Superior and Limit Inferior.................... 39 2.7 Cauchy Sequence............................. 41 2.8 Series................................... 42 Contents vi 2.8.1 Sequence of Partial Sums.................... 43 2.8.2 Oscillatory Series......................... 43 2.9 Dropping Finitely Many Terms..................... 44 2.10 Series of Nonnegative Terms....................... 47 2.11 The Comparison Test.......................... 48 2.12 The Number e.............................. 51 2.13 The Ratio Test.............................. 54 2.14 Cauchy's Root Test............................ 57 2.15 Absolute Convergence.......................... 58 2.16 Absolute Convergence.......................... 58 2.17 Dirichlet's Test for Series......................... 59 2.18 Alternating Series............................. 62 2.19 Grouping Terms in a Series....................... 62 2.20 Rearrangements of Series......................... 64 2.21 Addition and Multiplication of Series.................. 65 2.22 Power Series................................ 68 3 Continuity 73 3.1 Limits................................... 75 3.1.1 Limits (An informal view).................... 76 3.1.2 Formal Denition of Limit.................... 77 3.2 One Sided Limits............................. 81 3.3 Limits at ±∞............................... 83 3.4 Innite Limits............................... 84 3.5 Continuity................................. 85 3.6 Piecewise Continuous Functions..................... 87 3.7 Removable Discontinuity......................... 89 3.8 Bounded Functions............................ 91 3.9 The Intermediate Value Theorem.................... 94 3.10 Uniform Continuity............................ 94 3.11 Monotonic Functions........................... 97 3.12 Limits Inferior and Superior....................... 98 4 Dierentiability 102 4.1 Derivative................................. 102 4.1.1 Dierentiability Implies Continuity............... 104 4.2 One Sided Derivative........................... 108 4.3 Dierentiable Function.......................... 109 4.4 Extreme Values of a Function...................... 109 4.5 Rolle's Theorem.............................. 111 4.6 The Mean Value Theorem........................ 111 4.7 Generalized Mean Value Theorem.................... 112 4.8 Lipschitz Continuity........................... 113 4.9 Indeterminate Forms........................... 115 Contents vii 4.10 Indeterminate forms (0)(∞)....................... 117 4.10.1 Indeterminate form ∞ − ∞................... 118 4.10.2 0 ∞ Indeterminate forms 0 , 1 , ∞ 0................. 118 4.11 Taylor's Theorem............................. 120 5 Riemann Integration 126 5.1 Riemann Sums.............................. 126 5.2 Riemann Integral............................. 127 5.3 Upper and Lower Integrals........................ 130 5.4 Fundamental Theorem of Calculus................... 146 5.4.1 Anti-derivative of a Function.................. 149 5.5 Integration by Parts........................... 150 5.6 Integration by Substitution....................... 151 List of Figures 1.1 (a) Subset (b) Union of two sets (c) Intersection of two sets, (d) Disjoint sets................................ 4 2.1 Convergence of the sequence....................... 30 3.1 Tangent line as a limit of secant line.................. 75 3.2 Area under the curve as limit...................... 75 3.3 Limit x→2................................ 76 3.4 Limit x → 1 when f (x) is not dened at x = 1............ 76 3.5 Limit x → 0 when f (x) is not dened at x = 0............ 77 3.6 Limit x → 0................................ 77 3.7 Limits at ±∞............................... 83 3.8 Graph of the function.......................... 88 4.1 The tangent lines............................. 104 4.2 Extreme values of a function....................... 110 5.1 Area under the curve........................... 126 Chapter 1 The Real Number System 1.1 Basic Set Theory In this section, we are going to provide some basic terminology required to under- stand the forthcoming concepts about the real number system. 1.1.1 Universal Set The understanding that the members of all sets under consideration in any given context come from a specic collection of elements, called the universal set. For this chapter of the course our universal set will be the set of real numbers. If an element x is in A, we write x ∈ A and say that x is a member of A. If an element x is not in A, we write x∈ / A and say that x is not a member of A. How to dene a set? {4, 5, 6, 7}. 1.1.2 Set Builder Notion If P is a property that is meaningful and unambiguous for elements of a set S, then we write {x ∈ S : P (x)}, for the set of all elements x in S for which the property P is true. If every element of a set A also belongs to a set B, we say that A is a subset of B and write A ⊂ B. Proper subset: We say that A is proper subset of B if there exist at least one element of B which is not in A. Equal sets: Two sets are said to be equal if they contain the same number of elements. Examples: Consider the set of natural numbers N = {1, 2, 3,...}. 1.1. Basic Set Theory 3 {x ∈ N : x2 − 3x + 2 = 0} = {1, 2}. {x ∈ N : x2 − 4x + 2 = 0}. The set of even numbers {2k : k ∈ N}. The set of odd numbers {2k − 1 : k ∈ N}. Let S and T be sets. S contains T , and we write S ⊃ T or T ⊂ S, if every member of T is also in S. In this case, T is a subset of S. S−T is the set of elements that are in S but not in T. S equals T , and we write S = T , if S contains T and T contains S; thus, S = T if and only if S and T have the same members. Let S and T be sets. S strictly contains T if S contains T but T does not contain S ; that is, if every member of T is also in S , but at least one member of S is not in T. The complement of S, denoted by Sc, is the set of elements in the universal set that are not in S. The union of S and T, denoted by S ∪T, is the set of elements in at least one of S and T Let S and T be sets. The intersection of S and T , denoted by S ∩ T , is the set of elements in both S and T If S ∩ T = ∅ (the empty set), then S and T are disjoint sets A set with only one member x0 is a singleton set , denoted by {x0 }. Let S and T be sets. The intersection S and T , denoted by S ∩ T , is the set of elements of in both S and T If S ∩ T = ∅ (the empty set), then S and T are disjoint sets A set with only one member x0 is a singleton set , denoted by {x0 }. 