Introductory Real Analysis Dover Books On Mathematics Pdf
This course unit introduces students to the concepts of mathematics that are the building blocks of mathematical reasoning and mathematical proofs. The course unit handles concepts such as logic, methods of proof, sets, functions, real number properties, sequences and series, limits and continuity and differentiation. Real analysis provides students with the basic concepts and approaches for internalizing and formulation of mathematical arguments. The course unit is aimed at: • Providing learners with the knowledge of building mathematical statements and constructing mathematical proofs. • Giving learners an insight on the concepts of sets and the relevant set theories that are vital in the development of mathematical principles. • Demonstrating to learners the concepts of sequences and series with much emphasis on the bound and convergence of sequences and series. • Providing students with the knowledge of limits, continuity and differentiation of functions that will serve as an introduction to calculus.
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REAL ANALYSIS 1
UNDERGRADUATE LECTURE NOTES
NDUNGO ISSA
MOUNTAINS OF THE MOON UNIVERSITY
ndungoissa@mmu.ac.ug
+256776428589
ABOUT THE AUTHOR
The Author is Issa Ndungo, currently teaching at Mountains of the Moon
University, Uganda. The author passed through Musasa Primary School
for Primary Education, Mutanywana Secondary School for Ordinary
Level, Bwera Secondary School for Advanced Level, Mountains of the
Moon University for Bachelor of Science with Education Degree
(Mathematics/Economics) and Mbarara University of Science and
Technology for Master of Science Degree (Pure Mathematics). The
Author also holds a Certificate in Monitoring and valuation and a
Certificate in Financial Management.
The Author has taught in Secondary Schools such as Kamengo SS-Fort portal, Mutanywana SS-
Kasese, Munkunyu SS-Kasese and Kyarumba Islamic SS-Kasese. In addition to this work, the
Author has written Lecture Notes in Linear Algebra and Calculus. He has also written Mathematics
related peer reviewed journal articles.
Published Research Work by the Author
1. Ndungo. I, Biira. M (2018), Teacher quality factors and pupils' achievement in mathematics
in primary schools of Kyondo, sub-county, Kasese District, Uganda. Acad. J. Educ. Res. 6(7):
191-195. (DOI: 10.15413/ajer.2018.0117)
2. Ndungo. I, Mbabazi. A (2018). Institutional and communication factors affecting students'
decisions to choose a university: The case of Mountains of the Moon University, Uganda.
Acad. J. Educ. Res. 6(10): 257-262. (DOI: 10.15413/ajer.2018.0128 )
3. Ndungo. I, Sarvate. D (2016), GDD (n, 2, 4; λ1, λ2 ) with equal number of even and odd blocks.
Discrete Mathematics. 339:1344-1354. (http://doi.org/10.1016/j.disc.2015.11.004)
4. An Assessment of the Impact of the Leadership Training Program on Pupils' Leadership
Skills in Hakibale Sub-county Kabarole District: In MMU year book vol. 8 2017.
(mmu.ac.ug/wp-content/MMUYEARBOOK/Year%20Book%202017.pdf)
5. Existence of group divisible designs with two groups, block size four with equal number of
even and odd blocks: MSc. dissertation 2016
(https://pdfs.semanticscholar.org/8135/1d40c8cc0c936133716e23e124eb635e60d3.pdf)
These lecture notes were prepared to serve and facilitate the teaching of Bachelor of Science with
Education students at Mountains of the Moon University. The notes are also relevant to students
offering Computer Science.
The Author welcomes the readers to the content of the course unit that introduces the mathematical
logical thinking and proving for the existence of certain mathematical concepts. The contribution
of different Authors whose work was used in compiling this hand book is appreciated. Students are
encouraged to read these notes carefully and internalize the necessary concepts for use in the
subsequent course units. (For any error please notify the Author at ndungoissa@mmu.ac.ug).
Wishing you all the best, enjoy Real Analysis.
PREFACE
This course unit introduces students to the concepts of mathematics that are the building blocks of
mathematical reasoning and mathematical proofs. The course unit handles concepts such as logic,
methods of proof, sets, functions, real number properties, sequences and series, limits and continuity
and differentiation. Real analysis provides students with the basic concepts and approaches for
internalising and formulation of mathematical arguments. The course unit is aimed at:
• Providing learners with the knowledge of building mathematical statements and constructing
mathematical proofs.
• Giving learners an insight on the concepts of sets and the relevant set theories that are vital in
the development of mathematical principles.
• Demonstrating to learners the concepts of sequences and series with much emphasis on the
bound and convergence of sequences and series.
• Providing students with the knowledge of limits, continuity and differentiation of functions
that will serve as an introduction to calculus.
By the end of the course unit, students should be able to:
• Construct truth tables to prove mathematical statements or propositions
• Use relevant methods of proof in constructing proofs of simple mathematical principles
• Operate sets, proof basic set principles and have ability to explain the set concepts such as
closure of a set, boundary point, open set and neighborhood of a point.
• State and prove the axioms of real numbers and use the axioms in explaining mathematical
principles and definitions.
• Construct proofs of theories involved in sequences such as convergent, boundedness, and
Cauchy properties as well as showing understanding of the connection between bondedness
and convergent.
• Obtain the limit of a function, construct relevant proofs for the existence of limits and perform
algebra on limits.
• State and prove the rules of differentiations and show understanding of the application of the
concept of differentiation and the connection between limits, continuity and differentiation.
The course will be delivered through: (1) three hours of lecture per week every Friday 8:30am-
11:30am (12) a combination of lectures, discussions and presentations. Students will be given
lecture notes on each unit but students will be required to make use of the university E-library
for personal reading when answering the assignments.
The course is will be assessed through: (1) Course Work Assessment (class exercises, assignments
& tests (2) End of semester examination. The pass mark for this course unit is 50%.
Table of Contents
ABOUT THE AUTHOR I
PREFACE II
LOGIC AND METHODS OF PROOF 1
1.1 PROPOSITIONS 1
1.2 CONNECTIVES 1
EXERCISE 1.1 2
1.3 T RUTH TABLES AND TRUTH VALUES 2
EXERCISE 1.2 4
1.4 T AUTOLOGY, CONTRADICTIONS AND E QUIVALENCE 4
EXERCISE 1.3 4
1.5 O PEN SENTENCE AND QUANTIFIES 5
EXERCISE 1.4 5
1.6 NEGATION OF A QUANTIFIER 5
1.7 O VERGENERALIZATION AND C OUNTEREXAMPLES 6
1.8 METHODS OF PROOF IN MATHEMATICS 6
EXERCISE 1.5 8
EXERCISE 1.6 9
SETS AND FUNCTIONS 10
2.1 DEFINITIONS ABOUT SETS 10
2.2 I NTERVAL AND INEQUALITIES 11
2.3 O PERATIONS ON SETS 11
EXERCISE 2.1 11
2.4 I NDEXED FAMILIES OF SETS 13
EXERCISE 2.2 13
2.5 FUNCTIONS 13
EXERCISE 2.3 14
2.6 CARDINALITY: T HE SIZE OF A SET 14
2.7 S ETS OF REAL NUMBERS 14
2.8 O RDER ON SETS AND ORDERED SETS 14
2.9 B OUNDED SETS 15
EXERCISE 2.4 15
2.10 NEIGHBORHOODS 15
2.11 T YPES OF POINTS FOR SETS 16
EXERCISE 2.5 16
2.12 O PEN SETS AND T OPOLOGY 16
2.13 CLOSED SET 17
EXERCISE 2.6 17
REAL NUMBERS AND THEIR PROPERTIES 18
3.1 REAL NUMBERS 18
3.2 AXIOMS OF REAL NUMBERS 18
3.2.1 T HE FIELD AXIOMS 18
EXERCISE 3.1 20
3.2.2 T HE ORDER AXIOM 20
3.2.3 T HE COMPLETENESS AXIOM 21
3.3 T HE ARCHIMEDEAN PROPERTY OF REAL NUMBERS 22
3.4 T HE E UCLIDEAN SPACE 22
SEQUENCES AND SERIES 24
4.1 S EQUENCES 24
EXERCISE 4.1 24
4.2 B OUNDED SEQUENCES 24
4.3 CONVERGENT AND DIVERGENT SEQUENCES 25
EXERCISE 4.2 26
EXERCISE 4.3 29
4.4 ALGEBRA OF L IMITS OF SEQUENCES 29
EXERCISE 4.4 30
4.5 MONOTONE S EQUENCE 30
4.6 S UBSEQUENCES 32
4.7 CAUCHY S EQUENCES 32
EXERCISE 4.5 34
4.8 I NFINITE SERIES 35
4.9 GEOMETRIC SERIES 36
4.10 PROPERTIES OF INFINITE SERIES 37
4.11 CONVERGENT CRITERION FOR SERIES 37
4.11.1 L IMIT OF TERM TEST FOR DIVERGENCE 37
4.11.2 I NTEGRAL TEST 37
4.11.3 P-SERIES TEST 39
EXERCISE 4.6 39
4.11.4 COMPARISON TEST 39
4.11.5 ALTERNATING TEST 41
4.11.6 ABSOLUTE CONVERGENCE 42
4.11.7 RATIO TEST 42
4.11.8 ROOT TEST 42
LIMITS AND CONTINUITY 43
5.1 L IMIT OF A FUNCTION 43
EXERCISE 5.1 45
5.2 ALGEBRA OF LIMITS 46
EXERCISE 5.2 47
5.3 CONTINUITY OF FUNCTIONS 48
EXERCISE 5.3 50
5.4 T HE INTERMEDIATE VALUE THEORY 50
5.5 UNIFORM CONTINUITY 51
EXERCISE 5.4 51
5.6 DISCONTINUITY 51
DIFFERENTIATION 52
6.1 DERIVATIVE OF A FUNCTION 52
6.2 RULES OF DIFFERENTIATION 53
6.4 GENERALIZATION OF MEAN VALUE THEOREM (MVT) 54
6.5 L' HOSPITAL RULE 54
EXERCISE 6.1 54
REFERENCES 55
LOGIC AND METHODS OF PROOF
1.1 Propositions
A proposition (statement) is a sentence that is either true or false (but not both).
