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Real Analysis
(For Students at AIMS)
Introduction: The subject real analysis is one of the fundamental and core 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. Currently a thorough
understanding of the basic concepts in real analysis has become essential for the study of
advanced topics science and social sciences.
We will take up the following topics:
I. The Real Number System.
1.1. Sets and operations on sets
1.2 Functions
1.4 Mathematical Induction and the Well -Ordering Principle.
1.5 The Least Upper Bound Property and some of its consequence.
1.6 Countable and Uncountable sets.
2. Sequence of real numbers
2.1 Convergent sequences
2.2 Limit Theorems
2.3 Monotone Sequences.
2.4. Limit Superior and inferior of a Sequence
2.6 Cauchy sequences
3. Structure of point set in the set of real numbers
3.1Open and Closed sets
3.2Compact sets
.
4. Continuous Functions
4.1 Basic properties of continuous functions
4.2 Uniform Continuity
5. Sequence of Functions
5.1 Pointwise and uniform convergence of functions.
5.2 Uniform convergence and Continuity
6. The Riemann Integral
6.1 Definition of the Riemann Integral.
6.2 Set of measure zero
6.3.A necessary and sufficient condition for Riemann Integrability.
1
Notations used in this note
I = the set of positive integers or natural numbers = { 1,2,3, ... }
Z = the set of all integers = { . . . -3, -2 , -1 , 0 , 1 , 2, 3 ...}
Q = the set of rational numbers { p/q : p ∈ Z and q ∈ I }
R = the set of real numbers.
1. The Real Number System.
1.1. Sets and operations on sets
1.2 Functions
1.4 Mathematical Induction and the Well -Ordering Principle.
1.5 The Least Upper Bound Property and some of its consequence.
1.6 Countable and Uncountable sets.
1.1. Sets and Operations on Sets.
Sets are constantly encountered in mathematics. A set is conceived simply as a collection of
definable objects. The objects in a set are called elements. The notation x ∈ A means that x is an
element of the set A ; the notation x∉A means x is not an element of A. The set containing no
element is called the empty set or null set and will be denoted by φ .
A set can be described by listing its elements, usually within braces { }.For example
A = {-1 ,2 , 5, 4 }
describes the set consisting of the elements -1, 2, 5, and 5. More generally, a set A may be defined
as the collection of all elements x in some larger collection satisfying a given property. Thus the
notation
A = {x : p(x) }
defines A to be the set of all objects x having the property P(x). This is usually read as " A equals
the set of all elements x such that P(x)." For example
A = {x : x is a real number and 1 < x < 5}
is the set of all real numbers that lie between 1 and 5. For this example, 3.73 ∈ A , 6 ∉ A We will
also use the notation A = {x ∈ X : P(x) } to indicate that only those x that are elements of X are
being considered.
We say a set A is subset of a set B ,denoted A ⊆ B ,if every element of A is an element of B. The set
A is a proper subset of B, denoted A ⊂ B if A is a subset of B ,but A ≠ B. By definition
the empty set φ is a subset of every set
2
Two sets A and B .are equal denoted A = B if A ⊆ B and B ⊆ A.
Operation on sets:
The basic set operations are the union ,intersection and complement.
Definition:. If A and B are sets , the union of A and B ,written A ∪ B is the set of elements that
are in either A or B. Symbolically
A ∪ B = {x : x ∈ A or x ∈ B }
The intersection of A and B denoted, A ∩ B, is the set of elements that belong to both A and B ; i.e
A ∩ B = { x : x ∈ A and x ∈ B }
Two sets A and B are disjoint if A ∩ B = φ .
The relative complement of A in B ,denoted B \ A ( or B – A )is the set of elements in B which are
not in A .Thus
B \A = { x : x ∈ B and x ∉ A }.
If A is a subset of a fixed set X ( called the universal set) ,then X\A is referred to as the
complement of A and is denoted by Ac .
We state some of the basic properties of set operations:
1.A ∩ B = B ∩ A ; A ∪ B = B ∪ A
( Commutative property)
2.A ∩ (B ∩ C) = (A ∩ B) ∩ C ; A ∪ (B ∪ C) = (A ∪ B) ∪ C
(Associative property)
3. A ∩ (B ∪ C ) = ( A ∩ B) ∪ (A ∩ C) (Distributive property of intersection over union)
4. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ ) (Distributive property of union over intersection)
5.(A ∪ B)c = Ac ∩ Bc ,
( A ∩ B)c = Ac ∪ Bc ( De Morgan's laws)
6 A\ (B ∪ C)
= (A\B) ∩ (A\C)
= (A\B ) ∪ (A\C) ∩
7 A\ (B ∩ C)
We provide the proof of (3) and leave the remaining as exercises.
Suppose x ∈ A ∩ (B ∪ C) .Then x ∈ A and x ∈ B ∪ C.Since x B ∪ C., x ∈ B or x ∈ C. If x ∈ B
then x ∈ A ∩ B and hence x ∈ (A ∩ B) ∪ (A ∩ C) .Similarly ,if x ∈ C. then x ∈ A ∩ C and again
x ∈ A ∩ B) ∪ (A ∩ C) . This proves that
A ∩ (B ∪ C ) ⊆ (A ∩ B) ∪ (A ∩ C) .
To complete the proof we need to show the reverse inclusion, i.e,
( A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C ) ,thereby proving equality.
3
Let x ∈ ( A ∩ B) ∪ (A ∩ C)..Then by definition x ∈ A ∩ B or x ∈ A ∩ C. If
x ∈ A ∩ B then x ∈ A and x ∈ B. Since x ∈ B , x ∈ B ∪ C. Thus x ∈ A ∩ (B ∪ C).
Similarly, if x ∈ A ∩ C, then x ∈ A ∩ (B ∪ C).
If A is any set , the set of all subsets of A is denoted by P(A) and is referred to as the power set of A
Exercises 1.1
1.Let A be the set of all letters in the word " trivial,'' A = {a, i, l, r ,t, v}. Let B be the set of letters in
the word '' difficult'' Find A ∪ B, A ∩ B , A\B , B\A.
2. Prove (A ∪ B) ─( A ∩ B) = (A ─ B) ∪ (B─A)
3.True or false ( that is ,prove true for all sets,B,C ,or give an example to show false)
a) (A ∪ B) ─C = A ∪ (B ─C).
b) (A ∪ B) ─A = B
1.2
c)( A ∩ B) ∪ ( B ∩ C) ∪ (A ∩ C) = A ∩ B ∩ C..
Functions
Definition .If A and B are sets then the Cartesian product of A and B ,denoted by A X B is the set
of all order pairs (a, b) where a ∈ A and b ∈ B.That is,
A X B = { (a ,b) : a ∈ A and b ∈ B }.
Example . The Cartesian product of the set of real numbers with it self gives the set of all order
pairs of real numbers. We call this set the plane.
Definition. Let A and B be any two sets. A function f from A into B is a subset of AXB with the
property that each x ∈ A is the firs component of precisely one order pair (x,y) in f.; that is ,for
every x in A there is y in B such that (x,y)is in f, and if (x,y ) and (x,z ) are elements of f , then
y = z . The set A is called the domain of f, denoted Domf. The range of f, denoted Range f, is the
set
Range f = { y ∈ B : (x,y) ∈ f for some x ∈ A }.
If the rang f = B ,then the function f is said to be onto or sujective.
If f is a function from A to B and (x,y) ∈ f ,then the element y is called the value of the function f
at x or the image of x under f.and we write
y = f(x)
If f is a function from A to B we write
f : A→ B .
If B = R, the set of real numbers, then f is said to be a real –valued function.
4
Definition. Let f : A → B .
a) if E ⊆ A ,then f(E) , the image of E under f ,is defined by
f(E) = { f(x) : x ∈ E}.
b) If H ⊆ B , the inverse image of H, denoted f − 1(H), is defined by
f − 1(H) = { x ∈ A : f(x) ∈ H}.
If H = {y} ,we will write f − 1(y) instead of f − 1({y}). Thus for y ∈ B,
f − 1(y) = { x ∈ A : f(x) = y }.
Examples
a) Let A = { -3, -2, -1,0, 1} ,B = Z the set of integers, and f : A → Z the
function given by
f = { (-3, 2), (-2,-2) , (-1, 4), (0, -6), (1, 4) }.
Consider the subset E = {-1, 0,1} of A. Then
f(E) = {f(-1), f(0), f(1) } = {4, -6 }
.
If H = {0, 1,2, 3,4} , then
f − 1(H) = { x ∈ A: f(x) ∈ H} = {-3, -1,1}.
b) Consider the function g : Z → Z given by g(x) = x2 , and let E = {-1,-2,-3,. . .}.Then
g (E) = {(-n)2 : n ∈ N } = {1, 4, 9, . . .}. On the other hand,
g-1 (g(E)) = Z \ {0}. We see that E ⊂ g-1 (g(E)).
Theorem Let f be a function from A to B. If C and D are subsets of A ,then
a) f (C ∪ D ) = f(C ) ∪ f(D)
b) f(C ∩ D) ⊆ f(C ) ∩ f(D ).
Proof. To prove (a), let y ∈ f (C ∪ D).Then y = f(x) for some x in C ∪ D. Thus x ∈ C
or x ∈ D. Hence f(x) ∈ f(C) or f(x) ∈ f(D).Therefore y ∈ f(C) ∪ f(D).
Thus
f(C ∪ D) ⊆ f(C ) ∪ f(D)
Since Both C and D are subsets of C ∪ D , we have f(C) and f(D) are subsets of
f(C ∪ D), the reverse inclusion also holds, thereby proving equality..
Since f(C ∩ D) is a subset of both f(C) and f(D) , the relation stated in (b) holds.
To see that equality need not hold in (b) , consider the function g(x) = x2 ,Domg = Z.
If C = {-1, -2, -3,...} and D = {1,2,3,...}, then f(C) = f(D) = { 1,2, 3, . . .},but
C ∩D = φ .
Thus,
5
f (C ∩ D) = f( φ ) = φ ≠ f(C) ∩ f(D) = {1 , 2, 3, . . . }.
Theorem:. Let f be a function from A to B If C and D are subsets of B, then
(a) f −1 (C ∪ D) = f −1 (C) ∪ f −1 (D)
(b) f −1 (C ∩ D) = f −1 (C) ∩ f −1 (D)
(c) f −1 (B\C) = A\f −1 (C)
Proof: Left for exercises.
Inverse Functions.
Definition : A function f from A to B is said to be one- to –one or injective if whenever
x 1 ≠ x 2 , then
f(x 1 ) ≠ f(x 2 ).
Alternately ,a function f is one – to one if whenever (x 1 , y ) and ( x 2 , y) are elements of f
then x 1 = x 2 .From the definition it follows that f i s one – to –one if and only if f −1 (y)
consists of atmost one element of A for every y in B. If f is onto b ,then f −1 (y) ≠ φ . Thus f is
one- to –one and onto B, then f −1 (y) consists of exactly one element x in A and
g = { (y, x) ∈ B XA : f (x) = y }
defines a function from B to A. We denote g by f −1 .
Definition :If f is one- to –one function from A onto B, let
f −1 = { (y, x) ∈ B XA : f (x) = y }.
The function f −1 from B onto A is called the inverse function of f. Furthermore for each y in B,
x = f −1 ( y) iff f(x) = y.
Note that if f is one – to –one function from A to B, then f −1 is a function from Range f onto A
Example Let h(x) = 2x + 3 , Dom h = R. The function h is clearly one-to –one and onto R Thus
if,
1
x = h −1 (y) we have y = h(x) = 2x + 3 . Consequently h −1 (y) = x = ( y – 3).
2
Composition of functions
Definition . If f is a function f from A to B and g is a function from B to C, then the function
6
gof : A → C ,defined by (gof) (x) = g(f(x)) is called the composition of g with f. That is ,
gof = { (x, z) ∈ AX C : z = g(f(x)) }.