1.2. Number Theory 4 S T S T S T S ∪ T = shaded region (a) (b) S T S T S ∩ T = shaded region S∩T=∅ (c) (d) Figure 1.1: (a) Subset (b) Union of two sets (c) Intersection of two sets, (d) Disjoint sets The set of natural numbers The set of prime numbers Fundamental Theorem of arithmetic Diophantines A little bit about number theory The set of natural numbers: N = {1, 2, 3,...}. 1.2 Number Theory Number theory is a branch of mathematics that studies the properties of, and the relationships between, particular types of numbers. The set of natural numbers N. The prime numbers. 1.3. Principle of Mathematical Induction 5 The primes are the building blocks of the positive integers. Fundamental Theorem of Arithmetic: Every positive integer can be uniquely written as the product of primes in nondecreasing order. How may prime numbers are there? (2500 years ago, Euclid provided the proof ) Many dierent approaches have been used to determine whether an integer is prime.For example, in the nineteenth century, Pierre de Fermat showed that p divides 2p − 2 whenever p is prime. The search for integer solutions of equations is another important part of number theory. An equation with the added provision that only integer solutions are sought is called diophantine, after the ancient Greek mathematician Diophantus. Example: an + bn = cn , n ∈ Z. The set of nonnegative integers: Z+ = {0, 1, 2, 3,...}. The set of nonpositive integers: Z− = {0, −1, −2, −3,...}. The set of integers: Z = {0, ±1, ±2, ±3,...}. Some properties of the natural numbers: 2+4=4+2 m+3=p+3 then m = p. 4 + (2 + 7) = (4 + 2) + 7. 1.3 Principle of Mathematical Induction Let P (n)be a mathematical statement, where n∈N (or Z+ ). If P (1) is true. P (n) is true implies P (n + 1) is true. Then P (n) is true for all n∈N (or Z+ ). Examples: 1.3. Principle of Mathematical Induction 6 n(n+1)(2n+1) 12 + 22 + 32 +... + n2 = 6. n2 (n+1)2 13 + 23 + 33 +... + n3 = 4. Prove that if (x + 1/x) is an integer then (xn + 1/xn ) is also an integer for any positive integer n. Theorem: Suppose that m, n, p ∈ Z+. Then m + n = n + m, (commutativity) (m + n) + p = n + (m + p), (associativity) if m + n = p + n, then m=p (cancelation) if m + n = 0, then m=n=0 Proof : Proof of rst property. Step I: Dene U = {m ∈ Z+ : 0 + m = m + 0}. Step II: Dene V = {n ∈ Z+ : (m + 1) + n = (m + n) + 1, for all m ∈ Z+ }. Step III: Dene W = {n ∈ Z+ : m + n = n + m, for all m ∈ Z+ }. m + (n + 1) = (m + n) + 1 (1.1) Step I: Dene U = {m ∈ Z+ : 0 + m = m + 0}. Notice that 0 ∈ U. Suppose m∈U and consider 0 + (m + 1) = (m + 1) and we have (0 + m) + 1 = m + 1. Thus m + 1 ∈ U , and so U = Z+. Step II: Dene V = {n ∈ Z+ : (m + 1) + n = (m + n) + 1, for all m ∈ Z+ }. 0 ∈ V , suppose n ∈ V , then (m + 1) + (n + 1) = ((m + 1) + n) + 1 by (1.1) = ((m + n) + 1) + 1, since n∈V = (m + (n + 1)) + 1 by (1.1). Hence we have V = Z+. Step III: Dene W = {n ∈ Z+ : m + n = n + m, for all m ∈ Z+ }. Using the rst step we have 0 ∈ W. Suppose n ∈ W. Then m + (n + 1) = (m + n) + 1 by (1.1) = (n + m) + 1, since n∈W = (n + 1) + m, from step II. Hence W = Z+. Theorem: Suppose that m, n, p ∈ Z+. Then 1.3. Principle of Mathematical Induction 7 m.n = n.m, (commutativity) (m.n).p = n.(m.p), (associativity) if m.n = p.n and n ̸= 0, then m=p (cancellation) if m.n = 0, then m=0 or n = 0. 0.n = 0 and 1.n = n. 1.3.1 An initial segment An initial segment I of N is a nonempty subset of N with the property that if n∈I and m≤n then m ∈ I. Example: Consider the set {1, 2, 3,..., 20}. Consider the set {1, 2, 3,..., n}. Consider the set {2, 4, 6, 8, 10}. Proposition: If I is an initial segment of N then either I = N or there exists n ∈ N such that I = In = {m ∈ N : m ≤ n}. Proof : It follows immediately from the denition of an initial segment that if m∈ /I and n ≥ m then n ∈ / I. I ̸= N, then N\I is non-empty. Let m0 be its least element. If Suppose, if possible, that m0 = 1. If n ∈ N, then n ≥ 1, so that n ∈ / I and I is the empty set. Thus m0 > 1, and so there exists n ∈ N such that m0 = n + 1. Then n ∈ I , and so In ⊆ I. But if p > n then p ≥ n + 1 = m0 , and so p ∈ / I. Thus I ⊆ In. Recall a one to one and onto mapping: A mapping (function) f :A→B is said to be one to one if f (x1 ) = f (x2 ) ⇒ x1 = x2. A mapping (function) f : A → B is said to be onto if for every y ∈ B there exists at least an x∈A such that f (x) = y. A mapping (function) is said to be bijective if the mapping is one to one and onto. 1.4. Finite and Innite Set 8 1.4 Finite and Innite Set A set A is nite if either A is empty or there exists n ∈ N and a bijective mapping c : In → A. Thus the nite sequence (c1 ,..., cn ) lists the elements of A, without repetition. A set is innite if it is not nite. Proposition: If g : Im → In is an one to one (injective) mapping then m ≤ n. Proof : The proof is by induction on m. The result is trivially true if m = 1. Suppose that it holds for m, and that f : Im+1 → In is injective. Then m + 1 > 2, so that f (Im+1 ) contains at least two points, and so n = k + 1, for some k ∈ N. Let τ : In → In be the mapping that transposes f (m + 1) and n and leaves the other elements of In xed. Then τ ◦ f : Im+1 → In is injective, and τ (f (Im )) ⊆ Ik. By the inductive hypothesis, m ≤ k, and so m + 1 ≤ k + 1 = n. Corollary: If A is a non-empty nite set, there exists a unique n∈N for which there exists a bijection c : In → A. Proof : Suppose that c : In → A and c′ : In′ → A are bijections. Then c −1 ◦ c′ : In′ → In is a bijection, and so n′ ≤ n. Similarly, n ≤ n′. Remark: The number n is known as cardinality of A and is written as | A |. Proposition: Suppose that A is a nite set, and that f :A→B is a bijection. Then B is nite, and | B |=| A |. Proof : For if C : I|A| → A is a bijection, then the mapping f ◦ C : I|A| → B is the bijection. 