Examples of propositions are:
(1) John was born on 20th February 2019. (2) 3+6 = 11. (3) is irrational.
Examples of non-propositions are:
(1) What is the date today? (2) How old are you? (3) This statement is true.
There are two types of statements/propositions:
1. Atomic/simple propositions: These cannot be divided into smaller propositions.
We use capital letters to denote simple propositions.
Examples include: (1) John's leg is broken (2) 5 is a prime number (3) is irrational
2. Compound propositions: These can be broken down into smaller propositions. They are
constructed by using connectives.
Examples include: (1) (2) is odd whenever is an odd integer
1.2 Connectives
Connectives are symbols used to construct compound statements/propositions from simple
statements.
The most commonly used connectives are:
Examples
1) P Q means P and Q. 2) PQ means if P then Q. 3) PQ means Por Q
Example 1.1
Write the following compound statements into symbolic form
1) Jim is a lawyer but he is not a crook
Solution
Let P = Jim is a Lawyer and Q = Jim is a crook
Then we have
2) Although our Professor is young but he is knowledgeable
Let R = our Professor is young and S = our Professor is knowledgeable
Then we have .
3) If you don't attend class, then you either read a book or you will fail the exam.
Let P = you attend class Q = you read a book and R = you pass the exam
Then we have
NB. Brackets are used for punctuations.
4) If Lucy has a credit in MAT1 or has a credit in MAT2 and MAT3, then she does MAT7.
Let P = Lucy has a credit in MAT1, Q= Lucy has a credit in MAT2, R= Lucy has a credit in
MAT3 and S = Lucy does MAT7
Then we have
5) The lights are on if and only if John or Mary is at home.
Let R = the lights are on, S= John is at home and T = Mary is at home
Then we have
Exercise 1.1
Write the following compound propositions in symbolic form
1. If I go home and find lunch ready then I will not go to the restaurant
2. Either you pay your rent or I will kick you out of the apartments
3. Jolly will leave home and will not come back
4. If I go to Kampala, I will bring for you biscuits and bread.
1.3 Truth tables and truth values
A truth table is a convenient device to specify all possible truth values of a given atomic or compound
propositions. We use truth tales to determine the truth or falsity of a compound statement based on
the truth or falsity of its constituent atomic propositions.
We use T or 1 to indicate that a statement is true and F or 0 to indicate that the statement is false. So
1and 0 are called truth values.
Truth tables for connectives
1. Conjunction
Let P and Q be two propositions, the proposition is called the conjunction of P and Q.
is true if and only if both P and Q are true.
2. Disjunction
Let P and Q be two propositions, the proposition is called the disjunction of P and Q. The
proposition PVQ is true if and only if at least one of the atomic propositions is true. It is only false
when both atomic propositions are false.
3. Implication
Let P and Q be two propositions, the proposition PQ is called the implication of P and Q. The
proposition PQ simply means that P implies Q. P is called the hypothesis (condition or
antecedent) and Q is called the conclusion (consequence). There are many ways of stating that P
implies Q.
- If P then Q
- Q if P
- P is sufficient for Q
- Q is necessary for P
- P only if Q
- Q whenever P
P Q is only false when P is true and Q is false.
4. Biconditional
Let P and Q be two propositions, the proposition PQ is called the biconditional of P and Q. It
simply means that PQ and QP. It is called biconditional because it represents two conditional
statements. There are many ways of stating that P implies Q.
- P if and only if Q (P iff Q)
- P implies Q and Q implies P
- P is necessary and sufficient for Q
- Q is necessary and sufficient for P
- P is equivalent to Q
The proposition PQ is true when P & Q are both true and if P & Q are both false
5. Negation
Let P be a proposition, the proposition meaning "not P" is used to denote the negation of P.
If P is true then is false and vice versa.
Example 1.2
Let P and Q be propositions. Construct the truth table for the proposition (PÉ…Q) (PVQ)
Exercise 1.2
Let P, Q and R be propositions. Construct the truth tale for the proposition (PÉ…Q)VR.
1.4 Tautology, contradictions and Equivalence
Definition: A compound proposition is a tautology if it is always true irrespective of the truth values
of its atomic propositions. If on the other hand a compound statement is always false regardless of
its atomic propositions, we say that such proposition is a contradiction. Theorems in mathematics
are always true and are examples of tautologies.
Example 1.3
The statement PV P is always true while the statement PÉ… P is always false
Exercise 1.3
1. Let P, Q and R be proposition. Use truth tables to prove that:
a) (PÉ…Q) PV Q
b) (PVQ)PÉ… Q
c) P Q PVQ
d) (P Q)PÉ… Q
e) PÉ…(QVR)(PÉ…Q)V(PÉ…R)
f) (PVQ)VR PV(QVR)
Hint: To prove for example a), show that (PÉ…Q)PV Q is a tautology (always true).
2. Use mathematical logic to discuss the validity of the following argument:
"If girls are beautiful, they are popular with boys. Ugly girls are unpopular with boys.
Intellectual girls are ugly. Therefore, beautiful girls are not intellectual.''
1.5 Open sentence and quantifies
Definition: An open sentence (also called a predicate) is a sentence that contains variable(s) and
whose truth or falsity depend(s) on the value(s) assigned for the variable(s). We represent open
sentences with capital letters followed by the variable(s) in the bracket. Eg P(x),
Definition: The collection of all allowed values of the variable(s) in an open sentence is called the
universe if discourse.
Examples of open sentence are:
1) 2) It has 4 legs 3)
Universal quantifier- : To say that P(x) is true for all x in the universe of discourse we write
( . is called the universal quantifier.
means
Existential quantifier ( ): To say that there is (at least one) x in the universe of discourse for which
P(x) is true we write ( is called the existential quantifier.
means
NB: Quantifying an open sentence makes it a proposition.
Exercise 1.4
1. Translate the following statements using word notation
a) b) ( c) ( d) (
2. Write the following statements using qualifiers
a) For each real number x>0, x2 +x-6 = 0
b) There is a real number x>0 such that x2 +x-6 = 0
c) The square of any real number is non-negative
d) For each integer x, there is an integer y such that x+y = -1
e) There is an integer x such that for each integer y, x+y = -1
1.6 Negation of a quantifier
To negate a statement that involves the quantifiers , change and then negate
the open sentence.
Examples
1) [( ] (
2) [all birds can fly] [there is at least one bird that does not fly]
3) [(]
1.7 Overgeneralization and Counterexamples
Overgeneralization occurs when a pattern searcher discovers a pattern among finitely many cases and
then claims that the pattern is general (when in fact it is not). To disprove a general statement such as
, we must exhibit one x for which P(x) is false. i.e. .
Examples
1) Statement ( )(x 2) is false, x = ½ is a counterexample since ½ ½)2 .
