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Chapter I The Real Line System Readings: Elementary Analysis: The Theory of Calculus. (chapter 1) Introduction to Real analysis (chapter 2) The set N of Natural Numbers: We denote the set {1,2,3,……} of all numbers by N. The set N has the following properties: N1. 1 N . N2. If n N , then its successor n 1 N . N3. 1 is not the successor of any element in N. N4. If n and m in N have the same successor, then n = m. N5. A subset of N which contains 1 and which contains n+1 whenever it contains n, must equals N. (i.e. If B N such that (i) 1 B (ii) If n 1 Bn B , then B = N) Properties N1~N5 are known as the Peano Axioms. Axiom N5 is the basis of mathematical induction. The set Q of Rational Numbers: Rational numbers are numbers of the form 1 m where m , n Z and n n 0. The set Q of rational numbers is a very “nice” algebraic system until one tries to solve equations like x2 = 2. No rational number satisfies this equation (we’ll prove this later), but there is a number that satisfies this equation. 2 , , e Q ( Do you know why?). So there are “gaps” in Q. Definition 1.1: A number is called an algebraic number if it satisfies a polynomial equation an xn+an-1 xn-1+..........+a1 x +ao = 0 where ao , a1, ………, an are all integers, an 0 , and n 1 . Notes: (i) Rational numbers are algebraic numbers. If r m , then r satisfies the equation n nx-m=0. (ii) 2 is an algebraic number. Theorem 1.2: (The Rational Zeros Theorem) Suppose that ao , a1, ………, an are integers and that r is a rational number satisfying the equation an xn+an-1 xn-1+..........+a1 x +ao = 0 n 1 , an 0 , ao 0 , and r (1) where p , p and q are integers having no common factor and q q 0 . Then q divides an and p divides ao. 2 In other words: The only rational candidate for solutions of (1) have the form r p where q divides an and p divides ao. q Proof: r p satisfies equation (1). Then q n p p an ...............a1 ao 0 q q Multiplying by qn, we get an pn+an-1 pn-1q+an-2 pn-2q2+……………+a1 pqn-1+aoqn= 0 an pn=-q[an-1 pn-1+an-2 pn-2q+……………+a1 pqn-2+aoqn-1] q divides an pn But p and q have no common factor, therefore q divides an . Also, aoqn =-p[an pn-1+an-1 pn-2q+……………+a1 qn-1] p divides ao . 2 can’t represent a rational number. Example 1.3: By theorem 1.2 the only rational number that could possibly solve the equation x 2- 2 = 0 are ±1, ±2 (since an = 1 and ao = -2 p = ±1, ±2 and q = ±1). We can substitute each of the four numbers into the equation x2- 2 = 0 to eliminate them as possible solutions. But Example 1.4: 3 2 is a solution, therefore 2 can’t be a rational number. 6 can’t be a rational number. By theorem 1.2 the only rational numbers that could possibly solve the equation 3 x 3- 6 = 0 are ±1, ±2, ±3, ±6 (since an = 1 and ao = -6 p = ±1, ±2, ±3, ±6 and q = ±1). We know that none of these numbers is a solution of the equation x3- 6 =0. But 3 6 is a solution, therefore 3 6 can’t be a rational number. 1 2 Example 1.5: b (2 5 ) does not represent a rational number. 3 b is a solution of the equation x6 -6x4 +12x2 -13 = 0. ( an = 1 and ao = -13 p = ±1, ±13 and q = ±1). By theorem1.2 the only possible rational solutions are ±1, ±13. We know that none of these numbers is a solution. But b is a solution, therefore b can’t be a rational number. H.W. 1 : See Homework problems for chapter I The Algebraic and Order Properties of R: Algebraic Properties of R: A1. a +b = b +a a, b R . A2. (a +b) +c = a +(b +c) a, b, c R . A3. a +0 = 0 +a = a a R . A4. a R there is an element a R such that a +(-a ) = (-a ) +a = 0. M1. a .b = b .a a, b R . M2. (a .b) .c = a .(b .c) a, b, c R . M3. a .1 = 1 .a = a a R . 4 M4. a R(a 0) there is an element 1 1 1 R such that a .( ) = ( ) .a = 1. a a a D. a .(b +c) = (a .b) + (a .c) (b +c) .a = (b .a) + (c .a) a, b, c R Definition 1.6: A system that has more than one element and satisfies the nine above properties is called a field . The Order Properties of R: O1. Given a and b, either a ≤ b or b ≤ a O2. If a ≤ b and b ≤ a , then a = b O3. If a ≤ b and b ≤ c , then a ≤ c O4. If a ≤ b , then a + c ≤ b + c O5. If a ≤ b and 0 ≤ c , then a . c ≤ b . c Definition 1.7: A field with an ordering satisfying O1 ~ O5 is called an ordered field. Thus R is an ordered field. Theorem 1.8: If a R is such that 0 ≤ a < ε for every ε > 0, then a = 0. 1 2 Proof: Suppose to the contrary that a > 0. Then take o a 0 o a which is a contradiction. Therefore a = 0. 