1.5. Field 9 1.5 Field A eld is a set F, together with two laws of composition, addition (+) and multi- plication (.) such that for all a, b, c ∈ F the following properties holds a+b=b+a and ab = ba (commutative laws). (a + b) + c = a + (b + c) and (ab)c = a(bc) (associative laws). a(b + c) = ab + ac (distributive law). There are distinct members 0 and 1 such that a+0 = a and a1 = a for all a. For each a∈F there is an element −a ∈ F such that a + (−a) = 0, and if a ̸= 0, there is an element 1/a such that a(1/a) = 1. Remark: The left distributive law also holds. Examples: Let z2 = {0, 1} such that 0 + 0 = 1 + 1 = 0; 0 + 1 = 1 + 0 = 1 and 0.0 = 0.1 = 1.0 = 0, 1.1 = 1. Is the set of integers a eld? 1.6 Dedekind Innite Set Dedekind dened a set A to be innite if there is an injective map j:A→A which is not onto (surjective); such sets are now called Dedekind innite. Example: Show that N is Dedekind innite. The mapping f : N → N dened by f (n) = 2n is injective, and is not surjective. Corollary: The set of natural numbers N is innite set. Countable sets: A set A is countable if it is nite or if there is a bijection c : N → A; otherwise it is uncountable. 1.7. The Set of Rational Numbers 10 Remark: A set is countable if it is empty or if there is a bijection from an initial segment of N onto A. The function c is called an enumeration of A. A set is countably innite if it is innite and countable. Thus A is countably innite if and only if the elements of A can be listed, or enumerated, as an innite sequence (c1 , c2 ,...), without repetition. If A is countable (countably innite) and j : A → B is a bijection, then B is countable (countably innite). Not every set is countable. It was Cantor who rst showed, in 1873, that there are dierent sizes of innite set, showing that the set of real numbers is uncountable. Proposition: Let A be a non-empty set. Then the following are equivalent. (a) A is countable. (b) There exists a surjection f : N → A. (c) There exists an injection g : A → N. Theorem: The set N×N is countable. Proof : Dene the mapping f :N×N→N by f (k, l) = 2k 2l. Use above proposition to prove that N×N is countable. 1.7 The Set of Rational Numbers Construction of set of rational numbers: Let Z∗ = Z\{0} be the set of nonzero integers. Dene a relation on Z× Z∗ by setting (p, q) ∼ (r, s) if ps = qr. Proposition: The relation (p, q) ∼ (r, s) is an equivalence relation on Z × Z∗. Proof : Transitive: Suppose (p, q) ∼ (r, s) and (r, s) ∼ (t, u), we need to show that (p, q) ∼ (t, u). Consider pusr = (ps)(ru) = (qr)(ts) = qtsr Q is abelian group under addition: We dene addition on Q by p r ps + qr + =. q s qs ( )( ) p We dene the multiplication as q r s = pr qs. Proposition: Let Q∗ = Q/{0/1}. Then, (Q∗ ,.) and (Q∗ , +) are abelian group. 1.8. Ordered Set 11 The set of rational numbers is dened as m Q={ : m, n ∈ Z, n ̸= 0}. n √ Example: Find solution of the equation p = 2 in the set of rational numbers if possible. Solution: See Lecture. Example: Let A = {p ∈ Q : p2 < 2}. We will show that for every p ∈ A, we can nd a rational number q such that p < q. To do this, we associate with each rational p>0 the number −2 p2 2p + 2 q =p− =. (1.2) p+2 p+2 Then 2(p2 − 2) q2 − 2 =. (1.3) (p + 2)2 If p is in A then p2 − 2 < 0, (1.2) shows that q > p and (1.3) shows that q 2 < 2. Thus q is in A. B = {p ∈ Q : p2 > 2}. We Let will show that for every p ∈ B, we can nd a rational number q such that q < p. If p is in B then p2 − 2 > 0, (1.2) shows that 0 a. In other words, a is a lower bound of B , but β is not if β > a. This means that a = inf B. Theorem: There exists an ordered eld R which has the least-upper-bound prop- erty Moreover, R contains Q as a subeld. 1.12 The Completeness Axiom If a nonempty set of real numbers is bounded above, then it has a supremum. The above property is called completeness, and we say that the real number system is a complete ordered eld. Theorem: If a nonempty set S of real numbers is bounded above, then sup S is the unique real number β such that x≤β for all x in S; if ε > 0 (no matter how small), there is an x0 in S such that x0 > β − ε. Proof : We rst show that β = sup S has rst and second properties. Since β is an upper bound of S, it must satisfy the rst property. Since any real number a less than β can be written as β−ε with ε = β − a > 0, second property is just another way of saying that no number less than β is an upper bound of S. Hence, β = sup S satises rst and second properties. Now we show that there cannot be more than one real number with rst and second properties. Suppose that β1 < β2 and β2 has the second property ; thus, if ε > 0, there is an x0 in S such that x0 > β2 − ε. Then, by taking ε = β2 − β1 , we see that there is an x0 in S such that x0 > β2 − (β2 − β1 ) = β1 , so β1 cannot have the rst property. Therefore, there cannot be more than one real number that satises both parts. 1.13. The Archimedean Property of R 15 1.13 The Archimedean Property of R Theorem: If ρ and ε are positive, then nε > ρ for some integer n. Proof : The proof is by contradiction. If the statement is false, ρ is an upper bound of the set S = {x = nε, n ∈ Z}. Therefore, S has a supremum β (Why?), by denition of least upper bound property of real numbers. Therefore, nε ≤ β for all integers n. (1.4) Since n+1 is an integer whenever n is, (1.4) implies that (n + 1)ε ≤ β and therefore nε ≤ β − ε for all integers n. 1.14 Dense Set in R A set D is said to be dense in the set of real numbers if every open interval (a, b) contains a member of D. Theorem: The rational numbers are dense in R, that is, if a and b are real numbers with a < b, there is a rational number p/q such that a < p/q < b. Recall the The Archimedean property Theorem: If ρ and ε are positive, then nε > ρ for some integer n. Proof of the theorem: The Archimedean property with ρ = 1 and ε = b − a, there is a positive integer q such that q(b − a) > 1. There is also an integer j such that j > qa. This is obvious if a ≤ 0, and it follows from Archimedean property with ε = 1 and ρ = qa if a > 0. Let p be the smallest integer such that p > qa. Then p − 1 ≤ qa, so qa < p ≤ qa + 1. Since 1 < q(b − a), this implies that qa < p < qa + q(b − a) = qb, so qa < p < qb. Therefore, a < p/q < b. 1.15. The Set of Rational Numbers is not Complete 16 1.15 The Set of Rational Numbers is not Complete The rational number system is not complete; that is, a set of rational numbers