2) For all real numbers x and y |x+y| = |x|+|y|. This statement is false, the counterexample is
when x =1 and y = -1 since |1+-1|≠|1|+|-1| or 0 ≠ 2.
3) For all prime numbers p, 2p+1 is prime. This statement is false, the counterexample is p = 7.
1.8 Methods of proof in Mathematics
A proof is a process of establishing the truth of an assertion, it is a sequence of logical sound
arguments which establishes the truth of a statement in question.
1) Direct Method
Suppose P Q, in this method we assume that P is true and proceed through a sequence of logical
steps to arrive at the conclusion that Q is also true.
Example 1.4
a) Show that if m is an even integer and n is an odd integer then m+n is an odd integer.
Assume that m is an even integer and n is an odd integer.
Then m = 2k and n = 2 +1 for some integer k and .
Therefore for some integer d = k+. Since d is
an integer then m+n is an odd integer.
b) Show that if n is an even integer, then is also an even integer.
Assume that n is an even integer, then n = 2k for some integer k.
Now n2 = (2k)2 = 4k2 = 2(2k)2 = 2c2 for some integer c = 2k. Since c is an integer the is an even
integer.
2) Contrapositive Method
Associated with the implication is the logical equivalent statement Q , the
contrapositive of the conditional statement . So, one way to prove the conditional
statement is to give a direct proof of the contrapositive statement Q . The first step
is to write down the negation of the conclusion then follow it by a series of logical steps that this
leads to the negation of the hypothesis of the original conditional statement.
Example 1.5
a) Show that if is even integer then n is an even integer.
We will show that if n is odd then is odd. Assume that is odd. Then for some
integer .
Now for some integer
. Since is an integer then is an odd integer. Thus, by contrapositive if is even
then n is itself even.
b) Show that if 2n is odd integer then n is an odd integer.
By contrapositive, we show that if n is even integer then 3n is an even integer.
Now n = 2p and 3n =3*2p =2(3p) = 2c for some integer c = 3p. Since c is an integer then 3n is
even and so by contrapositive, if 3n is odd then n is odd
3) Contradiction Method
Assume that P is true and Q is false i.e. PQ is true. Then show a series of logical steps that leads
to a contradiction or impossibility or absurdity. This will mean that the statement PQ must have
been fallacious and therefore, its negation must be true. Since P Q] [PÉ… , it
follows that P Q] [PÉ… and hence PQ must be true.
Definition: A real number r is said to be rational if there are integers n and m (m≠0) such that
r =
with greatest common divisor between [n, m] = 1. We denote the set of rations by . A real
number that is not rational is said to be irrational.
Example 1.6 (proofs by contradiction)
a) Show that is irrational. That is there is no integer p and q such that =
or there is no
rational number such that
By contradiction, we assume that is rational.
So, we have =
. And so =
thus 2q2 = p2 ………………….. (*)
From (*), observe that p2 is even and so p is even and can be written as p = 2c for some integer c.
So (*) becomes: 2q2 =(2c)2 = 4c2 and so q2 = 2c2 ………………….. (**)
Similarly, from (**) it is observed that q2 is even and so q is itself even. Therefore, both p and q are
even integers; implying that they both have a common factor of 2. This contradict the fact that the
GCD [p, q] =1 and thus is not rational but rather it is irrational.
b) Show that if 3n is an odd integer then n is an odd integer.
We will use contradiction. Assume that 3n is an odd integer and n is even integer.
Then 3n = 2k+1 and n = 2d for some integers k and d.
Therefore 3n =2k+1 = 2(3d) = 2c for some integer c. Since c is an integer then 3n is an even integer
which indicates that 3n is both odd and even which is impossible. Thus, if 3n is odd n cannot be
even but rather n is odd
c) If r is rational and x is irrational then prove that r+x and rx are irrational.
Let r+x be rational (by contradiction), then r+x =
with q, GCD [p, q] = 1.
Since r is rational, then r =
and then x =
-
=
which is rational. This indicates that x is
both rational and irrational which is absurd and so r+x is irrational.
Similarly, suppose rx is rational, then rx =
for some integers p, q; q and GCD [p, q] =1. Since
r is rational then r =
and then rx =
which is rational. This indicate that x is both rational and
irrational; which is impossible and thus the assumption that rx is ration is false so rx is irrational
Exercise 1.5
Use the method of contradiction to prove the following:
a) If n is a positive integer which is not perfect square then is irrational.
b) is irrational c) is irrational
4) Mathematical induction Method
It is a method that proves statements of the type holds. Here A(n) is statement
depending on n. e.g.
Definition
Let no N and let A(n) be statement depending on nN such that:
(1) A(n) is true.
(2) For every k if A(k) is true then A(k+1) is also true. That is A(k) .
Then A(n) is true for n.
Steps for mathematical induction
1- Initial stage# verify that A(no) is true
2- Induction hypothesis# A(n) is true for some n say n = k
3- Induction step# deduce (from step1 and 2) that A(n) is true for n = k+1
4- Conclusion# by the PMI, A(n) is true for all integers n
Example 1.7
1) Prove that "7 is a divisor of "
Step 1: For n =1 we have True
Step 2: is divisible by 7 for some n = k
Step 3: set n = k+1, step 2 implies that j: =7j
=
=
= 9 (7j) +(9-2)*
= 7(9 j+2k ) which is divisible by 7
Step 4: It follows from the PMI that 7 is a divisor of for all n
2) Prove that for all n , 1+4+9+ …+ n 2 =
Solution
Check that the statement holds for no=1; 1=
= 1
A(n) holds for some n = k that is 1+4+9+ …+ k 2 =
Check that the statement holds for n = k+1
That is 1+4+9+ …+ k2 + (k+1)2 =
+(k+1)2
=
k+1)
Which is the RHS of A(k+1). Therefore, the statement is true for n = k+1
It follows by PMI that A(n) holds for all n
Exercise 1.6
Prove the following statements by PMI
i) 1+2+3+…+n = ½ (n2+n) for all n
ii) 1+3+5+…. + (2n-1) = n2 for all n
SETS AND FUNCTIONS
2.1 Definitions about sets
Definition: A set is a well-defined collection of objects that share a certain property or properties.
The term well-defined means that the objects contained in the set are totally determined and so any
given object is either in the set or not in the set.
We use capital letters to denote sets and small letters to denote the elements in the set. A set can be
described by listing its members (roster method) or by defining a rule or a function that describes its
elements. E.g. B = {x for some k}.
Examples of sets include set of integers, set of rational numbers, set of counting numbers etc.
Definition: Let A and B be sets. We say that:
a) B is a subset of A (or is contained in A) denoted by BA if every element of B is an element
of A. i.e. .
b) A = B if ) i.e .
c) If B is a subset of A and A then B is a proper subset of A written as B
d) We say that a set is empty if it contains no element e.g. {x: x2+1 =0} is empty since
x2 +1=0 has no solution in .
Proposition 2.1
1) If A is empty then AB for any set B
2) All empty sets are equal
3) For any set A, Aif Aand B, then A
Theorem 2.1
Given any two sets A and B, if A = B then (A
Proof
Let x, since A = B then x and so we have
A = B ( (x ]
( (x
( (x ]
2.2 Interval and inequalities
Let then only and only one of the statements a<b, a>b and a = b is true.
Finite intervals
Open interval (a, b) or {x
Closed interval [a, b] or {x
Half-closed interval [a, b), (a, b] or {x, {x respectively.
Infinite intervals
Open interval (a, ), () or {xrespectively
Closed interval [a, ( or {x, {x respectively
Half-closed interval [a, b), (a, b] or {x, {x respectively.
- and can not be boundaries on the subset of since they are not members of .
2.3 Operations on sets
Definition: Let A be a set. The power of set A denoted by P(A) is the set whose elements are all
the subsets of A. that is P(A) = {B: BA}.
Example: Let A = {x, y, z}. P(A) = {, {x}, {y}, {z}, {x, y}, {x, z}, {y, x}, {x, y, z}}
Definition: Let A and B be subsets of a universal set
a) The union of A and B denoted by AUB is the set of all elements in that are either
in A or B or both. i.e. AUB = {x: (x.
b) The intersection of A and B denoted by AB is the set containing all elements that
are both A and B. i.e. AB = {x: (x .
c) Sets A and B are said to be disjoint or mutually exclusive if AB =.
d) The complement of A relative to B denoted by B-A or B\A is a set of elements of B
that are not in A. i.e. B-A = {x : (x.
e) The complement of A denoted by Ac or AI is the set of all elements in U that are not
in A. i.e. Ac = {x : .
f) The symmetric difference between A and B denoted by AB is a set given by
A B = {(B-A)V(A-B).