5 Absolute Value and the Real Line: Definition 1.9: The absolute value of a real number a, denoted by |a|, is defined by: a if a 0 if a if a 0 a 0 a 0 Theorem 1.10: (i) |ab| = |a| |b| a, b R (ii) |a|2 = |a2| a R (iii) If c > 0, then |a| ≤ c if and only if –c ≤ a ≤ c (iv) -|a| ≤ a ≤ |a| a R Theorem 1.11: (Triangular Inequality) If a, b R , then |a + b| ≤ |a| + |b|. Proof: -|a| ≤ a ≤ |a| and -|b| ≤ b ≤ |b|. Adding we get (a b) ab a b ab a b by theorem 1.10(iii). Corollary 1.12: If a, b R , then (i) a b a b (ii) a b a b 6 Proof: (i) a = a - b + b a a b b a b b a b a b b=b–a+a b b a a b a a a b b a a b Therefore, a b a b a b a b a b by theorem 1.10(iii) (ii) a b a (b) a b a b Therefore, a b a b . Definition 1.13: The distance between two elements a and b in R is dist (a , b) = a b . Definition 1.14: Let a R and ε > 0. Then the ε-neighborhood of a is the set V a x R : x a [ x V a x a a x a ] Theorem 1.15: Let a R . If x V (a) for every ε > 0, then x = a. 7 Proof: x V (a) 0 xa 0 x a 0 by theorem 1.8 x a. H.W. 2: See Homework problems for chapter I ___________________________________________________________________ The Completeness property of R: Suprema and Infima: Definition 1.16: Let S be a nonempty subset of R. (a) If S contains a largest element so (i.e., so S and s so s S ), then we call so the maximum of S and we write so = max S. (b) If S contains a smallest element s1 (i.e., s1 S and s s1 s S ), then we call s1 the minimum of S and we write s1 = min S. Example 1.17: (a) Every finite nonempty subset of R has a maximum and a minimum. max{ n Z : 4 n 100} 100 min{ n Z : 4 n 100} 3 (b) max [a , b] = b min [a , b] = a (a , b) has no maximum or minimum. [a , b) has no maximum but its minimum is a. 8 (c) The sets Z and Q have no maximum or minimum. N has a minimum which is 1. (d) min{ r Q : 0 r 2} 0 . But the set has no maximum. Note: Always min S S and max S S . Definition 1.18: Let S be a nonempty subset of R. (a) The set S is said to be bounded above if there exists a number u R such that s u s S . Each number u is called an upper bound of S. (b) The set S is said to be bounded below if there exists a number w R such that s w s S . Each number w is called a lower bound of S. (c) A set is said to be bounded if it is both bounded above and below. A set is said to be unbounded if it is not bounded. Example 1.19: (a) The sets (a , b), (a , b], [a , b) and [a , b]are bounded above, the number b and any number larger then b is an upper bound. Also they are bounded below, the number a and any number less then a is a lower bound. (b) The sets Z, Q, N and R are unbounded. But N is bounded below, the number 1 and any number less then 1 is a lower bound. 9 (c) The set {r Q : 0 r 2} is bounded. The number then 2 and any number larger 2 is an upper bound, the number 0 and any number less then 0 is a lower bound. (d) The set {n( 1) : n N} is unbounded. It is not bounded above. n Note: If the set S is bounded above (or below) then the upper bounds (or the lower bounds) don’t have to belong to S. Definition 1.20: Let S be a nonempty subset of R. (a) If S is bounded above, then a number u is said to be a supremum (or a least upper bound) of S if it satisfies the conditions: (i) u is an upper bound of S. (ii) If v is any upper bound of S, then u ≤ v. We write u = sup S. (b) If S is bounded below, then a number w is said to be an infimum (or a greatest lower bound) of S if it satisfies the conditions: (iii) w is a lower bound of S. (iv) If t is any lower bound of S, then t ≤ w. We write w = inf S. 10 Example 1.21: (a) sup [a , b] = sup (a , b) = sup (a , b] = sup [a , b) = b. inf [a , b] = inf (a , b) = inf (a , b] = inf [a , b) = a. (b) sup {r Q : 0 r 2} = 2. inf {r Q : 0 r 2} = 0. (c) inf {n( 1) : n N} = 0. n Notes: (i) If a set S has a maximum, then max S = sup S. Also if it has a minimum, then min S = inf S. (ii) sup S and inf S need not belong to the set S. (iii) There can only be only one supremum and only one infimum. (iv) Not every subset of R has a supremum or an infimum. Lemma 1.22: Let S be a nonempty set, and u = sup S. Then the following statements are equivalent: (i) If v is any upper bound of S, then u ≤ v. (ii) If z < u, then z is not an upper bound of S. (iii) If z < u, then sz S such that z < sz . (iv) If ε > 0, then s S such that u – ε < sε . 