may be bounded above (by rational numbers), but not have a rational upper bound less than any other rational upper bound, that is, that set does not have a rational supremum. Recall: Theorem: If a nonempty set S of real numbers is bounded above, then sup S is the unique real number β such that x≤β for all x in S; if ε > 0 (no matter how small), there is an x0 in S such that x0 > β − ε. Consider the set A = {p ∈ Q : p2 < 2}. √ If p ∈ A, then p< 2. Then using the fact that there is a rational number between every two real numbers √ √ implies that if ε>0 there is a rational number √ r0 such that √ 2 − ε < r 0 < 2, so using above Theorem implies that 2 = sup A. However, 2 is irrational ; that is, it cannot be written as the ratio of integers. √ Therefore, if r1 is any rational upper bound of A, then 2 < r1. Since there is a rational number between every two real numbers, there is a rational number √ r2 such that 2 < r2 < r1. Since r2 is also a rational upper bound of A, this shows that A has no rational supremum. ExampleProduct of a rational and irrational number is irrational. Solution: See lecture. Example: Sum of a rational and an irrational number is irrational. Proof : See lecture. Theorem: The set of irrational numbers is dense in the reals ; that is, if a and b are real numbers with a < b, there is an irrational number t such that a < t < b. Proof : Since between every two real numbers there is a rational number, therefore there are rational numbers r1 and r2 such that a < r1 < r2 < b. (1.5) Let 1 t = r1 + √ (r2 − r1 ). 2 Then t is irrational (why?) and r1 < t < r2 , so a < t < b, from (1.5). 1.16. Applications of Properties of Field 17 1.16 Applications of Properties of Field Proposition: The axioms of addition imply the following (a) If x+y =x+z then y=z (Cancellation law). (b) If x+y =x then y=0 (Uniqueness of the additive identity). (c) If x+y =0 then y = −x Uniqueness of the additive inverse). ( (d) −(−x) = x. Proof : If x + y = x + z, then from the axioms of eld we have y = 0 + y = (−x + x) + y = −x + (x + y) = −x + (x + z) = (−x + x) + z = 0 + z = z. This proves (a). Take z=0 in (a) to obtain (b). Take z = −x in (a) to obtain (c). Since −x + x = 0, (c) with −xin place of x gives (d). Proposition: The axioms of multiplication imply the following (a) If x ̸= 0 and xy = xz then y=z Cancellation law). ( (b) If x ̸= 0 and xy = x then y = 1 (Uniqueness of the multiplicative identity). (c) If x ̸= 0 and xy = 1 then y = 1/x Uniqueness of the additive inverse). ( (d) If x ̸= 0 then 1/(1/x) = x. Proposition: The eld axioms imply the following statements for any x, y, z ∈ F (a) 0x = 0. (b) If x ̸= 0 and y ̸= 0 then xy ̸= 0. (c) (−x)y = −(xy) = x(−y). (d) (−x)(−y) = xy. The manipulative properties of the real numbers, such as the relations (a + b)2 = a2 + 2ab + b2 , (3a + 2b)(4c + 2d) = 12ac + 6ad + 8bc + 4bd, (−a) = (−1)a, a(−b) = (−a)b = −ab, a c ad + bc + = (b, d ̸= 0), b d bd 1.16. Applications of Properties of Field 18 all follow from the axioms of eld. Theorem: For every real x > 0 and every integer n>0 there is one and only one positive real y such that y n = x. Proof : The uniqueness is clear, since if there are two positive numbers y1 and y2 then 0 < y1 < y2 ⇒ y1n < y2n. Let E t such that tn < x. be the set containing of all positive real numbers The set E is not empty, as if t = x/(x + 1) then 0 ≤ t < 1 and t < t < x. n If t > 1 + x then t ≥ t > x, so that t ∈ / E. Thus t ∈ E is an upper bound of E. n Then by least upper bound property there exists y ∈ R such that y = sup E. We need to prove that yn = x we will show that each of the inequalities yn < x and yn >x leads to a contradiction. Recall the identity bn − an = (b − a)(bn−1 + bn−2 a +... + an−1 ) yields the identity bn − an = (b − a)nbn−1 , when 0 < a < b. n Assume y < x, choose h so that 0 x, and t∈ / E. It follows that y−k is an upper bound of E. But y − k < y, which contradicts the fact that y is the least upper bound of E. Theorem: If a and b are any two real numbers, then |a + b| ≤ |a| + |b|. (1.6) Proof : There are four possibilities: 1.17. The Extended Real Number System 19 (a) If a≥0 and b ≥ 0, then a + b ≥ 0, so |a + b| = a + b = |a| + |b|. (b) If a≤0 and b ≤ 0, then a + b ≤ 0, so |a + b| = −a + (−b) = |a| + |b|. (c) If a≥0 and b ≤ 0, then a + b = |a| − |b|. (d) If a≤0 and b ≥ 0, then a + b = −|a| + |b|. Eq. 1.6 holds in cases (c) and (d), since { |a| − |b| if |a| ≥ |b|, |a + b| = (1.7) |b| − |a| if |b| ≥ |a|. Corollary: If a and b are any two real numbers, then |a − b| ≥ |a| − |b| , and |a + b| ≥ |a| − |b|. (1.8) Proof : Replacing a by a−b in (1.6) yields |a| ≤ |a − b| + |b| ⇒ |a − b| ≥ |a| − |b|. Interchanging a and b here yields |b − a| ≥ |b| − |a|, which is equivalent to |a − b| ≥ |b| − |a|, (1.9) since |b − a| = |a − b|. Since { |a| − |b| if |a| > |b|, |a| − |b| = |b| − |a| if |b| > |a|. (??) and (1.9) imply (1.8). Replacing b by −b in (1.8) yields ( ??), since | − b| = |b|. 1.17 The Extended Real Number System A nonempty set S of real numbers is unbounded above if it has no upper bound, or unbounded below if it has no lower bound. It is convenient to adjoin to the real number system two ctitious points, +∞ (which we usually write more simply as ∞) and −∞, and to Dene the order relationships between them and any real number x by − ∞ < x < ∞. (1.10) We call ∞ and −∞ points at innity. 1.17. The Extended Real Number System 20 If S is a nonempty set of reals, we write sup S = ∞ (1.11) to indicate that S is unbounded above, and inf S = −∞ (1.12) to indicate that S is unbounded below. If S = {x : x < 2}, then sup S = 2 and inf S = −∞. If S = {x : x ≥ −2}, then sup S = ∞ and inf S = −2. If S is the set of all integers, then sup S = ∞ and inf S = −∞. A member of the extended reals diering from −∞ and ∞ is nite ; that is, an ordinary real number is nite. However, the word nite in nite real number is redundant and used only for emphasis, since we would never refer to ∞ or −∞ as real numbers. The real number system with ∞ and −∞ adjoined is called the extended real number system , or simply the extended reals. We must dened arithmetic operations with ±∞. A member of the extended reals diering from −∞ and ∞ is nite ; that is, an ordinary real number is nite. However, the word nite in nite real number is redundant and used only for emphasis, since we would never refer to ∞ or −∞ as real numbers. The arithmetic relationships among ∞, −∞, and the real numbers are dened as follows. If a is any real number, then a + ∞ = ∞ + a = ∞, a − ∞ = −∞ + a = −∞, a a = = 0. ∞ −∞ If a > 0, then a ∞ = ∞a = ∞, a (−∞) = (−∞) a = −∞. If a < 0, then a∞ = ∞a = −∞, a(−∞) = (−∞)a = ∞. 