Exercise 2.1
Let A and B be subsets of a universal set U = {all counting numbers}. Let A = {1,2,4,5,6, 12,13}
and B = {3,5,6,7,8,910}. Write down the elements for each of the following sets.
a) P = {x: (x
g) Q={x : (x
b) R= {x : (x
c) S = {x :
Proposition 2.2
Let A, B, and C be subsets of a universal set U.
a) i) AUB =BUA and ii) A = B (Commutative law)
b) i) (A (Associative law)
c) i) A and ii) AUA = A (Idempotent law)
d) i) AU(B and ii) A (B (Distributive law)
e) i) (AUB)I = AI B I and ii) (A B) I = AI BI (Demorgan's law)
f) A-(BUC) = (A-B) (A-C) (Distributive law)
We prove d) ii), e) i) and f). the rest can be proved in the same way by the reader.
d) ii) A (B . Let x A (B
x A (B (x
(x
[(x
. Thus A (B
e) i) (AUB)I = AI B I
Let x (AUB)I , then x
Therefore
x(AUB)I
Therefore (AUB)I = AI BI
f) A-(BUC) = (A-B) (A -C)
Let x A-(BUC)
x A-(BUC)
. Therefore A-(BUC) = (A-B) (A -C)
2.4 Indexed families of sets
Instead of using the alphabetical letters, due to existence of large collection of sets, we usually index
sets using some convenient indexing sets.
Suppose is a set and for each there corresponds one and only one subset of a universal set
U. The collection { is called an indexing family of sets or indexed collection of sets.
Definition:
Let { be an indexed family of subsets of the universal set U.
a) The union of the family { denoted by is the set of all those elements of U
which belong to at least one of the . That is = { for some } or
= { .
b) The intersection of the family { denoted by is the set of all those elements
of U which belong to all the . That is = { for each } or
{
Exercise 2.2
Let { be an indexed family of subsets of the universal set U and let B be a subset of U.
Prove that:
a) =
b) =
c) =
Definition: Let X and Y be sets, a function from X to Y denoted by f is a rule that assigns
to each X a unique element .
Definition: Let X and Y be sets, a function from X to Y denoted by f is said to be:
a) Injective (one-to-one) if for each there is at most one X such that .
Equivalently, is injective if for all , f implies that . That is
b) Surjective (onto) if for every there is an X such that f . That is
.
c) Bijective: If it is both injective and surjective
Definition
Let X, Y and Z be sets, f and g be functions. The composition of f and denoted by
is the function defined by . As illustrated in the figure below.
Theorem 2.2
Let and such that , then:
a) If f and g are onto then so is the composition function .
b) If f and g are one-to-one then so is the composition function
c) If is one-to-one then so is f.
d) If is onto then so is g.
Proof
a) Let . Since g is onto there is a such that g(y) = z since f is onto there is a X
such that f(x) = y. Therefore . Hence is onto
b) Let be in X such that . Then
]
since g is one-to-one
since f is one -to-one
So is one-to -one
c) Let be elements of X such that then
Since is one-to-one it follows that thus f is one-to-one
d) Let we must produce a such that g(y) = z. Since is onto there is an X such that
(x) = . Let then g(y) = z which proves that g is onto.
Exercise 2.3
1) Let be a bijection. Then is a bijection
2) Let and be bijections then
2.6 Cardinality: The size of a set
Definition: Two sets A and B are said to have same cardinality, denoted by |A| = |B| if there is a
one-to -one function from A onto B.
Sets with same cardinality are said to be equipotent or equinumerous.
2.7 Sets of real numbers
The sets of real numbers have unique description for example:
Set of counting numbers, integers, real numbers, complex numbers, rational numbers, whole
numbers, even numbers etc.
2.8 Order on sets and ordered sets
Order of a set
Let S be a non-empty set. An order on a set S is a relation denoted by "<" with the following
properties:
1) If and then one and only one of the statements x<y, x>y and x = y is true.
2) If x, y, z and if x<y, y<z then x<z.
Ordered set
A set S is said to be ordered if an order is defined on S.
2.9 Bounded sets
Let S be an ordered set and . If there exist a such that then we say that E
is bounded above and b is known as an upper bound of E.
Let S be an ordered set and . If there exist a such that then p is the lower
bound of E.
Least upper bound (supremum)
Suppose S is an ordered set, and E is bounded above. Suppose there exist a such that:
i) is an upper bound of E ii) if g< then, g is not an upper bound of E. Then is called a least
upper bound of E or supremum of E written as supE = . i.e. is the least member of the set of
upper bounds of E.
Greatest lower bound (infimum)
Suppose S is an ordered set, and E is bounded below. Suppose there exist a such that:
i) k is a lower bound of E ii) if g>k then, g is not a lower bound of E. Then k is called a greatest
lower bound of E or infimum of E written as inf E = k. i.e. k is the greatest member of the set of
lower bounds of E.
Note:
1) That the supremum and the infimum are not necessarily members of E.
E.g. Let E be a set of numbers of the form
where n is a counting number. {E =
}.
then the supE =1 which belongs to set E and the inf E = 0 which does not belong to set E.
2) If the sup E exist and is not in S then set S does not contain the least upper bound property.
3) An ordered set which has the least upper bound property also has the greatest lower bound
property.
Exercise 2.4
a) State the least upper bound property and greatest lower bound property of a set S.
b) Let be a non-empty subset of an ordered set, suppose a is lower bound of E and b is an
upper bound of E. Prove that
2.10 Neighborhoods
A neighborhood of a point is an interval of the form ( where for any
positive real number. Thus, the neighborhood consists of all points distance less than from p.
( = { }. E.g. (1.2, 2.8) is a neighborhood of 2.
2.11 Types of points for sets
We shall consider sets that are subsets of or subset of extended . Note that does not include
while extended includes .
Definition (interior point of S)
Consider , a point is said to an interior point if it has a neighborhood U laying entirely
outside S. i.e. with .
Definition (exterior point of S)
Consider , a point is said to an exterior point if it has a neighborhood U laying entirely
inside S. i.e. with .
Definition (boundary point of S)
Consider , a point is said to a boundary point of S if it is neither interior or exterior
point of S. Thus, p is a boundary point S if its neighborhood intersects both S and Sc.
Example 2.1
1) For the set A = [- decide which of the following are true or which is false.
a) -6 is an interior point of A (T)
b) 6 is an interior point of A ( F )
c) 9 is a boundary point of A (T)
d) 5 is a boundary point of A ( F )
Interior, exterior and boundary of a set
The set of all interior points of a set S is denoted by So and is called the interior of S, the set of all
boundary points of S is denoted by is called the boundary of S while the set of all points of S is
the exterior of S and is denoted by Sext.
Thus, the extended line is split up into 3 parts: = .
Example 2.2
For the set A = [-
i) = [- ii) = {4,5,9,7,6} iii) = [4, 5)
iv) The interior of the complement of =
Exercise 2.5
1) For the set B = (- decide which of the following are true or which is
false.
a) -6 is an interior point of B b) -5 is an interior point of B c) 5 is a boundary point of B
d) 7 is a boundary point of B e) 4 is an interior point of B
2) For the set D = {-4, 8} find , , and
2.12 Open sets and Topology
We say that a set is open if it does not contain any of its boundary points. Eg (2, 3) (5, 9) is open.
Note that every point of an open set is an interior point. Thus, if a set is open it means that S o = S.
This means that a union of open sets is open.
The collection of all open subsets of is called the topology of .
Theorem 2.3
If two sets A and B are open then is open.
Proof
Take a point , then p is in both A and B. Since and A is open there is a
neighborhood U of p which is a subset of A i.e. .
Similarly, there is a neighborhood V of p which is a subset of B i.e. .
But then is a neighborhood of p which is a subset of both A and B, so . Thus,
every point in is open.
2.13 Closed set
A set S is said to be closed if it contains all its boundary points. i.e.
e.g. B = [4, 8] U [9,
so, B is closed
A = [4, 5) is not closed.
N.B: If a set is open then its complement is closed and the converse is true.
Exercise 2.6
Prove the following theorems:
a) If a set is open then is complement is closed
b) The closure of a set is closed
c) The closure of a set is the smallest closed set containing S.
d) Every closed and bounded subset of is compact. Conversely every compact subset of
is closed and bounded. (Heine Borel theory)
REAL NUMBERS AND THEIR PROPERTIES
3.1 Real numbers
Definition: Real numbers are numbers that appear on a number line. They form an open set written
as (∞, -∞); ±∞ are not included in the set and so they are not real numbers.
