SMA206_NOTES

SMA 206: INTRODUCTION TO ANALYSIS
Lecture Notes
First Edition
By
Dr. Bernard Mutuku Nzimbi, PhD
School of Mathematics, University of Nairobi
P.o Box 30197, Nairobi, KENYA.
Copyright c 2009 Benz, Inc. All rights reserved.

Contents
Preface iv
Acknowledgements v
Dedication vi
1 THE REAL NUMBERS SYSTEM 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 ALGEBRAIC AND ORDER PROPERTIES OF R . . . . . . . . . . . . 3
1.2.1 FIELD AXIOMS . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2.2 ORDER AXIOMS . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 OTHER PROPERTIES OF R AND ITS SUBSETS . . . . . . . . . . . . 10
1.3.1 Properties of Integers . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.3.2 Properties of Rationals and Irrationals . . . . . . . . . . . . . . . 10
1.3.3 Properties of the Positive Real Numbers . . . . . . . . . . . . . . 12
2 THE UNCOUNTABILITY OF R 22
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2 COUNTABLE SETS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.3 THE UNCOUNTABILITY OF R . . . . . . . . . . . . . . . . . . . . . . 29
2.3.1 INTERVALS ON THE REAL LINE . . . . . . . . . . . . . . . . 30
2.3.2 Nested Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.3.3 Nested Interval Property . . . . . . . . . . . . . . . . . . . . . . . 31
i

3 STRUCTURE OF THE METRIC SPACE R 38
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.2 The notion of a metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.2.1 Examples of Metrics . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.3 Neighbourhoods, Interior points and Open sets . . . . . . . . . . . . . . . 41
3.3.1 Neighborhoods in a metric space . . . . . . . . . . . . . . . . . . 41
3.4 Limit Points and Closed sets . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.5 Properties of open and closed sets in R . . . . . . . . . . . . . . . . . . . 48
3.6 Relatively Open and Closed Sets . . . . . . . . . . . . . . . . . . . . . . 49
3.7 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.8 Tutorial Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
4 BOUNDED SUBSETS OF R 54
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4.2 Upper Bounds, Lower Bounds of a subset of R . . . . . . . . . . . . . . . 54
4.2.1 Supremum and Inﬁmum of a subset of R . . . . . . . . . . . . . . 55
4.3 The Completeness Property of R . . . . . . . . . . . . . . . . . . . . . . 58
4.4 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
5 SEQUENCES OF REAL NUMBERS 62
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
5.2 Convergence of a sequence . . . . . . . . . . . . . . . . . . . . . . . . . . 63
5.2.1 Criterion of Convergence . . . . . . . . . . . . . . . . . . . . . . . 63
5.2.2 Bounded Sequences . . . . . . . . . . . . . . . . . . . . . . . . . 67
5.3 Subsequences and the Bolzano-Weierstrass Theorem . . . . . . . . . . . . 68
5.4 Monotonic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
5.5 Limit Superior and Limit Inferior of a sequence . . . . . . . . . . . . . . 72
5.6 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
5.7 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
6 LIMITS AND CONTINUITY OF FUNCTIONS IN R 79
6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
6.2 Limit of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
ii

6.3 Some results on Limits of Real-valued Functions . . . . . . . . . . . . . . 82
6.3.1 Limit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
6.3.2 Limits at Inﬁnity . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
6.4 Continuous Functions in R . . . . . . . . . . . . . . . . . . . . . . . . . . 84
6.5 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
6.6 Points of Discontinuity of a Function . . . . . . . . . . . . . . . . . . . . 90
6.6.1 Right and Left Limits . . . . . . . . . . . . . . . . . . . . . . . . 90
6.6.2 Types of Discontinuities . . . . . . . . . . . . . . . . . . . . . . . 91
6.7 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
7 PROPERTIES OF CONTINUOUS FUNCTIONS IN R 100
7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
7.2 Boundedness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
7.3 Location of Roots Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 101
7.4 Bolzano’s Intermediate Value Theorem (IVT) . . . . . . . . . . . . . . . 102
7.4.1 Applications of the IVT ( Existence and location of real roots of
polynomial equations) . . . . . . . . . . . . . . . . . . . . . . . . 104
7.5 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
8 THE RIEMANN INTEGRAL 107
8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
8.2 Partitions of an Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
8.3 Lower and Upper Riemann Sums . . . . . . . . . . . . . . . . . . . . . . 111
8.4 Upper and Lower Riemann Integrals . . . . . . . . . . . . . . . . . . . . 113
8.5 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
8.5.1 Criterion for Riemann Integrability . . . . . . . . . . . . . . . . . 115
8.5.2 Some Classes of Riemann Integrable Functions . . . . . . . . . . . 117
8.6 Integral as a Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
8.7 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
Bibliography 127
iii

Preface
The study of mathematical analysis is indispensable for a prospective student of pure or
applied mathematics. It has great value for any undergraduate student who wishes to
go beyond the routine manipulations of formulas to solve standard problems, because
it develops the ability to think deductively, analyze mathematical situations, and ex-
tend ideas to a new context. The subject of analysis is one of the fundamental areas of
mathematics, and is the foundation for the study of many advanced topics, not only in
mathematics, but also in engineering and the physical sciences. A thorough understand-
ing of the concepts of analysis has also become increasingly important for the study
of advanced topics in economics and the social sciences. Topics such as Fourier series,
measure theory and integration are fundamental in mathematics and physics as well as
engineering, economics, and many other areas.
The only absolute prerequisites for mastering the material in the book are an interest
in mathematics and a willingness occasionally to suspend disbelief when a familiar idea
occurs in an unfamiliar guise. But only an exceptional student would proﬁt from reading
the book unless he/she has previously acquired a fair working knowledge of the processes
of elementary calculus.
This book is a development of various courses designed for second year students of math-
ematics, humanities and third year students of education at the University of Nairobi,
whose preparation has been several courses in calculus and analytical geometry.
iv

Acknowledgements
I would like to thank all my students who provided help and encouragement when
I was writing this edition. Their suggestions helped to remove many mistakes and
ambiguities, thus improving the exposition. I am indebted to my professors at the
University of Nairobi and Syracuse University who introduced me to and sharpened
my skills in Mathematical Analysis. The remaining errors, ambiguities and misleading
comments remain the responsibility of the author.
v

Dedication
To all my students for their patience, support and encouragement.
vi

Chapter 1
THE REAL NUMBERS SYSTEM
1.1 Introduction
We discuss the essential properties of the real number system R. We exhibit a list of
fundamental properties associated with R and show how further properties can be de-
duced from them. We begin this chapter by studying the decomposition of the real line
into the following subsets:
1.1 The Natural Numbers, N
N = {1, 2, 3, ...}.
This set is also called the set of counting numbers.
Deﬁnition 1.1 A non-empty set X is said to be closed with respect to a binary operation
∗ if for all a, b ∈ X, we have a ∗ b ∈ X.
Note that N is closed with respect to the usual addition and usual multiplication but
ont usual subtraction.
1

1.2 The Whole Numbers, W
W = {0, 1, 2, 3, ...}. Note that W = {0} ∪ N.
Note that W is closed with respect to usual addition and multiplication but not under
subtraction.
1.3 The Integers, Z
Z = {..., −3, −2, −1, 0, 1, 2, 3, ...}. Note that Z = −N ∪ W.
This system guarantees solutions to every equation x + n = m with n, m ∈ W. Clearly,
Z consists of numbers x such that x ∈ N or x = 0 or −x ∈ N. Z is closed w.r.t + and
×.
Note also that N ⊂ W ⊂ Z.
1.4 The Rational Numbers, Q
A rational number r is one that can be expressed in the form r = a
b
, for a, b ∈ Z, b = 0
and (a, b) = 1, where (a, b) denotes the greatest common divisor of a and b.
Deﬁnition 1.2 The set of rationals, denoted by Q, is given by
Q = {a
b
: a, b ∈ Z, b = 0, (a, b) = 1}.
With this system, solutions to all equations nx + m = 0 with m, n ∈ Z, and n = 0 can
be uniquely found: i.e. x = −n−
1m = −m
n
.
Examples: 2, 0, 1
2
, − 5
900
.
Note that N ⊂ W ⊂ Z ⊂ Q.
2

1.5 The Irrational Numbers, QC
An irrational number s is one that is not rational, i.e. s cannot be expressed as
s = a
b
, a, b ∈ Z, b = 0 and (a, b) = 1.Note that the sets of rationals and irrationals are
complements of each other.
Examples:
√
2,
√
3, Π.
Remark:
√
p , where p is a prime number is always an irrational number. This
result will be proved towards the end of this chapter.
1.6 The Real Numbers, R
The set of reals is the union of the set of rationals with the set of irrationals, i.e.
R = Q QC
. Graphically, R is represented by the real number line and called the real
number system.
1.2 ALGEBRAIC AND ORDER PROPERTIES OF R
We now introduce the ”algebraic” properties, often called the ”ﬁeld” axioms that are
based on the two binary operations of addition and multiplication.
We start with a given set S whose elements will be called numbers and consider the
following axioms for this set:
1.2.1 FIELD AXIOMS
(I) Addition Axioms: There is an addition operation ”+” such that for all numbers
x, y, z ∈ S the following hold:
1. x + y = y + x [ Commutativity ]
2. x + (y + z) = (x + y) + z [ Associativity ]
3. There is a number 0 such that x + 0 = x [ Existence of zero ]
4. For each x ∈ S there exists a number denoted −x such that
3

x + (−x) = 0; one writes y − x = y + (−x). [ Existence of Additive inverse]
(II) Multiplication Axioms: There is a multiplication operation ”.” such that for all
x, y, z ∈ S:
5. x.y = y.x [Commutativity]
6. x.(y.z) = (x.y).z [Associativity]
7. There is a number 1 such that 1.x = x [Existence of unity or unit element]
8. For each x = 0, ∃ a number x−1
such that x.x−1
= 1; one writes y.x−1
= y
x
. [
Existence of Reciprocals ]
9. x.(y + z) = x.y + x.z [ Distributivity ]
10. 1 = 0 [ Non-triviality ]
Any set or ”number system” with operations + and . obeying these rules is called a field.
For example, the rational numbers, Q, the reals, R are fields. The set of integers, Z is
not a field since the reciprocal of an integer (other than ±1) is not an integer.
The identities 0 and 1 are defined in W. Addition and multiplication are also defined in
W. However, the Existence of additive inverse and the Existence of reciprocals axioms
do not hold in W.
Axiom 10 outlaws the trivial field consisting of the single element 0.
Axioms 1 and 2 hold along with 5, 6, 7, 9, and 10 in N.
Z obeys axioms 1 through 7, 9, and 10.
Q and R obey all these axioms.
From the field axioms, one can deduce the usual properties for manipulation of alge-
braic equalities, such as deriving the identity (a − b)2
= a2
− 2ab + b2
and the laws of
exponents.
The real numbers also come equipped with a natural ordering. We usually visualize
them arranged on a line.
4

1.2.2 ORDER AXIOMS
(III) There is a relation ”≤” such that
11. For each x we have x ≤ x. [Reflexivity]
12. If x ≤ y and y ≤ x, then x = y. [Antisymmetry]
13. If x ≤ y and y ≤ z, then x ≤ z. [Transitivity]
14. For every pair of numbers (x, y), either x ≤ y or y ≤ x. [Linear ordering]
15. If x ≤ y then x + z ≤ y + z for every z. [Compatibility of ≤ and +].
16. If 0 ≤ x and 0 ≤ y then 0 ≤ xy. [Compatibility of ≤ and .]
Remark
Properties 11 and 13 state that the relation ”≤” is a partial ordering. Property 14 says
that every two numbers are comparable. This is described by saying that ≤ is a linear
ordering or a total ordering.
Definition 1.3 A system obeying all 16 properties listed above is called an ordered field.
Examples: Q and R are ordered fields.
W is well-ordered by ≤.
Remark: By definition, x < y shall mean that x ≤ y and x = y.
The ”order properties” of R refer to the notions of positivity and inequalities between
real numbers. Properties 11, 12 and 14 combine to give the following observation:
The Law of Trichotomy
If x and y are elements of an ordered field, then exactly one of the relations
x < y, x = y or x > y holds.
Remark:
5

There are other systems besides real numbers in which some of these axioms play a role.
For example, axioms 1 through 9 excluding 5 and 8 define a ring. Axioms 1 through 4
define a commutative group.
We use some of the axioms to prove the following result:
Proposition 1.1 . In an ordered field the following properties hold:
(i). Unique identities:
If a + x = a for every a, then x = 0.
If a.x = a for every a, then x = 1.
(ii). Unique inverses:
If a + x = 0, then x = −a.
If ax = 1, then x = a−1
.
(iii). No divisors of zero:
If xy = 0, then x = 0 or y = 0.
(iv). Cancellation Laws for addition:
If a + x = b + x, then a = b.
If a + x ≤ b + x ≤, then a ≤ b.
(v). Cancellation Laws for multiplication:
If ax = bx and x = 0, then a = b.
If ax ≥ bx and x > 0, then a ≥ b.
6