1.18. Principle of Mathematical Induction 21 We also dene ∞ + ∞ = ∞∞ = (−∞)(−∞) = ∞ and −∞ − ∞ = ∞(−∞) = (−∞)∞ = −∞. Finally, we dene |∞| = | − ∞| = ∞. It is not useful to dene ∞ − ∞, 0 · ∞, ∞/∞, and 0/0. They are called inde- terminate forms , and left undened. 1.18 Principle of Mathematical Induction The rigorous construction of the real number system starts with a set N of undened elements called natural numbers, with the following properties. The set of natural number N satisfy the following: N is nonempty. Associated with each natural number n there is a unique natural number n′ called the successor of n. There is a natural number n that is not the successor of any natural number. The set of natural number N satisfy the following: Distinct natural numbers have distinct successors; that is, if n ̸= m, then n′ ̸= m′. The only subset of N that contains n and the successors of all its elements is N itself. Theorem: Let P1 , P2 ,... , Pn ,... be propositions, one for each positive integer, such that P1 is true; for each positive integer n, Pn implies Pn+1. Then Pn is true for each positive integer n. Proof : Let M = {n ∈ N and Pn is true}. From rst axiom of Peano's, 1 ∈ M, and from second axiom, n+1 ∈ M whenever n ∈ M. Therefore, M = N, by fourth axiom. 1.18. Principle of Mathematical Induction 22 Example: Let Pn be the proposition that n(n + 1) 1 + 2 + ··· + n =. 2 Solution: Then P1 is the proposition that 1 = 1, which is certainly true. If Pn is true, then adding n+1 to both sides of the equation yields n(n + 1) (1 + 2 + · · · + n) + (n + 1) = + (n + 1) 2 ( ) n = (n + 1) +1 2 (n + 1)(n + 2) (n + 1)(n + 2) = = , 2 2 Example: For each nonnegative integer n, let xn be a real number and suppose that |xn+1 − xn | ≤ r|xn − xn−1 |, n ≥ 1, where r is a xed positive number. Show that |xn − xn−1 | ≤ rn−1 |x1 − x0 | if n ≥ 1. Solution: For n = 1, 2, and 3, we nd that |x2 − x1 | ≤ r|x1 − x0 |, |x3 − x2 | ≤ r|x2 − x1 | ≤ r2 |x1 − x0 |, |x4 − x3 | ≤ r|x3 − x2 | ≤ r3 |x1 − x0 |. It is important to verify P1 , since Pn may imply Pn+1 even if some or all of the propositions P1 , P2 ,..., Pn ,... are false. Let Pn be the proposition that 2n − 1 is divisible by 2. If Pn is true then Pn+1 is also, since 2n + 1 = (2n − 1) + 2. However, we cannot conclude that Pn is true for n ≥ 1. In fact, Pn is false for every n. 1.18. Principle of Mathematical Induction 23 1.18.1 Principle of Mathematical Induction For Z Theorem: Let n0 be any integer (positive, negative, or zero). Let Pn0 , Pn0 +1 ,... , Pn ,... be propositions, one for each integer n ≥ n0 , such that Pn0 is true ; for each integer n ≥ n0 , P n implies Pn+1. Then Pn is true for every integer n ≥ n0. Proof : For m ≥ 1, let Qm be the proposition dened by Qm = Pm+n0 −1. Then Q1 = Pn0 is true by rst part. If m ≥ 1 and Qm = Pm+n0 −1 is true, then Qm+1 = Pm+n0 is true by second part with n replaced by m + n0 − 1. Therefore,Qm is true for all m ≥ 1 by Mathematical induction Theorem with P replaced by Q and n replaced by m. This is equivalent to the statement that Pn is true for all n ≥ n0. Example: Prove the proposition Pn given by 3n + 16 > 0, n ≥ −5. Example: Let Pn be the proposition that n! − 3n > 0, n ≥ 7. If Pn is true, then (n + 1)! − 3n+1 = n!(n + 1) − 3n+1 > 3n (n + 1) − 3n+1 (by the induction assumption) = 3 (n − 2). n Therefore, Pn implies Pn+1 if n > 2. By trial and error, n0 = 7 is the smallest integer such that P n0 is true; hence, Pn is true for n ≥ 7, by Theorem ??. Theorem: Let n0 be any integer (positive, negative, or zero). Let Pn0 , Pn0 +1 ,... , Pn ,... be propositions, one for each integer n ≥ n0 , such that Pn0 is true ; for n ≥ n0 , Pn+1 is true if Pn0 , Pn0 +1 ,... , Pn are all true. 1.19. Generalization of Union and Intersection 24 Then Pn n ≥ n0. is true for Example: Let S = {x ∈ R : 0 < x < 1}, T = {x ∈ (0, 1) : x is rational}, and U = {x ∈ (0, 1) : x is irrational}. Then S ⊃ T and S ⊃ U , and the inclusion is strict in both cases. The unions of pairs of these sets are S ∪ T = S, S ∪ U = S, and T ∪ U = S, and their intersections are S ∩ T = T, S ∩ U = U, and T ∩ U = ∅. 1.19 Generalization of Union and Intersection : If F is an arbitrary collection of sets, then ∪{S : S ∈ F} is the set of all elements that are members of at least one of the sets in F , and ∩{S : S ∈ F} is the set of all elements that are members of every set in F. The union and intersection of nitely many sets S1 ,..., Sn are also written as ∪n ∩n k=1 Sk and k=1 Sk. The union and intersection of an innite sequence {Sk }∞ of sets are written ∪∞ ∩∞ k=1 as k=1 Sk and k=1 Sk. Example: If F is the collection of sets Sρ = {x : ρ < x ≤ 1 + ρ}, where 0 < ρ ≤ 1/2, then ∪ {Sρ : Sρ ∈ F} = {x : 0 < x ≤ 3/2} ∩ {Sρ : Sρ ∈ F } = {x : 1/2 < x ≤ 1}. Example: If, for each positive integer k, the set Sk is the set of real numbers that can be written as x = m/k for some integer m, ∪∞ ∩∞ then S k=1 k is the set of rational numbers and k=1 Sk is the set of integers. If a and b are in the extended reals and a < b, then the open interval (a, b) is dened by (a, b) = {x : a < x < b}. The open intervals (a, ∞) and (−∞, b) are semi-innite if a and b are nite, and (−∞, ∞) is the entire real line. ε-neighborhood: If x0 is a real number and ε > 0, then the open interval (x0 − ε, x0 + ε) is an ε-neighborhood of x0. If a set S contains an ε-neighborhood of x0 , then S is a neighborhood of x0 , and x0 is an interior point of S. The set of interior points of S is the interior of S , denoted by S. If every point 0 of S is an interior point (that is, S = S ), then S is open. 0 1.19. Generalization of Union and Intersection 25 A set S is closed if Sc is open. Example: An open interval (a, b) is an open set, because if x0 ∈ (a, b) and ε ≤ min{x0 − a, b − x0 }, then (x0 − ε, x0 + ε) ⊂ (a, b). R = (−∞, ∞) is