Real numbers are divided into two types, rational numbers and irrational numbers.
Rational Numbers:
Any number that can be expressed as the quotient of two integers (fraction).
Any number with a decimal that repeats or terminates.
Subsets of Rational Numbers:
Integers: rational numbers that contain no fractions or decimals {…,-2, -1, 0, 1, 2, …}
Whole Numbers: all positive integers and the number 0 {0, 1, 2, 3, … }
Natural Numbers (counting numbers): all positive integers (not 0) {1, 2, 3, … }
Irrational Numbers:
Any number that cannot be expressed as a quotient of two integers (fraction).
Any number with a decimal that is non-repeating and non-terminal (doesn't repeat and doesn't end).
Examples of irrational numbers include: Ï€, √2, √3 etc.
3.2 Axioms of Real numbers
3.2.1 The Field axioms
Definition: A field is a set together with two binary operations +: (called addition)
and (called multiplication) such that for all x, y, z the following are satisfied.
1. Closure law: x, y then x+y and .
2. Commutative law: and
3. Associative law: and ,
4. Existence of inverse:
a) Additive inverse-For any ,
b) Multiplicative inverse-For any ,
5. Existence of identity:
a) Additive identity-For any , 0
b) Multiplicative identity- For any ,
6. Distributive law: For
Theorem 3.1
Let then,
a) If , then
b) If then
c) If , then
d)
Proof
a) Suppose
Thus
b) Take in a)
c) Take in a)
d) Let
If
Then
Thus,
Theorem 3.2
Let
a) If and then,
b) If and then,
c) If and then,
d) If then,
Proof
a) Suppose
Since
Thus,
b) Suppose
From in a) put and get
c) Take
in a)
d) Since
then c) gives
Exercise 3.1
Use the field axioms to prove the following propositions.
a)
b)
c)
d) If then,
3.2.2 The order axiom
An ordered field is a field on which an order relation < is defined such that:
i) (Trichotomy)-for every exactly one of the following holds: .
ii) (Transitivity)- for all
iii) For all furthermore if then .
Theorem 3.3
The following propositions holds for any ordered field.
i) If then and vice versa
ii) If and then
iii) If and then
iv) If then in particular, 1>0
v) If then
Proof
i) If then . So
If then . So
ii) Since we have
which means that . Also
Therefore
iii) Since
. Also
Therefore
iv) If then
If then
i.e. If then Since 12 = 1, then 1>0
v) If and then But
Like wise
as . If we multiply both sides of the inequality by the positive quantity
we obtain
. i.e.
3.2.3 The completeness axiom
Definition: An ordered field is said to be complete if every subset S of which is bounded
above has the least upper bound.
Proposition: A nonempty subset S of an ordered field can have at most one least upper bound.
Proof
Suppose λ and are both least upper bounds of S. Then by the definition of least upper bound, we
have thus
Theorem 3.4 (Characterization of supremum)
Let S be a nonempty subset of an ordered field and . Then if and only if:
i) M is an upper bound for S
ii) For any with ε>0, there is an element such that .
Proof
Assume that M is a supremum for S. i.e. . Then, by definition, M is an upper bound for S.
If there is an with for which for all , then is an upper bound
for S which is smaller than M, a contradiction.
For the converse, assume that i) and ii) hold. Since S is bounded above, it has a supremum, A (say).
Since M is an upper bound for s, we must have that . If then with there is
an element such that i.e. which is absurd.
Therefore and so M is the supremum of .
Theorem 3.5
Let A and B be nonempty subsets of which are bounded above. Then the set
is bounded above and
Proof (Left to the reader)
3.3 The Archimedean Property of Real numbers
Theorem 3.6 (Archimedean Property)
The set of natural numbers is not bounded above.
Proof
Assume that is bounded above. By the completeness axiom, exists. Let then with
there is an element such that This implies that which is
impossible. Thus is not bounded above
Corollary:
The Archimedean property implies the following
1) For every real number there exist an integer such that
2) Given any number there exist an integer such that
3) If and are two positive real numbers there exist a natural number such that
4) If then there exists an such that
Theorem 3.7 (Density of rational in reals)
If and and then there exists a rational number such that That is in between
any two distinct numbers there is a rational number. Or is dense in .
Proof (lest to the reader).
3.4 The Euclidean space
Definition: Let and be vectors in The inner product or scaler product of and is
defined as
And the norm of is defined by
The vector with the above inner product and norm is called Euclidean .
Theorem 3.8
Let then:
i)
ii) (Cauchy Schwarz's inequality)
Proof
i) Since
therefore
ii) For we have:
=
Now put
(certain real number)
which holds if
i.e.
Theorem 3.9
Suppose then:
a) (Triangle inequality)
b)
Proof
a) Consider
Thus =
b) We have
The triangle inequality suggests that
Thus
SEQUENCES AND SERIES
4.1 Sequences
Definition: A sequence is a function whose domain is the set of natural numbers. If is such a
sequence, let denote the value of the sequence at . In this case we denote the
sequence by
of simply .
An infinite sequence is an unending set of real numbers which are determined according to some rule.
A sequence is normally defined by giving a formula for the term.
Examples
1)
is the sequence
2) is the sequence
3) is the sequence
We can also use recursive formulas e.g.
were and , then the terms
of the sequence are
.
Remark
(1) The order of the terms of the sequence is an important part of the definition of the sequence.
For example, the sequence is not the same as the sequence
(2) There is a distinction between the terms of a sequence and the values of a sequence. A
sequence has infinitely many terms while its values may or may not be finite.
(3) It is not necessary for the terms of a sequence to be different. For example,
is a particularly good sequence.
Exercise 4.1
Write down the first five terms of the following sequences
a)
b)
c)
Definition: A sequence is said to be:
(1) Bounded above if there is such that .
(2) Bounded below if there is such that .
(3) Bounded if it bounded below and bounded above; otherwise it is unbounded .
It is easy to see that a sequence is bounded if and only if there is a positive real number such
that for all .
Examples
a) The sequence
is bounded since
for all .
b) The sequence is not bounded above and is not bounded below.
c) The sequence
is bounded below by but is not bounded above.
4.3 Convergent and divergent sequences
Convergence of a sequence is concerned with the behaviour of the sequence as increases.
Definition: A sequence is said to converge to real number if given there exists a
natural number (which depends of ) such that for all .
Or
If converges to , then we say the is the limit of as increases without bound and we
write
or as .
Note that if a sequence does not converge to a real number, it is said to diverge.
Definition: A sequence is said to diverge to denoted by as if for any
particular real number there is an such that for all
Similarly, diverges to denoted by as if for any particular real number
there is an such that for all
Example 4.1
1. Show that a sequence converges to zero if and only if the sequence converges to zero.
Solution
Assume that the sequence converges to zero. Then given , there exists a natural number
(which depends on ) such that .
Now for all we have That is the sequence converges to zero.
For the converse, assume that the sequences converges to zero. That is , there exists a
natural number (which depends on ) such that:
. It follows that the sequence converges to zero
2. Show that
Solution
Let be given. We can find a such that
. By Archimedean property,
there is an such that
Thus if then we have that
.
That is
.
3. Show that
Solution
Let be given. We need to find an such that
Noting that:
and
We have
.
Now by Archimedean property, there is a such that
. Therefore, for all we
have:
.
Thus
4. Show that the sequence diverges.
Solution
Assume that the sequence converges to a number say . Then with
, there is an such that
.
In particular,
Therefore, for all ,
, which is impossible.
Thus diverges.
5. Show that the sequence diverges.
Solution
Assume that the sequence converges to some real number . Then with there exists a number
such that for all .
Now if is odd, then we have . Hence .
And if is even we have . Hence . A contradiction.
Exercise 4.2
1. Show that if and then
2. Suppose that is a sequence such that for all . Show that as
if and only iff
Theorem 4.1
Let and be sequences of real numbers and let if for some positive real number and
some we have for all and if
then
.
Proof
Let be given. Since as , there exist an such that
for all
Let , then for all we have
.
Thus
.
Example 4.2
Show that
Solution
Since
for each , there is a nonnegative real number such that
. Thus,
by binomial theorem we have:
Therefore,
hence
or
for all
Now since
and
we have by theorem 4.1 that
Theorem 4.2 (Uniqueness of limits)
Let be a sequence of real number. If
and
, then .
