Introduction to Real Analysis 4th Edition Bartle Solutions Manual
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CONTENTS
Chapter 1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 2 The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Chapter 3 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Chapter 4 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .28
Chapter 5 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Chapter 6 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .43
Chapter 7 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
Chapter 8 Sequences of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
Chapter 9 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Chapter 10 The Generalized Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
Chapter 11 A Glimpse into Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
Selected Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
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, CHAPTER 1
PRELIMINARIES
We suggest that this chapter be treated as review and covered quickly, without
detailed classroom discussion. For one reason, many of these ideas will be already
familiar to the students — at least informally. Further, we believe that, in practice,
those notions of importance are best learned in the arena of real analysis, where
their use and significance are more apparent. Dwelling on the formal aspect of
sets and functions does not contribute very greatly to the students’ understanding
of real analysis.
If the students have already studied abstract algebra, number theory or com-
binatorics, they should be familiar with the use of mathematical induction. If not,
then some time should be spent on mathematical induction.
The third section deals with finite, infinite and countable sets. These notions
are important and should be briefly introduced. However, we believe that it is
not necessary to go into the proofs of these results at this time.
Section 1.1
Students are usually familiar with the notations and operations of set algebra,
so that a brief review is quite adequate. One item that should be mentioned is
that two sets A and B are often proved to be equal by showing that: (i) if x ∈ A,
then x ∈ B, and (ii) if x ∈ B, then x ∈ A. This type of element-wise argument is
very common in real analysis, since manipulations with set identities is often not
suitable when the sets are complicated.
Students are often not familiar with the notions of functions that are injective
(= one-one) or surjective (= onto).
Sample Assignment: Exercises 1, 3, 9, 14, 15, 20.
Partial Solutions:
1. (a) B ∩ C = {5, 11, 17, 23, . . .} = {6k − 1 : k ∈ N}, A ∩ (B ∩ C) = {5, 11, 17}
(b) (A ∩ B) \ C = {2, 8, 14, 20}
(c) (A ∩ C) \ B = {3, 7, 9, 13, 15, 19}
2. The sets are equal to (a) A, (b) A ∩ B, (c) the empty set.
3. If A ⊆ B, then x ∈ A implies x ∈ B, whence x ∈ A ∩ B, so that A ⊆ A ∩ B ⊆ A.
Thus, if A ⊆ B, then A = A ∩ B.
Conversely, if A = A ∩ B, then x ∈ A implies x ∈ A ∩ B, whence x ∈ B.
Thus if A = A ∩ B, then A ⊆ B.
4. If x is in A \ (B ∩ C), then x is in A but x ∈ / B ∩ C, so that x ∈ A and x is
either not in B or not in C. Therefore either x ∈ A \ B or x ∈ A \ C, which
implies that x ∈ (A \ B) ∪ (A \ C). Thus A \ (B ∩ C) ⊆ (A \ B) ∪ (A \ C).
1
, 2 Bartle and Sherbert
Conversely, if x is in (A \ B) ∪ (A \ C), then x ∈ A \ B or x ∈ A \ C. Thus
x ∈ A and either x ∈ / B or x ∈ / C, which implies that x ∈ A but x ∈ / B ∩ C,
so that x ∈ A \ (B ∩ C). Thus (A \ B) ∪ (A \ C) ⊆ A \ (B ∩ C).
Since the sets A \ (B ∩ C) and (A \ B) ∪ (A \ C) contain the same elements,
they are equal.
5. (a) If x ∈ A ∩ (B ∪ C), then x ∈ A and x ∈ B ∪ C. Hence we either have
(i) x ∈ A and x ∈ B, or we have (ii) x ∈ A and x ∈ C. Therefore, either
x ∈ A ∩ B or x ∈ A ∩ C, so that x ∈ (A ∩ B) ∪ (A ∩ C). This shows that
A ∩ (B ∪ C) is a subset of (A ∩ B) ∪ (A ∩ C).
Conversely, let y be an element of (A ∩ B) ∪ (A ∩ C). Then either (j) y ∈
A ∩ B, or (jj) y ∈ A ∩ C. It follows that y ∈ A and either y ∈ B or y ∈ C.
Therefore, y ∈ A and y ∈ B ∪ C, so that y ∈ A ∩ (B ∪ C). Hence (A ∩ B) ∪
(A ∩ C) is a subset of A ∩ (B ∪ C).
In view of Definition 1.1.1, we conclude that the sets A ∩ (B ∪ C) and
(A ∩ B) ∪ (A ∩ C) are equal.
(b) Similar to (a).
6. The set D is the union of {x : x ∈ A and x ∈ / B} and {x : x ∈
/ A and x ∈ B}.