Note .If f and g are function ,the Dom gof = { x : x ∈ Dom f and f(x) ∈ Dom g }.
Example . If f(x) == 1 + x and g(x) = x2 ,then
( gof) (x) = g(f(x)) = g( 1 + x ) = ( 1 + x )2 = 1 + x
Since the Dom f = { x ∈ R : x ≥ -1 }and Dom g = R, we conclude that Dom gof = Dom f.
even though the equation (gof)(x) = 1 + x is defined for real numbers.
We also have (fog) (x) = f(g(x)) = f (x2 ) = 1 + x 2 .Since for every real number x,x2 is in the
domain of f , the Dom fog = R.
Suppose f is a one-to –one function from A to B .Then f − 1 is a function from the Range f on to A
Then
( f o f −1 )(y) = y for every y in the Range f and (f − 1 of)(x) = x for every x in A.
Real –valued functions
If f : A → R we call f a real – valued function.That is , a function f is called a real – valued
function if its rang is a subset of R.
We define the sum,difference , product , and quotient of real – valued functions.
Definition. If f : A → R and g : A → R we define f +g ,f - g , fg , f/g and cf ( c a real
number) to be the functions given by
a) (f +g)(x) = f(x) + g(x)
( x ∈ A).
b) (f – g) (x) = f (x) – g(x)
( x ∈ A),
c ) (fg)(x) = f(x).g(x)
( x ∈ A),.
f
f ( x)
d )  ( x) =
,for all x in A for which g(x) ≠ 0,
g ( x)
g
e) (cf)(x) = c f(x).
For real numbers a and b ,define max (a ,b ) to be the maximum of a and b ; and min(a ,b) the
minimum of a and b and if a =b we set max (a , b) = min(a, b) = a = b. This allows us to define
max(f , g) and min(f , g) for real – valued functions.
7
Definition. If f : A → R and g : A → R, the max (f ,g) is the function defined by
max( f ,g ) (x) = max (f(x) , g(x))
( x ∈ A).,
and min(f ,g) is the function defined by
min(f , g)(x) = min (f(x) , g(x) )
Example . Let f(x) = x2
( x ∈ A).,
, and g(x) = x3, (- ∞ < x < ∞ ) . Then
 x 2 , (−∞ < x ≤ 1)

max(f , g) (x) =  x 3 , (1 ≤ x < ∞
min (f , g )(x) =
Definition . If
 x 2 , 1 ≤ x < ∞
 3
 x , − ∞ < x ≤ 1
f : A → R, then f is the function defined by
f (x) = f (x)
( x ∈ A)
For are real numbers a and b, we have
max(a ,b) = (│a-b│+ a + b) / 2
min(a , b) = ( - │a-b│+ a + b) /2
From this it follows
max (f, g) = (│ f -g│+ f +g) / 2
and min(f , g) = (- │f- g│+ f + g) /2.
Definition . Let X be set and A ⊆ X , then χ A ,called the characteristic function of A is defined
as
χ A (x) = 1 ( x ∈ A)
χ A (x) = 0 ( x ∈ A ' ).
We list some properties of the characteristic functions where A and B are subsets of X
•
For sets A and B, A =B iff χ A = χ B .
8
•
χ A∪ B = max( χ A , χ B )
•
•
•
χ A∩ B = min( χ A , χ B ) = χ A χ B
χ A−B = χ A - χ B
provide A ⊆ B,
χ φ = 0 (the function identically zero)
Exercises.
1. Let A = {(x , y) ∈ RXR : x2 + y2 = 1 } .
a ) Is A a function ? Explain your answer.
b). Let B = {( x, y) ∈ A: y ≥ 0 }.Is B a function? Explain your answer.
2 .Let f = {(x ,y): x ∈ R and y = x3 + 1}.
a). Let A = {x : -1 ≤ x ≤ 2 }.Find f(A) and f −1 (A).
b) Show that f is one –to –one from R onto R.
c) Find the inverse function f −1 .
3 .Let f : A → B and F ⊂ A.
a) Prove that f(A)\ f(F) ⊆ f(A\F).
b) Give an example for which f(A) \f(F) ≠ f(A\F).
4. Let f : A → B and g : B → A be functions satisfying (gof)(x) = x for all x ∈ A. Show that
f is one-to –one function. Must f be onto?
5.If f : A → B and g : B → C are on- to-one functions ,show that (gof) −1 = f −1 og −1
on Range (gof).
6.Let f(x) = 2x (- ∞ < x < ∞ ).Can you find functions g and h which satisfy the two equations
gof = 2gh and hof = h2 ─ g2 ?
7. Let f : X → Y , A ⊆ X , B ⊆ Y .Show that
a) f -1(f (A)) ⊇ A.
b) f(f -1(B)) ⊆ B.
c) Show by example equality does not hold in (a).
d) Sate a sufficient and necessary condition for which equality holds in (a)
.3. Mathematical induction and the Well-ordering principle.
Consider the following statement. For each natural number n,
n(n + 1)
2
How can we prove that the statement is true for all positive integers n?
Mathematical Induction provides a very useful tool in establishing that such an identity is valid for
all positive integers n.
1 + 2 +3 + . . . + n =
9
1. 3.1 Theorem (Principle of Mathematical Induction).For each natural number n , let P(n) be a
statement. If
a) P(1 ) is true , and
b) P(k + 1) is true whenever P(k) is true,
then P(n ) is tue for all natural number n.
The proof of this theorem depends on the Well-ordering Principle, which is usually taken as a
postulate or axiom for the positive integers.
1.3.2 Well- Ordering Principle. Every non-empty subset of the set of positive integers has a
smallest element .That is, if A is a non- empty subset of N then there exist n in A such that n ≤ k
for all k in A.
Proof : Let S = { n ∈N: P(1) is true}.S ≠ φ , as 1∈S by (a). Claim S = N.
Suppose S ≠ N . Then N \S ≠ φ . By the well-ordering principle N\S has a smallest element say
n. and n ≠ 1. Thus n – 1 is a positive integer and not in N\S. Hence n -1∈ S and consequently
P(n-1) is true. By (b) P(n-1 +1 ) is true , i.e, P(n ) is true .Hence n is S ,which is a contradiction.
Therefore S = I
Examples. Use mathematical induction to prove that
a). 1 + 2 +3 + . . . + n =
is valid for all n ∈ I
n(n + 1)
2
Proof. a)Let P(n) , n ∈ I , be the statement
1 + 2 +3 + . . . + n =
n(n + 1)
2
1(1 + 1)
, which is true , i.e., P(1) is true.
2
b)Assume P(k) is true for k ≥ 1, i.e.,
when n = 1 we get 1 =
k (k + 1)
,
2
We need to show P(k + 1) is true; that is ,
1 + 2 +3 + . . . + k =
1 + 2 +3 + . . . + ( k + 1) =
(k + 1)((k + 1) + 1)
.
2
Now
10
1 + 2 +3 + . . . + ( k + 1) = 1 + 2 +3 + . . . + k + ( k + 1)
k (k + 1)
=
+ (k +1)
(
By induction hypothesis)
2
k (k + 1) + 2(k + 1)
2
(k + 1)(k + 2)
=
2
(k + 1)((k + 1) + 1)
=
2
.Thus the statement is true for k +1,i.e., P(k+1) is true and hence by the principle of mathematical
induction for all n ∈ N.
=
b). If h > -1 , then
(1 + h) n ≥ 1 + nh for all n∈ N.
Proof. when n =1, we have (1 + h)1 = 1 + 1.h .Thus since equality holds, the inequality is valid.
Assume the inequality is true when n = k ≥ 1.We need to show the inequality is valid for k +1
;i.e., (1 + h)k +1 ≥ 1 + (k +1)h
Now (1 + h)k +1 = (1 +h)k (1 +h) ≥ (1 + k h) (1 +h) (by assumption and the fact that (1 + h ) >
0
= 1 + h + kh +kh2 ≥ 1 +(k +1 )h ( since kh2 ≥ 0 )
Thus the inequality holds for n = k + 1, and by th alle mathematical induction for all n ∈ I.
Although the statement of Theorem 1.3.1 starts with n = 1,the result is still true if we start
with any integer no .In this case the principle of mathematical induction is stated as follows:
If for each n ∈ Z , n ≥ no , P(n) is a statement satisfying
a) P( no ) is true, and
b ) (k + 1 ) is true whenever P(k) is true, k ≥ no .
Then P(n ) is true for all n ∈ I , n ≥ no.
There is a second version of the principle of mathematical induction that is also quite useful.
Theorem. ( Second Principle of mathematical Induction)
For each n ∈ N ,let P(n) be a statement about the positive integer n. If
11
a) P(1) is true ,and
b) for k > 1 ,P(k) is true whenever P(j) is true for all positive integers j < k,
then P(n) is true for all n ∈ I.
Proof. Exercise.
Exercises.1.3
1. use mathematical induction to prove the following statements
a) 1 + 3 + 5 + . . . (2n-1) = n2
n(n + 1)(2n + 1)
b) 12 + 22 + 32 + . . . n2 =
6
n
c) 2 > n for all n
Index families of sets.
In section 1.1 we defined the union and intersection of two sets. We now extend these definitions to
a large collection of sets.
Definition. If A is a set ,by a sequence in A we mean a function from I into A.
For each n in I,let xn = f(n). We denote the sequence in A by {xn} ∞ n =1 or simply { xn}.
Definition .Let A and X be non-empty sets. An indexed subset of X ( or family of subsets of X ) is
a function from A in to X ( or power set of X). The set A is called an index set .
If f: A → X ( or power set of X) ,the for each i in A ,we let f(i) = xi (or Ei ).As for sequences ,we
denote this function by { xi } i∈A .( or { Ei i∈A }. If A = I the notion of indexed set coincides with
the notion of a sequence. In this case { En } n∈N is called a sequence of subsets of X .We call
{Ei } i∈A an indexed family of subsets of X.
Examples: a) The sequence { Nn} ∞ n =1 , where Nn = { 1, 2,3, . . .n} is a sequence of subsets of I..
b)For each n in I ,set In = { x ∈ R : 0 < x <
1
}. Then { In } ∞n =1 . is a sequence of subset of R
n
c)For each x , 0 < x < 1 ,let
Ex = { r ∈ Q : 0 ≤ r < x}
Then { Ex } x∈( 0, 1) is an indexed family of subset of Q. In this example the index set
is the open interval (0 , 1).
Definition. Suppose {E α } α ∈A is an indexed family of subsets of X. The union of the family of
set {E α } α ∈A is defined to be
12
∪ Eα = { x ∈ X : x ∈ Eα for some α ∈ A}.
α ∈A
The intersection of the family {E α } α ∈A is defined as
∩ E α = { x ∈ X : x ∈ Eα for all α ∈ A}.
α ∈A
If A = I we use the notation
∞
∞
n =1
n =1
∪ En and ∩ En
Examples.
a) Let Nn = {1 ,2, 3, . . .n} for each n in N. We have
∞
∞
n =1
n =1
∪ En = I and ∩ En = {1}
b) For each n in I ,let In = { x ∈ R : 0 < x <
Claim .
1
}.
n
∞
∩ In = φ .
n =1
Suppose not ,then there exist x in R such that x ∈ In for all n . Since x > 0, there is
a positive integer n such that
1
< x , i.e . x ∉ In .
n
which is a contradiction.
For the union , since In ⊆ I1 for all n =1,2,3, ...