Proof
Let be given. Then there exist natural numbers and such that
for all and
for all .
Let , then for all we have:
Thus, and since holds for every , we have and so
. Thus, a sequence converges to only and only one limit (the limit of a sequence is
unique)
Proposition 4.1
A sequence converges to if and only if for each , the set is
finite.
Theorem 4.3
Every convergent sequence of real numbers is bounded.
Proof
Let be a sequence of real numbers which converges to , then with there exists an
such that
By the triangle inequality, we have that: for all
Let . Then for all That is the sequence is
bounded.
The converse of Theorem 4.3 is not necessarily true. There are sequences which are bounded but do
not converge. E.g. the sequences is bounded but not convergent.
Theorem 4.4 (Squeeze theory on limits)
Suppose that , and are sequences such that for all . If
, then
Proof.
Let be given. Then there exist such that:
for all and for all
That is for all and
Let . Then for all we have
and consequently for all .
That is
Example 4.3
1. Show that
Solution
Since
and
it follows that
2. Show that for every with
Solution
Without loss of generality, assume that and Since then there is a positive real
number such that
.
Then
for some then
this implies that
and so,
Since
= 0 =
. We have by the squeeze theorem, that
Exercise 4.3
1) Show that for any ,
2) Show that
3) Show that
4) Find the term of the sequence
Theorem 4.5
Let be a subset of which is bounded above. Then there exists a sequence in such that
.
Proof
Let . By the characterization of supremum in Theory 3.4, for each there exists
such that
. Since
, we have by squeeze theorem that
4.4 Algebra of Limits of sequences
Theorem 4.6
Let and be sequences of real numbers which converges to and respectively. Then
i)
ii)
iii)
if for all and
Proof
i) Let be given. Then there exist and in such that:
for all and
for all .
Let . Then for all we have
Hence,
ii) Let be given. Now,
.
Since is convergent, it is bounded. There exist a such that for all .
Thus
Let . Then
Since and and , Then there exist and in such that:
for all
and
for all .
Let . Then for all we have:
Therefore,
iii) (left to the reader)
Exercise 4.4
Show that if the sequence converges to , then the sequence
converges to
Definition: Let be a sequence of real numbers we say that is:
a) Increasing if for each ,
b) Strictly increasing if for each ,
c) Decreasing if for each ,
d) Strictly decreasing if for each ,
e) Monotone if is increasing of decreasing
f) Strictly monotone if is strictly increasing or decreasing
Remark
An increasing sequence is bounded below by a decreasing sequence is bounded below by
it therefore follows that an increasing sequence is bounded if and only if it is bounded above and a
decreasing sequence is bounded if an only if it is bounded below.
Examples
1. The sequence is increasing
2. The sequence is decreasing
3. The sequence is strictly increasing
4. The sequence is strictly decreasing
Theorem 4.7
Let a sequence be a bounded sequence.
i) If is monotonically increasing then it converges to its supremum.
ii) If is monotonically decreasing then it converges to its infimum.
Proof
Let and and take
i) Since there exists such that .
Since is increasing then
for all
for all
ii) Since there exists such that .
Since is decreasing then
for all
for all
Theorem 4.8
A monotone sequence converges if and only if it is bounded.
Proof
We already proved in Theorem 4.3 that every convergent sequence is bounded. To prove the converse
let be a bounded increasing sequence and let Since S is bounded above it has a
supremum, say. We claim that
Let be given, by the characterization of
supremum, there exist such that for all . Thus,
for all
The proof for the case when the sequence is decreasing is similar.
Example 4.4
Show that
is a convergent sequence
Solution
We show that
is monotone and bounded. Its convergence will ben follow from theorem 4.8.
Monotonicity: Let
then,
This for all so the sequence
is monotone decreasing.
Another proof for monotone: Consider
,
for all . Thus, is
decreasing on . Therefore i.e.
for all
Boundedness:
is bounded below by 1. So
is a convergent sequence by Theorem 4.8.
4.6 Subsequences
If the terms of the sequence are contained in other sequences then is a subsequence of
.
Definition:
Let be a sequence of real numbers and let be a sequence of natural numbers
such that . Then the sequence is called a subsequence of That is a
subsequence of sequence is strictly increasing function .
Example:
Let be the sequence
then
and are subsequences of
Theorem 4.9
Let be a sequence which converges to . Then any subsequences of converges to .
Proof
Let be a subsequence of and let be given. Then there is an such that
for all . Thus when we have that and so for all
. Thus,
.
Theorem 4.10 (Bolzano Weierstrass theorem for sequences)
Every bounded infinite sequence of real numbers has a convergent subsequence.
Proof (is left to the reader)
4.7 Cauchy Sequences
Definition: A sequences is said to be Cauchy if given any , there exists a such
that
for all .
Or
Or is a Cauchy sequence if
.
Example 4.5
Show that the sequence
is a Cauchy sequences.
Solution
For all ,
.
Therefore if then,
Let be given. Then there is an such that
. Thus, for all we have
. Hence, is Cauchy.
Theorem 4.11
Every convergent sequence is a Cauchy sequence.
Proof
Assume that converges to . Then given there exists an such that
for
all . Now for all we have that:
Thus is a Cauchy sequence.
Theorem 4.12
Every Cauchy sequence is bounded.
Proof
Let , then there exists such that for all . Choose a and
observe that
Let .
Then for all and therefore, is bounded.
Theorem 4.13
Every Cauchy sequence of real numbers converges.
Proof
Let be Cauchy, by Theorem 4.12, is bounded and therefore by Bolzano Weierstrass
theorem, has a subsequence that converges to some real number .
We claim that converges to . Let be given. Then there exist , such that:
for all and
for all
Let Then for all we have:
Therefore
and so is convergent.
Theorem 4.11 and theorem 4.13 combined together gives the Cauchy's convergent criterion for
sequences: "A sequence of real numbers converges if and only if it is Cauchy."
Example 4.6
1) Use Cauchy criterion to show that the sequence
converges.
Solution
We need to show that the sequence
in Cauchy. To that end and
. then, for all
with
. Now there is an
such that
. Thus, for all we have
.
Thus
is a Cauchy sequence and so it converges.
2) Show that the sequence
diverges.
Solution
It suffices to show that is not a Cauchy sequence. Now for with , we have:
>
In particular if we take we get
thus, is not a Cauchy sequence and so it diverges.
Exercise 4.5
Show that every subsequence of a bounded sequence is bounded.
4.8 Infinite series
When the individual terms of a sequences are summed a series of real numbers is obtained. If
is an infinite sequence then:
is an infinite series. Then numbers
are terms of the series.
Definition: For the infinite series
the partial sum is given by:
.
If the sequence of partial sum converges to then, the series
converges. The limit is
called the sum of the series.
so
.
If diverges then
diverges.
Example 4.7
a) The series
has the following partial sums
Because
. It follows that the series converge and its sum is 1.
b) The partial sum of the series
is given by
because the
. Thus, the series converge and its sum is 1.
c) The series
diverges because and the sequence of partial sums
diverges.
NB. The series in Example 4.7b) is a telescoping series of the form
.
Because the sum of a telescoping series is given by , it follows that a telescoping
series will converge if and only if approaches a finite number as . Moreover, if the series
converges its sum is
.
Writing a series in telescoping form
Find the sum of the series
.
Solution
Using partial fractions, we write
.
The partial sums is
so the series
converge and its sum is
.
4.9 Geometric series
A geometric series is given by
. Provided , is
called the ratio.
Theorem 4.14
A geometric series with ratio diverges if and if the series converges to the sum
Proof
It is easy to see that the series diverge if . If then
. Let us multiply this equation by to yield:
Subtracting from we get:
, with
When , it follows that as that is
. Thus, the series diverge
When , it follows that 0 as that is
. Thus, the series converge
and the sum is
.
Example 4.8
a) The geometric series
has
and . Because
then the series converge and its sum is
.
b) The geometric series
has
. Since then the series
diverge.
c) Use a geometric series to write
as a ratio of two integers.
Solution
For the repeated decimal
, we write
For this series we have
and
so the series converge and the sum is given by
4.10 Properties of infinite series
Theorem 4.15
Let and be convergent series and let , and be real numbers. If and
, the following series converged to the indicated sums.
1)
2)
4.11 Convergent criterion for series
4.11.1 Limit of term test for divergence
We first provide a proposition whose contrapositive gives the desired test criterion for divergence.