open, and therefore ∅ = Rc is closed. However, ∅ The entire line is also open. for to deny this is to say that ∅ contains a point that is not an interior point, which is absurd because ∅ contains no points. Since ∅ is open, R(= ∅ ) is c closed. Thus, R and ∅ are both open and closed. They are the only subsets of R with this property A deleted neighborhood of a point x0 is a set that contains every point of some neighborhood of x0 except for x0 itself. For example, S = {x : 0 < |x − x0 | < ε} is a deleted neighborhood of x0. We also say that it is a deleted ε-neighborhood of x0. Theorem: The following statements are true for arbitrary collections, nite or innite, of open and closed sets. The union of open sets is open. The intersection of closed sets is closed. Proof : Let G be a collection of open sets and S = ∪{G : G ∈ G}. Ifx0 ∈ S , then x0 ∈ G0 for some G0 in G , and since G0 is open, it contains some ε-neighborhood of x0. Since G0 ⊂ S , this ε-neighborhood is in S , which is consequently a neighborhood of x0. Thus, S is a neighborhood of each of its points, and therefore open, by denition. Let F be a collection of closed sets and T = ∩{F : F ∈ F}. Then T c = ∪F c : F ∈ F and, since each F c is open, T c is open, from the rst part. Therefore, T is closed, by denition. Examples: If −∞ < a < b < ∞, the set [a, b] = {x ∈ R : a ≤ x ≤ b} is closed, since its complement is the union of the open sets (−∞, a) and (b, ∞). We say that [a, b] is a closed interval. 1.19. Generalization of Union and Intersection 26 Examples: The set [a, b) = {x ∈ R : a ≤ x < b} is a half-closed or half-open interval If −∞ < a < b < ∞, as is (a, b] = {x ∈ R : a < x ≤ b}; however, neither of these sets is open or closed. (Why not?) Examples: Semi-innite closed intervals are sets of the form [a, ∞) = {x ∈ R : a ≤ x} (−∞, a] = {x ∈ R : x ≤ a}, wherea is nite. They are closed sets, since their complements are the open intervals (−∞, a) and (a, ∞), respectively. Let S be a subset of R. Then x0 is a limit point of S if every deleted neighborhood of x0 contains a point of S. x0 is a boundary point S if every neighborhood of x0 contains at least one of point in S and one not in S. The set of boundary points of S is the boundary of S, denoted by ∂S. The closure of S , denoted by S , is S = S ∪ ∂S. Let S be a subset of R. Then x0 is an isolated point of S if x0 ∈ S and there is a neighborhood of x0 that contains no other point of S. Let S be a subset of R. Then x0 exterior to S if x0 is is in the interior of Sc. The collection of such points is the exterior of S. Let S = (−∞, −1] ∪ (1, 2) ∪ {3}. Then The set of limit points of S is (−∞, −1] ∪ [1, 2]. ∂S = {−1, 1, 2, 3} and S = (−∞, −1] ∪ [1, 2] ∪ {3}. Let S = (−∞, −1] ∪ (1, 2) ∪ {3}. Then 3 is the only isolated point of S. The exterior of S is (−1, 1) ∪ (2, 3) ∪ (3, ∞). 1.20. Open Coverings 27 For n ≥ 1, let [ ] ∞ ∪ 1 1 In = , and S= In. 2n + 1 2n n=1 Then The set of limit points of S is S ∪ {0}. ∂S = x : x = 0 or x = 1/n (n ≥ 2) and S = S ∪ {0}. S has no isolated points. The exterior of S is [ ∞ ( )] ( ) ∪ 1 1 1 (−∞, 0) ∪ , ∪ ,∞. 2n + 2 2n + 1 2 n=1 1.20 Open Coverings A collection H of open sets is an open covering of a set S if every point in S is contained in a set H belonging to H; that is, if S ⊂ ∪{H ∈ H : H ∈ H}. Example: The set S1 = [0, 1], is covered by the family of open intervals {( ) } 1 1 H1 = x − ,x + 0 0 and |s − s′ | is independent of ε, we conclude that |s − s′ | = 0; that is, s = s′. A sequence {sn } is said to be divergent to ∞ if for any real number a, sn > a for large n and written as lim sn = ∞ n→∞ Similarly, lim sn = −∞ n→∞ if for any real number a, sn < a for large n. not regard {sn } as convergent unless limn→∞ sn is nite, as required by Denition of limit. To emphasize this distinction, we say that {sn } diverges to ∞ (−∞) if limn→∞ sn = ∞ (−∞). Example: The sequence {n/2 + 1/n} diverges to ∞, since, if a is any real number, then n 1 + >a if n ≥ 2a. 2 n Therefore, we write ( ) n 1 lim + =∞ n→∞ 2 n Example: The sequence {n − n2 } diverges to −∞, since, if a is any real number, then √ −n2 + n = −n(n − 1) < a if n>1+ |a|. Therefore, we write lim (−n2 + n) = −∞. n→∞ Example: The sequence {(−1)n n3 } diverges, but not to −∞ or ∞. 2.2 Bounded Sequence A sequence {sn } is bounded above if there is a real number b such that sn ≤ b for all n, bounded below if there is a real number a such that sn ≥ a for all n, or bounded if there is a real number r such that |sn | ≤ r for all n. 2.3. Monotonic Sequences 32 Example: If sn = [1+(−1)n ]n, then {sn } is bounded below (sn ≥ 0) but unbounded above. {−sn } is bounded above (−sn ≤ 0) but unbounded below. If sn = (−1)n , then {sn } is bounded. If sn = (−1)n n, then {sn } is not bounded above or below. Theorem: A convergent sequence is bounded. Proof : By taking ε = 1, we see that if limn→∞ sn = s, then there is an integer N such that |sn − s| < 1 if n ≥ N. Therefore, |sn | = |(sn − s) + s| ≤ |sn − s| + |s| < 1 + |s| if n ≥ N, and |sn | ≤ max{|s0 |, |s1 |,... , |sN −1 |, 1 + |s|} for all n, so {sn } is bounded. Theorem: A sequence {sn } converges to s if and only if every neighborhood of s contains sn for all but nitely many n. 2.3 Monotonic Sequences A sequence {sn } is nondecreasing if sn ≥ sn−1 for all n, or nonincreasing if sn ≤ sn−1 for all n. A monotonic sequence is a sequence that is either nonincreasing or nondecreas- ing. If sn > sn−1 for all n, then {sn } is increasing , while if sn < sn−1 for all n, {sn } is decreasing. Theorem: If {sn } is nondecreasing, then limn→∞ sn = sup{sn }. If {sn } is nonincreasing, then limn→∞ sn = inf{sn }. Recall: The denition of supremum and inmum of a set. Proof : Let β = sup{sn }. If β < ∞, then by denition if ε > 0 then β − ε < sN ≤ β for some integer N. Since sN ≤ sn ≤ β if n ≥ N, it follows that β − ε < sn ≤ β if n ≥ N. 2.4. Some Special Sequences 33 This implies that |sn − β| < ε if n ≥ N, limn→∞ sn = β , by denition. so If β = ∞ and b is any real number, then sN > b for some integer N. Then sn > b for n ≥ N , so limn→∞ sn = ∞. For the proof of the second part try yourself. Example: If s0 = 1 and sn = 1 − e−sn−1 , then 0 < sn ≤ 1 for all n, by induction. Since sn+1 − sn = −(e−sn − e−sn−1 ) if n ≥ 1. The mean value theorem implies that sn+1 − sn = e−tn (sn − sn−1 ) if n ≥ 1, (2.1) where tn is between sn−1 and sn. Since s1 − s0 = −1/e < 0, it follows by induction from (2.1) that sn+1 − sn < 0 for all n. Hence, {sn } is bounded and decreasing, and therefore convergent. Remark: Let {xn } and {sn } be two sequences. If 0 ≤ xn ≤ sn for n ≥ N, where N is some xed number, and if sn → 0, then xn → 0. 