∞
∪ In = I1 = { x ∈ R : 0 < x < 1}.
n =1
Theorem (Distributive Laws): If E λ , λ ∈ A and E are subsets of X, then
a)
(
E ∩ ∪ Eλ
λ∈ A
)


b) E ∪  ∩ E λ  =
 λ∈A 
=
∪ ( E ∩ Eλ )
λ∈ A
∩ ( E ∪ Eλ )
λ∈ A
Proposition (De Morgan's Laws) : If {E λ } λ∈A is a family of subsets of X , then
13


a)  ∪ E λ 
λ




b)  ∩ E λ 
λ


c
c
∩ ( Eλ )
=
λ∈ A
c
∪ Eλ
=
c
λ
Theorem: Let f :X → Y .
a) If { E α } α ∈A is a family of subsets of X , then


f  ∪ Eα  = ∪ f ( Eα )
α
λ




f  Ι Eα  ⊆ ∩ f ( Eα )
α
 α 
b) If {B λ } is a family of subsets of Y, then
f
−1


 ∪ Bλ  = ∪ f
λ
λ




f −1  ∩ Bλ 
λ



=
−1
( Bλ )
∩f
λ
−1
(B λ )
Exercises .
∞
∞
n =1
n =1
1. Find ∪ An and ∩ An for each of the following sequence of sets {An}.
a). An = { x ∈ R : -n < x < n }
1
b) An = { x ∈ R : - < x < 1 }
n
1
1
c) An = { x ∈R : - < x < 1 + }
n
n
2.. For each x ∈(0 , 1),let Ex = { r ∈Q : 0 ≤ r < x }. Prove that
∩ Ex = { 0 } and
x∈( 0 ,1)
∪ E x = { r ∈Q : 0 ≤ r < 1 }.
x∈( 0 ,1)
14
1.4. The Least Upper Bound Property.
1.4.1. Upper bound of a set.
Definition. A subset E of R is bounded above if there exists a real number M such that x ≤ M for
every x ∈ E. Such a number M is called an upper bound of E.
The concepts bounded below and lower bounded are defined similarly. A set E is bounded if E is
bounded both above and below.
Examples. a) A = ( 1 , 4] be the half open interval. I is bounded below by any number r ≤ 1 and
bounded above by any real number s ≥ 4. one can guess that the number 1 is the largest of all the
lower bounds and 4 is the smallest of all the upper bounds of A
b I = { 1 , 2, 3 , . . . } .This set is bounded below by any number r ≤ 1.Our intuition
tells us that N is not bounded above.
Definition .Let E be a non- empty subset of real numbers.
a) A real number c is called the least upper bound or supremum of E if
i) c is an upper bound of E , and
ii) if m satisfies m < c ,then m is not an upper bound of E.
Condition (ii) is equivalent to c ≤ m for any upper bound m of E. If c is the least upper bound of
E , we write
c = supE
A real number d is called the greatest lower or infimum of E if
i) d is a lower bound of E , and
ii) k satisfies k > d , then k is not a lower bound of E.
Condition (ii) is equivalent to k ≤ d for any lower bound k of E.
We write
d = inf E
Remarks. i) If c = sup E then for any ε > 0 there exists x ∈ E such that
x>c- ε .
ii)If d = inf E, then for any ε > 0 there exists y ∈ E such that x < d - ε .
iiii ) sup ( E) = -inf (-E).
iv) If a ∈ E is an upper bound of E , then supE = a.
We state the least upper bound property, which is one of the fundamental properties of the real
numbers and which forms the foundation of many the results in analysis.
The Least Upper Bound Property of R.
15
Every non- empty subset of R that is bounded above has a supremum in R.
Theorem. Every non- empty sunset of R that is bounded above below has an infimumin R.
Definition. If E is a non –empty subset of R ,we set
supE = ∞ if E is not bounded above ,and
inf E = - ∞ if E is not bounded below.
For the empty set φ , every real number is an upper bound ,For this reason we set
sup φ = - ∞ . Similarly ,inf φ = ∞
Also for the symbols - ∞ and ∞ we adopt the convention that - ∞ < x < ∞ for every real
number x.
We now state some of the basic consequences of the Least Upper bound Property.
Theorem . (Archimedean property)
If x and and y are real numbers and x > 0 , then there exists a positive integer n such that
nx > y.
Proof. If y < 0 ,then the result is true for all n . Thus assume y > 0. Suppose the assertion is
not true ,i.e, for all n ,nx ≤ y.
Let
S = { nx : n ∈ IN}.
Then y is an upper bound of S. Since S is nonempty ,by the least upper bound property
S has a least upper bound in R. let sup S = t. As x is in S and x > 0 we have t – x < t.
.Therefore ,
t –x is not an upper bound of S and thus there is an element in S say mx such that
t – x < mx .
Which implies t < (m – 1) x. Contradicting the fact that t is an upper bound of S. therefore ,
there exist a positive integer n such that nx > y.
Special case of the above theorem is the case x = 1 ,which gives n > y.
Corollary .Given ε > 0 , there exists a positive integer n such that
1
< ε.
n
1
Proof. Take y = and x = 1 in the above theorem.
ε
Theorem . If x and y are real numbers and x < y , then there exists a rational number r such that
16
x < r < y.
Proof. Assume first x ≥ 0.Since y –x > 0 , there is a positive integer n such that
n(y –x ) > 1..
Consider the set A = { k ∈ I : k > nx } , which non- empty ,by the above theorem. Thus by the
well ordering principle A has a smallest element ,say m. Consequently
,
m -1 ≤ nx < m.
Therefore ,
nx < m ≤ nx + 1 < ny ,
dividing by n ,we get
m
< y.
n
Suppose x < 0. Then – x > 0 .and there is a positive integer n such that n > -x.
Thus , we have 0 < n +x < n + y. By the first case the a rational number t such that
x <
n+x <t <n+y,
therefore, x < t –n < y and r = t – n is a rational number.
Examples a ) Let E = (0, 1) , the half open interval. Prove that sup E = 1 and infE = 0.
Solution. Since x ≤ 1, 1 is an upper bound of E .Let t be any real number such that t < 1.
We need to show t is not an upper bound of E. If t ≥ 0 then there is a rational r such that
t < r < 1.
Since r is in (0, 1) ,t is not an upper bound of E. If t < 0 then, clearly t is not an upper bound of E.
Therefore, sup E = 1. Similarly inf E = 0.
b)E = { 1, ½ , 1/3 ,...} .Prove sup E =1 and inf E = 0.
Solution. Since 1/n ≤ 1 for all n, 1 is an upper bound .But 1 ∈ E ,and hence
supE = 1.
Let t be any real number such that t > 0.We need to show t is not a lower bound of E.
Since t > 0, there is a positive integer n such that 1/ n < t. As 1/n is in E t is not a lower
bound of E . Therefore, inf E = 0.
c) Prove that the set of natural numbers, I is not bounded above.
Solution. Suppose N is bounded above. the by the Least upper bound property , N has a
least upper bound in R, say t. Then for all n in I n ≤ t .Since for all n ; n + 1 is also inIN
, we have n + 1 ≤ t . Therefore, n ≤ t -1 for all n. Thus t -1 is an upper bound of i
smaller than t, contradicting the definition of t.
17
Exercises.
1.Find the supremum and infimum of the each of the following sets.
a) A = {1 , ½ , ¼ , 1/8, . . . }.
b) B = { cos n
π
: n = 1 , 2, 3 , . . .}
4
c) C = { ( 1 – ( -1)n )n : n = 1,2, 3, . . }
d) D = { x ∈ R : x2 < 4 }
2+n
: n = 1,2 , . . .}
e) E = {
n
2) Let A and B subset s of R Define A + B = { a + b : a ∈ A and b ∈ B },and
AB = { ab : a ∈ A and b ∈ B }.
a) If A and B are non empty and bounded above ,prove that
sup( A + B ) = supA + sup B
b) If A and B are nonempty subsets of the positive real numbers that are bounded
above, prove that sup(AB) = ( supA) ( supB).
c) Give an example of two non-empty sets A and B for which
sup(AB) ≠ (supA) (supB)
3).Let E be a nonempty subset of R that is bounded above , and let U be the set of all
upper bounds of E. Prove that supE = inf U.
4) If A and B are non-empty subset of R with A ⊂ B , prove that
inf B ≤ inf A ≤ supA ≤ supB.
5) Let f ,g be real valued functions defined on a non- empty set X satisfying Range f
and range g are bounded subsets of R. Prove each of the following :
a) sup{ f(x) + g(x) : x ∈X} ≤ sup{f(x) : x ∈ X } + sup{ g(x) : x ∈ X }
b)inf { f(x) : x ∈ X } + inf { g(x) : x ∈ X } ≤ inf{ f(x) + g(x) : x ∈ X }
Countable and uncountable sets.
Definition: Two sets A and B are said to be equivalent ,denoted A ~ B if there exists a one –to
–one function from A onto B.
The notion of equivalence of sets satisfies the following:
a) A ~ A ( reflexive )
b) If A ~ B , then B ~ A . ( symmetric)
c) If A ~ B and B ~ C ,then A ~ C. ( transitive).
Example . a) Any two set having five elements are equivalent.
b ) If E = { 2n : n ∈I } ,then E ~ I .The mapping f : I → E defined by
f(n ) = 2n is one- to –one and onto .
18
Definition. For each positive integer n, let Nn = { 1, 2, 3, . . . n}
Let A be any set ,we say :
a) A is finite if A ~ Nn for some n, or if A = φ .
b) A is infinite if A is not finite.
c) A is countable if it is finite or if A~ I.
d) A is uncountable if nor countable.
e) If A is countably infinite (denumerable) if A~ I.
Example .a).The set of integers Z is countable.
f: I → Z defined by
n
(n − 1)
( n odd)
, (n even) , and f(n) = −
f(n ) =
2
2
is one –to – one mapping of N onto Z
b)Let S = {12 , 22 , 32 , . . .} . The function g (n ) = n2 is one-to-one mapping of I onto S.
Thus S is countable.
If a set A is countably infinite then there exists a one – to -one function f from I onto A.
Thus
if we set
f(1 ) = a 1 , f (2) = a 2 ; . . . , f (n ) = a n , .
then A = { a 1 , a 2 , . . . a n , ...} = . { a n : n = 1,2, 3,. . .}, which is an enumeration
of A and write { a n }n =1
Similarly if A is non-empty finite set then there is a positive integer n such that
A = { a1 , a 2 , . . . a n }
Theorem:The image of a countable set is countable
Proof. Let A be countable and f . : A → B .We need to show f(A) ; the image of A is
countable. Since A is countable A = { a n }n =1 .Set f(an) = bn . ,n =1 ,2, 3, ....
Thus the range can be listed. as : b1 , b2 , b3 , Κ ( we remove those already listed).
Theorem;. Every subset of a countable set is countable.
Proof: Let E = {x n } be a countable set, and A a subset of E. If A is empty, by
definition A is countable. If A is not empty, choose x in A. Define a function
f : I → A by
 x n if x n ∈ A
f(n ) = 
 x if x n ∉ A
Then f is on to A and by the above proposition A is countable.
Corollary .I x I is countable.
Proof Let f : I x I → I be defined by f( n , m) = 2n .3m . f is injective , hence
equivalent to a subset of I. Thus I x I is countable
19
Theorem: If { En } ∞ n =1
is a sequence of countable sets and
∞
S = ∪ En .
n =1
Then S is countable
Proof . Since En is countable for each n in I, we can write
En = { xn,k : k = 1,2,. . .}.
Let h: I x I → S be defined by h (n, m ) = xn,m .
The function h is onto and hence S is countable.
Corollary: Q is countble.
Proof. For each n in I let
En = {
m
: m ∈ Z }.
n
Then En is countable, and since Q =
∞
∪ En ,by the above proposition Q is countable.
n =1
Theorem:. The interval ( 0 , 1) is uncountable.
1
: n = 2,3,. . .} is infinite subset of (0 ,1) ,the set
n
is not finite. Thus we enumerate or list its elements as { x 1 , x 2 , x 3 , . . . }.
Each x i has a non terminating decimal representation, i.e,
x i = .x i ,1 x i , 2 x i ,3 ...x i ,i ...