7. Here An = {n + 1, 2(n + 1), . . .}.
(a) A1 = {2, 4, 6, 8, . . .}, A2 = {3, 6, 9, 12, . . .}, A1 ∩ A2 = {6, 12, 18, 24, . . .} =
{6k :k ∈ N} = A5 .
(b) An = N \ {1}, because if n > 1, then n ∈ An−1 ; moreover 1 ∈ / An .
Also An = ∅, because n ∈ / An for any n ∈ N.
8. (a) The graph consists of four horizontal line segments.
(b) The graph consists of three vertical line segments.
9. No. For example, both (0, 1) and (0, − 1) belong to C.
10. (a) f (E) = {1/x2 : 1 ≤ x ≤ 2} = {y : 14 ≤ y ≤ 1} = [ 14 , 1].
(b) f −1 (G) = {x : 1 ≤ 1/x2 ≤ 4} = {x : 14 ≤ x2 ≤ 1} = [−1, − 12 ] ∪ [ 12 , 1].
11. (a) f (E) = {x + 2 : 0 ≤ x ≤ 1} = [2, 3], so h(E) = g(f (E)) = g([2, 3]) =
{y 2 : 2 ≤ y ≤ 3} = [4, 9].
(b) g −1 (G) = {y : 0 ≤ y 2 ≤ 4} = [−2, 2], so h−1 (G) = f −1 (g −1 (G)) =
f −1 ([−2, 2]) = {x : −2 ≤ x + 2 ≤ 2} = [−4, 0].
12. If 0 is removed from E and F , then their intersection is empty, but the
intersection of the images under f is {y : 0 < y ≤ 1}.
13. E \ F = {x : −1 ≤ x < 0}, f (E) \ f (F ) is empty, and f (E \ F ) =
{y : 0 < y ≤ 1}.
14. If y ∈ f (E ∩ F ), then there exists x ∈ E ∩ F such that y = f (x). Since x ∈ E
implies y ∈ f (E), and x ∈ F implies y ∈ f (F ), we have y ∈ f (E) ∩ f (F ). This
proves f (E ∩ F ) ⊆ f (E) ∩ f (F ).
15. If x ∈ f −1 (G) ∩ f −1 (H), then x ∈ f −1 (G) and x ∈ f −1 (H), so that f (x) ∈ G
and f (x) ∈ H. Then f (x) ∈ G ∩ H, and hence x ∈ f −1 (G ∩ H). This shows
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CONTENTS
Chapter 1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 2 The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Chapter 3 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Chapter 4 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .28
Chapter 5 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Chapter 6 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .43
Chapter 7 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
Chapter 8 Sequences of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
Chapter 9 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Chapter 10 The Generalized Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
Chapter 11 A Glimpse into Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
Selected Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
This sample only, Download all chapters at: alibabadownload.com
, CHAPTER 1
PRELIMINARIES
We suggest that this chapter be treated as review and covered quickly, without
detailed classroom discussion. For one reason, many of these ideas will be already
familiar to the students — at least informally. Further, we believe that, in practice,
those notions of importance are best learned in the arena of real analysis, where
their use and significance are more apparent. Dwelling on the formal aspect of
sets and functions does not contribute very greatly to the students’ understanding
of real analysis.
If the students have already studied abstract algebra, number theory or com-
binatorics, they should be familiar with the use of mathematical induction. If not,
then some time should be spent on mathematical induction.
The third section deals with finite, infinite and countable sets. These notions
are important and should be briefly introduced. However, we believe that it is
not necessary to go into the proofs of these results at this time.
Section 1.1
Students are usually familiar with the notations and operations of set algebra,
so that a brief review is quite adequate. One item that should be mentioned is
that two sets A and B are often proved to be equal by showing that: (i) if x ∈ A,
then x ∈ B, and (ii) if x ∈ B, then x ∈ A. This type of element-wise argument is
very common in real analysis, since manipulations with set identities is often not
suitable when the sets are complicated.
Students are often not familiar with the notions of functions that are injective
(= one-one) or surjective (= onto).
Sample Assignment: Exercises 1, 3, 9, 14, 15, 20.
Partial Solutions:
1. (a) B ∩ C = {5, 11, 17, 23, . . .} = {6k − 1 : k ∈ N}, A ∩ (B ∩ C) = {5, 11, 17}
(b) (A ∩ B) \ C = {2, 8, 14, 20}
(c) (A ∩ C) \ B = {3, 7, 9, 13, 15, 19}
2. The sets are equal to (a) A, (b) A ∩ B, (c) the empty set.
3. If A ⊆ B, then x ∈ A implies x ∈ B, whence x ∈ A ∩ B, so that A ⊆ A ∩ B ⊆ A.
Thus, if A ⊆ B, then A = A ∩ B.