Proposition 4.1 (limit of term for a convergent series)
If
converges then
Proof
Assume that
then because and
it follows that
That is
and so
Proposition 4.2 (limit of term test for divergence)
If
then
diverges. (Contrapositive of proposition 4.1)
Example 4.9
1) For the series
we have:
so, the limit of the term is not zero so the series diverge.
2) For the series
we have
so, the series diverge.
4.11.2 Integral test
Proposition 4.3
If is positive, continuous and decreasing for and then:
and
either both diverge of converge.
Proof
Begin by partitioning the interval into unit interval as illustrated on Figure 4.1 a and
4.1b. The total area of the inscribed rectangles and the circumscribed rectangles are as follows:
inscribed rectangles
circumscribed rectangles
The exact area under the graph i from to lies between the inscribed area and the
circumscribed area.
Using the partial sum, we write (1) as
, assuming that
to it follows that for
Consequently is bounded and monotonic and by Theorem 4.7, it converges. So converges.
For the other direction proof;
Assume that the improper integral
diverges, then
approaches infinity as
and the inequality
implies that diverges and so diverges.
Example 4.10
Apply the integral test to the series
Solution
is positive and continuous for . We find
for and so
is decreasing. satisfies the conditions for the integral test.
So
so the series
diverge.
Inscribed rectangles
Circumscribed rectangles
4.11.3 P-series test
Proposition 4.4
A series of the form
1) Converge if and 2) Diverges if
Example 4.11
Discuss the convergence and divergence of
a) Harmonic series and b) p-series with
Solution
a) By the p-series test, it follows that for the harmonic series
,
diverges.
b) It follows from the p-series test that the series
, so
the series converges.
Exercise 4.6
1) Use the integral test to determine the divergence and convergence of the following series
a)
b)
c)
d)
2) Explain why the integral test does not apply to the following series
a)
b)
c)
3) use the p-series test to determine the convergence of divergence of the following series.
a)
b)
4.11.4 Comparison test
Direct comparison
Proposition 4.5
Let for all
1. If
converges then
converges. 2. If
diverges then
diverges
Proof
To prove the first property, let
and let . Because
the sequences is nondecreasing and bounded above by so it must converge. Because
it follows that converges.
The second property is logically equivalent to the first.
Example 4.7
1) Determine the convergent and divergent of the series
Solution
The series
resembles
converging geometric series. Term by term comparison
yields:
so, since
converges then
also converges.
2) Determine the convergent and divergent of the series
.
Solution
The series
resembles
divergent p-series. And
which does
not meet the requirement for divergence. We also compare with
divergent harmonic series.
and by the direct comparison test, the given series converge.
Limit comparison
Proposition 4.6
Suppose that , and
where is finite and positive. Then the two series
and either both diverge of converge.
Proof
Because and
there exists such that
for .
This implies that . So, by the direct comparison test, the convergence of
implies the convergence of . Similarly, the fact that
can be used to show
that the convergence of implies the convergence of .
NB: Some examples of p-series to use in comparison tests for given series are in the table below
The table above suggests that when choosing a series for comparison one disregards all but the
highest powers of in both the numerators and the denominator.
Example 4.8
1) Show that the general harmonic series
diverge.
Solution
By comparing with
divergent harmonic series, we have
because this limit
is finite and positive then, the given series diverges.
2) Determine the convergence of divergence of
Solution
Compare the series with
convergent p-series
Because
then by limit comparison test the given series converge.
3) Determine the convergence of divergence of
.
Solution
Compare with
divergent series. Now that the series diverge, by term test from the limit
so, the given series diverge.
4.11.5 Alternating test
Proposition 4.7
Let the alternating series and converges if the following
conditions are satisfied.
1)
2) for all .
Proof
Consider the alternating series . For this series, the partial sum (where )
has all nonnegative terms and
therefore the sequence is a nondecreasing sequence, we can also write:
which implies that for
every integer . So, is bounded, nondecreasing and converges to some value Because
and we have
Because both and converges to the same limit , it follows that also converges to .
Consequently, the given alternating series converge.
4.11.6 Absolute convergence
Proposition 4.8
If a series converges then the series converge.
Because for all , the series
converges by comparison with
the convergent series . Furthermore because we write e
where both series on the right converge.
So, it follows that converges.
The converse of proposition 4.8 is not true. For example, the alternating harmonic series
converge by alternating series test. Yet the harmonic series diverge. This type of convergence is
called conditional.
Note: (1) is absolutely convergent converges
(2) conditionally converges if converges but diverges
4.11.7 Ratio test
Proposition 4.9
Let be a series with nonzero terms:
1) converges if
2) diverges if
3) The ratio test is inconsistent if
4.11.8 Root test
Proposition 4. 10
Let be a series,
1) converges absolutely if
2) diverges absolutely if
3) The root test is inconsistent if
LIMITS AND CONTINUITY
5.1 Limit of a function
Definition:
Let be a function defined on an open interval containing (expcept possibly at ) and let be a
real number. is called a limit of at if given , there exists a (depending on and
) such that: for all satisfying .
We write
.
Or 1) Suppose is defined for all real numbers where , then is the limit of
as tends to if, given there exist a real number such that whenever
and
.
2) Suppose is defined for all real numbers where , then is the limit of
as tends to denoted by and
if, given there exist a real number
such that whenever .
Example 5.1
1) Show that
.
Let be given. We need to produce a >0 whenever
Now we have
Consider all which satisfy the inequality . Then, for all such we have
Thus . Choose
. Then, whenever
we have that .
2) Show that and
.
Let be given. We need to find a such that: for all
satisfying .
Note that since we are interested in
the value of near , we may consider those values of satisfying the inequality
i.e. . For all these values we have that . Therefore, if we
have that . Choose
, then, working backwards,
we have that: for all satisfying
3) Show that and
.
Solution
Let be given. We need to find a such that:
for all satisfying
. Consider values of which satisfy
i.e.
.
Recognizing that as the distance from to we have
. Thus,
Choose
. Then whoever we have that
.
4) Show that
where
does not exists.
Solution
Assume that the limit exists and
. Then with there is a such that
for all satisfying . Taking
, we have that
and
so,
…………………….. (*)
On the other hand, if
, we have
so
………………………. (**)
Therefore, there is no real number that simultaneously satisfy equations (*) and (**). So,
does not exit.
5) Show that
Solution
Let be given. We need to find a , such that:
for all satisfying
. We have
Which proves that that
6) Consider the function given by
Show that if , then that
does not exist.
Solution
Assume that there is an such that that
. Then with
, there exist a
such that for all satisfying .
If then, we have
whenever, .
If then, we have
whenever, .
Since the set contains both rational and irrationals, we have that
, which is absurd. And so
does not exist.
7) Given
, find such that whenever
Solution
Given . To find notice that
Choose
, then implies that
. So .
Exercise 5.1
Use definition of limits to prove that:
a)
b)
c)
Theorem 5.1
Let be defined in some open interval containing except possibly at The
if and only if for every sequence such that
we have
.
Prove
Assume that
and be a sequence such that
. Then given ,
there exists a and a such that fo r all satisfying and
for all Now, since for all Thus
for all Therefore,
For the converse, assume that for every such that
we have
Claim that
. If the claim were false, then there would exist an such that for every
with , we have . Let and take
, then we can find
such that
and
Clearly is a sequence in with the property that
and for all
Thus that
this is a contradiction.
Theorem 5.2 (uniqueness of limits)
Let be a function which is defined on some open interval containing , except possibly at . If
and
then
Proof
If , let
. Then there is a and such that
whenever and and
whenever and
. Let . Then whenever, we have
, which is impossible and so .
5.2 Algebra of limits
Theorem 5.3
Let . Suppose that and are real valued functions defined on some open interval
containing , except possibly at itself, and that
and
then,
a)
b)
c)
provided and
Proof
a) Let be given. Then there exists and such that:
whenever, and and
whenever, and
Let Then, whenever and we have:
Thus
A similar argument shows that
b) With , there exists a such that whenever and
.
This implies that whenever and
.
Now with we have
Given there exists and such that
whenever with and
whenever with
Let . Then, whenever with we have
=
Thus
c) It is enough to show that
provided for all and . Since
,
>0. Therefore, there exists a such that
whenever
and .
Now for all satisfying we have
that is
for all satisfying . It now
follows that for all satisfying
Given , there exist such that
whenever and
.
Let . Then, whenever with we have
Thus,
and from part b) of theorem 5.3,
Theorem 5.4
Let . Suppose that and are real valued functions defined on some open interval
containing , except possibly at itself and that . If
and
then .
