PREFACE
This manual is oered as an aid in using the fourth edition of Introduction to Real
Analysis as a text. Both of us have frequently taught courses from the earlier
editions of the text and we share here our experience and thoughts as to how to
use t

Chapter 3 Sequences
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3. Y = (X + Y ) X .
4. If zn := xn yn and lim(xn ) = x = 0, then ultimately xn = 0 so that yn = zn /xn .
5. (a) (2n ) is not bounded since 2n > n by Exercise 1.2.13.
(b) The sequence is not bounded.
(b) 0, since |(1)n /(n + 2)| 1/n,

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1/n
20. If L < r < 1 and := r L, then there exists K such that |xn L| < =
1/n
r L for n > K , which implies that xn < r for n > K . Then 0 < xn < rn for
n > K , and since 0 < r < 1, we have lim(rn ) = 0. Hence lim(xn ) = 0.
21. (a)

Chapter 3 Sequences
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5. have y2 = p + p > p = y1 . Also yn > yn1 implies that yn+1 =
We
p + yn > p + yn1 = yn , so (yn ) is increasing. An upper bound for (yn )
is B := 1 + 2 p because y1 B and if yn B then yn+1 < p + B =
1 + p < B . If y := lim(yn ), t

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order is inherited from the order of the given sequence. The distinction between
a sequence and a set is crucial here.
The Bolzano-Weierstrass Theorem 3.4.8 is a fundamental result whose importance cannot be over-emphasized. It will

Chapter 3 Sequences
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Sample Assignment: Exercises 1, 2, 3, 5, 8, 9.
Partial Solutions:
1. If the set cfw_xn : n N is not bounded above, choose nk+1 > nk such that
xnk k for k N.
(b) xn := n, yn := n.
2. (a) xn := n, yn := n,
3. Note that |xn 0| < if and

Chapter 3 Sequences
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14. Show that bk a1 /k for k N, whence b1 + + bn a1 (1 + + 1/n).
15. Evidently 2a(4) a(3) + a(4) 2a(2) and 22 a(8) a(5) + + a(8) 22 a(4),
etc. The stated inequality follows by addition. Now apply the Comparison
Test 3.7.7.
16. Clear

Chapter 3 Sequences
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11. Show that lim(1)n xn ) = 0.
12. Choose n1 1 so that |xn1 | > 1, then choose n2 > n1 so that |xn2 | > 2, and,
in general, choose nk > nk1 so that |xnk | > k .
13. (x2n1 ) = (1, 1/3, 1/5, . . .).
14. Choose n1 1 so that xn1 s 1, t

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given sequence. For a series to be convergent, the given terms must approach
0 suciently fast. Unfortunately there is no clear demarcation line between
the convergent and the divergent series. Thus it is especially important for the

CHAPTER 4
LIMITS
In this chapter we begin the study of functions of a real variable. This and
the next chapter are the most important ones in the book, since all subsequent
material depends on the results in them. In Section 4.1 the concept of a limit of

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Section 4.2
Note the close parallel between this section and Section 3.2. While the proofs
should be read carefully, the main interest here is in the application of the theorems
to the calculation of limits.
Sample Assignment: Exerc

CHAPTER 5
CONTINUOUS FUNCTIONS
This chapter can be considered to be the heart of the course. We now use all the
machinery that has been developed to this point in order to study the most important class of functions in analysis, namely, continuous functio

Chapter 4 Limits
31
f (x c), we infer that lim (f (x) f (c) = lim f (x c) = lim f (z ) = 0, so that
xc
xc
z 0
lim f (x) = f (c).
xc
13. (a) g (f (x) = g (x + 1) = 2 if x = 0, so that lim g (f (x) = 2, but g ( lim f (x) =
x0
x0
g (f (0) = g (1) = 0. Not eq

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9. (a) If |f (x) f (c)| < for x V B , then |g (x) g (c)| = |f (x) f (c)| < for
x V (c) A.
(b) Let f = sgn on B := [0, 1], g = sgn on A := (0, 1] and c = 0.
10. Note that |x| |c| |x c|.
11. Let c R be given and let > 0. If |x c| < /K

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Bartle and Sherbert
8. Note that |f (x) L| < for x > K if and only if |f (1/z ) L| < for
0 < z < 1/K .
9. There exists > 0 such that |xf (x) L| < 1 whenever x > . Hence |f (x)| <
(|L| + 1)/x for x > .
10. Modify the proof of Theorem 4.3.11 (using Denit

Chapter 5 Continuous Functions
35
5. The function g is not continuous at 1 = f (0).
6. Given > 0, there exists 1 > 0 such that if |y b| < 1 , then |g (y ) g (b)| < .
Further, there exists > 0 such that if 0 < |x c| < , then |f (x) b| < 1 .
Hence, if 0 < |

Chapter 4 Limits
29
7. Let b := |c| + 1. If |x| < b, then |x2 + cx + c2 | 3b2 . Hence |x3 c3 |
(3b2 )|x c| for |x| < b.
8. Note that x c = ( x c)( x + )/( x + c) = (x c)/( x + c).
c
Hence, if c = 0, we have | x c| (1/ c)|x c|, so that we can take :=
c.

14
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Bartle and Sherbert
(b) Let u := sup S and b < 0. If x S , then (since b < 0) bu bx so that
bu is a lower bound of bS . If v bx for all x S , then x v/b (since b < 0),
so that v/b is an upper bound for S . Hence u

CHAPTER 3
SEQUENCES
Most students will nd this chapter easier to understand than the preceding one
for two reasons: (i) they have a partial familiarity with the notions of a sequence
and its limit, and (ii) it is a bit clearer what one can use in proofs t

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 tha

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5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
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

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left to the reader by means of a phrase such as: The proof is by Mathematical
Induction. Since may students have only a hazy idea of what is involved, it may
be a good idea to spend some time explaining and illustrating what constitu

Chapter 1 Preliminaries
16.
17.
18.
19.
20.
21.
22.
23.
24.
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.
If f (a) = f (b), then a/ a2 + 1 = b/ b2 + 1, from which it follows that

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Bartle and Sherbert
2. Part (b) Let f be a bijection of Nm onto A and let C = cfw_f (k ) for some
k Nm . Dene g on Nm1 by g (i) := f (i) for i = 1, . . . , k 1, and g (i) :=
f (i + 1) for i = k, . . . , m 1. Then g is a bijection of Nm1 onto A\C . (Why?

CHAPTER 2
THE REAL NUMBERS
Students will be familiar with much of the factual content of the rst few sections,
but the process of deducing these facts from a basic list of axioms will be new
to most of them. The ability to construct proofs usually improve

Chapter 1 Preliminaries
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
5
The sum is 1 + 3 + + (2n 1) = n2 . Note that k 2 + (2k + 1) = (k + 1)2 .
If n0 > 1, let S1 := cfw_n N : n n0 + 1 S Apply 1.2.2 to the set S1 .
If k < 2k , then k + 1 < 2k + 1 < 2k + 2k = 2

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4.
5.
6.
7.
8.
9.
(c) Add 3 to both sides and factor to get x2 4 = (x 2)(x + 2) = 0. Now
apply 2.1.3(b) to get x = 2 or x = 2.
(d) Apply 2.1.3(b) to show that (x 1)(x + 2) = 0 if and only if x = 1 or
x = 2.
Clearly a = 0 satises a a