Solution Manual For
Introduction To Real Analysis by William F. Trench
business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based Questions:
Contents
Chapter 1 The Real Numbers 1
1.1 The Real Number System 1
1.2 Mathematical Induction 4
1.3 The Real Line 13
Chapter 2 Differential Calculus of Functions of One Variable 17
2.1 Functions and Limits 17
2.2 Continuity 24
2.3 Differentiable Functions of One Variable 30
2.4 L’Hospital’s Rule 36
2.5 Taylor’s Theorem 43
Chapter 3 Integral Calculus of Functions of One Variable 53
3.1 Definition of the Integral 53
3.2 Existence of the Integral 56
3.3 Properties of the Integral 61
3.4 Improper Integrals 66
3.5 A More Advanced Look at the Existence
of the Proper Riemann Integral 77
Chapter 4 Infinite Sequences and Series 79
4.1 Sequences of Real Numbers 79
4.2 Earlier Topics Revisited With Sequences 87
4.3 Infinite Series of Constants 89
4.4 Sequences and Series of Functions 100
4.5 Power Series 107
,Chapter 5 Real-Valued Functions of Several Variables 116
5.1 Structure of Rn 116
5.2 Continuous Real-Valued Function of n Variables 121
5.3 Partial Derivatives and the Differential 123
5.4 The Chain Rule and Taylor’s Theorem 130
Chapter 6 Vector-Valued Functions of Several Variables 141
6.1 Linear Transformations and Matrices 141
6.2 Continuity and Differentiability of Transformations 146
6.3 The Inverse Function Theorem 152
6.4 The Implicit Function Theorem 160
Chapter 7 Integrals of Functions of Several Variables 170
7.1 Definition and Existence of the Multiple Integral 170
7.2 Iterated Integrals and Multiple Integrals 187
7.3 Change of Variables in Multiple Integrals 207
Chapter 8 Metric Spaces 217
8.1 Introduction to Metric Spaces 217
8.2 Compact Sets in a Metric Space 224
8.3 Continuous Functions on Metric Spaces 226
business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based Questions:
, CHAPTER 1
THE REAL NUMBERS
1.1 THE REAL NUMBER SYSTEM
l.l.l. Note that |a — b| = max(a, b) — min(a, b).
(a) a + b + |a — b| = a + b + max(a, b) — min(a, b) = 2 max(a, b).
(b) a + b — |a — b| = a + b — max(a, b) + min(a, b) = 2 min(a, b).
(c) Let a = a+b+2c +|a—b |+ˇa +b—2c +|a—b|ˇ. From (a), a = 2 [max(a, b) + c + |max(a, b) —c |] =df
β. From (a) with a and b replaced by max(a, b) and c, β 4 max (max(a, b), c)
4 max(a, b, c).
(d) Let a a b 2c a b ˇa b 2c a b ˇ. From (b), a 2 [min(a, b) c min(a, b) c ] df
β. From (a) with a and b replaced by min(a, b) and c, β 4 min (min(a, b), c)
4 min(a, b, c).
l.l.2. First verify axioms A-E:
Axiom A. See Eqns. (1.1.1) and (1.1.2).
Axiom B. If a 0 then (a b) c b c and a(b c) bc, so (a b)
c a (b c). Similar arguments apply if b 0 or c 0. The remaining case is
a b c l. Since (l l) l 0 l l and l (l l) l 0 l, addition
is associative. Since
0, unless a = b = c = l,
(ab)c a(bc)
l, if a = b = c = l,
multiplication is associative.
Axiom C. Since
0, if a = 0,
a(b c) ab ac
b + c, if a = l,
the distributive law holds.
Axiom D. Eqns. (1.1.1) and (1.1.2) imply that 0 and l have the required properties.
business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based Questions:
, Axiom E. Eqn. (1.1.1) implies that —0 = 0 and —l = l; Eqn. (1.1.2) implies that l/l = l.
To see that the field cannot be ordered suppose that 0 < l. Adding l to both sides yields
0 l < l l, or l < 0, a contradiction. On the other hand, if l < 0, then l l < 0 l,
so 0 < l, also a contradiction.
, ,
l.l.3. If 2 is rational we can write 2 = m/n, where either m is odd or n is odd. Then
so n= =
2 m is even; thus, m = 2m where m is an integer. Therefore, 4m = 2n ,
2 2 2 2
m 2 2n , so
2m and 1 1
n is also even, a contradiction.
, ,
l.l.4. If p is rational we can write p = m/n where either m or n is not divisible
by p. Then m = pn , so m is divisible by p; thus, m = pm 1 where m 1 is an integer.
2 2
Therefore, p m = pn , so n = pm and n is also divisible by p, a contradiction.
2 2 2 2 2
l.l.6. If S is bounded below then T is bounded above, so T has a unique supremum, by
Theorem 1.1.3. Denote sup T = —a. Then (i) if x e S then —x ≤ —a, so x ≥ a; (ii) if
‹ > 0 there is an xO e T such that —xO < —a — ‹, so xO > a + ‹. Therefore, there is
an a with properties (i) and (ii). If (i) and (ii) hold with a replaced by a1 then —a1 is a
supremum of T , so a1 = a by the uniqueness assertion of Theorem 1.1.3.
l.l.7. (a) If x e S, then inf S ≤ x ≤sup S, and the transitivity of ≤implies (A), with
equality if and only if S contains exactly one point.