11 Exercise: Obtain the four equivalent statements above for w = inf S. The Completeness Property of R: Every nonempty set of real numbers that has an upper bound also has a supremum in R. Corollary 1.23: Every nonempty subset of R that has a lower bound also has an infimum in R. Proof: Let –S = { s : s S } . S is bounded below, therefore there exists a real number t such that t s s S t s s S t is an upper bound of –S. By the completeness property of R, the set –S has a supremum u = sup (-S). Then by definition 1.20: (i) s u s S (ii) If v is any upper bound of –S, then u v . (i) u s s S (ii) If -v is any lower bound of S, then v u . Therefore, -u = inf S. i.e. sup (-S) = -inf S . H.W. 3: See Homework problems for chapter I 12 Theorem 1.24: (The Archimedean Property): If x R , then nx N such that x < nx . Proof: Assume that n x n N . x is an upper bound of N. Therefore by the completeness property, N has a supremum u R . u - 1 < u u – 1 is not an upper bound of N m N such that u – 1 < m u<m+1 But m 1 N , and this contradicts the fact that u is an upper bound of N. Hence, nx N such that x < nx . Corollary 1.25: If t > 0, then nt N such that 0 1 t. nt 1 Proof: inf : n N = 0. n 1 t > 0 t is not a lower bound of the set : n N . Thus nt N such that n 0 1 t. nt 13 Density of Rational Numbers in R: Theorem 1.26: (The Density Theorem): If x and y are real numbers with x < y, then a rational number r Q such that x < r < y. (In other words: Between any two real numbers there is a rational number). Proof: Assume that x > 0. -x + y > 0 by corollary 1.24 n N such that 1 yx n nx + 1 < ny m N such that m 1 nx m nx > 0 m nx 1 Therefore, m < ny Now, nx < m < ny x m y. n Thus the rational number r m satisfies x < r < y. n What if x 0 ? (Exercise) Corollary 1.27: If x and y are real numbers with x < y, then an irrational number z such that x < z < y. Proof: x 2 y 2 By theorem 1.26, r Q such that xr 2 y. Then z r 2 is irrational and satisfies x < z < y. 14 x 2 r y 2 The Extended Real Numbers. It is often convenient to extend the system of real numbers by the addition of two elements +∞ and -∞. This enlarges the set R to the set R {,} . The set R R {,} is called the set of extended real numbers. One use of the extended real numbers is in the expression “sup S ”. If S is a nonempty set of real numbers with an upper bound, we define sup S to be the least upper bound of S. If S has no upper bound, we write sup S = +∞. Then sup S is defined for all nonempty sets. If S is bounded below then inf S is defined to be the greatest lower bound of S. If S has no lower bound, we write inf S = -∞. 15 Finite and Infinite Sets: Definition 1.28: (a) The empty set φ is said to have 0 elements. (b) If n N , a set S is said to have n elements if there is a bijection (1-1 and onto mapping) from the set Nn = {1,2,3,……..,n} onto S. (c) A set is said to be finite if it is either empty or it has n elements. (d) A set S is said to be infinite if it is not finite. i.e., if S has an infinite number of elements. Example 1.29: (a) The sets N, Z, Q, and R are all infinite sets. (b) All intervals are infinite sets. (c) The set {r Q : 0 r 2} is an infinite set. (d) The set {n Z : 10 n 20} is a finite set. Theorem 1.30: Suppose that S and T are sets such that T S . (i) If S is a finite set, then T is a finite set. (ii) If T is an infinite set, then S is an infinites set. Theorem 1.31: The finite union of finite sets is also finite. (i.e., if A1 , A2 , ……,An are all finite n sets, then B Ai is also finite). i 1 16 Countable Sets: Definition1.32: (a) A set S is said to be denumerable if there exists a bijection of N onto S. And we write S = {s1 , s2 , s3 ,………}. (b) A set S is said to be countable if it is either finite or denumerable. (c) A set S is said to be uncountable if it is not countable Sets Finite Sets (Countable) Infinite Sets Denumerable Sets (Countable) Uncountable Sets Theorem 1.33: (i) The union of two denumerable sets is denumerable.( countable) (ii) N × N is denumerable.( countable) (iii) The set Q of rational numbers is denumerable.( countable) Theorem 1.34: Suppose that S and T are sets such that T S . (i) If S is a countable set, then T is a countable set. (ii) If T is an uncountable set, then S is an uncountable set. 17 Theorem 1.35: The countable union of countable sets is countable. (i.e. if A1 , A2 , A3…… are all countable sets, then B Ai is also countable). i 1 Example 1.36: (a) The set E {2n : n N } of even natural numbers is denumerable since the mapping f (n) = 2n for n N is a bijection of N onto E. Similarly, the set O {2n 1 : n N } is denumerable. (why?) (b) The set Z of integers is denumerable since the mapping f : N → Z defined by 0 n f ( n) 2 1 n 2 if n 1 if nE if nO is a bijection of N onto Z. (c) All intervals are uncountable sets. (d) The set {r Q : 0 r 2} is a countable set. (d) The set {n Z : 10 n 20} is a countable set. Exercise: Is the countable union of finite sets also finite? 18