Proof .Suppose it is countable. Since {
where for each k , x i ,k ∈ {0, 1,2, . . .9}. We now define a new number
b = .b 1 b 2 b 3 ...
as follows : bi = 3 if xi,i ≤ 5 and bi = 4 if xi,i ≥ 6..Then b is in(0,1) and since
bi ≠ 0 or 9 ,b is not one of the real numbers with two decimal expansions. Also,
since for each i , bi ≠ xi,i , we have b ≠ xi for all i. Thus b ∉ (0, 1). This is a
contradiction.
Theorem:If A is the set of all sequences whose elements are 0 or 1 , then A is
uncountable.
Proof. Exercise.
Exercises.
1.Let O denote the set of positive odd integers ,prove that O ~ I.
2. a) if a , b are real numbers ,a < b , prove that (a ,b) ~ (0 , 1).
20
b) Prove that (0 , 1) ~ (0 , ∞ ).
3.Show that the function f : (0 , 1) → R given by f(x) =
x
is one- to –one and
1 − x2
onto R
4. If A and B are countable sets , prove that A x B is countable. Use induction to prove
that for any positive integer n , A1 x A2 x . . .x An is countable when ever each set Ai is
countable.
5.a) For each n ∈ I ,prove that the collection of all polynomials in x of degree less than or
equal to n with rational coefficients is countable.
b)Prove that the set of all polynomials in x with rational coefficients is countable.
6.Prove that the set of irrational real numbers is uncountable.
Chapter 2
Sequence of Real Numbers
Definition. A sequence in R is a function f : I → R. For each n ∈ I, an = f(n) is called
the nth term of the sequence f. We denote the sequence f by {an } ∞n =1 or simply {an }.
Definition. Let {an } be a sequence of real numbers and L a real number. We say the limit
of the sequence {an } is L if for every ε >0 there is a positive integer N =N( ε ) such
that
an − L < ε
for all n ≥ N
In this case we say the sequence converges to L and write
lim an = L., or an → L ( n → ∞ ) .
n →∞
We call L the limit of the sequence {an } .
We say the sequence {an } is convergent if there is a real number L such that
lim an = L,.
n →∞
Otherwise it is said to be divergent.
Examples. a) Let an =
to 0.
1
1
, ( n =1 , 2 , 3 , . . .). Prove that {
} converges
n +1
n +1
Solution. Let ε > 0 be given. We need to find a positive integer N such that
21
1
- 0│ < ε , for all n ≥ N.
n +1
│
By the corollar of Archimedean there exists a positive integer N such that
1
< ε .Thus for all n ≥ N , we have
N
1
1
1
< ε .
- 0│=
<
n +1
n +1
N
│
b) Let sn = n for each for each positive integer n. Prove that {sn } has no limit .
Solution : Suppose sn → L .Then for each ε > 0 there is a positive integer N
such that
│ sn - L│ < ε
for all n ≥ N.
In particular for ε = 1 we obtain
│ n - L│ < 1. for all n ≥ N. Hence n < L + 1 for all n ≥ N.
Thus n < L +1 for all positive integers , which contradicts the set of positive
integers is not bounded above.
2n
c) Let an =
1
2
, for n =1 ,2 3,. . .
n + 4n
Show that an → 2.
Solution. Let ε > 0 be given .we need to find a positive integer N such that
2n
│
for all n ≥ N.
- 2│ < ε
1
n + 4n 2
Now
│
2n
n + 4n
1
2
- 2│ =
8n
n + 4n
Thus chose N so that N so that
8
N
│
1
2
2n
n + 4n
1
2
1
2
1
2
≤
8
1
n2
< ε . It is easy to check that if n ≥ N ,
- 2│ < ε .
22
Subsequence .
Definition : A subsequence S of {n } ∞n =1 is a function from I into I such that
S( i ) < S( j ) if i < j (i ,j ∈ I).
If we set S ( i ) = ni we denote S by { ni } i∞=1 and we have n 1 < n 2 < n 3 < . . . < n k < . ..
Examples. a) Let S : I → I defined by S( i ) = 2i .Then {2i} is a subsequence of {n}.
{2i} = 2 ,4, 6, 8, . . .
c) T : I → I where T(i) = 2i – 1 is a subsequence of {n} ,i.e. {2i – 1 } i∞=1 is a
subsequence of { n }.
Definition . Let f = { xn } be sequence and g = { ni } i∞=1 a subsequence of I ; then the
composition fog is called a subsequence of {xn }.
Thus (fog)(i) = f(g(i)) = xn i ( i = 1, 2, ...). and fog = { xn i } i∞=1 .note that . n 1 < n 2 < n 3 <
. . . < n k < . . ..
Example.
a) Let f be the sequence : 1, 0,1 ,0, 1, 0, . . . and g ( i ) = ni = 2i – 1.Then
(fog)( i ) = f(g(i)) = f( ni ) = = xn i
xn 1
=
x1 = 1 , xn 2 = x3 = 1, . . .
fog = { xn i } = 1 , 1, 1, 1 , . . .
Theorem: (Uniqueness) If the sequence {an } has a limit then it is unique .i.e.,if
lim an = L
n →∞
and
lim an = M then L = M.
n →∞
Proof. Assume L ≠ M . Then L − M > 0 . Let ε =
1
L − M .By the hypothesis
2
lim an = L there exists N1 ∈ I such that
n →∞
an − L < ε
(n ≥ N1 )
Similarly, since lim an = M there exists N2 ∈ I such that
n →∞
an − M < ε
(n ≥ N2 )
Let N = max (N1 , N2 )..Then since N ≥ N1 and N ≥ N2 we have
L−M
=
(a n − M ) + ( L − a n )
≤ an − M + an − L < 2 ε = L − M ,
which is a contradiction. Hence L = M.
23
Theorem. If the sequence {an } is convergent to L ,then any subsequence of {an } converges to L.
Proof .exercise
This theorem is useful in proving that a given sequence is divergent.
Consider the sequence {an } where an = (-1)n .Let { a 2 n } = 1 ,1 ,1,1, ...
and {a 2 n −1 } = -1 , -1 ,-1, -1,....which are subsequences of { (-1)n } converging to 1 and -1
respectively , hence the sequence is divergent.
Divergent sequences.
Definition. Let {an } be a sequence of real numbers. We say that {an } approaches to infinity as
n tends → ∞ if for any real number M > 0 there exists a positive integer N such that
( n ≥ N ).
an > M
In this case we write an → ∞ as n → ∞ or lim an = ∞ .
n →∞
Definition .Let {an } be a sequence of real numbers .We say {an } approaches to minus infinity
written
lim an = - ∞
n →∞
If for each positive real number M there exists a positive integer N such that
an < - M
( n ≥ N).
1
, n =1, 2 ,3 . . .. Show that lim an = - ∞ .
n →∞
n
Solution.Let M > 0 be given .We need to find a positive integer N such that
Example. Let an = log
an < - M
( n ≥ N ).
Now
an
< -M
1
log < - M implies log n > M . Thus n > eM .Choose a positive integer N such that
n
1
N > eM .It is easy to see check that for all n ≥ N , log < -M.
n
Bounded sequences.
Definition. We say that the sequence {a n } is bounded above if the range of {a n }, i .e,
if { an : n =1, 2, 3, . . . } is bounded above.
Similarly, we say {an } is bounded below if{ an : n =1, 2, 3, . . . } is bounded below. The
sequence { an } is bounded if it is both bounded above and below,i.e, if there is a real number
M such that a n ≤ M for all n.
Note. I f a sequence diverges to + ∞ or - ∞ , the sequence is not bounded.A sequence that
diverges to infinity must be bounded below .
24
Theorem: If an ≥ 0 for all n and {an } is convergent , then lim an ≥ 0 .
n →∞
Theorem. If the sequence {an } converges , then it is bounded.
Exercise
The converse is not in general true. Consider the sequence {(-1)n }.
Monotone sequences.
Definition. Let {an } be a sequence of real numbers. If a 1 ≤ a 2 ≤ a3 ≤ . . . , then {an } is called
nondecreasing or increasing . If a 1 ≥ a 2 ≥ a3 ≥ . . . , then {an } is called non increasing
or decreasing.
A monotone sequence is sequence which is either nondecreasing or nonincreasing (or both) .
The definition is equivalent to : if a n ≤ a n +1 for all n the sequence {an } is nondecreasing, and
if a n ≥ a n +1 for all n .
,then {an } is nondecreasing.
Examples. a) let an = 2 b). Let bn =
1
2
n −1
. Since
1
1
≤ n −1 for each n ,{an } is nondecreasing.
n
2
2
n +1
1
1
(n + 1) + 1
=
= bn +1 for each n,
. Since bn = 1 + > 1 +
n
n
n +1
n +1
{bn } is nonincreasing.
Theorem:. A nondecreaising sequence which is bounded above is convergent.
Proof. Let {an } be a nondecreasing sequence which is bounded above.
Then the set S = { an : n =1 , 2, 3, . . .} is bounded above. By the completeness property S has a
least upper bound in R. Let sup S = L. Claim lim an = L . Let ε > 0 be given. Then
n →∞
L - ε is not an upper bound of S. Hence there exists a positive integer N
L- ε
such that
< aN .
Since the sequence is nondecreasing , we have
L- ε
< a N < an ≤ L < L + . ε
an − L < ε
for all n ≥ N. Thus,
( n ≥ N).
 1  n 
Corollary:. The sequence 1 +   is convegent.
 n  
25
n
 1
1 +  . By Binomial Theorem ,
 n
Proof. Let an =
n 1
an = 1 +  
1  n
== 1 + n .
1
n
2
 n  1 
+    + ...
 2  n 
+
n(n − 1) 1
2! n 2
 n  1 
  
 n  n 
+ ...+
n
n(n − 1)(n − 2)...(n − k + 1) 1
1
+. . . + n
k
k!
n
n
For each k =1,2,3,.. n the (k+1) term is
1
1
2
k −1
n(n − 1)(n − 2)...(n − k + 1) 1
= (1 − )(1 − ). . .(1 −
).
k
k!
k!
n
n
n
n
Consider a n +1 .As in the above for each k = 1,2,3, . . . (n+1) the (k+1) term is
2
1
1
k −1
! (1 −
)(1 −
) . . .(1 −
).
k!
n +1
n +1
n +1
We see that an ≤ a n +1 . Thus {an } is nondecreasing . Moreover,
1
1
1
1
an < 1 + 1 + + 2 + . . . + n −1 = 1 +
= 3 for all n.
1
2
2
2
1−
2
Hence { an } converges.. We denote lim an by e..
n →∞
Theorem. A nondecreasing sequence which is not bounded above diverges to infinity.
Proof. Exercise
Theorem. A nonincreasing sequence which is bounded be below is convergent.
proof .Exercise
Theorem .A nonincreasing sequence which is not bounded below diverges to minus infinity.
Operation on convergent sequences.
If {an } and {bn } are sequences and c is a constant ,then we form the following sequences:
• { an + bn } = a1 + b1 , a 2 + b2 , . . ., an + bn , . . . (addition)
(subtraction)
• { an - bn } = a1 − b1 , a 2 − b2 , . . ., an - bn , . . .
= ca 1 , ca 2 , . . . , can , . . . ( constant multiple)
• {can }
= a1b1 , a 2 b2 , . . ., an bn , . . .
(multiplication)
• {anbn }
26
Theorem. If {an } and {bn } are sequences such that an → L and bn → M , then
a n − bn → L − M .
c a n → cL (c a constant) , a n + bn → L + M ,
Theorem.a) If 0 < x < 1, then {xn } converges to zero.
b) If 1 < x < ∞ , then {xn } diverges to infinity.
Proof. a) xn+1 = x xn < xn if 0 < x < 1 .Thus {xn } is a decreasing sequence and bounded
below by 0.Hence it converges. Let xn → L.To show L = 0.
With c = x , we have
L = limxn+1 = limx.xn = x limxn = xL.