Prove (Left to the reader)
Theorem 5.5
Let and be real valued function which are defined on some open interval containing , except
possibly at and that for all if
,
, then
.
Exercise 5.2
1. Prove Theorem 5.4
2. Prove Theorem 5.5
3. Find the limits of
a)
b)
5.3 Continuity of functions
Definition:
Continuity at appoint: A function is continuous at if the following conditions are met.
i) is defined
ii)
exists
iii)
Continuity on an open interval: a function is said to continuous at an open interval if its
continuous at each point in the interval. A function that is continuous on the entire real line
is everywhere continuous.
Or 1) Let and . Then function is said to be continuous at if given
there exist a such that whenever and .
2) Let and . Then function is said to be continuous at if for each
of there is of such
that whenever .
Example 5.2
1. Show that is continuous on
Solution
Let be given and . We need to produce a such that:
whenever .
Now since we need the behavior of near , we
may restrict our attention to those real numbers that satisfy the inequality
therefore, for all those real numbers we have
. Now take
. Then,
Thus, is continuous at . Since was arbitrary chosen from , it follows that is continuous on
.
2. Show that the function
is not continuous at 0.
Solution
Let be given. We need to produce a such that:
whenever . Now
Choose Then, implies that
Thus, is continuous at 0.
3. Show that the function given by
is discontinuous
at every real number.
Solution
Assume that is continuous at some number . Then, given there exists a such
that whenever . Since rational numbers and irrational numbers are
dense in , the interval contains both rational and irrationals. If and
then whenever ,
On the other hand, if and then whenever .
But there is no real number which simultaneous satisfy the inequalities and
.
Therefore is discontinuous at every .
4. Show that
is continuous at 1.
Solution
Let be given. We need to find a such that whenever .
Since we are interested in the values of for which
. These values satisfy the inequality
. Now for all which satisfy
we have
. Chose
. Then, whenever
we have that
Theorem 5.6
Let and . Then is continuous at if and only if for every sequence
such that
we have that
Proof.
Suppose that is continuous at and that a sequence in such that the
. Given
, there exist a and an such that whenever and
for all .
Therefore for all . That is
For the converse, assume that for every sequence such that
we have
and that is not continuous at . Then, there exist an such that for every
with we have . For , let
and then we can
find such that
and .
Clearly, is a sequence in with the property that
and for all
. That is
a contradiction.
Example 5.3
Find the limit of the sequences
, if it exists
Solution
Since
and the function is continuous on , if follows from Theorem
5.6 that
That is the sequence
converges to zero.
Exercise 5.3
Show that the function
is continuous at
Theorem 5.7
Let and be functions with a common domain and let . If and are continuous
at then so are the functions (i) (ii) for each (iii) .
Theorem 5.8
Let be a function which is continuous at . Suppose that is a function which is
continuous at the point Then the composition function is also continuous at .
5.4 The intermediate value theory
Theorem 5.9 (Intermediate Value Theorem)
If is a continuous function on a closed interval and then for each number
between and there is a point such that .
Proof
For definiteness, assume that .
Let . Then since . Thus exists as a real number in
, by Theorem 4.5 there exists a sequence in such that
Since
for each we have that , and so is continuous at . This implies that
.
As for each , we deduce that and so, It now remains to show
that since , and ,
for each . Also, since we have that
. Therefore, there exists an such that
. Hence for each we have that
i.e.
. this implies that for all ,
and
. Thus
for all . By the continuity of we obtain that hence .
Theorem 5.9 (Fixed point theorem)
If is continuous on a closed interval and for each then has a
fixed point. i.e. there is a point such that
Definition:
Let and . The function is said to be uniformly continuous on if given any
there exist a such that whenever and .
The most important point to note here is that does not depend on any particular point of the
domain - the same works for all paints of
Example 5.4
1) Show that the function is uniformly continuous on.
Solution
Let be given. We must produce such that whenever and
. Since we may choose . Then for all with
we have Thus is uniformly continuous on .
2) Show that is not uniformly continuous on .
Solution
Let be given. We must show that for every there exist such that
and Choose with
and
.
Then and
. Thus is not continuous on .
3) Show that the function is continuous on .
Solution
Let be given. Then for all we have
. Choose
. Then for all with we have
Therefore is uniformly continuous on .
Exercise 5.4
Show that
is not uniformly continuous on
Theorem 5.10
If is uniformly continuous on , then it is continuous on .
5.6 Discontinuity
If is a point in the domain of definition of the function at which is not continuous we say that
is discontinuous at or has a discontinuity at .
If the function is defined on an interval, the discontinuity is divided into two:
(1) Let be defined on . If is discontinuous at point and if and exist the
is said to have a discontinuity of the 1st kind or simple discontinuity.
(2) Otherwise the discontinuity is said to be of the 2nd kind.
DIFFERENTIATION
6.1 Derivative of a function
Definition:
Let be defined and real valued on . For any point , form a quotient
and
define
provided the limits exists.
is called derivative of .
Observation
1. If is defined at point then is differentiable at .
2. exists if and only iff for a real number a real number such that
whenever .
3. then, we have:
.
4. is differentiable at if and only if is removable discontinuity of the function
.
Example 6.1
1. A function defined by
This function is differentiable at because:
2. Let , (n is an integer), . Then:
Implying that is differentiable everywhere and
Theorem 6.1
Let be defined on , if is differentiable at a point , then is continuous at .
Proof
We want that
where and .
Now
Implying that
which shows that is continuous at
Note that the converse of Theorem 6.1 does not hold.
Example 6.2
Let be defined by
Then
=
And
=
Since then does not exist.
6.2 Rules of differentiation
Theorem 6.2 (Rules of differentiation)
Suppose and are differentiable on and are differentiable at a point on [a, b], then ,
and
are differentiable at and
i) ii)
iii)
6.3 Local Maximum
Let be a real valued function defined on a metric space we say that has a local maximum at
point if there exists such that with .
Local minimum is defined otherwise.
Theorem 6.3
Let be defined on , if has a local maximum at a point and if exists then
. (the analogy for local minimum is of course also true)
Proof
Choose such that .
Now if then,
. Taking limits at we get
--------------------------------------------------(1)
If then,
. Again, taking limits as we get
--------------------------------------------------(2)
Combining (1) and (2) we get that
6.4 Generalization of Mean Value theorem (MVT)
Theorem 6.4
If and are continuous real valued functions on closed interval , then there is a point
at which
Theorem 6.5 (Lagrange's MTV)
Let be (i) Continuous at
(ii) Differentiable on
Then a point such that
6.5 L'hospital rule
Theorem 6.6
Suppose and exist and . Then
Proof
=
Exercise 6.1
1. Prove the Lagrange's Mean value theorem
2. Let be defined for all real numbers and suppose that
Prove that is constant.
3. If in then prove that is strictly increasing in and let be its inverse
function, prove that the function is differentiable and that
,
4. Suppose is defined and differentiable for every and as , put
Prove that as
5. If then compute and show that does not exist.
6. Suppose is defined in the neighborhood of a point and exist. Use Lagrange's
mean value theorem to show that
End
Congratulation upon finishing this course unit. As you prepare to take your examination, I wish you
the very best.
References
1. Gelbaum, B.R. (1992). Problems in Real and Complex Analysis, Springer-Verlag, New
York.
2. Hart, F. M. (1987), A Guide to Analysis, The Macmillan Press Ltd, New York.
3. Kolmogorov, A.N. and Formin, S.V. (1975). Introductory Real Analysis, Dover
Publications, Inc., New York.
4. Rudin, W. (1987). Real and Complex Analysis, McGraw-Hill Book Company, New York.
5. Shilov, G.E. (1996). Elementary Real and Complex Analysis, Dover Publications, Inc.,
New York.
6. Edwards, C.H. (1994). Advanced Calculus of Several Variables, Dover Publications, Inc.,
New York
7. Marsden, J.E. and Hoffman, M.J. (1993). Elementary Classical Analysis, 2nd ed., W.H.
Freeman and Company, New York
8. Marsden, J.E. and Tromba A.J. (2003). Vector Calculus, 5th ed., W.H. Freeman and
Company, New York.
9. Larson. E. (2010). Calculus (9th Ed). Cengage Learning.
ResearchGate has not been able to resolve any citations for this publication.
Calculus (9 th Ed). Cengage Learning
- E Larson
Larson. E. (2010). Calculus (9 th Ed). Cengage Learning.
Introductory Real Analysis Dover Books On Mathematics Pdf
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