(vi). 0.x = 0 for every x.
(vii). −(−x) = x for every x.
(viii).−x = (−1)x for every x.
(ix). If x = 0, then x−1
= 0 and (x−1
)−1
= x.
(x). If x = 0 and y
.
= 0, then xy = 0 and (xy)−1
= x−1
y−1
.
(xi). If x ≤ y and 0 ≤ z, then xz ≤ yz.
If x ≤ y and z ≤ 0, then yz ≤ xz.
(xii). If x ≤ 0 and y ≤ 0, then xy ≥ 0.
If x < 0 and y ≥ 0, then xy ≤ 0.
(xiii). 0 < 1.
(xiv). For any x, x2
≥ 0.
Proof
(i). Suppose x + a = a. Then
x = x + 0 = x + (a + (−a)) = (x + a) + (−a) = a + (−a) = a + (−1)a = (1 + (−1))a =
0.a = 0.
Likewise, suppose ax = a for all a. Then
x = x.1 = x(a.a−1
) = (a.x)a−1
= a.a−1
= 1.
(ii). Suppose a + x = 0. Then
7

−a = −a + 0 = −a + (a + x) = (−a + a) + x = 0 + x = x, and so −a = x, as desired.
Likewise, if ax = 1, then a−1
= a−1
.1 = a−1
(ax) = (a−1
a)x = 1.x = x.
(iii). It suﬃces to assume that x = 0 and prove that y = 0. Multiply xy by 1
x
and apply
Associativity of multiplication, Existence of reciprocals, and Existence of unit axioms
to get:
1
x
(x.y) = ((1
x
).x).y = 1.y = y. Since xy = 0, (1
x
)(xy) = 1
x
.0 = 0. Thus y = 0.
(iv). Suppose a + x = b + x. Then
a = a + 0 = a + (−x + x) = −x + (a + x) = −x + (b + x) = (−x + x) + b = 0 + b = b.
Likewise, suppose a + x ≤ b + x. Then (a + x) + (−)(b + x) ≤ 0. That is, (a − b) + (x +
(−x)) ≤ 0, i.e. (a − b) + 0 ≤ 0, i.e. (a − b) ≤ 0. Adding b both sides, we have a ≤ b.
(v). Suppose that ax = bx and x = 0. Then ax + −(bx) = 0. That is, (a + (−b))x = 0.
Since x = 0, x−1
exists and (a + (−b))x.x−1
= 0.x−1
= 0. That is, a + (−b) = 0 i.e.
a = b.
Likewise, suppose that ax ≥ bx and x > 0. Then ax − bx ≥ 0. That is, (a + (−b))x ≥ 0.
Since x > 0 ⇒(a-b)≥ 0 ⇒a≥ b.
(vi). 0.x = (0 + 0)x = 0.x + 0.x, and so 0 = 0.x + (−0.x) = (0.x + 0.x) + (−0.x) =
0.x + (0.x + (−0.x)) = 0.x + 0.x = 0.x.
(viii). x + (−1)x = 1.x + (−1)x = (1 + (−1))x = 0.x = 0 by (vi). Thus, (−1).x = −x
by (ii).
(x). Suppose x = 0, y = 0 but xy = 0. Then since 0x = 0 by (vi), we have that
1 = (1
y
)(1
x
).xy = (1
y
)(1
x
)0 = 0, contradiction to Proposition 1.1.1 axiom 10. Hence,
xy = 0. The proof of (xy)−1
= x−1
y−1
is left as an exercise.
(xiii). Suppose 1 ≤ 0. Then 1 + (−1) ≤ 0 + (−1) and so 0 ≤ −1. Using property 16:
since 0 ≤ −1 and 0 ≤ −1, we get 0 ≤ (−1)(−1) = −(−1) = 1. Therefore, 1 ≤ 0 and
8

0 ≤ 1 and so 1 = 0 by property 12, in contradiction to property 10. Hence 0 < 1.
(xiv). Consider two cases: If x ≥ 0, then x2
= x.x ≥ 0, by axiom 16. If x < 0,
then x2
= (−(−x))(−(−x)) = (−1)2
(−x)2
, by (vii) and (viii). But (−1)2
= 1, since
0 = (−1)(−1 + 1) = (−1)2
+ (−1).1 = (−1)2
− 1. Thus, x2
≥ 0. ♣
Remark: The purposes of the axioms of an ordered field is to isolate the key properties
we need for manipulation of algebraic equalities and inequalities.
Example 1 Using the axioms and properties of an ordered field given in this section,
prove that a2
− b2
= (a − b)(a + b).
Solution
By the distributive law, (a − b)(a + b) = (a − b).a + (a − b).b. Using the commutativity
and the distributive law again, along with a − b = a + (−b):
(a − b).a + (a − b).b = a.(a − b) + b.(a − b) = a2
+ a.(−b) + b.a + b.(−b).
Now,
a.(−b) = a.(−1).b = (−1)ab = −(ab) by Proposition 1.1.1 (viii), associativity and com-
mutativity. Similarly, b.(−b) = −b2
. Thus, (a−b)(a+b) equals a2
−a.(−b)+b.a+b.(−b) =
a2
− (ab) + ba − b2
= a2
− ab + ab − b2
( by axiom 5)
= a2
− b2
(by axioms 3 and 4). ♣
Example 2 In an ordered field prove that if 0 ≤ x < y, then x2
≤ y2
.
Solution If 0 ≤ x < y, then 0 ≤ x ≤ y, and so by Proposition 1.1 (xi), x2
≤ yx. By the
same reasoning, x ≤ y⇒xy≤ y2
. Thus
x2
≤ yx = xy ≤ y2
, and so x2
≤ y2
. We now need to exclude the possibility that x2
equals y2
. But if x2
= y2
, then x2
− y2
= 0 (add −y2
to each side).
(x − y)(x + y) = 0 (by Example 1).
By Proposition 1.1(xii), we have 0 ≤ x and y > 0. Now x+y = 0, since x+y = 0 would
imply that y = (−x) ≤ 0, so that y ≤ 0, which is impossible by the Law of Trichotomy.
By the Cancelation Law for multiplication, (x − y)(x + y) = 0, i.e. x − y = 0, i.e. x = y.
9

But we are given x < y, and so this case is excluded as desired. ♣
1.3 OTHER PROPERTIES OF R AND ITS SUBSETS
1.3.1 Properties of Integers
Definition 1.4 Let m ∈ Z. Then m is said to be even if it can be expressed as m = 2n,
for some n ∈ Z.
Definition 1.5 Let m ∈ Z. Then m is said to be odd if m = 2n + 1, for some n ∈ Z.
Proposition 1.2 (i). Let m ∈ Z. Then m is even iff m2
is even.
(ii). Let m ∈ Z. Then n is odd iff m2
is odd.
Proof
(i). (⇒) Let m ∈ Z be even. Then m = 2n for some n ∈ Z ⇒m2
= 4n2
= 2(2n2
).
Hence m2
is divisible by 2, hence m2
is even.
(⇐) Conversely, let m2
be even and assume to the contrary that m is odd. Then
m = 2n + 1 for some n ∈ Z.
Therefore, m2
= (2n + 1)2
= 4n2
+ 4n + 1 = 2(2n2
+ 2n) + 1, where 2n2
+ 2n ∈ Z. Thus
m2
is odd. This contradicts the fact that m2
is even.
Therefore, m must be even whenever m2
is even.
(ii). (⇒) Suppose m is odd. Then m = 2n + 1, for some n ∈ Z. So, m2
= (2n + 1)2
=
4n2
+ 4n + 1 = 2(2n2
+ 2n) + 1, where 2n2
+ 2n ∈ Z. Hence m2
is odd.
(⇐) Conversely, let m2
be odd and assume to the contrary that m is even. Then m = 2n,
for some n ∈ Z. Therefore, m2
= 4n2
= 2(2n2
), where 2n2
∈ Z. Thus m2
is even, a
contradiction to our hypothesis that m2
is odd. Therefore m must be odd whenever m2
is odd. ♣
1.3.2 Properties of Rationals and Irrationals
Proposition 1.3 : Q is ”dense” in itself: If x and y are in Q, with x < y, then there
exists an element z ∈ Q such that x < z < y.
10

Proof
Choose z = x+y
2
. ♣
Theorem 1.4 [Archimedian Property of R] If x, y ∈ R and x > 0, y > 0 and
x < y, then there exists a positive integer n such that nx > y.
Theorem 1.5 The Density Theorem If x and y are any real numbers with x < y,
then there exists a rational number r ∈ Q such that x < r < y.
Proof
Without loss of generality (WLOG) assume that x > 0. Since y − x > 0, ∃n ∈ N such
that 1
n
< y − x. Therefore we have nx + 1 < ny. Since x > 0, we have nx > 0, there
exists m ∈ N with m−1 ≤ nx < m. Therefore, m ≤ nx+1 < ny, whence nx < m < ny.
Thus, the rational number r = m
n
satisfies x < r < y. ♣
Proposition 1.6 QC
is dense in R: If x and y are real numbers with x < y, then there
exists an irrational number z such that x < z < y.
Proof
Applying the Density Theorem to the real numbers x√
2
and y√
2
, we obtain a rational
number r = 0 such that x√
2
< r < y√
2
. If we let z = r
√
2, then clearly z is irrational and
satisfies x < z < y. ♣
Remark:
If we start to mark the rational numbers on the number line, we find that they are
scattered densely along the line and seem to be filling it up. However, we know they
do not; for example
√
2 is missing. That is, there exist at least one irrational real
number, namely
√
2. There are ”more” irrational numbers than rational numbers in the
sense that the set of rational numbers is countable, while the set of irrational numbers
is uncountable. Concepts of countability and uncountability of sets will be studied in
Chapter 2.
This forms the basis of the following proposition:
Proposition 1.7
√
2 is irrational.
11

Proof
We need to show that there does not exist an r ∈ Q such that r2
= 2.
We prove by contradiction. Assume to the contrary that
√
2 is rational. Then by
definition,
√
2 = a
b
, a, b ∈ Z, b = 0, (a, b) = 1.
(*)
Squaring both sides of (*), we have
2 = a2
b2 or a2
= 2b2
.
Therefore a2
is even, and hence a is even (by Proposition 1.2). Since a is even, a = 2k,
for some k ∈ Z. Hence, a2
= 4k2
. But a2
= 2b2
= 4k2
. That is b2
= 2k2
⇒ b2
is even
and hence b is even (by Proposition 1.2). This means that 2 is a common factor for a
and b, a contradiction since (a, b) = 1 was our assumption. Hence
√
2 is irrational. ♣
Proposition 1.8
√
3 is irrational.
Proof
Assume to the contrary that
√
3 is rational. Then
√
3 = p
q
, with p, q ∈ Z, q = 0 and
(p, q) = 1. Therefore, 3 = p2
q2 or p2
= 3q2
which implies that 3 divides p2
and hence will
divide p (by Proposition 1.2). That is, p = 3k, for some k ∈ Z. Therefore, p2
= 9k2
. But
p2
= 3q2
, which implies that 3q2
= 9k2
or q2
= 3k2
. Thus 3 divides q2
and hence q (by
Proposition 1.2). Thus m and n have a common factor 3. This leads to a contradiction
of our assumption. Hence,
√
3 is irrational. ♣
Exercise. Prove that the
√
p is irrational for any prime number p.
(Hint: Use a similar proof as above).
1.3.3 Properties of the Positive Real Numbers
We now define a nonempty subset P of R called the set of positive real numbers (some-
times denoted R+
) that satisfies the following properties:
(i). If a, b ∈ P, then a + b ∈ P.
12

(ii). If a, b ∈ P, then ab ∈ P.
(iii). If a ∈ R , then exactly one of the following holds:
a ∈ P, a = 0, − a ∈ P [Trichotomy Property]
Remark
Property (iii) is the Trichotomy Property because it divides R into three distinct types
of elements. It states that the set {−a : a ∈ P} of negative real numbers has no elements
in common with the set P of positive real numbers, and , moreover, the set R is the
union of three disjoint sets.
Definition 1.6 If a ∈ P, we write a > 0 and say that a is a positive (or a strictly
positive) real number.
If a ∈ P ∪ {0}, we write a ≥ 0 and say that a is a nonnegative real number. Similarly,
if −a ∈ P, we write a < 0 and say that a is negative (or strictly negative) real number.
If − a ∈ P ∪ {0}, we write a ≤ 0 and say that a is a nonpositive real number.
Remark We now use the above definitions to prove the following theorem:
Theorem 1.9 Let a, b ∈ R.
(a). If a > b and b > c, then a > c.
(b). If a > b then a + c > b + c.
(c). If a > b and c > 0 , then ca > cb. If a > b and c < 0, then ca < cb.
Proof
(a). If a − b ∈ P and b − c ∈ P, then by the order properties of a field, this implies that
(a − b) + (b − c) = a − c belongs to P. Hence a > c.
(b). If a − b ∈ P, then (a + c) − (b + c) = a − b is in P. Thus a + c > b + c.
13

(c). If a − b ∈ P and c ∈ P, then ca − cb = c(a − b) is in P. Thus ca > cb when c > 0.
On the other hand, if c < 0, then −c ∈ P, so that cb − ca = (−c)(a − b) is in P. Thus
cb > ca when c < 0. ♣
Theorem 1.10 (a). If a ∈ R and a = 0, then a2
> 0.
(b). 1 > 0
(c). If n ∈ N, then n > 0.
Proof
(a). By the Trichotomy Property, if a = 0, then either a ∈ P or −a ∈ P. If a ∈ P, then
by the order property 3.(ii), a2
= a.a ∈ P. Also, if − a ∈ P, then a2
= (−a)(−a) ∈ P.
We conclude that if a = 0, then a2
> 0.
(b). Since 1 = 12
, it follows from (a)that 1 > 0.
(c). We use Mathematical Induction: The assertion for n = 1 is true by (b). If we
suppose the assertion is true for the natural number k, then k ∈ P, and since 1 ∈ P,we
have k + 1 ∈ P by order property (i). Therefore, the assertion is true for all natural
numbers. ♣
Remark
The product of two positive numbers is positive. However, the positivity of a product
of two numbers does not imply that each factor is positive.
Theorem 1.11 If ab > 0, then either
(i). a > 0 and b > 0, or
(ii). a < 0 and b < 0.
Proof
Note that ab > 0 implies that a = 0 and b = 0. From the Trichotomy Property, either
a > 0 or a < 0. If a > 0, then 1
a
> 0, and therefore b = (1
a
)(ab) > 0. Similarly, if a < 0,
then 1
a
< 0, so that b = (1
a
)(ab) < 0. ♣
14

Corollary 1.12 If ab < 0, then either
(i). a < 0 and b > 0 or
(ii). a > 0 and b < 0.
We apply the above results in working with inequalities.
Inequalities
The order properties can be used to ”solve” certain inequalities.
Examples
(a). Determine the set A of all numbers x such that 2x + 3 ≤ 6.
Solution
x ∈ A iff 2x + 3 ≤ 5 iff 2x ≤ 3 iff x ≤ 3. Therefore A = {x ∈ R} : x ≤ 3
2
}.
(b). Determine the set B = {x ∈ R : x2
+ x > 2}
Solution
Note that x ∈ B ⇔ x2
+ x − 2 > 0 ⇔ (x − 1)(x + 2) > 0. Therefore, we either have
(i). x − 1 > 0 and x + 2 > 0 or we have
(ii). x − 1 < 0 and x + 2 < 0.
In case (i), we must have both x > 1 and x > −2, which is satisfied iff x > 1. In case
(ii), we must have both x < 1 and x < −2, which is satisfied iff x < −2. We conclude
that B = {x ∈ R : x > 1} ∪ {x ∈ R : x < −2}.
(c). Determine the set C = {x ∈ R : 2x+1
x+2
< 1}.
15

Note that C = {x ∈ R : 2x+1
x+2
− 1 < 0} = {x ∈ R : 2x+1−(x+2)
x+2
< 0} =
{x ∈ R : x−1
x+2
< 0}
Therefore, we have either
(i). x − 1 < 0 and x + 2 > 0 or
(ii). x − 1 > 0 and x + 2 < 0.
In case (i), we must have both x < 1 and x > −2, which is satisfied iff −2 < x < 1. In
case (ii), we must have both x > 1 and x < −2, which is never satisfied. We conclude
that C = {x ∈ R : −2 < x < 1}.
Exercise
1. Let a ≥ 0 and b ≥ 0. Prove that a < b ⇔ a2
< b2
⇔
√
a <
√
b.
Definition 1.7 If a and b are positive real numbers, then their arithmetic mean is
1
2
(a + b) and their geometric mean is
√
ab.
The Arithmetic-Geometric Mean Inequality for a and b is
√
ab ≤ 1
2
(a + b), with equality occurring if and only if a = b. Note that if a > 0, b > 0,
and a = b, then
√
a > 0,
√
b > 0, and
√
a =
√
b. Therefore, by a previous result,
(
√
a −
√
b)2
> 0. Expanding the square, we obtain a − 2
√
ab + b > 0, whence it follows
that
√
ab < 1
2
(a + b).
16

The general Arithmetic-Geometric Mean Inequality for the positive real numbers
a1, a2, ..., an
is
(a1a2...an)
1
n ≤
a1 + a2 + ... + an
n
with equality iﬀ a1 = a2 = · · · = an.
Solved Problems
1. Show that if t is irrational then any number s is given by s = t
t+1
is also irrational.
Solution
Assume to the contrary that s is rational. Then we can write s = m
n
, m, n ∈ Z, n =
0, (m, n) = 1.
Therefore, t
t+1
= m
n
.
i.e. nt = m(t + 1) or nt = mt + m. That is, (n − m)t = m or t = m
n−m
.
Since Q is closed under addition and multiplication, it follows that m
n−m
is rational and
hence t is rational, a contradiction since it is known to be irrational. Hence, t
t+1
is
irrational.
2. What is meant by saying that a number r is rational? Show that if s =
√
n + 1 −
√
n − 1 for any integer n ≥ 1, then r is irrational.
Solution
Let s =
√
n + 1 −
√
n − 1 for n ≥ 1. Assume that s is rational. Then s =
√
n + 1 −
√
n − 1 = a
b
, (a, b) = 1.
Therefore,
a2
b2 = n + 1 + (n − 1) − 2(
√
n + 1)(
√
n − 1) = 2n − 2
√
n + 1 −
√
n − 1
= 2(n −
√
n + 1
√
n − 1).
That is a2
b2 is even. Hence a
b
is even. That is a and b have the number 2 as a common
factor, a contradiction. Hence s is irrational.
17

3. Given that a and b are ratinals with b = 0 and s is an irrational number such that :
a − bs = t, show that t is irrational. Hence show that
√
2−1√
2+1
is irrational.
Proof
t = a − bs, b ∈ Q, s ∈ QC
.
Assume that t is rational.
Then t = p
q
with p, q ∈ Z, q = 0, (p, q) = 1.
Therefore p
q
= a − bs or p = q(a − bs),i.e. bqs = aq − p.
Therefore, s = aq−p
bq
. Since Q is closed under + and . , we have aq−p
bq
is rational. Hence
s is rational, a contradiction. Hence t is irrational.
Now
√
2−1√
2+1
= (
√
2−1)(
√
2−1)
(
√
2+1)(
√
2−1)
= 3 − 2
√
2.
Since 2 and 3 are rationals and
√
2 is irrational, we have by the above result that 3−2
√
2
can be expressed in the form 3 − 2
√
2 = a − bs.
Hence it is irrational.
4. Let x and y be positive real numbers. Show that :
(a). x + y is also positive
(b). x < y iﬀ x2
< y2
(c). x < y implies 1
y
< 1
x
Solution
(a). x, y > 0. So 0 = 0 + 0 < x + y. That is 0 < x + y. Hence x + y is also positive.
(b). Let x < y. Multiply each side by x > 0 to get x2
< yx. Also multiply each side by
y > 0 to get xy < y2
. Therefore, x2
< yx < y2
. Thus x < y ⇒ x2
< y2
.
Conversely, let x2
< y2
. That is x2
− y2
< 0, or (x + y)(x − y) < 0. Dividing each side
by x + y > 0, gives x − y < 0. That is x < y. Thus x2
< y2
⇒ x < y.
NB: This result may not hold if we are not told ”x > 0 and y > 0” .
(c). x < y, x > 0, y > 0, so xy > 0 and so is 1
xy
. Since x < y, we have that x 1
xy
< y 1
xy
.
That is 1
y
< 1
x
.
18

5. Prove that
(a). If x and y are negative then x + y is also negative.
(b). If 0 < x < y and 0 < w < z then xw < yz.
(c). If x ∈ R, and 1 < x, i.e. x = 1 + h, h > 0, the 1 + nh < xn
for each positive
integer n.
Proof
(a) x < 0, y < 0. Let x = −p, for p > 0, y = −q, for q > 0. Therefore,
x + y = −p + (−q) = (−1)(p + q) < 0, since p + q > 0. Hence x + y < 0 and thus
negative.
(b). 0 < x < y and 0 < w < z. Since 0 < x and x < y, then 0 < y. Now since 0 < w,
we have xw < yw.
Also, since 0 < y, we have wz < yw. Also, since 0 < y, we have wy < zy. Therefore
xw < yw = wy < zy. That is xw < zy.
(c). Since x = 1 + h, we have
xn
= (1 + h)n
= 1 + nh + n(n−1)
2!
h2
+ ...
Ignoring the terms involving h2
and higher terms we have that
1 + nh < xn
.
6. Show that:
(a). If x > 0, then −x < 0 and conversely.
(b). If x, y ∈ R are such that x < y, then there exists an irrational number r such
that x < r < y.
19

Proof
(a). Since x > 0,
−x = 0 + (−x) < x + (−x) = 0. That is, −x < 0. Hence −x is negative.
Conversely, if x < 0, then 0 = x + (−x) < 0 + (−x) = −x, i.e 0 < −x.
Hence −x is positive.
(b). Given any pair of real numbers x and y such that x < y we have that since rationals
are dense in R, or are everywhere on the real line, we should be able to ﬁnd a rational
number between x and y no matter how close x and y are.
In particular, there is a rational number, say s such that
x√
2
< s < y√
2
. That is x < s
√
2 < y. Now let r = s
√
2. Then r is an irrational number
such that x < r < y.
7. Bernoulli’s Inequality: If x > −1, then
(1 + x)n
≥ 1 + nx, for all n ∈ N.
(**)
Proof By Mathematical Induction:
The case n = 1 yields equality, so the assertion is valid in this case. Next, we assume
the validity of the inequality (**) for k ∈ N and will deduce it for k + 1.
The assumptions that (1 + x)k
≥ 1 + kx and that 1 + x > 0 imply that
(1 + x)k+1
= (1 + x)k
(1 + x) ≥ (1 + kx)(1 + x) = 1 + (k + 1)x + kx2
≥ 1 + (k + 1)x.
Thus, inequality (**) holds for n = k + 1. Therefore, it holds for all n ∈ N.
Tutorial Exercises
1. If a ∈ R satisﬁes a.a = a, prove that either a = 0 or a = 1.
2. (a). Show that if x, y are rational numbers, then the sum x + y and the product xy
are rational numbers.
(b). Prove that if x is a rational number and y is an irrational number, then the sum
x + y is an irrational number. If in addition, x = 0, then show that xy is an irrational
number.
3. Give an example to show that if x and y are irrational numbers, the sum x + y and
the product xy need not be irrational.
4. Prove that
√
2 +
√
3 is irrational.
20

5. Prove that there is no rational number whose square is 12.
6. Suppose that x ∈ R and 0 < x. Show that there is an irrational number between 0
and x.
21

Chapter 2
THE UNCOUNTABILITY OF R
2.1 Introduction
We analyze subsets of the real line, R to determine those that are countable and those
that are uncountable.
Definition 2.1 Let A and B be any two non-empty sets. If there is a function f which
maps A onto B such that f is one-to-one (i.e. f is a 1-to-1 correspondence or a bijec-
tion), then A and B are said to be equivalent or equinumerous or A and B are said to
have the same cardinality.
We thus write A ∼ B.
Remark
When we count the elements in a set, we say ”one, two, three,...”, stopping when we
have exhausted the set. From a mathematical perspective, what we are doing is defining
a bijective mapping between the set and a portion of the set of natural numbers. If the
set is such that the counting does not terminate such as the set of natural numbers, then
we describe the set as being infinite (see definition below).
Definition 2.2 The empty set ∅ is said to have 0 elements.
Definition 2.3 Let A be any non-empty set. Then we have that:
(i). A is called a finite set if it has n elements for some positive integer n. That is
A ∼ Jn for some positive integer n, where the set Jn denotes the set {1, 2, 3, ..., n} for
22

n ∈ N.
(ii). A is called an infinite set if it is not finite.
(iii). A is called countable or countably infinite if A ∼ N.
(iv). A is called uncountable if it is neither countable nor finite.
(v). A is called at most countable if it is finite or countable.
Properties of finite and infinite sets
(a). If A is a set with m elements and B is a set with n elements and if A ∩ B = ∅, then
A ∪ B has m + n elements.
(b). If A is a set with m ∈ N elements and C ⊆ A is a set with 1 element, then AC is
a set with m − 1 elements.
(c). If C is an infinite set and B is a finite subset of C, then C is an infinite set.
Proof
(a). Let f be a bijection of Jm onto A, and let g be a bijection of Jn onto B. We define
h on Jm+n by
h(x) = {
f(i) i = 1, 2, ..., m
g(i − m) i = m + 1, ..., m + n
We leave it as an exercise to show that h is a bijection from Jm+n onto A ∪ B. ♣
Parts (b) and (c) are left as exercises.
2.2 COUNTABLE SETS
These are an important type of infinite sets.
Definition 2.4 A set S is said to be denumerable or countably infinite if there exists
a bijection of N onto S.
Definition 2.5 A set S is said to be countable if it is finite or denumerable.
23

Examples
(a). The set E = {2n : n ∈ N} of even numbers is denumerable(countable), since the
mapping f : N → E deﬁned by f(n) = 2n for n ∈ N is a bijection of N onto E.
Similarly, the set O = {2n − 1 : n ∈ N} of odd natural numbers is denumerable. Deﬁne
a bijection g : N → O by g(n) = 2n − 1.
(b). The set Z of all integers is countable(denumerable).
To construct a bijection of N onto Z , we map 1 onto 0, we map the set E of even nat-
ural numbers onto the set N of positive integers, and we map the set O of odd natural
numbers onto the negative integers.
(c). The union of two disjoint denumerable(countable) sets is denumerable(countable).
Indeed, if A = {a1, a2, ...} and B = {b1, b2, ...}, we can enumerate the elements of A∪B as
a1, b1, a2, b2, ...
Theorem 2.1 The set N × N is countable.
Proof
N × N consists of all ordered pairs (m, n), where m, n ∈ N.
We can enumerate these pairs as:
(1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1), (1, 4), ...
according to increasing sum m + n, and increasing m.
.(1, 4) .(2, 4) . .
.(1, 3) .(2, 3) .(3, 3) .
.(1, 2) .(2, 2) .(3, 2) .(4, 2)
.(1, 1) .(2, 1) .(3, 1) .(4, 1)
shows a bijection f : N × N → N. ♣
Theorem 2.2 Suppose S and T are sets and that T ⊆ S.
24

(a). If S is a countable set, then T is a countable set
(b). If T is an uncountable set, then S is an uncountable set.
Examples
1. N is countable.
Consider the identity map i : N → N, i.e i(n) = n ∀n ∈ N. Then i is one-to-one
and onto. Thus i is a one-to-one correspondence from the set N onto itself. Hence N is
countable.
2. Consider the set of integers Z = {0, ±1, ±2, ...}.
Define f : N → Z by
f(n) =
n
2
if n is even
−(n−1)
2
if n is odd
Then f is 1-to-1 and onto Z. Hence Z is countable.
Remark
In Example 2 above, Z is equivalent to its subset N. Clearly, N and Z do not have the
same number of elements.
Theorem 2.3 Let A be a countable set and E be any infinite subset of A. Then E is
also countable.
Proof
Since A is countable, we can arrange its elements in a sequence say,
x1, x2, x3, ...
of distinct elements. Let n1 be the least integer such that xn1 ∈ E. Having selected xn1
we find the smallest number n2 > n1 such that xn2 ∈ E and so on.
In this way we construct a sequence
n1, n2, ...,
and have the elements
xn1 , xn2 , ...
25

all belonging to E.
We now have the correspondence
1 −→ xn1
2 −→ xn2
3 −→ xn3
. .
. .
. .
r −→ xnr
Thus the mapping f : N → E deﬁned by f(r) = xnr is bijective. Therefore, E ∼ N.
Hence E is countable. ♣
Theorem 2.4 Let (En )∞
n=1 be a sequence of countable sets. Then S =
∞
∪ En
n = 1
is
also countable.
Proof 1(Less formal but intuitive proof)
Since Er is countable for r = 1, 2, 3, ..., we can arrange Er in a sequence as xr1 , xr2 , ... of
distinct elements. It now follows that we can arrange elements of S in an array as follows:
x11, x12, x13, x14, . . .
x21, x22, x23, x24, . . .
x31, x32, x33, x34, . . .
.
.
.
where the rth
row in the array above represents the elements Er (r = 1, 2, ...). This
inﬁnite array contains all elements of S. We can now re-arrange this array in a sequence
by considering diagonals as follows
26

1st
diagonal 2nd
diagonal 3rd
diagonal
x11 , x21, x12, x31, x22, x13, . . .
, etc
If any two of the sets Er have a common element, then this would be repeated in the
sequence above. This means that we can find a subset say T of N such that T is equiva-
lent to S. Clearly T is at most countable and hence S is at most countable. Otherwise
S is countable since its elements can be arranged in a sequence as shown above. ♣
Proof 2 (Alternative Proof)
For each n ∈ N, let ϕn be a surjection of N onto En. We define
ψ : N × N → S by
ψ(n, m) = ϕn(m).
We claim that ψ is a surjection. Indeed, if a ∈ S, then there exists a least n ∈ N such
that a ∈ En, whence there exists a least m ∈ N such that a = ϕ(m).
Therefore, a = ψ(n, m).
Since N × N is countable, it follows that there exists a surjection
f : N → N × N whence ψ ◦ f is a surjection of N onto S. Hence S is countable. ♣
Theorem 2.5 Let A be a countable set and Bn denote the set of all n-tuples. Thus
Bn = {(a1, a2, ..., an) : ai ∈ A} for i = 1, 2, ..., n; where ai need not be distinct.
Then Bn is countable.
Proof(By Mathematical Induction)
Let n = 1. Then B1 = A and since A is countable, it follows easily that B1 is countable.
Now assume that Br−1 is countable. We show that Br is countable where r ≥ 2. Note
that every element of Br is of the form (b, a), where b ∈ Br−1 and a ∈ A.
Now, keep b fixed and let a vary over A. Then the set of all such elements (b, a) is
equivalent to A and hence is countable. But b ∈ Br−1 and Br−1 is countable by the
induction hypothesis.
Therefore, we have a countable number of countable sets which is countable. Thus Br
is also countable. Hence Bn is countable for all n ∈ N. ♣
27

Corollary 2.6 The set Q of all rational numbers is countable.
Proof(Method 1)
We ﬁrst note that every rational number can be expressed in the form a
b
where a, b ∈
Z, b = 0 and (a, b) = 1.
Consider the ordered pair (a, b) and identify it with a
b
, i.e. the map
ψ : (a, b) → a
b
is a one-to-one correspondence. But the set {(a, b) : a, b ∈ Z} = Bn with
n = 2 and hence is countable by the previous theorem. Thus we have that B2 ∼ Q.
Hence Q is also countable. ♣
Proof (Alternative proof)
Observe that the set Q+
of positive rational numbers is contained in the enumeration
1
1
, 1
2
, 2
1
, 1
3
, 2
2
, 3
1
, 1
4
, ...
which is another ”diagonal mapping”
1
1
2
1
3
1
4
1
. . .
1
2
2
2
3
2
4
2
. . .
1
3
2
3
3
3
4
3
. . .
1
4
2
4
3
4
4
4
. . .
. . .
. . .
. . .
The set Q+
So there exists a surjection of N onto N × N:
f : N → N × N.
If g : N × N → Q+
is a mapping that sends the ordered pair (m, n) into the rational
numbers having the representation m
n
, then g is a surjection onto Q+
.
Therefore, the composition g ◦ f is a surjection of N onto Q+
and therefore Q+
is a
countable set.
Similarly, the set Q−
of negative rational numbers is countable. Hence,
Q = Q−
∪ {0} ∪ Q+
28

is countable. ♣
Remark
Since Q contains N, it must be denumerable since N is.
This argument that Q is countable was first given in 1874 by Georg Cantor (1845−1918).
He was the first mathematician to examine the concept of infinite set in rigorous detail.
He also proved that the set of real numbers R is an uncountable set.
2.3 THE UNCOUNTABILITY OF R
Theorem 2.7 Let A be the set of all infinite sequences whose terms consist of only 0
and 1. Then A is uncountable.
Proof
Consider E as a countable subset of A. Enumerate E as a sequence:
s1, s2, ..., sn, ...
We construct an infinite sequence S as follows: The nth
member of S is 1 if the nth
member of sn is 0 and vice versa for n = 1, 2, 3, ...
Thus we have that:
S =
1 if sn = 1
0 if sn = 0
where sn is any member of E.
Clearly, s differs from every member of E. Thus s is not in E and yet s ∈ A.
Hence E is a proper subset of A.
Thus every countable subset of A is a proper subset of A. In this case A must be un-
countable for if it was countable then it would be a proper subset of itself. This is an
absurdity. Hence the result. ♣
Remark
Every real number when expressed in binary uses only the digits 0 and 1. This means
that every real number can be viewed as one of the sequences of A. Thus A constitutes
29

the set of real numbers. Hence R is uncountable. The set QC
of irrational numbers is
uncountable.
Proof
We know that Q is countable. Now assume QC
is also countable. Then this implies
that R = Q ∪ QC
is countable since a union of countable sets is again countable. But
R = Q ∪ QC
is uncountable by the theorem above. This leads to a contradiction. Hence
QC
is uncountable.♣
2.3.1 INTERVALS ON THE REAL LINE
The order relation on R determines a natural collection of subsets called intervals.
Definition 2.6 Bounded Intervals
If a, b ∈ R satisfy a < b, then
(a, b) = {x ∈ R : a < x < b} is the open interval between a and b.
[a, b] = {x ∈ R : a ≤ x ≤ b} is the closed interval between a and b.
[a, b) = {x ∈ R : a ≤ x < b}
(a, b] = {x ∈ R : a < x ≤ b}
are the half-open (or half-closed) intervals between a and b.
Definition 2.7 Unbounded Intervals
The infinite open intervals are:
(a, ∞) = {x ∈ R : x > a}
(−∞, b) = {x ∈ R : x < b}
The infinite closed intervals are:
[a, ∞) = {x ∈ R : x ≥ a}
(−∞, b] = {x ∈ R : x ≤ b}
30

Remark
It is often convenient or customary to think of the entire R as an inﬁnite interval, and
write R = (−∞, ∞).
Note that −∞ and ∞ are not elements in R, but only convenient symbols.
2.3.2 Nested Intervals
Deﬁnition 2.8 A sequence of intervals In, n ∈ N is nested if the following chain of
inclusions holds
I1 ⊇ I2 ⊇ I3 ⊇ ... ⊇ In ⊇ In+1 ⊇ ...
Figure 2.1: Nested intervals
Example If In = [0, 1
n
], for n ∈ N, then In ⊇ In+1 for each n ∈ N, so this sequence of
intervals is nested.
2.3.3 Nested Interval Property
Theorem 2.8 [Nested Interval Property]
If In = [an, bn], n ∈ N is a nested sequence of closed and bounded intervals, then there
exists a number ξ ∈ In for all n ∈ N.
31

Application of the Nested Interval Property We use the Nested Interval Property
to prove that the set R of real numbers is an uncountable.
Theorem 2.9 The set R of real numbers is not countable.
Proof
It suffices to prove that the unit interval I = [0, 1] is an uncountable set. This implies
that the set R is an uncountable set, for if it were countable, then the subset I would
also be countable. We prove by contradiction.
Assume that I is countable. Then we can enumerate the set as I = {x1, x2, ..., xn, ...}.
We first select a closed subinterval I1 of I such that x1 ∈ I1, then select a closed interval
I2 of I1 such that x2 ∈ I2, and so on. In this way, we obtain nonempty closed intervals
I1 ⊇ I2 ⊇ I3 ⊇ ... ⊇ In ⊇ ...
such that In ⊆ I and xn ∈ In for all n.
The Nested Intervals Property implies that there exists a point ξ ∈ I such that ξ ∈ In
for all n. Therefore ξ = xn for all n ∈ N, so the enumeration of I is not a complete listing
of the elements of I, as claimed. Hence, I is an uncountable set. Since I is equivalent
to R (see Exercise below), it follows that R is uncountable. ♣
Exercise: Find a one-to-one correspondence f : R −→ [0, 1].
Cardinality of subsets of R
If two sets A and B are equivalent, then they have the same cardinality or the same
cardinal number.
Definition 2.9 If A is finite, then cardinality of A is the number of elements in A.
Remarks
The cardinal number of a countable set A is denoted by the symbol ℵ0 and is called
aleph zero or aleph nought or aleph null and written
Card A = ℵ0.
Since every infinite subset of a countable set is also countable, it follows that the count-
able infinity is the smallest infinity among infinities of all orders.
32

It therefore follows that infinity of an uncountable set like R is of a higher order than
that of a countable set like Q of rationals.
Example
Given the following limits
1. lim
n→∞
n2
= ∞
2. lim
n→∞
2n
= ∞
We note that the infinity generated by the limit in part (2) is of a higher order than the
infinity generated by the limit in part (1).
Definition 2.10 A set A is said to have a cardinal number less than of another set B
if A is equivalent to a proper subset of B but A is not equivalent to B.
Thus Card A < Card B.
Theorem 2.10 Let M be an infinite set and P(M) denotes the class of all subsets of
M. Then we have that:
Card M < Card P(M).
Proof
Let M = {a, b, c, ...}. Then in particular the singleton subsets
{a}, {b}, {c}, ... ∈ P(M).
Thus the mapping
a −→ {a}
b −→ {b}
. . .
is a one-to-one correspondence.
It follows that M is equivalent to a proper subset of P(M) that contains only single-
tons. Note that P(M) contains other subsets like {b, a}, {b, c}, {a, c}, etc. which are not
mapped to under this correspondence. Thus M is not equivalent to P(M).
33

Hence by definition, Card M < CardP(M). ♣
Example
If M is a finite set and thus has n elements then Card M = n. But we have
Card(P(M)) =
n
0
+
n
1
+
n
2
+ · · · +
n
n
= 2n
Clearly 2n
> n.
Thus the theorem is equally true for the case of finite sets.
Exercise
Given M = {x, y, z}, write down all the elements of P(M).
Remarks
The concept of countability of a set is equivalent to the concept of nextness of a set.
This is why every countable set can be enumerated as a sequence.
The cardinality of an infinite set is infinity and all those cardinalities which are infinity,
the one involving a countable set is the smallest (i.e. of least order).
SOLVED PROBLEMS
(1). Show that the set Z of integers is countable. Hence deduce that the set of all
negative whole numbers is countable.
Solution
We first show that the set Z is countable.
Define a mapping
f : N −→ Z as follows:
f(n) =
n
2
, if n is even
−(n−1)
2
, if n is odd
Clearly f is one-to-one and onto. Thus Z is equivalent to N. Hence Z is countable.
We now show that the set of all negative whole numbers is also countable.
Firstly, note that the required set here is an infinite subset of Z. By Theorem 2.3: An
34

infinite subset of a countable set is again countable, we can conclude that the set of all
negative whole numbers is also countable. ♣
(2). Show that the set of all polynomials with integral coefficients is countable.
Solution
A polynomial of degree n with integral coefficients can be expressed in the form
p(x) = a0 + a1x + a2x2
+ a3x3
+ . . . + anxn
,
with a0, a1, . . . , an as integers.
Now, the set of (n + 1) tuples {(a0, a1, . . . , an) : ai ∈ Z} is denoted by Bn+1 and is
countable as seen in Theorem 2.5.
Thus the collection of all polynomials Pn of degree n with integral coefficients can be
put in a one-to-one correspondence with the set Bn+1. In this case, the mapping
f : Pn −→ Bn+1
is one-to-one and onto. Hence the collection of such Pn is also countable. But n is any
positive integer, i.e. P1, P2, . . . , Pn, are countable sets. Thus P =
∞
∪ Pn
n = 1
is also
countable.
Hence the set of all polynomials of any degree with integral coefficients is countable. ♣
(3). It is well known that a real root to f(x) = 0 when f(x) is a polynomial with rational
coefficients, is called an algebraic number and that the set of all algebraic numbers is
countable.
Given that a real number is called transcendental if it is not algebraic, determine
whether the set of all transcendental numbers is countable or uncountable.
Solution
Let A be the set of all transcendental numbers and B be the set of all algebraic numbers.
35

Then we have that R = A ∪ B.
Now, B is countable and R is known to be uncountable. Assume A is countable. Thus
A∪B is countable, since the union of countable sets is again countable. Thus A∪B = R
is countable. This is a contradiction, since R is known to be uncountable. Hence the set
of all transcendental numbers is uncountable. ♣
4. Prove that every subset of a countable set is countable.
Proof
Follows easily from Theorem 2.3.
Alternative Proof Let E = {xn} be a countable set, and let A be a subset of E. If A
is empty, A is countable by definition. If A is not empty, choose x ∈ A. Define a new
sequence {yn} by setting
yn =
xn, if xn ∈ A
x, if xn ∈ A
Then A is the range of {yn} and is therefore countable.
5. Let A be a countable set. Prove that the set of all finite sequences from A is also
countable.
Proof
Since A is countable, it can be put into a one-to-one correspondence with a subset of the
set N of natural numbers. Thus it suffices to prove that the set of all finite sequences
of natural numbers is countable. Let {2, 3, 5, 7, 11, · · · , pk · · · } be the sequence of prime
numbers. Then each n in N has a unique factorization of the form
n = 2x1
3x2
· · · pxk
k ,
where xi ∈ N0 = N ∪ {0} = W and xk > 0.
Let f be the function on N that assigns to the natural number n the finite sequence
{x1, · · · , xk} from N0. Then S is a subset of the range of f. Hence S is countable by
Problem 4 above.
Tutorial Problems
36

1. Show that the set Ω = N − {2, 4, · · · , 2n, · · · } is countable, where N denotes the set
of natural numbers.
2. Prove that the set N × N is countable by identifying a bijection f : N × N −→ N.
3. Show that the set S = {12
, 22
, 32
, · · · } of the squares of the positive integers is count-
able.
4. Let A and B be sets such that A is countable and B is uncountable. Prove that
B − A is uncountable.
5. Prove that the set Q of rational numbers is countable by identifying a bijection from
a countable set to Q.
6. Let A and B be countable sets. Prove that Aand B are equivalent.
7. Prove that the set of all polynomials in x with rational coeﬃcients is countable.
8. Prove that (0, 1) ∼ (a, b). [Hint: f : (0, 1) −→ (a, b) deﬁned by f(x) = a + x(b − a)
is a bijection of (0, 1) onto (a, b)]
9.(a). Prove that (0, 1) ∼ (0, 1]. [This problem is not easy! Hint: Consider the function
on (0, 1) that for each n ∈ N, n ≥ 2, maps 1
n
to 1
n−1
, and is the identity mapping
elsewhere]
(b). Prove that (0, 1) ∼ [0, 1] and hence deduce that [0, 1] ∼ R.
37

Chapter 3
STRUCTURE OF THE METRIC
SPACE R
3.1 Introduction
The system of real numbers has two types of properties. The ﬁrst type which consists
of the algebraic, dealing with addition and multiplication, etc was studied in Chapter
one. In this Chapter we concentrate on another aspect of the real numbers-the concept
of distance, which is fundamental in classical analysis. The latter properties are called
topological or metric. The results of this chapter will come in handy in the rest of the
chapters in this course. For instance, the classical deﬁnition of continuity:
f : R −→ R is continuous at x ∈ R, if given > 0, then for some δ > 0,
|f(x) − f(y)| < whenever |x − y| < δ, for y ∈ R
can be crudely restated (using | x − y | as a measure of distance between x and y) as:
”f is continuous at x if f(y) is near to f(x) whenever y is near enough to
x”.
In this chapter, we furnish R with a geometric structure which provides for the concept
of distance between any two given elements of R. We endow R with a metric (which is
an abstraction of a distance function) and hence refer to it as a metric space.
38

We will discuss the concepts of an − neighborhood of a point, open and closed sets and
later apply the results to convergence of sequences and continuity of functions defined
on metric spaces. We will define the notions of ”convergence of a sequence” and ”limit
of a set” in terms of −neighbouhoods.
3.2 The notion of a metric
Definition 3.1 A metric on a non-empty set X is a function
d : X × X −→ R
that satisfies the following properties:
(i). d(x, y) ≥ 0 for all x, y ∈ X. [positivity]
(ii). d(x, y) = 0 iff x = y [definiteness or nondegeneracy]
(iii). d(x, y) = d(y, x) for all x, y ∈ X [symmetry]
(iv). d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X [triangle inequality]
A set X equipped with a metric d, and denoted (X, d) is called a metric space. That is,
a metric space is a set X with a metric defined on it.
Remark:
There are always many different metrics for a given set X.
3.2.1 Examples of Metrics
1. The usual or familiar or standard metric for R is defined by
d(x, y) =| x − y |, for all x, y ∈ R.
We show that d is a metric by running all the axioms of a metric:
(i). d(x, y) =| x − y |≥ 0, for all x, y ∈ R.
39

(ii). d(x, y) = 0 ⇐⇒ |x − y| = 0
⇐⇒ x − y = 0
⇐⇒ x = y, ∀ x, y ∈ R
(iii). d(y, x) = |y − x| = | − (x − y)|
= | − 1||x − y|
= |x − y|
= d(x, y) , ∀ x,y∈ R
(iv). follows from the triangle inequality for absolute values because we have
d(x, y) = |x − y| = |(x − z) + (z − y)|
≤ |x − z| + |z − y| = d(x, z) + d(z, y)
d(x, z) = |x − z| = |x − y + y − z|
≤ |x − y| + |y − z|
= d(x, y) + d(y, z), ∀ x, y, z ∈ R
Thus d is a metric on R and hence (R, d) is a metric space.
2. The discrete metric. If X is a non-empty set, deﬁne d by
d(x, y) =
1 if x = y
0 if x = y
Exercise: Verify that d in (2) above is a metric on X.
Solution
Note that the ﬁrst three properties follow easily. The triangle inequality does not hold
if d(x, y) = 1 and d(x, z) = d(y, z) = 0. However, this would only be possible for
x = y = z. Hence, d(x, y) cannot be equal to 1. This proves that
d(x, y) ≤ d(x, z) + d(y, z), ∀ x, y, z ∈ X.
Remark
40

We note that if (X, d) is a metric space, and T ⊆ S, then d defined by d (x, y) := d(x, y),
for all x, y ∈ T gives a metric on T, which we generally denote by d and say that (T, d)
is a metric space. For instance, the standard metric on R is a metric on the set Q of
rational numbers, and thus (Q, d) is also a metric space.
In general, suppose that (X, d) is a metric space and A is a non-empty subset of X. If
x and y are in A, d(x, y) is the distance between x and y in the metric space (X, d),
and clearly d generates a notion of distance between points in the set A. However (if
A = X), d is not a metric for A because a metric for A is a function on A × A while d
is a function on X × Y . This defect can be remedied as follows:
Let dA be the restriction of d to A × A. Then it is easy to verify that dA is a metric for
A, called the relative metric induced by d on A. The metric space (A, dA) is called the
subspace of (X, d) generated by A. Despite the formalism, the idea is very simple; the
distance between two points in (A, dA) is precisely the distance between them in (X, d).
However, despite the simplicity of the idea, some care is required when working with
relative metrics; a subspace may have properties quite different from the original space.
We now present some basic definitions and theorems about metric spaces.
3.3 Neighbourhoods, Interior points and Open sets
There are special types of sets that play a distinguished role in analysis. These are the
open and closed sets in R. To expedite this discussion, it is convenient to have sound
grip of the notion of a neighbourhood of a point.
This is the basic notion needed for the introduction of limit concepts.
3.3.1 Neighborhoods in a metric space
Definition 3.2 Let (X, d) be a metric space. Then for > 0, the open −neighbourhood
of a point x0 in X is the set
V (x0) = {x ∈ X : d(x0, x) < }
41

Other names are: the open −ball centre x0, an open disc with centre x0 and
radius or simply a neighbourhood of x0, or a sphere, if precision is not required.
In R with its usual metric, an open sphere centred at p radius is the set
S (p) = {x ∈ R : |x − p| < }.
This consists of all those real numbers x which satisfy the inequality
− < x − a < ⇐⇒ p − < x < p +
Remark We note that spheres in various metric spaces can look quite different from
those in Euclidean space. It should be particularly noticed that the spheres in a given
sphere may be quite unlike those in a subspace.
Definition 3.3 Let x0 ∈ (R, d) be a fixed element and r > 0 be a real number, where d
is the usual metric on R. Then the set given by
N(x0, r) = {x ∈ R : |x − x0| < r}
is called the open neighbourhood (nbhd) of x0 centered at x0 with radius r.
42

Graphically, N(x0, r) looks like
Figure 3.1: N(x0, r)
This nbhd is the open interval (x0 − r, x0 + r).
Deﬁnition 3.4 Let A be a subset of R. A point x ∈ A is said to be an interior point
of A if there exists a neighbourhood N(x, r) for some r > 0 such that N(x, r) ⊂ A.
That is, N(x, r) is properly contained in A.
Example
Let A ⊆ R be given by A = {x ∈ R : 0 < x < 1}. Then an element like 1
4
∈ A is an
interior point of A since N(1
4
, 1
8
) ⊂ A. But the element 1 is not an interior point of A
since there does not exist a neighbourhood N(1, r) such that N(1, r) ⊂ A.
Deﬁnition 3.5 Let A be a subset of R. The set of interior points of A is called the
interior of A and is denoted by A◦
or int(A).
That is
int(A) = {x : x is an interior point of A}.
Example
Let A = {x ∈ R : 0 < x ≤ 1}. Then int(A) = {x ∈ R : 0 < x < 1}
43

Remark Clearly, int(A) ⊆ A.
Definition 3.6 A point x ∈ R is said to be a boundary point of A ⊆ R if every nbhd
N(x, r), r > 0 of x contains points in A and points in AC
.
Definition 3.7 The boundary of a set A ⊆ R, usually denoted by ∂A, or B dary(A)
is the collection of all the boundary points of A.
Remark Clearly x ∈ ∂A iff for all r > 0 N(x, r) ∩ A = ∅ and N(x, r) ∩ AC
= ∅.
Exercise: Show that a set A and its complement AC
have exactly the same boundary
points.
That is: ∂A = ∂AC
.
Definition 3.8 Let A be a subset of R. Then A is said to be open in R if every
element of A is an interior point of A. In other words, a subset A ⊆ R is said to be
open in R if for each x ∈ A, ∃ a nbhd N(x, r) of x radius r > 0 such that N(x, r) ⊂ A.
Remark Note that the interior of a set is always an open set. Examples
1. The set G = {x ∈ R : 0 < x < 1} is open.
Proof
For any x ∈ G we may take rx to be the smaller of the numbers x, 1 − x. It is left as an
exercise to show that if |u − x| < rx, then u ∈ G.
Alternative Proof
Follows easily since every member of G is an interior point of G.
2. The set R = (−∞, ∞) is open.
Proof
For any x ∈ R, we may take r = 1. That is R is an open set.
3. Generally any open interval I = (a, b) is an open set.
In fact, if x ∈ I, we can take rx to be the smaller of the numbers x − a, b − x.
44

Exercise: Show that (x − rx, x + rx) ⊂ I.
Similarly, the intervals (−∞, b) and (a, ∞) are open sets.
4. The set A ⊆ R given by A = {x ∈ R : 0 < x ≤ 1} is not open since the element
1 ∈ A but 1 is not an interior point of A.
5. It is easy to prove that:
(i) int(N) = ∅
(ii). int(Q) = ∅
(iii). int(QC
) = ∅
(iv). int({x}) = ∅
(v). int(R) = R
3.4 Limit Points and Closed sets
Definition 3.9 Let A be a subset of R. A point x ∈ R is said to be a limit point or
a cluster point or an accumulation point of A if every nbhd of x has at least one
element of A different from x or equivalently, if every nbhd N(x, r) of x has infinitely
many points.
Remarks
• Clearly, x is a limit point of A iff for every open nbhd, N(x, r) of x, we have
N(x, r) ∩ A = ∅.
• If x ∈ R, the definition demands that N(x, r) should contain at least one other point
of A.
45

Example: Let A ⊆ R be given by A = {x ∈ R : 0 < x < 1}. Then an element 1
2
∈ A is
a limit point of A; i.e. N(1
2
, r) ∩ A = ∅ for any r > 0. Also, the 1 ∈ R is a limit point of
A since N(1, r) ∩ A = ∅ for any r > 0.
Remarks
1. Note that a limit point of a set may or may not belong to the set.
2. If there is a member of a set which is not a limit point of the set, then it is called an
isolated point of the set.
Definition 3.10 The set of all the limit points of a set A, usually denoted by A is called
the derived set of A.
That is, A = {x : x is a limit point of A}.
Definition 3.11 A subset A of R is said to be closed if every limit point of A belongs
to A.
That is, A is closed if AC
is open in R. To show that A ⊆ R is closed, it suffices to show
that each point y of A has an open neighborhood N(x, r) disjoint from A.
Examples
1. The set I = [0, 1] is closed in R.
To see this, let y ∈ I; then either y < 0 or y > 1. If y < 0, we take y = |y|, and if y > 1,
take y = y − 1.
Exercise: Show that in either case, we have I ∩ (y − y, y + y) = ∅.
Alternatively, show that every limit point of I belongs to I. ♣
2. The set H = {x ∈ R : 0 ≤ x < 1} is neither open nor closed.
3. The set A = {x ∈ R : 0 < x ≤ 1} is not closed in R since 0 is a limit point of A but
0 ∈ A.
46

4. The empty set ∅ is open in R. In fact the empty set contains no points at all, so the
requirement in the definition is vacuously verified. The empty set is also closed since its
complement R is open.
Definition 3.12 Let A be a subset of R. Then the closure of A, denoted by A or cl(A)
is given by
A = A ∪ {x : x is a limit point of A} = A ∪ A .
Remark
Clearly A ⊆ A.
Example
Let A = {x ∈ R : 0 < x ≤ 1}. Then
A = A {0} = {x ∈ R : 0 ≤ x ≤ 1} = [0, 1].
Remark: We established that Q and QC
are dense in R. We give a more rigorous
definition of a dense set:
Definition 3.13 A subset A of R is said to be dense in R if every limit point of R is
also a limit of A.
That is if A = R.
Thus, we have Q = R, and QC = R.
That is if p is any real number, then every nbhd of p contains at least one rational
number and it also contains at least one irrational number.
Theorem 3.1 Let A be a subset of R. Then A is closed if and only if AC
is open.
Proof
(⇒) Assume A is closed and let x ∈ AC
. Then x cannot be a limit point of A for if it
is then x ∈ A, for A is closed. Thus there exists an open nbhd N(x, r) of x such that
N(x, r) ∩ A = ∅.
Thus, N(x, r) ⊂ AC
. Hence AC
is open.
(⇐) Conversely, assume that AC
is open and let x be any limit point of A. Then every
open nbhd N(x, r) of x is such that N(x, r) ∩ AC
= ∅. Thus x cannot be an interior
47

point of AC
. Since AC
is open (by assumption-and hence doesn’t contain all of its limit
points), x ∈ AC
.
This implies that x ∈ A. Thus every limit point of A belongs to A. Hence A is closed.
3.5 Properties of open and closed sets in R
Theorem 3.2 (a). The union of an arbitrary collection of open subsets in R is open.
(b). The intersection of any finite collection of open sets in R is open.
Theorem 3.3 (a). The intersection of an arbitrary collection of closed sets in R is
closed.
(b). The union of any finite collection of closed sets in R is also closed.
Examples
(1). Let Gn = (0, 1 + 1
n
), for n ∈ N. Then Gn is open for each n ∈ N.
However, the intersection
G =
∞
∩ Gn
n = 1
= (0, 1], which is not open in R.
Thus, the intersection of infinitely many open sets in R need not be open
(2). Let Fn = [1
n
, 1], for n ∈ N. Each Fn is closed, but the union
F =
∞
∪ Fn
n = 1
= (0, 1], which is not closed in R. Thus, the union of infinitely
many closed sets in R need not be closed.
Theorem 3.4 A subset of A ⊂ R is closed if and only if it contains all its limit points.
Proof
(⇒) Let A be a closed subset of R and let x be a limit point of A. We will show that
x ∈ A. For a contradiction suppose that x ∈ A. Then x ∈ AC
, an open set. Therefore,
there exists an open neighbourhood N(x, r) of x such that N(x, r) ⊂ AC
.
Consequently, N(x, r) ∩ A = ∅, which contradicts the assumption that x is a limit point
48

of A. Thus A must contain all of its limits points.
(⇐) Conversely, let A be a subset of R that contains all of its limit points. We show
that A is closed. It suffices to show that AC
is open. For if y ∈ AC
, then y is not a
limit point of A. It follows that ∃ an open nbhd N(y, r) of y that does not contain a
point of A. (except possibly y).
But since y ∈ AC
, it follows that N(y, r) ⊂ AC
. Since y is an arbitrary element of
AC
, we deduce that for every point in AC
, there exists an open nbhd that is entirely
contained in AC
. But this means that AC
is open in R. Therefore A is closed in R. ♣
Theorem 3.5 A subset of R is open if and only if it is the union of countably many
disjoint open intervals in R.
Remark It does not follow from the above theorem that a subset of R is closed iff it is
the intersection of a countable collection of closed intervals (why not? ). In fact, there
are closed sets in R that cannot be expressed as the intersection of a countable collection
of closed intervals in R. A set consisting of two points is one example(why? ).
3.6 Relatively Open and Closed Sets
One of the reasons for studying topological or metric concepts is to enable us to study
properties of continuous functions. In most instances, the domain of a function is not
all of R, but rather a proper subset of R. When discussing a particular function we will
always restrict our attention to the domain of the function rather than all of R. With
this in mind, we make the following definition.
Definition 3.14 Let X be s subset of R.
(a). A subset U of X is open in (or open relative to) X if for every p ∈ U, there
exists an r > 0 such that N(p, r) ∩ X ⊂ U.
(b). A subset C of X is closed in (or relative to) X if X − C is open in X
Example. Let X = [0, ∞) and let U = [0, 1). Then U is not open in R but is open in
X.(Why?)
The following theorem provides a simple characterization of what it means for a set to
be open or closed in X.
49

Theorem 3.6 Let X be a subset of R.
(a). A subset U of X is open in X if and only if U = X ∩ O for some open subset O of
R.
(b). A subset C of X is closed in X if and only if C = X ∩ F for some closed subset F
of R.
Remark
Clearly, open(closed) =⇒ relatively open ( relatively closed) but the converse is not
generally true.
3.7 Solved Problems
1. (a). Give the definition of an open subset of R and a closed subset of R.
(b). Let A ⊆ R be given by A = {x ∈ R : 1 ≤ x < 2}. Show that A is neither closed
nor open.
Solutions
(a). A subset A of R is said to be open if every member of A is an interior point of A.
But A is said to be closed in R if every limit point of A belongs to A.
(b). A = {x ∈ R : 1 ≤ x < 2} is not closed since 2 is a limit point of A but 2 ∈ A.
Also A is not open since 1 ∈ A but 1 is not an interior point of A.
2.(a). Construct a set of real numbers with only 3 limits.(Hint: Note that the set
A = {1
n
: n ∈ N} has only 0 as the limit point.)
(b). Let A ⊆ R. Prove that:
(i). A is open iff A = Int(A)
(ii). A is closed iff A = A.
Solution
(a). Given A = {1
n
: n ∈ N}, we note that
lim 1
n
= 0
n → ∞
. Now let
50

B = {1 + 1
n
: n ∈ N} and C = {2 + 1
n
: n ∈ N}. Similarly sets B and C have only one
limit point each. Thus S = A ∪ B ∪ C is a set whose limit points are 0, 1, and 2.
(b). A ⊆ R.
(i). A is open iﬀ A = Int(A).
(⇒) Assume A = Int(A). Then A is open because int(A) is always open.
(⇐) Conversely, assume that A is open. Then for any x ∈ A, we have that x is an
interior point of A. That is x ∈ A =⇒ x ∈ Int(A). Thus
A ⊆ Int(A)
(1)
But the inclusion
Int(A) ⊆ A
(2)
is immediate(obvious).
From (1) and (2) it follows that A = Int(A).
(ii). A is closed iﬀ A = A.
(⇒) Assume that A = A. Then A is closed since A is always closed.
(⇐) Conversely, assume that A is closed. Then every limit point of A belongs to A. But
x is a limit point of A means that x ∈ A.
Thus x ∈ A =⇒ x ∈ A.
That is A ⊆ A
(1)
But the inclusion
51

A ⊆ A
(2)
is obvious.
From (1) and (2) equality follows. That is A = A. ♣
3. Show that the set N of natural numbers is closed in R.
Solution
The complement of N is the union (−∞, 1) ∪ (1, 2) ∪ · · · of open intervals , hence open.
Therefore, N is closed since its complement is closed.
4. Show that the set Q of rational numbers is neither open nor closed.
Solution
Every nbhd of x ∈ Q contains a point not in Q.
5. Give an example of a set A ⊆ R such that int(A) = ∅ and A = R.
Solution
A = Q or A = QC
.
3.8 Tutorial Problems
1. Let A and B be subsets of a metric space X.
(a). Prove that
(i). int(A) ∪ int(B) ⊆ int(A ∪ B).
(ii). int(A) ∩ int(B) = int(A ∩ B).
(iii). (A ∪ B) = A ∪ B.
(iv). (A ∩ B) ⊆ A ∩ B.
52

(b). Give an example of two subsets A and B of R such that
(i). int(A) ∪ int(B) = int(A ∪ B).
(ii). (A ∩ B) = A ∩ B.
2. Prove that
(a). ∂A = A ∩ AC
(b). ∂A = A − int(A)
3. Find the boundary points of each of the following sets
(a). A = (a, b)
(b). A = {1
n
: n ∈ N}
(c). A = Q
(d). A = N
(e). A = R
4. (a). Prove that a set A ⊆ R is open if and only if A does not contain any of its
boundary points.
(b). Prove that a set A ⊆ R is closed if and only if A contains all of its boundary points.
53

Chapter 4
BOUNDED SUBSETS OF R
4.1 Introduction
In this chapter we will consider the concept of the least upper bound of a set and
introduce the least upper bound property of the real numbers R. We will show that
that this property fails for the rational numbers Q. We first define the notions of upper
bound and lower bound of a subset of real numbers.
4.2 Upper Bounds, Lower Bounds of a subset of R
Definition 4.1 A non-empty subset S of real numbers is said to be bounded above and
thus has an upper bound, say b if b ≥ x for all x ∈ S.
A non-empty subset S of R is said to be bounded below and thus has a lower bound,
say q if q ≤ x for all x ∈ S.
Remark
1. If b is an upper bound for S then any real number b > b is also an upper bound for
S.
In other words, if a set has an upper bound, then it has infinitely many upper bounds,
because b + 1, b + 2, ... are upper bounds of S.
54

2. If q is a lower bound for S, then any real number q < q is also a lower bound for S.
Thus, if a set has a lower bound, then it has infinitely many lower bounds, because
q − 1, q − 2, ... are lower bounds of S.
Definition 4.2 A set is said to be bounded if it is both bounded above and bounded
below. A set is said to be unbounded if it is not bounded.
Example The set S = {x ∈ R : x < 2} is bounded above; the number 2 and any other
number larger than 2 is an upper bound of S. This set has no lower bounds, so that the
set is not bounded below. Thus it is unbounded (even though it is bounded above !).
In the set of upper bounds of S and set of lower bounds of S, we single out their least
and greatest elements, respectively, for special attention.
4.2.1 Supremum and Infimum of a subset of R
Definition 4.3 Let S be a nonempty subset of R.
(a). If S is bounded above, then a number u is said to be the supremum (or the least
upper bound) of S if it satisfies the conditions:
(1). u is an upper bound of S, and
(2). if v is any other bound of S, then u ≤ v.
That is, if u < u, then u is not an upper bound of S. Then there exists u ∈ S such
that u < u < u.
Definition 4.4 If S is bounded below, then a number w is said to be the infimimum (or
greatest lower bound) of S if it satisfies the conditions:
(1’). w is a lower bound of S, and
(2’). if t is any lower bound of S, then t ≤ w. That is , if q is a lower bound of S and
if q > q, then q is not a lower bound for S. Thus, ∃ q ∈ S such that q > q > q.
Remark
1. It is not difficult to see that there can be only one supremum of a given subset S of
R. Thus we can refer to it as the supremum of S instead of a supremum). For suppose
that u1 and u2 are both suprema of S. If u1 < u2, then the hypothesis that u2 is a
supremum, then this implies that u1 cannot be an upper bound of S. Similarly, we see
55

that u2 < u1 is not possible. Therefore, we must have u1 = u2.
A similar argument can be given to show that the infimum of a set is unique.
If the supremum and infimum of a set exists, we will denote them by Sup S and Inf S,
respectively.
We observe that if u is an arbitrary upper bound of a non-empty set S, then
Sup S ≤ u (because SupS is the least upper bound of S).
2. Note that in order for a nonempty set S in R to have a supremum, it must have an
upper bound. Thus, not every subset of R has a supremum. Similarly, not every subset
of R has an infimum.
Four possibilities for a nonempty subset of R
A subset of R can have:
(1). both a supremum and an infimum,
(2). a supremum but no infimum,
(3). an infimum but no supremum,
(4). neither a supremum nor an infimum.
56

The ﬁgure below show the properties of a bounded set.
Figure 4.1: Bounds of S
Examples
1. Let A = {x ∈ R : 0 < x < 1}. Then
u = lubA = SupA = 1.
If we take any u < u, say u = 0.9, then ∃ u” say u” = 0.95 ∈ A and 0.9 < 0.95 < 1.
Thus u < u” < u.
2. Let B = {x ∈ R : 0 < x < 1} {2}. Then u = SupB = 2. If we take u = 1.5, then
u < b” = b.
Example
Let A = {x ∈ R : 0 < x < 1}. Then q = glbA = infA = 0. If we take any q > q say
q = 1
8
and 0 < 1
8
< 1
4
.
Thus q > q” = q.
Example Let B = {x ∈ R : 0 < x < 1} {−1}. Then q = infA = −1. If we let
q = −1
4
, say, then q” = −1 = q. Thus q > q” > q.
57

Remark
Below are equivalent definitions of sup S and inf S using > 0.
Definition 4.5 A real number b is said to be the least upper bound of S if for each
> 0, b − is not an upper bound for S. For ∃ b ∈ S such that b − < b < b.
Similarly, a real number q is said to be the greatest lower bound for S if for each > 0,
q + is not a lower bound for S, since ∃ q ∈ S such that q + > q > q.
Remarks
1. From the above definition, it follows that every nbhd of b = sup S or q = inf S has at
least one point of S. Thus b = sup S and q = inf S are limit points of S which need not
belong to S.
2. sup S ∈ S, inf S ∈ S.
3. In the special case when sup S ∈ S and inf S ∈ S, then sup S is called the maximum
element of S while inf S is called the minimum element of S.
Example Let A ⊆ R be given by
A = {x ∈ R : 0 < x ≤ 1}
Then inf A = 0 ∈ A. Hence A has no minimum element. But sup A = 1 ∈ A. Hence 1
is the maximum element of A.
Remark In general some sets of real numbers do not have sup or inf unless bounded
above or below, respectively. The completeness axiom guarantees existence of supremum
and infimum.
We will give an example to show that the claim of completeness axiom is only enjoyed
by set of real numbers but not other sets like those of rational or irrational numbers.
4.3 The Completeness Property of R
Proposition 4.1 (Least Upper Bound Property of R) Every nonempty subset of
real numbers that has an upper bound also has a supremum in R.
58

Remark
This property is called the Supremum Property of R. The analogous property for infima
can be deduced from the Completeness Property as follows:
Every nonempty subset of R that is bounded below has an infimum or greatest lower
bound in R. Every nonempty subset of R that is bounded below has an infimum or
greatest lower bound in R. Suppose that S is a nonempty subset of R that is bounded
below. Then the nonempty set S∗
= {−s : s ∈ S} is bounded above, and the supremum
property implies that u = SupS∗
exists in R.This can be re-stated as follows:
Proposition 4.2 (The Infimum or Greatest Lower Bound property of R) Ev-
ery nonempty subset of R that is bounded below has an infimum or greatest lower bound
in R.
Combining the immediate two propositions we have:
Theorem 4.3 (The Completeness Property of R): Every nonempty subset of real
numbers that has an upper bound also has a supremum in R and every nonempty subset
of R that is bounded below has an infimum or greatest lower bound in R.
Exercise:In the remark above, verify in detail that −u is the inf of S.
We demonstrate that not all subsets of R enjoy the Completeness Property.
Example
Let S = {y : y ∈ Q, 0 ≤ y, y2
< 2}. Then clearly 0 ∈ S so that S = ∅. Now suppose that
s ∈ S, then s2
< 2 and 2 − s2
is a positive rational number. Select a positive rational
number h such that h < 1 and h < 2−s2
2s+1
and let u = s + h.
Clearly h is rational and 0 < s < u. Also we can easily check that :
u2
= s2
+ 2sh + h2
< s2
+ 2sh + h < 2. Hence u ∈ S. That is for any given member of
S, there always exists a larger member of S. Hence S has no largest member.
This result shows that rational numbers do not have the completeness property.
4.4 Solved Problems
1. (a). Give an example of the following:
59

(i). a set S ⊆ R such that SupS = InfS.
(ii). a set S ⊆ R which has got inﬁmum, minimum and supremum but no maximum
element.
(b). Show that if the maximum element of a set S ⊆ R exists then it is unique.
Solution
(a). An example of
(i). a set S such that SupS = InfS.
Let S = {x}. Then SupS = x and InfS = x.
Hence SupS = InfS.
(ii). a set S of reals which has got inﬁmum, minimum, supremum but no maximum
element:
Let S = {x ∈ R : 0 ≤ x < 1}.
Then 0 = InfS and 0 ∈ S. Thus 0 ie equal to the minimum element in S. We also have
that 1 = SupS. But 1 ∈ S. Hence S has no maximum element.
(b). Let b = MaxS. We show that b is unique. Assume we also have b = MaxS. Then
in particular b is an upper bound of S and b is the lub of S.
That is
b < b
(1)
but we also have that b is an upper bound of S while b is the least upper bound of S.
That is
b < b
(2)
From (1) and (2), b = b .
60

Hence MaxS is unique. ♣
4.(a). State the Completeness property of R.
(b). Let E be a closed and bounded subset of R. Show that sup E and inf E belong
to E.
Solution
(a). A set of reals bounded above has the least upper bound in R and if it is bounded
below then it has the greatest lower bound.
(b). Let E be a closed and bounded subset of R. We show that SupE and InfE belong
to E.
Let b = SupE. Then for each > 0, b − is not an upper bound of E. Thus there
exists b ∈ E such that b − < b < b
This is equivalent to saying that every nbhd N(b, ) has at least one element of E. That
is b is a limit point of E. But E is closed, hence it contains all its limit points. Thus
b ∈ E. Thus SupE ∈ E.
Similarly, let q = InfE. Then for each > 0, q + is not a lower bound of E. Thus
there exists q ∈ E such that q < q < q + .
Thus q is a limit point of E. But E is closed. Hence q = InfE ∈ E. ♣
5. Deﬁne a set T by T = {z : z ∈ Q and 2 < z2
}. Show that T has no smallest element.
61

Chapter 5
SEQUENCES OF REAL
NUMBERS
5.1 Introduction
In this chapter we study the properties of a sequence of real numbers. In our study
of sequences we encounter our ﬁrst serious introduction to the limit process. We begin
the chapter by introducing the notion of convergence of a sequence of real numbers and
by proving the standard limit theorems for sequences normally encountered in calculus.
We use the least upper bound property of R to show that every bounded monotone
sequence of real numbers converges in R. We introduce the notion of subsequences and
subsequential limits and use these to provide a proof of the fact that every Cauchy
sequence of real numbers converges.
Deﬁnition 5.1 A sequence is an ordered set of numbers, say a1, a2, ... where each mem-
ber is followed by another according to a given rule. In this case we write the sequence as
{an}∞
n=1 = {a1, a2, ..., an, ...}
The members a1, a2, .. are called the terms of the sequence. The term an is called the nth
term.
Example
Let {an}∞
n=1 = {n2
} for all n ∈ N. Then {an} is a sequence of real numbers whose terms
62

are as follows:
an = 12
, 22
, 32
, ..., n2
, ...
The nth
term of this sequence is n2
.
Remark (1) Note that a sequence can also be defined as a function whose domain is
the set N of natural numbers,e.g. the sequence above, f(n) = n2
for all n ∈ N.
5.2 Convergence of a sequence
Definition 5.2 A sequence {xn}∞
n=1 in R is said to converge if there exists a point
x ∈ R such that for every > 0, there exists a positive integer n0 such that
xn ∈ N(x, ) for all n ≥ n0.
Remark
If this is the case, we say that {xn} converges to x or that x is the limit of the
sequence {xn}, and we write:
lim xn = x
n → ∞
or xn −→ x
If {xn} does not converge, then {xn} is said to diverge.
In the definition, the statement xn ∈ N(x, ) for all n ≥ n0 is equivalent to
|xn − x| < for all n ≥ n0
(*)
(*) gives the criterion of convergence of a sequence.
As a general rule, the integer n0 will depend on the given .
Remark
From (*), we can restate the criterion of convergence as follows:
5.2.1 Criterion of Convergence
A sequence {xn} of real numbers converges to a real number x if no matter how small
a positive real number is we should be able to find a natural number n0 depending on
such that the distance between the terms xn of the sequence and the limit x is always
63

less than provided the subscript n is greater than n0.
Example 1 Let {xn} = {1
n
}. We show that {xn}∞
n=1 converges to 0 in R.
Given > 0, there exists a positive integer n0 such that n0 > 1. Thus for all n ≥ n0,
|1
n
− 0| = |1
n
| < .
Therefore,
lim 1
n
= 0
n → ∞
.
In this example, the integer n0 must be chosen so that n0 > 1
.
Example 2 If x ∈ R, the sequence {xn} defined by xn = x for all x ∈ R is the constant
sequence of x. Since |xn − x| = 0 for all n ∈ N, we have
lim xn = x.
n → ∞
Example 3 Consider the sequence {2n+1
3n+2
}∞
n=1. We show that
lim 2n+1
3n+2
= 2
3
n → ∞
.
Since |2n+1
3n+2
− 2
3
| = 1
3(3n+2)
< 1
9n
, given > 0, choose n0 ∈ N such that n0 > 1
9
. Then for
all n ≥ n0, |2n+1
3n+2
− 2
3
| < .
Thus the given sequence converges to 2
3
.
Example 4 The sequence {xn} = {1 − (−1)n
}∞
n=1 diverges in R. To prove this, we first
note that for this sequence
|xn − xn+1| = 2 for all n. Suppose xn −→ x for some x ∈ R. Let 0 < < 1. Then by
definition of convergence, there exists an integer n0 such that
|xn − x| < for all n ≥ n0. But if n ≥ n0, then
2 = |xn − xn+1| ≤ |xn − x| + |x − xn+1| < 2 < 2.
This, however, is a contradiction. Thus our assumption that the sequence converges is
false; i.e, the sequence diverges.
64

Example 5 Prove that
lim {
√
n + 1 −
√
n} = 0
n → ∞
.
Solution
First we note that
|
√
n + 1 −
√
n| = (
√
n+1−
√
n)
1
(
√
n+1+
√
n)
(
√
n+1+
√
n)
= 1√
n+1+
√
n
< 1
2
√
n
. Given > 0, we want to
choose n0 such that 1
2
√
n
< for all n ≥ n0. This is easily verified to be the case if
n0 ∈ N is chosen so that n0 ≥ 1
4 2 . With this choice of n0, we now have
|
√
n + 1 −
√
n| < for all n ≥ n0 ♣
Example 6 Show from first principles that the sequence
{xn} = {1 + (−1)n 1
n2 } for all n ∈ N converges to 1 in R.
Solution
Let > 0 be given such that |xn − 1| < . Then we have that
|1 + (−1)n 1
n2 − 1| <
That is |(−1)n 1
n2 | <
That is |(−1)n
|| 1
n2 | <
That is 1
n2 < . That is n2
> 1
i.e. n > 1√ .
In this case if we take n0 = N( ) = [ 1√ ] + 1, where [x] denotes the largest integer less or
equal to x, then we have found the natural number depending on such that
|xn − x| < , ∀n ≥ n0( ).
Hence xn −→ 1. ♣
Remark
Note that in determining n0( ) we have to add 1 because [ 1√ ] could be 0.
Definition 5.3 A sequence xn in R is said to be bounded if there exists a positive con-
stant M such that |xn| ≤ M for all n ∈ N.
65

This deﬁnition is equivalent to saying that the range of {xn : n ∈ N} of the sequence
{xn} is a bounded subset of R or if the terms of {xn} are trapped between two given
real numbers.
Example 1 Let {xn} = {1
n
}. Then the range is given by
Range xn = {1, 1
2
, 1
3
, ...}
Clearly, 0 < xn < 1, ∀ n ∈ N.
Hence {xn} = {1
n
} ∀ n ∈ N is bounded.
Example 2
Deﬁne the terms of a sequence as:
xn =
1 if n is odd
0 if n is even
Then the range of xn = {0, 1}. Hence {xn} is bounded.
Example 3 Let xn = n2
for all n ∈ N. Then range xn = {12
, 22
, ...}. Clearly, {xn} is
not bounded. Thus it is unbounded.
Theorem 5.1 Let {xn} be a sequence of real numbers. If {xn} converges then its limit
is unique.
Proof
We prove by contradiction. We assume that the sequence converges but that its limit
is not unique. So suppose the sequence {xn} converges to two distinct points x, y ∈ R,
i.e. xn −→ x and xn −→ y, and x = y.
Thus, using the criterion for convergence we have that for each > 0 ∃ N1( ), N2( ) ∈ N
such that
|xn − x| < 2
∀ n ≥ N1( ).
Also
|xn − y| < 2
∀ n ≥ N2( ).
Let N = max{N1, N2}. Then
66

|xn − x| < 2
∀ n ≥ N and
|xn − y| < 2
∀ n ≥ N.
By the triangular inequality we have that
|x − y| = |x − xn + xn − y| ≤ |x − xn| + |xn − y| < 2
+ 2
= ∀ n ≥ N.
That is |x − y| < and > 0 is arbitrary. Thus x = y and hence limit of {xn} is unique
(if it exists).
5.2.2 Bounded Sequences
Theorem 5.2 Let {xn} be a convergent sequence of real numbers. Then {xn} is bounded.
Proof
Let xn −→ x. Then x ∈ R. Therefore, for each > 0, ∃ n0 ∈ N such that
|xn − x| < ∀ n ≥ n0. Since > 0 is arbitrary, WLOG(Without loss of general-
ity) we take = 1. Then we have, for this :
|xn − x| < 1 ∀n ≥ n0
Let r = min{|x1 − x|, |x2 − x|, ..., |xn0−1 − x|, 1}
Then it follows that:
|xn − x| < r ∀n ∈ N, where r > 0. But we have that
||xn| − |x|| < |xn − x| < r
i.e. ||xn| − |x|| < r, ∀n ∈ N.
In particular, |xn| − |x| < r ∀n ∈ N.
That is |xn| < r + |x|, ∀n ∈ N.
Let M = r + |x| > 0. Then |xn| ≤ M ∀n ∈ N. Hence {xn} is bounded. ♣
Remark We note that the theorem above asserts that:
convergence of {xn} =⇒ boundedness of {xn}.
But the converse is not true in general as the following example shows:
67

Example 1
Let {xn} be a sequence of real numbers defined by
xn =
1 if n is odd
0 if n is even
That is {xn} = {1, 0, 1, 0, ...}
Then {xn} is bounded. But xn is not convergent since it oscillates between 0 and 1.
Example 2 The sequence {xn} = {1 − (−1)n
}∞
n=1 is bounded, but the sequence does
not converge. The sequence is bdd since |xn| = |1 − (−1)n
| ≤ 2 ∀ n ∈ N.
Example 3 The sequence {n(−1)n
} is not bdd in R, and thus cannot converge.
5.3 Subsequences and the Bolzano-Weierstrass Theorem
We prove that every bounded sequence of real numbers has a convergent subsequence.
This is the sequential version of the Bolzano-Weierstrass Theorem.
Definition 5.4 Given a sequence {xn} in R, consider a sequence {nk}∞
k=1 of positive in-
tegers such that n1 < n2 < n3 < ... Then the sequence {xnk
}∞
k=1 is called a subsequence
of the sequence {xn}.
If the sequence {xnk
} converges, its limit is called a subsequential limit of the sequence
{xn}. Specifically, a point x ∈ R is a subsequential limit of the sequence {xn} if there
exists a subsequence {xnk
} of {xn} that converges to x. Also, we say that ∞ is a subse-
quential limit of {xn} if there exists a subsequence {xnk
} so that xnk
−→ ∞ as k −→ ∞.
Similarly for −∞.
Examples
1. Let {xn}∞
n=1 = {x1, x2, ...} be any sequence of real numbers. If we extract the terms
whose subscripts are even then we have a subsequence:
{xnk
}∞
k=1 = {xn1 , xn2 , ...}, where xn1 = x2, xn2 = x4, xn3 = x6, ...
Therefore {xnk
} = {x2, x4, x6, ...} is a subsequence of {xn}.
2. Consider the sequence {xn} = {(1 − (−1)n
)}.
If n is even, then xn = 0, and if n is odd, then xn = 2. Thus 0 and 2 are subsequential
68

limits of the given sequence.
Exercise Prove that 0 and 2 are the only two subsequential limits of the above sequence.
3. Consider the sequence {xn} = {(−1)n
+ 1
n
}.
Both 1 and −1 are subsequential limits. If n is even, i.e. n = 2k, then xn = x2k = 1+ 1
2k
,
which converges to 1. On the other hand, if n is odd, i.e. n = 2k + 1, then xn = x2k+1 =
−1 + 1
2k+1
,
which converges to −1.
This shows that −1 and 1 are subsequential limits.
4. Let {xn} = {1
n
} ∀ n ∈ N. Then {x2n} is a subsequence given by
{x2n} = {1
2
, 1
4
, 1
6
, ...}
We also have that {x2n+1} is a subsequence of {xn} given by {x2n+1} = {1, 1
3
, 1
5
, ...}.
Remarks
1. Note that if a subsequence {xnk
}∞
k=1 converges to x, then x is called a subsequential
limit.
2. Note that a sequence is a subsequence of itself.
3. Note that a sequence can have subsequential limit without being convergent:
Example
Let xn =
1 if n is odd
−1 if n is even
That is {xn} = {1, −1, 1, −1, ...}.
Then {xn} is not convergent since it oscillates between −1 and 1. However, the subse-
quence {x2n+1} = {1, 1, 1, ...} converges to 1. Thus 1 is a subsequential limit of {xn}.
Hence a subsequence can be convergent without the sequence being convergent.
Remark We state and prove a result that links the concept of a subsequence to the
convergence of the sequence.
Theorem 5.3 Let {xn} be a sequence of real umbers. Then {xn} converges to x if and
only if every subsequence of {xn} also converges to x.
Proof
(⇒) Firstly assume that xn −→ x. Let {xnk
}∞
k=1 be any subsequence of {xn}. As
xn −→ x we have:
69

For each > 0 ∃ n0( ) ∈ N such that
|xn − x| < ∀ n ≥ n0( ).
Take nk > n0( ). Then clearly
|xnk − x| < ∀ nk > n0( ).
Thus xnk −→ x.
Since xnk
was any subsequence, it follows that xn −→ x =⇒ every subsequence xnk of
xn converges to x.
(⇐) Conversely, assume that every subsequence of {xn} converges to x. Then xn −→ x
since xn is a subsequence of itself. ♣
Examples
1. Consider the sequence
xn =
1 if n is odd
0 if n is even
That is {xn} = {1, 0, 1, 0, ...}. Then the subsequence {x2n} = {0, 0, ...} converges to 0
but the subsequence {x2n+1} = {1, 1, 1, ...} converges to 1. Now 0 = 1. Hence {xn}
diverges in R since its subsequences do not converge to the same limit.
2. Consider the sequence {xn} = {1
n
}, for all n ∈ N. Then xn −→ 0. The subse-
quence of xn given by {x2n} = {1
2
, 1
4
, 1
6
, ...} also converges to 0. Also the subsequence
{x2n+1} = {1, 1
3
, 1
5
, ...} converges to 0.
That is every subsequence of this sequence converges to 0 since {xn} (which is a subse-
quence of itself) converges to 0.
Theorem 5.4 (Bolzano-Weierstrass) Every bounded infinite subset of R has a limit
point.
Remark
The conclusion of the Bolzano-Weierstrass theorem may fail if either hypothesis is re-
moved. For example, a finite set has no limit point. On the other hand, the st N of
natural numbers is an infinite unbounded subset of R with no limit points.
The following corollary is often called the sequential version of the Bolzano-Weierstrass
Theorem.
70

Corollary 5.5 (Bolzano-Weierstrass) Every bounded sequence in R has a conver-
gent subsequence.
5.4 Monotonic Sequences
Deﬁnition 5.5 Let {xn} be a sequence of real numbers. Then we say that:
(i). {xn} is monotonic increasing (in symbols xn ) if xn ≤ xn+1 ∀ n ∈ N.
(ii). {xn} is monotonic decreasing (in symbols xn ) if xn ≥ xn+1 ∀ n ∈ N.
We say that {xn} is monotonic if it is monotonic increasing or monotonic decreasing.
Examples
1. The sequence {xn} = {n2
} ∀ n ∈ N = {1, 4, 9, ...} is monotonic increasing.
2. The sequence {xn} = { 1
n2 } ∀ n ∈ N = {1, 1
4
, 1
9
, ...} is monotonic decreasing.
Remark We know that convergence of {xn} =⇒ boundedness of {xn} but the converse
is not true in general. But we now show that these two concepts are equivalent in the
case of monotonic sequences. That is
Convergence ⇐⇒ Boundedness if {xn} is monotonic.
Theorem 5.6 Let {xn} be a monotonic sequence of real numbers. Then {xn} is con-
vergent if and only if it is bounded.
Proof
We only prove the case when {xn} is monotonic increasing. For the other case, the proof
is similar. Let E = range xn. Since {xn} is bounded it follows that E is also bounded.
Thus E is a set of real numbers bounded above. Let x = lub E. Then xn ≤ x ∀ n ∈ N
and for each > 0, ∃ n0( ) ∈ N such that x − < xn0 ≤ x for otherwise x − would be
an upper bound.
By monotonicity of {xn} it follows that
x − < xn < x ∀ n ∈ N.
71

Therefore |xn − x| < ∀ n ∈ N.
Thus xn −→ x and hence boundedness =⇒ Convergence.
Remark
1. Note that if xn −→ ∞ or xn −→ −∞ as n −→ ∞, we say that {xn} diverges in R
because −∞ and ∞ are not real numbers.
2. If xn and not bounded above then xn −→ ∞ and hence {xn} diverges. Also if
xn and is not bounded below then xn −→ −∞ and hence diverges.
5.5 Limit Superior and Limit Inferior of a sequence
These two limit operations are very important because unlike the limit of a sequence,
the lim sup and lim inf of a sequence always exist. They come in handy in the study of
series of real numbers and power series.
Let {xn} be a sequence in R. For each k ∈ N, we deﬁne ak and bk as follows:
ak = inf{xn : n ≥ k},
bk = sup{xn : n ≥ k}.
From the deﬁnition, ak ≤ bk for all k.
Furthermore, sequences {ak} and {bk} satisfy the following:
ak ≤ ak+1 and bk ≥ bk+1 for all k.
The sequence {ak} is nondecreasing.
As a consequence the sequence {ak} is monotone increasing and the sequence {bk} is
monotone decreasing.
These sequences always have limits in R ∪ {−∞, ∞}.
72

Definition 5.6 Let {xn} be a sequence in R. The limit superior/supremum of {xn},
denoted
lim xn
n → ∞
or lim supxn, is defined as :
lim xn
n → ∞
=
lim bk
k → ∞
=
inf sup {xn : n ≥ k}
k ∈ N
.
The limit inferior/infimum of {xn}, denoted
lim xn
n → ∞
or lim infxn, is de-
fined as:
lim xn
n → ∞
=
lim ak
k → ∞
=
sup inf {xn : n ≥ k}
k ∈ N
.
Example
Let {xn} = {1 + (−1)n
}∞
n=1. Let xn = 1 + (−1)n
. Then xn = 2 if n is even, 0 otherwise.
Thus ak = 0 for all k and bk = 2 for all k.
Therefore
lim xn
n → ∞
= 2 and
lim xn
n → ∞
= 0.
Theorem 5.7 Let {xn} be a sequence of real numbers and let L be the set of all subse-
quential limits of {xn}.
That is L = {x : x is a limit of some subsequence {xnk} of {xn} }.
Then
lim xn
n → ∞
= sup L
lim xn
n → ∞
= inf L
Also limxn and limxn belong to L.
Remark Note that the theorem above asserts that for a given sequence {xn} there are
two subsequences such that one converges to limxn and another one to limxn.
73

Theorem 5.8 Let {xn} be a sequence in R. Then xn −→ x if and only if
limxn = limxn = x.
Proof
(⇒) We use the result xn −→ x iﬀ every subsequence of {xn} converges to x. Now,
let xn −→ x. Then L = {x}, where L is the set of all subsequential limits. Thus,
sup L = x = inf L.
That is limxn = limxn = x.
(⇐) Conversely, let limxn = limxn = x. Then sup L = inf L = x. Therefore L = {0}.
That is every subsequence of {xn} converges to x. Hence in particular, xn −→ x. ♣
Remark
Note that in the result above if limxn = limxn = +∞, then L = {+∞}.
Thus xn −→ +∞ and hence {xn} diverges. So the validity of the above result demands
that limxn and limxn should be real numbers.
Example Consider the sequence
{xn} = {1
n
} ∀ n ∈ N = {1, 1
2
, 1
3
, ...}.
Then
x1 = sup{1, 1
2
, 1
3
, ...} = 1
x2 = sup{1
2
, 1
3
, ...} = 1
2
x3 = sup{1
3
, 1
4
, ...} = 1
3
Thus
inf xk = 0
k ∈ N
=⇒ lim sup xn = 0.
We also have that :
x1 = inf{1, 1
2
, 1
3
, ...} = 0
74

SMA206_NOTES

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to SMA206_NOTES

Similar to SMA206_NOTES (20)

SMA206_NOTES