Conversely, if A = A ∩ B, then x ∈ A implies x ∈ A ∩ B, whence x ∈ B.
Thus if A = A ∩ B, then A ⊆ B.
4. If x is in A \ (B ∩ C), then x is in A but x ∈ / B ∩ C, so that x ∈ A and x is
either not in B or not in C. Therefore either x ∈ A \ B or x ∈ A \ C, which
implies that x ∈ (A \ B) ∪ (A \ C). Thus A \ (B ∩ C) ⊆ (A \ B) ∪ (A \ C).
1
, 2 Bartle and Sherbert
Conversely, if x is in (A \ B) ∪ (A \ C), then x ∈ A \ B or x ∈ A \ C. Thus
x ∈ A and either x ∈ / B or x ∈ / C, which implies that x ∈ A but x ∈ / B ∩ C,
so that x ∈ A \ (B ∩ C). Thus (A \ B) ∪ (A \ C) ⊆ A \ (B ∩ C).
Since the sets A \ (B ∩ C) and (A \ B) ∪ (A \ C) contain the same elements,
they are equal.
5. (a) If x ∈ A ∩ (B ∪ C), then x ∈ A and x ∈ B ∪ C. Hence we either have
(i) x ∈ A and x ∈ B, or we have (ii) x ∈ A and x ∈ C. Therefore, either
x ∈ A ∩ B or x ∈ A ∩ C, so that x ∈ (A ∩ B) ∪ (A ∩ C). This shows that
A ∩ (B ∪ C) is a subset of (A ∩ B) ∪ (A ∩ C).
Conversely, let y be an element of (A ∩ B) ∪ (A ∩ C). Then either (j) y ∈
A ∩ B, or (jj) y ∈ A ∩ C. It follows that y ∈ A and either y ∈ B or y ∈ C.
Therefore, y ∈ A and y ∈ B ∪ C, so that y ∈ A ∩ (B ∪ C). Hence (A ∩ B) ∪
(A ∩ C) is a subset of A ∩ (B ∪ C).
In view of Definition 1.1.1, we conclude that the sets A ∩ (B ∪ C) and
(A ∩ B) ∪ (A ∩ C) are equal.
(b) Similar to (a).
6. The set D is the union of {x : x ∈ A and x ∈ / B} and {x : x ∈
/ A and x ∈ B}.
7. Here An = {n + 1, 2(n + 1), . . .}.
(a) A1 = {2, 4, 6, 8, . . .}, A2 = {3, 6, 9, 12, . . .}, A1 ∩ A2 = {6, 12, 18, 24, . . .} =
{6k :k ∈ N} = A5 .
(b) An = N \ {1}, because if n > 1, then n ∈ An−1 ; moreover 1 ∈ / An .
Also An = ∅, because n ∈ / An for any n ∈ N.
8. (a) The graph consists of four horizontal line segments.
(b) The graph consists of three vertical line segments.
9. No. For example, both (0, 1) and (0, − 1) belong to C.
10. (a) f (E) = {1/x2 : 1 ≤ x ≤ 2} = {y : 14 ≤ y ≤ 1} = [ 14 , 1].
(b) f −1 (G) = {x : 1 ≤ 1/x2 ≤ 4} = {x : 14 ≤ x2 ≤ 1} = [−1, − 12 ] ∪ [ 12 , 1].
11. (a) f (E) = {x + 2 : 0 ≤ x ≤ 1} = [2, 3], so h(E) = g(f (E)) = g([2, 3]) =
{y 2 : 2 ≤ y ≤ 3} = [4, 9].
(b) g −1 (G) = {y : 0 ≤ y 2 ≤ 4} = [−2, 2], so h−1 (G) = f −1 (g −1 (G)) =
f −1 ([−2, 2]) = {x : −2 ≤ x + 2 ≤ 2} = [−4, 0].
12. If 0 is removed from E and F , then their intersection is empty, but the
intersection of the images under f is {y : 0 < y ≤ 1}.
13. E \ F = {x : −1 ≤ x < 0}, f (E) \ f (F ) is empty, and f (E \ F ) =
{y : 0 < y ≤ 1}.
14. If y ∈ f (E ∩ F ), then there exists x ∈ E ∩ F such that y = f (x). Since x ∈ E
implies y ∈ f (E), and x ∈ F implies y ∈ f (F ), we have y ∈ f (E) ∩ f (F ). This
proves f (E ∩ F ) ⊆ f (E) ∩ f (F ).
15. If x ∈ f −1 (G) ∩ f −1 (H), then x ∈ f −1 (G) and x ∈ f −1 (H), so that f (x) ∈ G
and f (x) ∈ H. Then f (x) ∈ G ∩ H, and hence x ∈ f −1 (G ∩ H). This shows