2.4 Some Special Sequences Theorem: If p > 0, then 1 lim = 0. n→∞ np Proof : Before presenting the proof let us recall the following: Recall the binomial theorem: x2 (1 + x)n = 1 + nx + x(n − 1) +.... 2! We have the following inequality 1 + nx ≤ (1 + x)n , x > 0. Theorem: If p > 0, then √ lim n p = 1. n→∞ Proof : We will discuss the three cases when p = 1, when p > 1 and when 0 < p < 1. √ If p > 1, put xn = n p − 1. Then xn > 0, and, by the binomial theorem 1 + nxn ≤ (1 + xn )n = p, 2.4. Some Special Sequences 34 so that p−1 0 < xn ≤. n Theorem: Show that √ n lim n = 1. n→∞ √ Proof : Take xn = n n − 1. Then xn ≥ 0, and, by the binomial theorem n(n − 1) 2 n = (1 + xn )n ≥ xn. 2 Hence √ 2 0 ≤ xn ≤ , n ≥ 2. n−1 Theorem: If p>0 and α is real, then nα lim = 0. n→∞ (1 + p)n Proof : Let k be an integer such that k > α, k > 0. For n > 2k , n(n − 1)...(n − k + 1) k nk pk (1 + p)n > p > k. k! 2 k! Hence nα 2k k! α−k 0< < n , n > 2k. (1 + p)n pk Since α − k < 0, nα−k → 0 by 1 lim = 0. n→∞ np Theorem: If |x| < 1, then lim xn = 0. n→∞ Theorem: Let limn→∞ sn = s and limn→∞ tn = t, where s and t are nite. Then lim (csn ) = cs (2.2) n→∞ if c is a constant; lim (sn + tn ) = s + t, n→∞ lim (sn − tn ) = s − t, n→∞ lim (sn tn ) = st, n→∞ sn s lim = n→∞ tn t 2.4. Some Special Sequences 35 if tn is nonzero for all n and t ̸= 0. Proof : We write sn tn − st = sn tn − stn + stn − st = (sn − s)tn + s(tn − t); hence, |sn tn − st| ≤ |sn − s| |tn | + |s| |tn − t|. Since {tn } converges, it is bounded. Therefore, there is a number R such that |tn | ≤ R for all n, and ( ??) implies that |sn tn − st| ≤ R|sn − s| + |s| |tn − t|. (2.3) By denition, if ε>0 there are integers N1 and N2 such that |sn − s| < ε if n ≥ N1 |tn − t| < ε if n ≥ N2. If N = max(N1 , N2 ), then both inequalities hold when n ≥ N , and the (2.3) implies that |sn tn − st| ≤ (R + |s|)ε if n ≥ N. This proves (2.3). We consider the special case where sn = 1 for all n and t ̸= 0; thus, we want to show that 1 1 =. lim n→∞ tn t First, observe that since limn→∞ tn = t ̸= 0, there is an integer M such that |tn | ≥ |t|/2 if n ≥ M. By denition with ε = |t|/2; thus, there is an integer M such that |tn − t| < |t/2| if n ≥ M. Therefore, |t| |tn | = |t + (tn − t)| ≥ ||t| − |tn − t|| ≥ if n ≥ M. 2 If ε > 0, choose N0 so that |tn − t| < ε if n ≥ N0 , and let N = max(N0 , M ). Then 1 1 |t − tn | 2ε − = ≤ 2 if n ≥ N. tn t |tn | |t| |t| hence, limn→∞ 1/tn = 1/t. Now we obtain (2.3) in the general case from (2.3) with {tn } replaced by {1/tn }. Example: Find the limit of the sequence 1 nπ 2(1 + 3/n) sn = sin +. n 4 1 + 1/n 2.5. Subsequence 36 Solution: We apply the applicable parts of Theorem as follows: 1 nπ 2 [limn→∞ 1 + 3 limn→∞ (1/n)] lim sn = lim sin + 1 + lim (1/n) n→∞ n n→∞ 4 limn→∞ n→∞ 2(1 + 3 · 0) = 0+ = 2. 1+0 Example: Find the limit of the sequence sn = limn→∞ (n/2)+log n √ 3n+4 n. Solution: (n/2) + log n 1/2 + (log n)/n lim √ = lim n→∞ 3n + 4 n n→∞ 3 + 4n−1/2 limn→∞ 1/2 + limn→∞ (log n)/n = limn→∞ 3 + 4 limn→∞ n−1/2 1/2 + 0 = 3+0 1 =. 6 2.5 Subsequence A sequence {tk } is a subsequence of a sequence {sn } if tk = s n k , k ≥ 0, where {nk } is an increasing innite sequence of integers in the domain of {sn }. We denote the subsequence {tk } by {snk }. Example: If { } { } 1 1 1 1 {sn } = = 1, , ,... , ,... , n 2 3 n then letting nk = 2k yields the subsequence { } { } 1 1 1 1 {s2k } = = , ,..., ,... , 2k 2 4 2k and letting nk = 2k + 1 yields the subsequence { } { } 1 1 1 {s2k+1 } = = 1, ,... , ,.... 2k + 1 3 2k + 1 Example: The sequence {sn } dened by ( ) 1 n sn = (−1) 1 + n does not converge, but {sn } has subsequences that do. For example, { } 1 {s2k } = 1 + 2k 2.5. Subsequence 37 lim s2k = 1. k→∞ Example: The sequence {sn } dened by ( ) n 1 sn = (−1) 1 + n does not converge, but {sn } has subsequences that do. For example, { } 1 {s2k+1 } = −1 − 2k + 1 lim s2k+1 = −1. k→∞ It can be shown that a subsequence {snk } of {sn } converges to 1 if and only if nk is even for k suciently large, or to −1 if and only if nk is odd for k suciently large. Otherwise, {snk } diverges. Theorem: If lim sn = s (−∞ ≤ s ≤ ∞), n→∞ then lim snk = s k→∞ for every subsequence {snk } of {sn }. Proof : By denition for every ε > 0, there is an integer N such that |sn − s| < ε if n ≥ N. Since{nk } is an increasing sequence, there is an integer K such that nk ≥ N if k ≥ K. Therefore, |snk − L| < ε if k ≥ K. Theorem: If {sn } is monotonic and has a subsequence {snk } such that lim snk = s, k→∞ then lim sn = s. n→∞ Recall the following theorem: If {sn } is nondecreasing, then limn→∞ sn = sup{sn }. If {sn } is nonincreasing, then limn→∞ sn = inf{sn }. Proof : Since {snk } is also nondecreasing in this case, it is sucient to show that sup{snk } = sup{sn } 2.5. Subsequence 38 Since the set of terms of {snk } is contained in the set of terms of {sn }, sup{sn } ≥ sup{snk }. Since {sn } is nondecreasing, there is for every n an integer nk such that sn ≤ snk. This implies that sup{sn } ≤ sup{snk }. Theorem: A point x is a limit point of a set S if and only if there is a sequence {xn } of points in S such that xn ̸= x for n ≥ 1, and lim xn = x. n→∞ Proof : By denition for each ε > 0, there is an integer N such that 0 < |xn −x| < ε if n ≥ N. Therefore, every ε-neighborhood of x contains innitely many points of S. This means that x is a limit point of S. x be a limit point of S. Then, for every For necessity, let integer n ≥ 1, the interval (x − 1/n, x + 1/n) contains a point xn (̸= x) in S. Since |xm − x| ≤ 1/n if m ≥ n, limn→∞ xn = x. Theorem: If {xn } is bounded, then {xn } has a convergent subsequence. If {xn } is unbounded above, then {xn } has a subsequence {xnk } such that lim xnk = ∞. k→∞ If {xn } is unbounded below, then {xn } has a subsequence {xnk } such that lim xnk = −∞. k→∞ Proof : Let S be the set of