Solutions Manual Introduction to Real Analysis 4th Edition By Robert G. Bartle, Donald R. Sherbert

Solutions Manual
Introduction to Real Analysis 4th Edition
By
Robert G. Bartle,
Donald R. Sherbert



( All Chapters Included - 100% Verified Solutions )




1

, 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).

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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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, Chapter 1 — Preliminaries 3

that f −1 (G) ∩ f −1 (H) ⊆ f −1 (G ∩ H). The opposite inclusion is shown in
Example 1.1.8(b). The proof for unions is similar.
√ √
16. If f (a) = f (b), then a/ a2 + 1 = b/ b2 + 1, from which it follows that a2 = b2 .
Since a and b must have the  same sign, we get a = b, and hence f is injective.
If −1 < y < 1, then x := y/ 1 − y 2 satisfies f (x) =√y (why?), √ so that f takes R
onto the set {y : − 1 < y < 1}. If x > 0, then x = x2 < x2 + 1, so it follows
that f (x) ∈ {y : 0 < y < 1}.
17. One bijection is the familiar linear function that maps a to 0 and b to 1,
namely, f (x) := (x − a)/(b − a). Show that this function works.
18. (a) Let f (x) = 2x, g(x) = 3x.
(b) Let f (x) = x2 , g(x) = x, h(x) = 1. (Many examples are possible.)
19. (a) If x ∈ f −1 (f (E)), then f (x) ∈ f (E), so that there exists x1 ∈ E such
that f (x1 ) = f (x). If f is injective, then x1 = x, whence x ∈ E. Therefore,
f −1 (f (E)) ⊆ E. Since E ⊆ f −1 (f (E)) holds for any f , we have set equality
when f is injective. See Example 1.1.8(a) for an example.
(b) If y ∈ H and f is surjective, then there exists x ∈ A such that f (x) = y.
Then x ∈ f −1 (H) so that y ∈ f (f −1 (H)). Therefore H ⊆ f (f −1 (H)). Since
f (f −1 (H)) ⊆ H for any f , we have set equality when f is surjective. See
Example 1.1.8(a) for an example.
20. (a) Since y = f (x) if and only if x = f −1 (y), it follows that f −1 (f (x)) = x and
f (f −1 (y)) = y.
(b) Since f is injective, then f −1 is injective on R(f ). And since f is surjec-
tive, then f −1 is defined on R(f ) = B.
21. If g(f (x1 )) = g(f (x2 )), then f (x1 ) = f (x2 ), so that x1 = x2 , which implies that
g ◦ f is injective. If w ∈ C, there exists y ∈ B such that g(y) = w, and there
exists x ∈ A such that f (x) = y. Then g(f (x)) = w, so that g ◦ f is surjective.
Thus g ◦ f is a bijection.
22. (a) If f (x1 ) = f (x2 ), then g(f (x1 )) = g(f (x2 )), which implies x1 = x2 , since
g ◦ f is injective. Thus f is injective.
(b) Given w ∈ C, since g ◦ f is surjective, there exists x ∈ A such that
g(f (x)) = w. If y := f (x), then y ∈ B and g(y) = w. Thus g is surjective.
23. We have x ∈ f −1 (g −1 (H)) ⇐⇒ f (x) ∈ g −1 (H) ⇐⇒ g(f (x)) ∈ H ⇐⇒ x ∈
(g ◦ f )−1 (H).
24. If g(f (x)) = x for all x ∈ D(f ), then g ◦ f is injective, and Exercise 22(a)
implies that f is injective on D(f ). If f (g(y)) = y for all y ∈ D(g), then
Exercise 22(b) implies that f maps D(f ) onto D(g). Thus f is a bijection of
D(f ) onto D(g), and g = f −1 .

Section 1.2
The method of proof known as Mathematical Induction is used frequently in real
analysis, but in many situations the details follow a routine patterns and are




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