(b) There are three cases: (i) If S is bounded below and unbounded above, then inf S = a
(finite) and sup S = ∞ from (13); (ii) If S is unbounded below and bounded above, then
inf S = —∞ from (14) and sup S = β (finite); (iii) If S is unbounded below and above
then sup S = ∞ from (13) and inf S = —∞ from (14). In all three cases (12) implies (A).
l.l.8. Let a = inf T and b = sup S. We first show that a = b. If a < b then a <
(a + b)/2 < b. Since b = sup S, there is an sO e S such that sO > (a + b)/2. Since
a = inf T , there is a tO e T such that tO < (a + b)/2. Therefore, tO < sO, a contradiction.
Hence a ≥ b. If a > b there is an x such that b < x < a. From the definitions of a and
b, x /e T and x /e S, a contradiction. Hence a = b. Let β = a(= b). Since a and b are
uniquely, defined so is β. If x > β then x Ø S (since β = sup S), so x e T . If x < β then
x Ø T (since β = inf T ), so x e S.
l.l.9. Every real number is in either S or T . T is nonempty because U is bounded. S is
nonempty because if u U and x < u then x S. If s S there is a u U such that
u > s, since s is not an upper bound of U . If t T then t u, since t is an upper bound
of U . Since u > s and t u, t > s. Therefore, S and T satisfy the conditions imposed
in Exercise 1.1.8, so there is a number β such that every number greater than β is an upper
bound of U and no number less that β is an upper bound of U . However, β is also an upper
bound of U (if not, there would be a uO e U such that uO > β, which is impossible, since
if uO > β then every number in (β, uO) is an upper bound of U ). Therefore, β = sup U .
l.l.l0. (a) Let a = sup S and β = sup T . If x = s + t then x ≤ a + β. If ‹ > 0 choose sO
— ‹/2 and tO > —
in S and tO in T such that sO > a β ‹/2. Then xO= sO+ tO > a+ β ‹
and a β sup(S T ) by Theorem 1.1.3. This proves (A). The proof of (B) by mean
of Theorem 1.1.8 is similar.
(b) For (A), suppose that S is not bounded above. Then sup S = sup(S + T ) = ∞. Since
sup T > —∞ if T is nonempty, (A) holds. The proof of (B) is similar.
business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based Questions:
Introduction To Real Analysis by William F. Trench
business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based Questions:
Contents
Chapter 1 The Real Numbers 1
1.1 The Real Number System 1
1.2 Mathematical Induction 4
1.3 The Real Line 13
Chapter 2 Differential Calculus of Functions of One Variable 17
2.1 Functions and Limits 17
2.2 Continuity 24
2.3 Differentiable Functions of One Variable 30
2.4 L’Hospital’s Rule 36
2.5 Taylor’s Theorem 43
Chapter 3 Integral Calculus of Functions of One Variable 53
3.1 Definition of the Integral 53
3.2 Existence of the Integral 56
3.3 Properties of the Integral 61
3.4 Improper Integrals 66
3.5 A More Advanced Look at the Existence
of the Proper Riemann Integral 77
Chapter 4 Infinite Sequences and Series 79
4.1 Sequences of Real Numbers 79
4.2 Earlier Topics Revisited With Sequences 87
4.3 Infinite Series of Constants 89
4.4 Sequences and Series of Functions 100
4.5 Power Series 107
,Chapter 5 Real-Valued Functions of Several Variables 116
5.1 Structure of Rn 116
5.2 Continuous Real-Valued Function of n Variables 121
5.3 Partial Derivatives and the Differential 123
5.4 The Chain Rule and Taylor’s Theorem 130
Chapter 6 Vector-Valued Functions of Several Variables 141
6.1 Linear Transformations and Matrices 141
6.2 Continuity and Differentiability of Transformations 146
6.3 The Inverse Function Theorem 152
6.4 The Implicit Function Theorem 160
Chapter 7 Integrals of Functions of Several Variables 170
7.1 Definition and Existence of the Multiple Integral 170
7.2 Iterated Integrals and Multiple Integrals 187
7.3 Change of Variables in Multiple Integrals 207
Chapter 8 Metric Spaces 217
8.1 Introduction to Metric Spaces 217
8.2 Compact Sets in a Metric Space 224
8.3 Continuous Functions on Metric Spaces 226
business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based Questions:
, CHAPTER 1
THE REAL NUMBERS
1.1 THE REAL NUMBER SYSTEM
l.l.l. Note that |a — b| = max(a, b) — min(a, b).
(a) a + b + |a — b| = a + b + max(a, b) — min(a, b) = 2 max(a, b).
(b) a + b — |a — b| = a + b — max(a, b) + min(a, b) = 2 min(a, b).