distinct numbers that occur as terms of {xn }. (For example, if {xn } = {(−1)n }, S = = {1, −1}; if {xn } {1, 12 , 1, 13 ,... , 1, 1/n,... }, S = {1, 12 ,... , 1/n,... }.) If S contains only nitely many points, then some x in S occurs innitely often in {xn }; that is, {xn } has a subsequence {xnk } such that xnk = x for all k. Then limk→∞ xnk = x, and we are nished in this case. If S is innite, then, since S is bounded (by assumption), the Bolzano- Weierstrass theorem implies that S has a limit point x. There is a sequence of points {yj } in S, distinct from x, such that lim yj = x. j→∞ 2.6. Limit Superior and Limit Inferior 39 Although each yj occurs as a term of {xn }, {yj } is not necessarily a subsequence of {xn }, because if we write yj = x n j. There is no reason to expect that {nj } is an increasing sequence as required in denition of subsequence. However, it is always possible to pick a subsequence {njk } of {nj } that is in- creasing, and then the sequence {yjk } = {snjk } is a subsequence of both {yj } and {xn }. 2.6 Limit Superior and Limit Inferior Theorem: If {sn } is bounded above and does not diverge to −∞, then there is a unique real number s such that, if ε > 0, sn < s + ε for large n (2.4) and sn > s − ε for innitely many n. (2.5) If {sn } is bounded below and does not diverge to ∞, then there is a unique real number s such that, if ε > 0, sn > s − ε for large n (2.6) and sn < s + ε for innitely many n. (2.7) Proof : {sn } is bounded above, there is Proof of the rst part. Since a number β such that sn < β for all n. Since {sn } does not diverge to −∞, there is a number α such that sn > α for innitely many n. If we dene Mk = sup{sk , sk+1 ,... , sk+r ,... }, then α ≤ Mk ≤ β , so {Mk } is bounded. Since {Mk } is nonincreasing (why?), it converges. Let s = lim Mk. (2.8) k→∞ If ε > 0, then Mk < s + ε for large k, and since sn ≤ Mk n ≥ k , s satises (2.4). for If (2.5) were false for some positive ε, there would be an integer K such that sn ≤ s − ε if n ≥ K. However, this implies that Mk ≤ s − ε if k ≥ K, 2.6. Limit Superior and Limit Inferior 40 which contradicts (2.8). Therefore, s has the stated properties. Now we must show that s is the only real number with the stated properties. If t < s, the inequality s−t s−t sn < t + =s− 2 2 cannot hold for all large n, because this would contradict (2.5) with ε = (s − t)/2. If s < t, the inequality t−s t−s sn > t − =s+ 2 2 cannot hold for innitely many n, because this would contradict (2.4) with ε = (t − s)/2. Therefore, s is the only real number with the stated properties. The numbers s and s dened in the previous Theorem are called the limit supe- rior and limit inferior , respectively, of {sn }, and denoted by s = lim sup sn and s = lim inf sn. n→∞ n→∞ We also dene lim sup sn = ∞ if {sn } is not bounded above, n→∞ lim sup sn = −∞ if lim sn = −∞, n→∞ n→∞ lim inf sn = −∞ if {sn } is not bounded below, n→∞ lim inf sn = ∞ if lim sn = ∞. n→∞ n→∞ Theorem: Every sequence {sn } of real numbers has a unique limit superior, s, and a unique limit inferior, s, in the extended reals, and s ≤ s. Examples:   ∞, |r| > 1, lim sup rn = 1, |r| = 1, n→∞  0, |r| < 1; and    ∞, r > 1,     1, r = 1, lim inf rn = 0, |r| < 1, n→∞     −1, r = −1,   −∞, r < −1. 2.7. Cauchy Sequence 41 Example: lim sup n2 = lim inf n2 = ∞, n→∞ n→∞ ( ) 1 lim sup(−1) 1 − n = 1, n→∞ n ( ) 1 lim inf (−1) n − n = −1, n→∞ n and lim sup [1 + (−1)n ] n2 = ∞, n→∞ lim inf [1 + (−1)n ] n2 = 0. n→∞ Theorem: If {sn } is a sequence of real numbers, then lim sn = s (2.9) n→∞ if and only if lim sup sn = lim inf sn = s. (2.10) n→∞ n→∞ Proof : s = ±∞, the equivalence of (2.9) and (2.10) follows immediately from If their denitions. If limn→∞ sn = s (nite), then denition of a convergent sequence implies that (2.4)(2.7) hold with s and s replaced by s. Hence, (2.10) follows from the uniqueness of s and s. For the converse, suppose that s = s and let s denote their common value. Then (2.4) and (2.6) imply that s − ε < sn < s + ε for large n, and (2.9) follows from Denition and the uniqueness of limn→∞ sn. 2.7 Cauchy Sequence A sequence {sn } of real numbers is called a Cauchy sequence if for every ε > 0, there is an integer N such that |sn − sm | < ε if m, n ≥ N. Theorem: If {sn } is a Cauchy sequence of real numbers, then {sn } is bounded. Proof : See Lecture Theorem (Cauchy's convergence criterion): A sequence {sn } of real numbers converges if and only if, for every ε > 0, there is an integer N such that |sn − sm | < ε if m, n ≥ N. 2.8. Series 42 Proof : Suppose that limn→∞ sn = s and ε > 0. There is an integer N such that ε |sr − s| < if r ≥ N. 2 Therefore, |sn − sm | = |(sn − s) + (s − sm )| ≤ |sn − s| + |s − sm | < ε, whenever n, m ≥ N. Therefore, the stated condition is necessary for convergence of {sn }. Recall a Cauchy sequence {sn } is bounded. So s and s are nite. Now suppose that ε>0 and N satises |sn − sm | < ε, n, m ≥ N. From denition of limit superior and limit inferior |sn − s| < ε, |sm − s| < ε for some integer m, n > N. Since |s − s| = |(s − sn ) + (sn − sm ) + (sm − s)| ≤ |s − sn | + |sn − sm | + |sm − s|. We have |s − s| < 3ε. Since ε is an arbitrary positive number, this implies that s = s, so {sn } converges. 2.8 Series If {an }∞ k is an innite sequence of real numbers, the symbol ∞ ∑ an n=k is an innite series ∑∞ , and an is the nth term of the series. We say that n=k an converges to the sum A, and write ∞ ∑ an = A, n=k if the sequence {An }∞ k dened by An = ak + ak+1 + · · · + an , n ≥ k, converges to A. 2.8. Series 43 2.8.1 Sequence of Partial Sums ∑∞ The nite sum An is the nth partial sum of n=k an. ∑∞ If {An }∞ k diverges, we say that n=k an diverges. ∑∞ In particular, if limn→∞ An = ∞ or −∞, we say that n=k an diverges to ∞ or −∞, and write ∞ ∑ ∞ ∑ an = ∞ or an = −∞. n=k n=k 2.8.2 Oscillatory Series A divergent innite series that does not diverge to ±∞ is said to oscillate , or be oscillatory. Example: Consider the series ∞ ∑ rn , −1 < r < 1. n=0 Here an = rn (n ≥ 0). The nth term of sequence of partial sum is 1 − rn+1 An = 1 + r + r2 + · · · + rn = , 1−r which converges to 1/(1 − r) as n → ∞. Thus, we write ∞ ∑ 1 rn = , −1 < r < 1.

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