Hence L = xL which implies L ( 1 – x) = 0.Since 1 –x > 0 ,we must have L = 0.
b)Left for exercise.
Corollary. Let {an } and {bn } are convergent sequences of real numbers with
an ≤ bn
(n ∈ I ).
If an → L and bn → M , then L ≤ M.
2
Lemma. I f an → L then { a n } converges to L2 .
Proof. Let ε > 0 be given .We need to find a positive integer N such that
a n − L2 < ε
2
(n ≥ N ).
Since {an } converges ,there is a positive number M such that a n ≤ M for all n.
and there is a positive number N such that
an − L <
ε
L +M
(n ≥ N ).
Hence for n ≥ N we have
2
a n − L2 =
a n + L a n − L ≤ (M + L ) a n − L < ε .
Theorem. If an → L and bn → M
Proof. Use the identity
then anbn → LM.
27
ab =
1
[(a + b) 2 − (a − b) 2 ]
.
4
Exercises.
1.If L and M are real numbers such that L ≤ M + ε for every ε > 0 , prove that L ≤ M .
2.If {an } is a sequence of real numbers and fi ,for every ε > 0 ,
an − L < ε
(n ≥ N ).
where N does not depend on ε , prove that all but a finite number of terms of (an } are equal to
L.
3. a) Find a positive integer N such that
2n
1
(n ≥ N ).
−2 <
n+3
5
2n
b) Prove lim
= 2.
n →∞ n + 3
4.If θ is a rational number prove that the sequence {sin(n θ π ) } has a limit.
5. Prove that if {an } converges to L then { a n } converge to L
Give an example that the converse is not generally true .
6.True or false ? If a sequence of positive numbers is not bounded then sequence diverges
to infinity.
7 If the sequence {an } is bounded , prove that for any ε > 0 there is a closed interval J of length
ε such that an ∈ J for infinitely many values of n.
8.If {an } is a sequence of real numbers and if
lim a 2 n = L and lim a 2 n −1 = L ,
n →∞
n ←∞
prove that an → L as n → ∞ .
10 n
9.If an =
,find N such that a n +1 < an
(n ≥ N ).Find lim an.
n!
2 and
10. Let a1 =
a n +1 = 2 . a n for n = 2 , 3, . . . Prove that
a) an ≤ 2 for all n.
b) a n +1 ≥ a n for all n
c) {an } is convergent.
d) that lim a n = 2.
n →∞
11.Suppose a1 > a 2 > 0 , and let a n +1 =
1
(a n + a n −1 )
2
for n = 2, 3 , 4 ,. . ..Prove that
a) { a 2 n −1 } ∞n =1 is nondecreasing.
b) { a 2 n } ∞n =1 is nonincreasing.
28
Limit Superior and Limit Inferior.
Let {an } be a sequence of real numbers that is bounded above. For each positive integer n
the set
An = { a k : k = n , n+1, n+ 2 , . . .} = { a n , a n +1 , a n + 2 , . . .} is bounded above .Hence by the
Least upper bound property , it has a least upper bound.
Let supAn = sup a k = bn , n = 1, 2, 3 , . . .
k ≥n
We see that since A n +1 ⊆ An , bn +1 ≤ bn Thus {bn } is nonincreasing sequence of real
numbers. and consequently {bn } either converges or diverges to minus infinity.
____
We call the limit of { bn } the limit superior of { an } and is denoted by lim sup a n (or lim a n )
n →∞
n
,i.e.
____
lim sup a n = lim bn . = lim a n .
n →∞
n →∞
n
____
Since lim bn = inf bn we can write lim a n = inf (sup a k ) .
n→∞
n
n
n
If { an } is not bounded above we write
k ≥n
lim sup a n = ∞ .
n →∞
Example. Let an = (-1 )n , n=1,2 , 3, .. Here
An = { (-1 )k : k = n , n+1, . . . } = {-1,1 }
Thus supAn = 1 for all n . Hence
lim sup a n = 1.
n →∞
Suppose { an } be a sequence of real numbers that is bounded below.
For each n, let
cn = inf a m = inf { a n , a n +1 , . . .}.
m≥ n
The sequence {cn } is a nondecreasing sequence of real numbers. Hence it either converges
or
diverges to infinity. We call the limit of {cn } the limit inferior of {an } and is denoted by
lim inf a n or lim a n i.e,,
n →∞
lim inf a n =
n →∞
_____
lim sup a n = sup(inf a k ) .
n →∞
n
k ≥n
In the above example lim inf (−1) n = 1.
n →∞
29
If { an } is not bounded below then we write lim inf a n = - ∞
n →∞
Example .let an = -n foe each n. Then bn = -n =sup{ -n , -(n+1) , . . .} for each n and
cn = = - ∞ = inf {-n , -(n+1) , ...}. Hence
lim sup a n = − ∞ =
n →∞
lim inf a n
n →∞
Theorem: If { an } is sequence of real numbers , then
lim inf a n ≤
n →∞
lim sup a n .
n →∞
Proof:.Exercise.
Theorem: If {an } is a convergent sequence of real numbers, then
lim sup a n == lim inf a n = lim a n .
n →∞
n →∞
n →∞
Proof: We proof the equality , lim sup a n =
n →∞
lim a n .
n →∞
Let L = lim a n .Let ε > 0 be given .we need to find a positive integer N such that
n →∞
sup a k − L
<ε
for all n ≥ N.
k ≥n
By hypothesis there is a positive integer N such that
an − L <
ε
for all n ≥ N.
2
This implies
L-
ε
2
< an < L +
ε
2
Hence for each n ≥ N , L +
(n ≥ N.).
ε
is an upper bound for { a n , a n +1 , a n + 2 , . . .} and L -
2
is an upper bound. Thus if n ≥ N we have
L - ε < bn = sup a k = sup { a n , a n +1 , a n + 2 , . . .}
k ≥n
≤ L +
ε
2
ε
2
< L + ε ,i.e,
30
<ε
sup a k − L
( n ≥ N)
k ≥n
Hence lim sup a n = L.
n →∞
Theorem: .If lim sup a n =
n →∞
lim inf a n . = L ∈ R, then lim a n = L.
n→∞
n →∞
Proof. Let ε > 0 be given .we need to find a positive integer N such that
an − L < ε
( n ≥ N ).
By hypothesis there is a positive integer N such that
sup{a n , a n +1 , a n + 2, . ..} − L < ε
for all n ≥ N . Hence if
and
inf{a n , a n +1 , a n + 2 , . . .} − L , < ε
n ≥ N,
L - ε < inf{ a n , a n +1 , . . .} ≤ an ≤ sup{ a n , a n +1 , . . .} < L + ε .
Exercises.
1.Let {an } be a sequence of real numbers. Prove that
lim sup(−a n ) = - lim inf a n .
n→∞
n →∞
2 Find the limit superior and limit inferior for the following sequences.
π
b) 1, 2 , 3, 1, 2, 3, 1, 2, 3, . . .
a) { (1 – 1/n) sin(n ) }
2
c) { n(1 + (-1)n)}
3. Let Let { an } be a bounded sequence of real numbers. If lim sup a n = L prove
n →∞
that : (a) Given ε > 0 there is a positive integer N such that an < L + ε
for all n ≥ N,
.,
(b) Given ε and n there is k ≥ n such that ak > L - ε .
4 .Prove that any bounded sequence of real numbers has a convergent subsequence.
(Hint : Apply problem 2 above).
5.Let {an } is a bounded sequence of real numbers and
a1 + a 2 + . . . + a n
n
a) Prove that lim sup σ n ≤ lim sup a n
σn =
n →∞
( n = 1,2,3, . . .).
n →∞
b) If lim a n = 0 then prove that lim σ n = 0
n →∞
n →∞
31
c ) Use (b) to prove that if lim a n = L ∈ R , then
n →∞
lim σ n = L.
n →∞
6. a) Let { an } and { bn } be bounded sequences of real numbers. Prove that
lim an + lim bn ≤ lim (an + bn ) ≤ lim an + lim bn .
b) Give an example that to show that equality need not in (a)
Cauchy sequences.
Definition. Let {an } be a sequence of real numbers.We call {an } a Cauchy sequence if for
every ε > 0 there exists an N ∈ I such that
an − am < ε
( n ,m ≥ N).
Theorem: If the sequence of real numbers {an } converges then it is a Cauchy sequence.
ε
Proof: Let L = lim a n .the given
> 0 ,there exists an N ∈ I such that
n →∞
an − L <
ε
2
Thus if n ,m ≥ N we have
. an − am =
so that
( n ≥ N)
( a n − L) + ( L − a m )
an − am <
ε
≤
an − L
+ L − am <
ε
2
+
ε
2
=
ε
( n ,m ≥ N).
Theorem: If {an } is a Cauchy sequence then it is bounded.
Theorem : If {an } is a Cauchy sequence , then {an } is convergent.
Proof. By hypothesis
{an } is bounded and hence
lim an and lim an are real numbers .
It is enough to prove lim an = lim an .Since lim an ≤ lim an , we need to prove the reverse
inequality.
Since {an } is a Cauchy sequence ,given ε > 0 there exists a positive integer N such that
an − am <
ε
(n ,m
≥
N)
In particular
32
ε
an − a N <
aN −
Thus
aN −
Consequently
ε
2
ε
2
(n
≥
N ).
< a n < a N + ε for all n
≥
N. This implies ,for n
≥
N
≤ inf{ a n , a n +1 , a n + 2 ,...} ≤ sup{a n , a n +1 , a n + 2 ,...} ≤ a N +
ε
2
sup{ a n , a n +1 , a n + 2 ,...} − inf{a n , a n +1 , a n + 2 ,...} ≤ ε .
sup{ a n , a n +1 , a n + 2 ,...} ≤ inf{a n , a n +1 , a n + 2 ,...} + ε
Taking limits on both sides we obtain,
≤
lim an
Since
ε
lim an +
ε.
is arbitrary ,we have
lim an
≤
lim an.
Exercises.
1 Nested – Interval Theorem.For each n let In = [ an , bn] be a nonempty closed bounded
interval of real numbers such that
I 1 ⊃ I 2 ⊃ I 3 ⊃ . . . ⊃ I n ⊃ I n +1 ⊃ . . . ,
and lim(bn − a n ) = lim(lenth of I n ) = 0.
n→∞
n →∞
∞
Prove that Ι I n contains precisely on point.
n =1
Show by an example that the assertion need not hold if the intervals are not closed or not
bounded.
2. Let a1 and a 2 be distinct real numbers. For n ≥ 3 , define inductively by
1
a n = (a n −1 + a n − 2 ).
2
Show that the {an } is a Cauchy sequence. Find the limit of {an }.
 1
Hint: a n +1 − a n =  − 
 2
n −1
(a 2 − a1 ) for n = 2, 3 ,4,. . .
Series of Real Numbers.
An expression of the form
a1 + a 2 + a3 + .Λ an + Κ
where an is a real number for each n is called an infinite series. It is denoted by
33
∞
∑a
n =1
n
.
an is called the nth term of the series.
we want to define what is mean by the sum of a series.
For each n ( n =1 , 2, 3, ...),let
n
∑a
s n = a1 + a 2 + a3 + .Λ an =
k =1
k
. s n is called the nth partial sum of the series and the
sequence { s n } ∞n =1 is called the sequence of nth partial sums of the series.
Definition: We say the series
∞
∑a
n =1
n
converges if the sequence { s n } of nth partial sums
converge in R. If lim s n = s , then s is called the sum of the series ,and write
n→∞
s =
∞
∑a
n =1
n
If the sequence { s n } diverges, then series
∞
∑a
n =1
n
is said to diverge.
Examples.
a) For r < 1,consider the geometric series
∞
∑r
For n ∈ I ,
k
.
k =1
sn = r + r 2 + r 3 + r 4 + Κ + r n .
Thus
(1- r) s n = r – rn-1,
and as a consequence ,
r − r n −1
.
1− r
sn =
Since r < 1 , lim r n = 0.Therefore
n →∞
∞
lim s n =
n→∞
r
, thus by definition
1− r
r
r <1 .
,
1− r
k =1
For r ≥ 1 the series diverges.
∑r
k
=
c) consider the series
∞
1
∑ k (k + 1)
.
k =1
34
1
1
1
= −
k (k + 1) k k + 1
Here a k =
Then
n
1 1
1 1
1
1
= ∑ ak = ( − ) + ( − + Κ + ( −
)
1 2
2 3
n n +1
k =1
1
= 1n +1
∞
1
= 1.
Thus lim s n = 1 and by definition
∑
n→∞
k =1 k ( k + 1)
sn
Theorem (the Cauchy criterion) The series
∞
∑a
n =1
converges if and only if given ε > 0,
n
there exists a positive integer N such that
m
∑a
k = n +1
k
< ε
for all m > n ≥ N.
Proof. The result follows from the relation
m
∑a
k = n +1
Corollary: If
k
∞
∑a
n =1
n
=
sm − sn .
converges ,then lim a n = 0
n→∞
Proof : Since a n = s n − s n −1 , the result follows from the Cauchy criterion.
Note.the condition lim a n = 0 is not sufficient for the convergence of
n →∞
The series
∞
1
∑n
n =1
∞
∑a
n =1
n
.
1
= 0.
n→∞ n
diverges ,yet lim
Theorem: Suppose a n ≥ 0 for all n. Then
∞
∑a
n =1
n
converges if and only if { s n } is bounded .
Proof: Exercise
Exercises.
1. Let
0 ≤ a n ≤ bn for each n. Prove that if
∞
∑b
n =1
n
converges, then
∞
∑a
n =1
n
converges.
35
Give an example to show that the converse is not true.
∞
1
2. Prove that the series ∑ 2
converges.
n =1 n + n
3.P rove that the series
∞
1
∑ n!
converges.
n =1
4. Prove the series
∞
n =1
5. If
∞
∑a
n =1
a)
∞
∑ ca
n =1
b)
n
n
∞
∑ (a
n =1
6 . If
n
and
∑b
n
2
converges. ( use the inequality
1
1
≤
2
n(n − 1)
n
=
1
1
− ).
n −1 n
both converge, prove each of the following:
converges for all real c.
+ bn ) converges.
∞
∑ (a
n =1
∞
n =1
1
∑n
n
+ bn ) converges, does this imply the series a1 + b1 + a 2 + b2 + Κ
converges?
Chapter 3
Open and closed set of real numbers.
Open sets of R.
The notion open sets is a generalization of open intervals.
Definition. A set O of real numbers is called open if for each x ∈ O there exists a δ > 0 such
that the interval ( x - δ , x + δ ) ⊆ O.
Example a) Each open interval is an open set , i.e , (a. b) , (a , ∞ ) and (- ∞ , a)
c) The empty set and the set R are open sets.
d) Any finite set is not open.
e) ) The set of rational numbers ,Q is not open.
We establish some basic properties of open sets.
Theorem: The intersection O1 ∩ O2 of two open sets O1 and O2 is open.
Proof:
Corollary: The intersection of any finite collection of open sets is open .
Theorem: The union of any collection of open sets is open.
Proof: Let { Oλ }λ∈J be any collection of open sets and U their union. Let x ∈ U .Then there
exists Oλ for some λ ∈ J such that x ∈ Oλ . Since O is open there is a δ > 0
such that x ∈ ( x − δ , x + δ ) ⊆ Oλ . Since Oλ ⊆ U , x ∈ ( x − δ , x + δ ) ⊆ U . Hence U is open.
Theorem: Every open set of real numbers is the union of a countable collection of open
36
intervals.
Proof: Let O be an open set. For each x in O there is a y such that the open interval (x,y) ⊆ and a
z with (z , x) ⊆ O since O is open.
Let a x = inf{z : (z , x) ⊆ O } and bx = sup{y : (x ,y) ⊆ O }.
Claim. i) a x and bx are not in O
iii) The open interval ( a x , bx ) ⊆ O
ii) a x < bx
Let Ix = ( a x , bx ).
iv) For x and y in O either Ix ∩ Iy = φ or Ix = Iy
Thus any two distinct intervals in the collection { Ix } x∈O are disjoint and O is the union of the
collection {Ix } .Each interval Ix contains a rational number say rx. Since the collection {rx } is
countable we can put {Ix} in one-to-one correspondence with the subset of rational numbers.
Thus {Ix } is countable.
We consider to study the notion of closed sets which is a generalization of closed intervals.
Definition: A real number x is called a point closure of a set E if for every δ > 0 there is a point y
in E such that y − x < δ .
From the definition every point of E is a point of closure of E.
The definition can be restated as follows: x is a point of closure of e if every open interval I
containing x contains a point of E, i.e I ∩ E ≠ φ . We denote the set of points of closure of E by
E . Thus E ⊆ E .
Examples.
a) Q = R b) Z = Z c) (a, b)
,
Theorem: If A ⊆ B , then A ⊆ B . Also A ∪ B = A ∪ B .
Proof: The first assertion follows from the definition. Since A ⊆ A ∪ B ,we have
A
⊆ A ∪ B . Similarly, B ⊆ A ∪ B . Hence A ∪ B ⊆ A ∪ B .
To prove the reverse inclusion , assume x ∉ A ∪ B and show that x ∉ A ∪ B .
Definition: A set F is closed if F = F .
Since we always have F ⊆ F , a set is closed if F ⊆ F , that is if F contains all of its points
of closure.
Examples: The empty set and R are closed sets. The closed interval [a , b] and [a, ∞ ) are closed
sets.
37
Theorem: For any set E the set E is closed ; i.e. E = E
Proof: Exercise.
Theorem: If A and two closed sets , then A ∪ B is closed.
Proof: A ∪ B = A ∪ B = A ∪ B.
Theorem: The intersection of any collection of closed sets is closed.
Note. The union of arbitrary collection of closed sets need not be closed.
1
1
Consider the following example. Let En = [a + , b - ] ( n = 1, 2 , 3 ... )
n
n
En is closed and
∞
Υ En = (a , b) , which is not closed . Why?
n =1`
Theorem: The complement of an open set is closed and the complement of a closed set is open.
Proof: Exercise.
The Heine- Borel theorem.
A collection of sets { Oλ } covers set F if F ⊆ Υ Oλ . The collection { Oλ } is called
λ
a covering of F. If in addition Oλ is for each λ , we call the collection { Oλ } an open cover of F.
Definition:A subset F of R is compact if every open cover of F has a finite subcover of F ; i.e.
if { Oλ } is an open cover of F, then there exists λ1 , λ 2 ,Κ , λ n such that
n
F ⊆ ∪ Oλi
i =1
Note that { Oλ1 , Oλ2 Κ Oλn } ⊆ { Oλ }.
Examples.
a) Every finite set is compact. Suppose F ={ x1 , x 2 , Κ x n } and { Oλ } is any open cover of F. Then
for each k ( k = 1 ,2, ...,n) there exists λ k such that x k ∈ Oλk . Hence { Oλk }n k =1
is a finite subcollection which covers F.The open interval (0, 1) is not compact. For each n (n = 2,
3,4 ,...,) let
1
).
n
The collection { On } ∞n =1 is an open cover of (0 ,1) , but no finite subcollection covers (0, 1).
Suppose on the contrary that a finite number say On1 , On2 , Κ Onk covers (0 , 1). Let N -=
On = ( 0 , 1 −
max{ n1 , n2 ,Κ , nk }. Then
1
1
for j = 1,2,3,...,k. Hence
≤
N nj
38
On j = (0 , 1 −
1
1
) ⊆ (0, 1 − ) , j = 1, 2,...,k
nj
N
Thus
k
(0, 1) = ∪ On j ⊆ (0 , 1j =1
1
),
N
which is a contradiction.
Some properties of compact sets.
Theorem: a) Every compact subset of R is closed and bounded
b)Every closed subset of a compact set is compact.
Proof: Let K be a compact subset of R. To show K is bounded. Consider the open cover
{ (-n , n )} n∈I of R ,and hence of K. Since K is compact there exists k1 , k 2 , Κ , k n in I such that
K⊆
n
∪ (.-kj , kj )
j =1
If N = max{ k1 , k 2 , Κ , k n }, then K ⊆ (-N , N). Thus K is bounded.
To show K is closed ,we need to show K c is open. Let x∈ K c be fixed(arbitrary).We need an
open interval containing x and contained in Kc .For each y in K
there exists δ y > 0 and ε y > 0 the intervals Uy = (x- δ y , x + δ y ) (and Vy = ( y- ε y , y + ε y ) are
dis joint.The collection { Vy } y∈K is an open cover of K.Since K is compact there exists
y1 , y 2 , y 3 ,Κ y n of elements in K such that the
collection { V y1 ,V y2 ,Κ , V yn } covers K. Let δ = min { δ y1 , δ y2 ,Κ , δ yn } and U = (x- δ , x + δ ).
Since U ∩ V y j = φ for each j (j = 1,2, 3,..., n) we have U ∩ K = φ . Thus U ⊆ K c and K c is
open .Hence K is closed
Let F be closed subset of the compact set K. To show F is compact. Let { Oλ } be an open cover
of F. Then the collection
{ Oλ } ∪ { Fc }
is an open cover of K. Since K is compact, a finite number of these will cover K, and hence
also F.
Examples. Since (0, 1) is not closed , it is not compact. [2 , ∞ ) is not bounded and hence
not compact.
Theorem (Heine- Borel) Every closed and bounded subset of R is compact.
Proof: Let F be a closed and bounded subset of R and let Ω = { Oλ } be an open cover of F. We
consider first the case F = [a , b] where a and b real numbers.
Let E = { x ∈ [a , b]; [a , x] can be covered by a finite number of sets of Ω } .
39
E ≠ φ since a ∈ Oλ for some λ .The set E is bounded above by b.Thus by the least upper bound
property ,E has a least upper bound in R. Let supE = c .Since b is an upper bound c ≤ b.
Cliam. i) c ∈ E, i.e. [a , c] is covered by a finite number of sets in Ω .
ii) c = b.
Since c ∈ [a , b] , c ∈ Oβ for some β .Since Oβ is open there is δ > such that
i)
(c- δ , c+ δ ) contained in O β . As c- δ is not an upper bound of E, there is an element y in E
such that c- δ < y ≤ c. Thus there is a finite subcollection Ω say { Oλk }n k =1
which covers E.
Hence the collection { Oλ1 , Oλ2 Κ , Oλn , Oβ } covers [a , c]. Therefore c ∈ [a , b].
ii)
is left as an exercise.
For the general case ,let F be any closed and bounded subset of R and Ω = { Oλ } an open cover of
F. Since F is bounded there is a positive real number M such that F is contained in the interval [M, M]. Consider the collection Ω ∪ { Fc } which is an open cover of R and hence of [-M , M] .
Apply the first case to [-M , M].
Exercises:
1.Let A = {1//n : n =1 , 2, 3, . . . }
a) Show that A is not compact.
b) Using the definition to prove that F = A ∪ {0} is compact .
2. If A and B are compact subsets of R prove that A ∪ B and A ∩ B are compact sets.
3 Find a countable infinite collection of compact sets such that their union is not compact
Continuous functions.
Definition. Let f : E → R be a function . We say f is continuous at x ∈ E if given ε > 0,there is a
δ > 0 such that
y ∈ E and y − x < δ implies
f ( y ) − f ( x) < ε .
We say f is continuous on a subset A of E if it is continuous at each point of A.
We state some of the most important result of continuous functions.
Theorem: Let F be a compact subset of R and f : F → R. Then f is bounded and
assumes its maximum and minimum on F; that is , there are points x 1 and x 2 in F such that
f(x 1 ) ≤ f(x) ≤ f(x 2 ) for all x in F.
Proof: We proof that f is bounded on F . Let x ∈ F be arbitrary. Since f is continuous, for
ε = 1 there is a δ x > 0 such that
y ∈ F and y − x < δ x implies f ( y ) − f ( x) < 1. Hence f ( y ) ≤ f ( x) + 1 for all y ∈ F and
y ∈ Ix = ( x- δ x , x + δ x ).
The collection { I x } x∈F is an open cover of F. Since F is compact there is a finite subcollection
I x , I x , I ,Κ , I x of { I x } x∈F which covers F.
1
2
x3
n
40
Let
M = 1 + max { f ( x1 ) , f ( x 2 ) ,Κ , f ( x n )}
If y is in F ,th en y ∈ I x j for some j .Hence f ( y ) ≤ f ( x j ) + 1 ≤ M. Since y is arbitrary,
we have f ( x) ≤ M for all x in F.
The second assertion is left for exercise.
Theorem: Let f be real-valued function defined on R. Then f is continuous iff for each open set O of
real numbers f-1 (O) is an open set.
Proof:( ⇒ ) Assume f is continuous and O is an open set. Let x be in f-1 (O). Then f(x) is in O.
Since O is open there is an ε > 0 such that (f(x) - ε , f(x) + ε ) ⊆ O. By the continuity of f there
is a δ > 0 such that f ( y ) − f ( x) < ε for y − x < δ . That is for each y in (x- δ , x+ δ )
f(y) ∈ (f(x) - ε , f(x) + ε ) ⊆ O. Hence y ∈ f-1( O ), for each y in (x- δ , x+ δ ). therefore,
( x- δ , x+ δ ) ⊆ O. The proof ( ⇐ ) is left for exercise.
Definition. A real –valued function f defined on a set E is said to be uniformly continuous if
on E if given ε > 0 there is a δ > 0 such that for all x, y in E with y − x < δ we have
f ( y ) − f ( x) < ε .
Note that δ depends only on ε . A uniform continuous is continuous.
Examples.
a) Let f(x) = x2, x ε [ 0 , 2] = E. We claim f is uniformly continuous on E. Let ε > 0.We
need to find δ > 0 such that for all x , y in E y − x < δ we have f ( y ) − f ( x) < ε .
Now
f ( y ) − f ( x) = y 2 − x 2 = y − x x + y ≤ 2 y − x < ε .
Take δ =
ε
2
.
1
, for x in ( 0 , 1). Show that f is not uniformly continuous on (0 , 1
x
Solution: Let ε = 1.Then for any δ > 0 we need to find two points x and y in
(0 , 1)such that y − x < δ but f ( y ) − f ( x) ≥ 1.
b) Let f(x) =
Let δ > 0 be given. There is a positive integer N > 2 such that
2
<δ .
N
1
2
and y =
. The numbers x and y are in (0 , 1);and y − x < δ but
N
N
N
f ( y ) − f ( x) =
> 1. Thus f is not uniformly continuous in (0, 1). Note f is
2
continuous on (0, 1).
Let x =
41
Theorem: If f is continuous on a compact subset F of real numbers, then f is uniformly
continuous on F.
Proof: Let ε > 0 be given. Then for each x in F there is a δ x > 0 such that for all y ∈ F and
y − x < δ x we have f ( y ) − f ( x) <
ε
.Let Ox = ( x -
δx
, x+
δx
) . Then the collection {
2
2
2
Ox } x∈F is an open cover of F. Since F is compact there a finite subcollection
1
{ O x1 , O x2 ,Κ , O xn } which covers F. Let δ = min { δ x1 , δ x2 , Κ , δ xn }.
2
Let x and y be any two points in F such that y − x < δ .We want to show
f ( y ) − f ( x) < ε .
δx
The point y belongs to O xi for some i, and hence y _ xi <
x − xi ≤
y − x + y _ xi < δ +
Hence
f ( y ) − f ( xi ) <
and
f ( x) − f ( xi ) <
∈ Therefore,
f ( y ) − f ( x) ≤
δx
i
2
i
2
. As a result we have
≤ δ xi .
ε
2
ε
2
.
f ( y ) − f ( xi ) + f ( x ) − f ( xi ) <
ε
2
+
ε
2
= ε,
showing f is uniformly continuous on F.
Exercises.
1. If and g are real-valued continuous on E , show that f+g ,fg f , max{f ,g} and min{f, g}
are continuous on E.
2. Let F : [0 , 1] → [0 , 1] be defined as follows: F(r) = 1/q , where r is a rational and
p
r=
, q > 0 p and q have no common factor, and F(x) = 0 if x is irrational .
q
42
a . Prove that { p/q : q > 0 and p/q ∈ [0 , 1] } is finite.
b) Prove that F is not continuous at any rational.
c) Prove that F is continuous at each irrational.
d) Show that F can be extended to a function g on R such that g is continuous at each
irrational but not continuous at any rational.
3.Let f be a function defined by setting
p
1
in lowest terms.
f(x) = x , if x is irrational and f(x) = psin if x =
q
q
At what points is f continuous?
4. Prove that the set of discontinuity of a monotone function is countable.
1
, x ∈ [1 , ∞ ) .Show that f is uniformly continuous on [1 , ∞ ).
5. Let f (x) =
1+ x2
6. Let f(x) = x , x ∈ [0 , ∞ ). Show that f is uniformly continuous on [0 , ∞ ).
7.If f be a continuous functions defined on R ,prove that for each c ,{x∈ R: f(x) ≥ c } is
a closed set.
8.Show that the function f (x) = sinx is uniformly continuous on R.
9. Suppose fis continuous real –valued function on R and that
liim f(x) = 0 = liim f(x).
x →∞
x → −∞
Prove that f is uniformly continuous on R.
Discontinuous function on R.
∞
Definition: The subset D of R is to be an Fσ if D = ∪ Fn , where each Fn is a closed subset
n =1
of R.
Example
a) A closed set is an is an Fσ .(Why)
b) An open interval (a , b) is an Fσ (Why?)
Our aim is to show if f : R → R is the set of points of discontinuity of f is an Fσ .
Definition . Let f : R → R.If J is any bounded interval in R, we define ω [f : J ] called
the oscillation of f over J as
ω [f : J ] = lub f(x) - inf f(x).
x∈J
x∈J
If a ∈ R we define ω [f ; a] called the oscillation of f at a to be
ω [f ; a] = inf ω [f : J ]
where the infimum is taken over all bounded open interval J containing a.
ω [f : J ] roughly measures the distance between the lowest point and the highest point of
the graph of f in J. Note both ω [f : J ] and ω [f ; a] are nonnegative.
43
Theorem: If f : R → R and a ∈ R, then the following statements hold:
1) If f is continuous at a , then ω [f ; a].
2) If f is not continuous at a, then ω [f ; a] > 0.
Proof. Exercise.
Theorem: Let f : R → R.For any r > 0 let Er be the set of all points a in R such that
ω [f ; a] ≥ 1/r. Then Er is closed.
Proof. let x be a point of closure of Er. We must show x ∈ r. Let J be any bounded open
interval containing x. Then by the definition of closure point J contains is a point y of Er
Hence ω [f : J] ≥ ω [f ;y] ≥ 1/r.Since J is arbitrary open interval containing x, it follows
that ω [f ; x] ≥ 1/r .Thus x ∈ Er .
Theorem: Let f : R → R. let D the set of points in R at which f is not continuous. Then
D is an Fσ .
Proof: Let x be in D. Then ω [f ; x] > 0. and thus there is a positive integer n such that
∞
ω [f ;x] ≥ 1/n. Let En = { x : ω [f ;x] ≥ 1/n} .Then we have D ⊆ ∪ En. The reverse
n =1
inclusion is obvious .Therefore D =
∞
∪ En
n =1
Sequence of functions.
Definition. Let { f n }∞n =1 be sequence of real-valued functions on a set E. We say that
{ f n }∞n =1 converges pointwise to the function f on E if for each x in E we have
lim f n ( x) = f ( x) .
n→∞
That is, the sequence { f n ( x) }∞n =1 of real numbers converges to f ( x) for each x in E. This means:
given x in E and ε > 0 there exists a positive integer N such that for all n ≥ N we have
f n ( x) − f ( x) < ε .
Note that N depends on x and ε .This dependence is denoted by N (x, ε ).
We also write f n → f pointwise on E. The function f is called the limit of { f n }∞n =1
Examples.
( 0 ≤ x ≤1 ). Then the sequence { f n ) ∞n =1 converges to f on [0 ,1] where
a) Let f n ( x) = x n
f ( x) = 0
( 0 ≤ x <1 ),
f (1) = 1.
x
b) Let g n ( x) =
( o ≤ x < ∞ ). Find g such that lim g n ( x) = g ( x) ( o ≤ x < ∞ ).
n→∞
1 + nx
Solution. If x = 0 , g n (0) = 0 for all n .Hence g (0) = 0 . If x > 0 , we have
0< g n ( x) =
x
x
1
≤
= . Hence
1 + nx
nx
n
lim g n ( x) = 0 for x > 0.
n →∞
44
Thus g ( x) = 0 for all x , o ≤ x < ∞ , i.e. the sequence {g n }∞n =1 converges to 0 (the function
identically zero).
c)Let En = [-n , n] and fn = χ En , the characteristic function on [-n , n].
Find the limit of {fn } .
Solution. Let x ∈ R. Then there is a positive integer N such that x < N. Hence for all n ≥ N
x ∈ [-n , n] and thus fn(x) = 1 (n ≥ N ) .Theref ore,
lim fn(x) = 1 for all x in R.
n →∞
Hence f(x) = 1 ,the constant function, is the limit of {fn } .
Uniform Convergence.
Definition. Let { f n }∞n =1 be a sequence of real-valued functions on a set E. We say { f n }∞n =1
converges uniformly to the function f on E if given ε > 0 there exists N∈ I such that
for all x in E and n ≥ N we have f n ( x) − f ( x) < ε .
Here N = N ( ε ) ,i.e .N depends only on ε but not on x.
It is clear that if { f n }∞n =1 converges uniformly to f on E, then { f n }∞n =1 converges pointwise to
f on E.
Examples.
sin(nx)
a) Let f n ( x) =
x in R. Show that { f n } converges uniformly to 0 ; the function
n
identically zero on R.
Solution. Let ε > 0 be given. We need to find an N in I such that for all x in R and n ≥ N
f n ( x) − 0 < ε .
f n ( x) − 0 =
that
sin(nx)
1
≤
for all x in R. By Archimedean, there is a positive integer N such
n
n
1
< ε . Hence for all x and n ≥ N we have
N
f n ( x) − 0
< ε
b) Let f n ( x) = x n
0( ≤ x ≤ 1 ). Show that { f n } does not converge uniformly to f on [0, 1]
where f = 0 on [0 , 1) and f(1) = 1, but converges uniformly on any interval [0 , r] , 0< r < 1.
Solution. Suppose the contrary .Then corresponding to ε = 1/2, there is a positive integer N
for all x in [0 , 1] and n ≥ N ,
x n − f ( x) < 1/2 .
In particular,
x N < 1/2
holds for all x in [0 , 1) since f ( x ) = 0 in [0 , 1).
45
Letting x → 1− , we get 1 ≤ ½. ; which is a contradiction.
On [0 , r] we have f ( x ) = 0 . Since rn → 0 , there is a positive integer N such that
rn < ε for all n ≥ N . Thus
0 ≤ f n ( x) ≤ rn < ε for all x in [0 , r] and n ≥ N .
Theorem:. Let { f n } be a sequence of continuous functions defined on a set E. If { f n } converges
uniformly to f on E then f is continuous on E.
Theorem(Cauchy criterion ). A sequence {fn} of real – valued functions defined on a set E
converges uniformly on E if and only if for every ε > 0 , there exists an integer N such that
│fn(x) - fm(x)│ < ε
(1)
for all x ∈ E and all n, m ≥ N.
Proof. ( ⇒ ) Assume {fn} converges uniformly on E ,say to a function f, Let ε > 0 be given.
there is a positive integer N such that
│fn( x) - f(x)│ < ε /2
for all x ∈ E and all n ≥ N.
Thus
│fn(x) - fm(x)│ ≤ │fn( x) - f(x)│ + │f( x) - fm (x)│ < ε /2 + ε /2 = ε
for all x ∈ E and all n ≥ N.
( ⇐ ) Asume that (1 ) holds.Then for each x in E the sequence {fn(x)} is a Cauchy sequence of real
numbers ,hence converges in R. Therefore
f(x) = lim fn(x)
n →∞
exists for every x in E.
We claim .fn → f uniformly on E. Let ε > 0 be given. By hypothesis there is a positive integer N
such that
│fn( x) - fm(x)│ < ε /2
for all x ∈ E and all n, m ≥ N. Fix m ≥ N. Then we have
│fm( x) - f(x)│ = lim fn( x) - fm(x)│ ≤ ε /2 < ε
n →∞
for all x ∈ E. Since the above holds for all m ≥ N ,the convergence is uniform.
Exercises.
1 Let f be a uniformly continuous real –valued function on R, and for each natural
number n ,let
46
1
fn(x) = f(x + )
n
Prove that {fn } converges uniformly to f on R.
xn
x in [0 , 1].
1+ xn
a) show that {fn } converge uniformly on [0 , ½].
b) Does {fn } converge uniformly on [0 , 1] ?
2 .Let fn (x) =
3) If {fn } is a sequence of functions which converge uniformly to the continuous function
f on ( (−∞, ∞) , prove that
1
lim fn(x + ) = f(x)
( − ∞ < x < ∞ ).
n →∞
n
x
x −n
4. Let f n (x) =
e
( 0 ≤ x < ∞ ).
n
a)) Prove that { f n } converge pointwise to 0 on [0 , ∞ ).
b) Show that the sequence converges uniformly on [0 , A] ,where A> 0and A a real number
but does not converge uniformly on [0 , ∞ ).
The Riemann Integral.
Sets of measure zero.
If J is an interval of real numbers we denote the length of J by J .
Definition.. The subset E of R is said to be of measure zero if for each ε > 0 there exists a
countable number of open intervals {In } which covers E and
∑
In < ε .
A set consisting of one point is a set of measure zero
The empty set is set of measure zero.
∞
Theorem: .If E1 E2 , E3 . . . are subsets of R each of measure zero ,then E= Υ En is also of
n =1
measure zero.
Proof: Let ε > 0 be given. Since En is a set of measure zero , for each n in I, there is a
∈
countable collection {In i } i =1 of open intervals which cover En and ∑ │In i │ < n .
2
i =1
The union of all such open intervals is countable and covers E , and the total sum of the
lengths of these intervals is less
ε
2
+
ε
2
2
+
ε
23
+ .... = ε .
Corollary: Every countable subset of R has measure zero.
47
Definition. A statement is said to hold at almost every point of [a, b](or almost every where
in [a ,b] ) if the set of points of [a ,b]at which the statement does not hold is a set of measure
zero.
Thus to say " fis continuous almost every where in [a , b] " means that if E is set of points of
[a , b] at which f is not continuous , then E is of measure zero.
Exercises
1.If a < b ,prove that [a , b ] cannot be covered by a finite number of open intervals whose
lengths add up to less than b – a..Use the Heine –Borel property of [a, b] to deduce that [a , b]
is not of measure zero.
2. If a < b , prove that (a , b) is of measure zero.
Definition of the Riemann integral.
. Let J be any bounded interval and f a bounded real-valued function on J. We write
M[f; J] = lub f(x). m[f ; J] = inf f(x) and ω [f ; J] = M[f; J] - m[f ; J]
x∈J
x∈J
If a is a point of J ,we write
ω [f ; a] = inf ω [ f ; I]
where the infimum is over all open subintervals I of J such that a ∈ I
By a partition of [a , b] (closed bounded interval) we mean a finite set { x o , x1 , Κ , x n } of
[a , b] such that a = x 0 < x1 < Κ ,< x n = b.If P1 and P2 are two subdivisions of [a , b] , we say
P2 is a refinement of P1 if P1 ⊆ P2 ( that is , P2 is obtained from P1 by adding more points
of subdivision ).
If P = { x o , x1 , Κ , x n } is a subdivision of [a ,b] , then the closed intervals I1 = [xo , x1] ,
I2 = [ x2 , x3] , . . ., In = [xn-1 , xn ] are the component intervals of P.
Definition. Let f be a bounded function on the closed and bounded interval [a , b] and let P be
any subdivision of [a , b].We define U (f ; P), called the upper sum for f corresponding to P as
U(f ; P) =
n
∑
M[f;Ik].│Ik│
k =1
where Ik , k = 1 ,2 ,3,... n.are the component intervals of P. Similarly, the lower sum L[f ; P] is
defined as
L(f ; p) =
n
∑
m[f ; Ik]..│Ik│.
k =1
Clearly U(f ; P ) ≥ L(f ; P).
Example .Let f(x) = x2 , x ∈ [ 2 , 4] and let P = { 2 ,2 .5 , 3 , 4} be a partition of [2 ; 4].
Here I1 = [2 ; 2.5] , I2 = [2.5 ; 3] and I3 = [3 ; 4] and M[f ; I1] = 25/4 , M[f ; I2] = 9 and
M[f ; I3] = 16. Thus ,U(f ; P) = 25/8 + 9/2 + 16/2 = 93/8.
Lemma Let f be a bounded function on [a , b].Then every upper sum is less than or equal to
every lower sum for e. That is , if P1 and P2 ant two partitions of [a , b] ,then
48
U(f : P1) ≥ (L f ; P2).
Proof. First we show that if P is a refinement of P1 then U (f ; P1 ) ≥ U (f ; P).
it is enough to prove that if where P is obtained P1 by adding one point of subdivision .let
I1 , I2 , . . , In be the components intervals of P1 and P has component intervals ,
I1 , I2 , . . Ik* , Ik**, . . . In where Ik = Ik* ∪ Ik** and │ Ik*│ + │ Ik** │ = │Ik│. Since Ik* ⊆
Ik we have M[f ; Ik* ] ≤ M[f : Ik] .Similarly, M[f ; Ik* *] ≤ M[f : Ik.. Hence
n
∑
U (f ; P ) =
M[f ; Ii].Ii│ + M[f; Ik*]│ Ik*│ + M[ f ; Ik** ] │ Ik** │
i≠k
i
≤
n
∑
M[f ; Ii].I + M[f ; Ik] ( │ Ik*│ + │ Ik** │ ) = U(f , P1).
i≠k
i
Similarly one proves L( f , P2 ) ≤ L ( f , P).
To complete the proof of the lemma , let P1 and P2 any subdivision of [a , b].Then
P = P1 ∪ P2 is a refinement of P1 and P2 .Hence
L(f , P2) ≤ L( f , P) ≤ U(f , P) ≤ U(f , P1),
which proves the lemma.
Let
U = { U(f, P ) : P is a partition of [a , b] } = The set of all upper sums of f over [a , b].
L = { L (f , P ) : P is a partition of [a , b] } = The set f all lower sums of over [a , b].
From the lemma it follows that every number t in L is a lower bound of U .Hence we have ,
t ≤ inf U. for all t in L.
Thus inf U is an upper bound for L and c0nsequently,
supL ≤ inf U.
Definition. Let f be abounded function on the closed and bounded interval [a , b].
:
a) The upper integral of f over [a ,b ] denoted by
−
b
∫
f(x)dx
a
is defined by
−
b
∫
f(x)dx = inf U
a
49
b) The lower integral of f over [a , b] ,denoted by
∫
b
a
−
f(x)
is defined
∫
b
a
−
f(x) dx = supL
Note that we are not attaching no meaning to dx.
It follows
∫
b
a
−
f(x) dx ≤
−
b
∫
f(x)dx =
a
We will show that equality holds for all continuous functions f on [a , b]. However , there exist
functions for which strict inequality holds
Example. Let A be the set of rational numbers on [0, 1]. and consider the characteristic
function χ A .For any interval J contained in [0 ,1] ,we have
M[ χ A ; J] = 1 and m[ χ A ; J] = 0.
Hence for any subdivision P of [0 ,1] we have U( χ A ; P) = 1 and L( χ A ; P) = 0.
Thus U = {1 } and L = { 0 } .Therefore , inf U = 1 and sup L = 0.,i.e.
−
1
∫
χ A = 1 and
0
∫
b
a
−
χ A = 0.
Definition .If f is a bounded function on the closed and bounded interval [a , b] we say f is
Riemann integrable on [a , b] if
−
b
∫
a
f(x)dx =
∫
b
a
−
f(x) dx
Theorem:. Let f be a bounded function on the closed and bounded interval [a , b].
Then f is Riemann integrable if and only if , for each ε > 0 , there exists a partition P of [a ,b]
such that
U (f ; P) < L(f; P) + ε .
50
Proof: ( ⇒ ) Assume f is Riemann integrable. Let ε > 0 be given. By the definition of the
upper integral there is a partition P1 of [a , b] such that
U(f ; P1) <
−
b
∫
f(x)dx +
ε /2.
(1)
a
Similarly , there is a partition P2 of [a ,b] such that
∫
b
a
−
ε /2.
f(x) dx < L(f ; P2 ) +
(2)
Let P = P1 ∪ P2 . P is a refinement of P1 and P2 .Applying the above lemma , we get
from (1) and (2)
Adding (1) &(2) and using the hypothesis that the upper and lower integrals are equal we
obtain
U(f ; P1) < < L(f ; P2 ) +
ε.
Let P = P1 ∪ P2 . P is a refinement of P1 and P2 . Thus applying the above lemma , we get
U ( f ; P) < L(f; P) +
ε ..
( ⇐ ) Suppose for a given
ε
> 0 there exists a partition P of [a , b] such that
ε .. To show f is Riemann integrable.
U ( f ; P) < L(f; P) +
Since
−
b
∫
≤ U(f ; P) and
f(x)dx
L(f ; P) ≤
a
∫
b
a
−
f(x) dx,
we have
−
b
∫
f(x)dx
a
Since
ε
<
∫
b
a
−
f(x) dx,
+
ε
is arbitrary, we get
−
b
∫
a
f(x)dx
≤
∫
b
a
−
f(x) dx.
As the reverse inequality is always hold, we obtain
51
−
b
∫
f(x)dx
=
a
∫
b
a
−
f(x) dx..
Theorem: A continuous function defined on a closed and bounded interval [a ,b] is Riemann
integrable.
Proof. Let f be a continuous function defined on [a , b].Then f is bounded. Let ε > 0 be
given. Since f is continuous on [a , b] ,f is uniformly continuous .Hence there is δ > 0 such
that for all x , y in [a, b] with │x - y│ < c we have
│f (x) – f(y)│ <
ε
b−a
.
b−a
b−a
< δ . ∆ Let =
.and
N
N
P = {a , a + ∆ , a + 2 ∆ , a + 3 ∆ , . . . a + N ∆ = b } be a partition of [a , b].and I 1 , I 2 ,
. . . , I N be the resulting component intervals .Since each component interval Ik is closed
and bounded , f assumes its supremum and infimum (actually maximum and minimum value)
at points in Ik .
Since the length of each component interval is less than δ , it follows .
Let N be a positive integer such that
M[f ; Ik ]  m[f ; Ik] <
ε
b−a
for each k = 1 , 2 , 3 , . ., N. Thus
U (f ; P)  L (f ; P ) < ε .
By the above theorem f is Riemann integrable .
Now we state a more general theorem for the existence of Riemann existence.
Theorem: Let f be bounded function on the closed and bounded interval [a , b].
Then f is Riemann integrable if and only if f is continuous almost every on [a, b] , i.e., the set
of points of discontinuity of f in [a ,b] is a set of measure zero .
52