(c) Let a = a+b+2c +|a—b |+ˇa +b—2c +|a—b|ˇ. From (a), a = 2 [max(a, b) + c + |max(a, b) —c |] =df
β. From (a) with a and b replaced by max(a, b) and c, β 4 max (max(a, b), c)
4 max(a, b, c).
(d) Let a a b 2c a b ˇa b 2c a b ˇ. From (b), a 2 [min(a, b) c min(a, b) c ] df
β. From (a) with a and b replaced by min(a, b) and c, β 4 min (min(a, b), c)
4 min(a, b, c).
l.l.2. First verify axioms A-E:
Axiom A. See Eqns. (1.1.1) and (1.1.2).
Axiom B. If a 0 then (a b) c b c and a(b c) bc, so (a b)
c a (b c). Similar arguments apply if b 0 or c 0. The remaining case is
a b c l. Since (l l) l 0 l l and l (l l) l 0 l, addition
is associative. Since
0, unless a = b = c = l,
(ab)c a(bc)
l, if a = b = c = l,
multiplication is associative.
Axiom C. Since
0, if a = 0,
a(b c) ab ac
b + c, if a = l,
the distributive law holds.
Axiom D. Eqns. (1.1.1) and (1.1.2) imply that 0 and l have the required properties.
business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based Questions:
, Axiom E. Eqn. (1.1.1) implies that —0 = 0 and —l = l; Eqn. (1.1.2) implies that l/l = l.
To see that the field cannot be ordered suppose that 0 < l. Adding l to both sides yields
0 l < l l, or l < 0, a contradiction. On the other hand, if l < 0, then l l < 0 l,
so 0 < l, also a contradiction.
, ,
l.l.3. If 2 is rational we can write 2 = m/n, where either m is odd or n is odd. Then
so n= =
2 m is even; thus, m = 2m where m is an integer. Therefore, 4m = 2n ,
2 2 2 2
m 2 2n , so
2m and 1 1
n is also even, a contradiction.
, ,
l.l.4. If p is rational we can write p = m/n where either m or n is not divisible
by p. Then m = pn , so m is divisible by p; thus, m = pm 1 where m 1 is an integer.
2 2
Therefore, p m = pn , so n = pm and n is also divisible by p, a contradiction.
2 2 2 2 2
l.l.6. If S is bounded below then T is bounded above, so T has a unique supremum, by
Theorem 1.1.3. Denote sup T = —a. Then (i) if x e S then —x ≤ —a, so x ≥ a; (ii) if
‹ > 0 there is an xO e T such that —xO < —a — ‹, so xO > a + ‹. Therefore, there is
an a with properties (i) and (ii). If (i) and (ii) hold with a replaced by a1 then —a1 is a
supremum of T , so a1 = a by the uniqueness assertion of Theorem 1.1.3.
l.l.7. (a) If x e S, then inf S ≤ x ≤sup S, and the transitivity of ≤implies (A), with
equality if and only if S contains exactly one point.
(b) There are three cases: (i) If S is bounded below and unbounded above, then inf S = a
(finite) and sup S = ∞ from (13); (ii) If S is unbounded below and bounded above, then
inf S = —∞ from (14) and sup S = β (finite); (iii) If S is unbounded below and above
then sup S = ∞ from (13) and inf S = —∞ from (14). In all three cases (12) implies (A).
l.l.8. Let a = inf T and b = sup S. We first show that a = b. If a < b then a <
(a + b)/2 < b. Since b = sup S, there is an sO e S such that sO > (a + b)/2. Since
a = inf T , there is a tO e T such that tO < (a + b)/2. Therefore, tO < sO, a contradiction.
Hence a ≥ b. If a > b there is an x such that b < x < a. From the definitions of a and
b, x /e T and x /e S, a contradiction. Hence a = b. Let β = a(= b). Since a and b are
uniquely, defined so is β. If x > β then x Ø S (since β = sup S), so x e T . If x < β then
x Ø T (since β = inf T ), so x e S.
l.l.9. Every real number is in either S or T . T is nonempty because U is bounded. S is
nonempty because if u U and x < u then x S. If s S there is a u U such that
u > s, since s is not an upper bound of U . If t T then t u, since t is an upper bound
of U . Since u > s and t u, t > s. Therefore, S and T satisfy the conditions imposed
in Exercise 1.1.8, so there is a number β such that every number greater than β is an upper
bound of U and no number less that β is an upper bound of U . However, β is also an upper
bound of U (if not, there would be a uO e U such that uO > β, which is impossible, since
if uO > β then every number in (β, uO) is an upper bound of U ). Therefore, β = sup U .
l.l.l0. (a) Let a = sup S and β = sup T . If x = s + t then x ≤ a + β. If ‹ > 0 choose sO
— ‹/2 and tO > —
in S and tO in T such that sO > a β ‹/2. Then xO= sO+ tO > a+ β ‹
and a β sup(S T ) by Theorem 1.1.3. This proves (A). The proof of (B) by mean
of Theorem 1.1.8 is similar.
(b) For (A), suppose that S is not bounded above. Then sup S = sup(S + T ) = ∞. Since
sup T > —∞ if T is nonempty, (A) holds. The proof of (B) is similar.
business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based Questions:
