REAL ANALYSIS: INTRODUCTION

Transcription

1 REAL ANALYSIS: INTRODUCTION DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY JAIPUR Contents 1. The real number system 1 2. Field Axioms 1 3. Order Axioms 2 4. Completeness Axiom Dedekind Cut Archimedean Theorem Forcing Principle 3 5. Dense Set 4 6. Nested Intervals 5 1. The real number system Real analysis (traditionally, the theory of functions of a real variable) is a branch of mathematical analysis dealing with the real numbers and real-valued functions of a real variable. Real numbers are represented graphically as points on a line called Real line or Real axis. R Set of real numbers R is a Complete Ordered Field. (R, +,.) is a Field. (R, +) is an abelian group. 2. Field Axioms (R,.) is an abelian group. Multiplicative inverse exists for all non-zero numbers. Multiplication is distributive over addition i.e. a.(b + c) = a.b + a.c Figure 1. Real Line 1

2 2 DR RITU AGARWAL 3. Order Axioms Order Property Law of trichotomy is the property of an order relation < on a set X that for any x and y, exactly one of the following holds: x < y, x = y, or x > y. (R, +,.) is a Field. Exactly one of the relations hold: x = y or x < y or x > y. This is an exclusive property of R If x < y then x + z < y + z for all z R If x > 0, y > 0 then xy > 0. Transitivity: If x > y and y > z then x > z. If x > 0, x is called positive real number and x R +. If x < 0, x is called negative real number and x R. 4. Completeness Axiom Upper bound A real number b is called an upper bound for set S if x b for every x in S. Supremum (or L.U.B.) Let S be a set of real numbers bounded above. A real number b is called a least upper bound or Supremum for S if it has following properties: (1) b is an upper bound for S. (2) No number less than b is an upper bound for S. In other words, for ɛ > 0, there exists x S such that b ɛ < x whenever b is supremum of S. Lower bound A real number a is called an lower bound for set S if a x for every x in S. Infimum (or G.L.B.) Let S be a set of real numbers bounded below. A real number a is called a Greatest Lower bound or Infimum for S if it has following properties: (1) a is an lower bound for S. (2) No number greater than a is a lower bound for S. In other words, for ɛ > 0, there exists x S such that a + ɛ > x whenever a is infimum of S. Completeness Property:Every non-empty set S of real numbers which is bounded above has a supremum in R, i.e. there is a real number b such that b = Sup S 4.1. Dedekind Cut. In mathematics, a Dedekind cut, named after Richard Dedekind, is a partition of the rational numbers into two non-empty sets A and B, such that all elements of A are less than all elements of B, and A contains no greatest element. Dedekind cuts are one method of construction of the real numbers. Theorem 4.1 (Approximation Theorem). Let a non-empty set S of real numbers has a supremum b. Then for every a < b, x S such that a < x b. Proof. By definition of supremum, x b for all x S. If we had x a for every x S, then a would be an upper bound for S smaller than l.u.b. Therefore x > a for at least one x in S. Theorem 4.2 (Comparison Property). Let S and T be non-empty and bounded above subsets of R such that s t for every s S and t T. Then Sup S Sup T.

3 MAT612-REAL ANALYSIS 3 Theorem 4.3 (Property of integers). The set Z + of integers is unbounded above. Proof. If Z + were bounded above and has a supremum, say a = Sup(Z + ). Then by approximation theorem, a 1 < n for some n Z + a < n + 1 for this n Since n + 1 Z +, it contradicts the fact that a = Sup(Z + ). Theorem 4.4 (Archimedean Property). If F is an ordered field then x F, n Z + such that n > x. Proof. Let there do not exists any n such that n > x then x would be an upper bound for Z + contradicting that Z + is unbounded Archimedean Theorem. Theorem 4.5 (Alternative statement). If x > 0 and y R then n Z + such that nx > y. Geometrically, any line segment (no matter how long) can be covered by a finite number of line segments of a given positive length, no matter how small. Theorem 4.6. A complete ordered field is Archimedean. Proof: Suppose F is complete ordered field but not Archimedean. Then for all x > 0, n Z + such that n x. The set Z + is a non-empty set in F with an upper bound x. Since F is complete, x 0 = Sup Z +. Now n Z +, n + 1 Z + n + 1 x 0 n x 0 1 x 0 1 is an upper bound for Z +. But x 0 is supremum. This is a contradiction. Hence the theorem Forcing Principle. Theorem 4.7 (Forcing Principle). Suppose F is an Archimedean ordered field and x, a, b F, then (1) If x ɛ ɛ > 0 then x 0. (2) If x a + ɛ ɛ > 0 then x a. (3) If x ɛ ɛ > 0 then x = 0. (4) If a b ɛ ɛ > 0 then a = b. Proof: (1) Suppose ɛ > 0, x 0 but x > 0 x 2 > 0. Taking ɛ = x 2, we have x ɛ x x 2 Which is a contradiction. Hence x 0 x 0. Theorem 4.8. If an ordered field F is complete, then x F such that x 2 = 2. Proof: Suppose F is an complete ordered field. Let us define the sets A and B as (Using Dedekind cut) A = {x > 0 : x 2 < 2} and B = {x > 0 : x 2 > 2}. Then A φ as 1 A and B φ as 2 B.Also A is bounded above. Since F is complete, let u = Sup A. Then u 1 since 1 A. Every positive number less then u lies in A and greater than u lies in B. Since

4 4 DR RITU AGARWAL Figure 2. Dedekind Cut (1) Let 0 < y < u then y < Sup A. a A such that y < a u y 2 < a 2 < 2 since a A y A (2) Let u < y then u < t < y for some t B. t > Sup A, t / A 2 < t 2 < y 2 y B Let ɛ > 0 and choose δ F such that 0 < δ < Min { ɛ u, 4u }. Then 0 < δ < u and 0 < δ < ɛ 4u. Now, u δ A and u + δ B. (u δ) 2 < 2 < (u + δ) 2. (u + δ) 2 < 2 < (u δ) 2 Also, (u δ) 2 < u 2 < (u + δ) 2. Therefore, we obtain 4uδ < u 2 2 < 4uδ u 2 2 < 4uδ = 4u ɛ 4u = ɛ. Since this holds for all ɛ >, by Forcing principle, u 2 = 2. Theorem 4.9 ( 2 is not rational). Show that there is no rational number whose square is 2. Proof: Let 2 is a rational number. Than p, q Z + such that 2 = p, q 0, (p, q) = 1 q ( ) 2 2 = p 2 = 2q 2 p q Since q is an integer, 2q 2 is also an integer. Also, since p 2 is divisible by 2, p is also divisible by 2.(Why??) Therefore let p = 2k for some k Z +. Then, q 2 = 2k 2 q is also divisible by 2. p and q have common factor 2. Which is a contradiction to our assumption. Q.E.D. Theorem 4.10 (Q is not complete). The ordered field Q of rational numbers is not complete. Proof: Since there is no rational number whose square is 2 and a complete ordered field must have an element whose square is 2. Therefore set Q of rational numbers is not complete. Example Consider the subset S of Q defined as: S = { 1, !, ! + 1 2!, ! + 1 3!, ! + 1 3! + 1 4!,...} Sup S=e (an irrational number)does not belong to S. 5. Dense Set Definition 1 (Dense set in ordered field). A set S is dense in an ordered field F if a < b in F, x S such that a < x < b.

5 MAT612-REAL ANALYSIS 5 Figure 3. A dense Julia set (see below for how to draw them). All the points in the plane belong to the Julia set. No matter where you zoom, interesting details will appear. Figure 4. Nested intervals. Definition 2 (Dense set in metric space). A set S is dense in a metric space X if S = X. A subset A of a set X is dense in X if for any point x in X, any neighborhood of x contains at least one point from A (i.e., A has non-empty intersection with every non-empty open subset of X) Theorem 5.1 (Denseness of the rationals). The rational numbers form a dense set in any Archimedean ordered field F. OR Between two rational numbers, there is a rational number Proof: Since F is an Archimedean ordered field, for b > a, a, b Q, b a > 0. Take x = b a and y = 1 in the Archimedean theorem. Then n Z + such that n(b a) > 1 nb na > 1 Since na and nb are one unit apart, there exists an integer m between na and nb. na < m < nb a < m n < b This shows that between two rational numbers a and b, there is a rational number. 6. Nested Intervals Definition 3 (Nested Intervals). A sequence of intervals I n, n N is nested if the following chain of inclusion hold I 1 I 2... I n I n+1... Example: I n = [0, 1/n] and n=1 = {0}. Let I n = [a n, b n ] i.e. I 1 = [a 1, b 1 ], I 2 = [a 2, b 2 ],... Then for I n to be nested interval, a n a n+1 and b n+1 b n for all n Theorem 6.1 (Nested interval Theorem). Consider a family of closed intervals, I 1 = [a 1, b 1 ], I 2 = [a 2, b 2 ],... If a n a n+1 and b n+1 b n for all n then there is an ξ R which is in every I n, that is, there is an ξ n=1 I n.

6 6 DR RITU AGARWAL Proof: Since the intervals are nested, I n I n+1 for all n N. a n b 1 for all n N {a n : n N} is non empty set bounded above. Let ξ = sup{a n : n N}. We claim that ξ b n for all n N. This is established by showing that for any particular n, the number b n is an upper bound for the set {a k : k N}. Consider two cases: (i) n k then I n I k a k b k b n. (ii) k < n then I n I k a k a n b n. a k b n for all k b n is an upper bound of the set {a k : k N} ξ b n n N. Since a n ξ b n n N ξ I n n N. Theorem 6.2. Consider a nested sequence of closed intervals, I n = [a n, b n ], n N such that the length b n a n of I n satisfy inf{b n a n : n N} = 0 then the number ξ contained in I n for all n N is unique. Exercise 1. Write a detailed note on Real numbers and its properties. Show that 2 is irrational. References (1) Tom M. Apostol, Mathematical Analysis, Narosa Publishing House, ISBN X