Intro to Analysis Notes

8/8/2019 Intro to Analysis Notes

1/111

Introduction to Analysis

Lecture Notes 2005/2006

Vitali LiskevichWith minor adjustments by Vitaly Moroz

School of MathematicsUniversity of Bristol

Bristol BS8 1TW, UK


2/111

Contents

1 Elements of Logic and Set Theory 11.1 Propositional connectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.1 Negation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.2 Conjunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.3 Disjunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.4 Implication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.5 Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Logical laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3.1 Subsets. The empty set . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.2 Operations on sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3.3 Laws for operations on sets . . . . . . . . . . . . . . . . . . . . . . . . 101.3.4 Universe. Complement . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4 Predicates and quantiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Ordered pairs. Cartesian products . . . . . . . . . . . . . . . . . . . . . . . . 141.6 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.7 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.7.1 Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.7.2 Injections and surjections. Bijections. Inverse functions . . . . . . . . 21

1.8 Some methods of proof. Proof by induction . . . . . . . . . . . . . . . . . . . 24

2 Numbers 272.1 Various Sorts of Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.1 Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.1.2 Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.1.3 Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.1.4 Cuts of the Rationals . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.2 The Field of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.3 Bounded sets of numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.4 Supremum and inmum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

II


3/111

CONTENTS

3 Sequences and Limits 403.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.2 Null sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.3 Sequence converging to a limit . . . . . . . . . . . . . . . . . . . . . . . . . . 413.4 Monotone sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.5 Cauchy sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.6 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.6.1 Series of positive terms . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4 Limits of functions and continuity 554.1 Limits of function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.2 Continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.3 Continuous functions on a closed interval . . . . . . . . . . . . . . . . . . . . 634.4 Uniform continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.5 Inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

5 Differential Calculus 685.1 Denition of derivative. Elementary properties . . . . . . . . . . . . . . . . . 685.2 Theorems on differentiable functions . . . . . . . . . . . . . . . . . . . . . . . 725.3 Approximation by polynomials. Taylors Theorem . . . . . . . . . . . . . . . 75

6 Series 786.1 Series of positive and negative terms . . . . . . . . . . . . . . . . . . . . . . . 78

6.1.1 Alternating series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 786.1.2 Absolute convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . 796.1.3 Rearranging series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.1.4 Multiplication of series . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

6.2 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

7 Elementary functions 877.1 Exponential function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

7.2 Logarithmic function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 897.3 Trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

8 The Riemann Integral 928.1 Denition of integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

8.1.1 Denition of integral and integrable functions . . . . . . . . . . . . . . 928.1.2 Properties of upper and lower sums . . . . . . . . . . . . . . . . . . . 94

8.2 Criterion of integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 958.3 Integrable functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 968.4 Elementary properties of integral . . . . . . . . . . . . . . . . . . . . . . . . . 97

8.5 Integration as the inverse to differentiation . . . . . . . . . . . . . . . . . . . 102

III January 28, 2006


4/111

8.6 Integral as the limit of integral sums . . . . . . . . . . . . . . . . . . . . . . . 1048.7 Improper integrals. Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

8.8 Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106


5/111

Chapter 1

Elements of Logic and Set Theory

In mathematics we always assume that our propositions are denite and unambiguous, so thatsuch propositions are always true or false (there is no intermediate option). In this respectthey differ from propositions in ordinary life, which are often ambiguous or indeterminate. Wealso assume that our mathematical propositions are objective, so that they are determinatelytrue or determinately false independently of our knowing which of these alternatives holds.

We will denote propositions by capital Roman letters.Examples of propositions1. A London is the capital of the UK.2. B Paris is the capital of the UK.3. C

3 > 5

4. D 3 < 5. etc.(We will use the sign in order to dene propositions.)A and D are true, whereas B and C are false, nonetheless they are propositions. Thus

with respect to our agreement a proposition may take one of the two values : true orfalse (never simultaneously). We use the symbol T for an arbitrary true proposition andF for an arbitrary false one.

Not every sentence is a mathematical proposition. For instance, the following sentencesare not mathematical propositions.

(i) It is easy to study Analysis.

(ii) The number 0.00001 is very small.

(iii) Is there a number whose square is 2?

(iv) x > 2.

About (i) it is impossible to judge whether it is true or false (depends for whom...). (ii) doesnot have precise sense. (iii) is a question, so does not state anything. And (iv) contains aletter x and whether it is true or false depends on the value of x.

At the same time, there are propositions for which it is not immediately easy to establishwhether they are true or false. For example,

E (1263728 + 15 15876 ) + 4 is a prime number.1


6/111

1.1. PROPOSITIONAL CONNECTIVES

This is obviously a denite proposition, but it would take a lot of computation actually todetermine whether it is true or false.

There are some propositions in mathematics for which their truth value has not beendetermined yet. For instance, in the decimal representation of there are innitely manydigits 7. It is not known whether it is true, but this is certainly a mathematical proposition.The truth value of this proposition constitutes an open question (at least the answer is notknown to the author of these notes). There are plenty of open questions in mathematics!

1.1 Propositional connectives

Propositional connectives are used to combine simple propositions into complex ones. They

can be regarded as operations with propositions.

1.1.1 Negation

One can build a new proposition from an old one by negating it. Take A above as an example.The negation of A (not A) will mean

A London is not the capital of the UK .We will use the symbol A to denote not A. Another example. For the proposition D {8 is a prime number }, its negation is D {8 is not a prime number }. Since we agreedthat we have one of the two truth values : truth or false, we can dene negation of a propositionA by saying that if A is true then A is false, and if A is false then A is true. This denitionis reected in the following table.

A AT F F T

This is called the truth table for negation.

1.1.2 Conjunction

Conjunction is a binary operation on propositions which corresponds to the word and inEnglish. We stipulate by denition that A and B is true if A is true and B is true, and Aand B is false if A is false or B is false. This denition is expressed in the following truthtable. We use the notation AB for the conjunction A and B .

A B ABT T T T F F F T F

F F F

2 January 28, 2006


7/111

CHAPTER 1. ELEMENTS OF LOGIC AND SET THEORY

The four rows of this table correspond to the four possible truth combinations of the propo-sition A and the proposition B . The last entry in each row stipulates the truth or falsity of

the complex proposition in question.Conjunction is sometimes called logical product .

1.1.3 Disjunction

Disjunction is a binary operation on propositions which corresponds to the word or inEnglish. We stipulate by denition that A or B is true if A is true or B is true, and Aor B is false if A is false and B is false. This denition is expressed in the following truthtable. We use the notation AB for the disjunction A or B .

A B A

BT T T T F T F T T F F F

Disjunction is sometimes called logical sum .

1.1.4 Implication

Implication is a binary operation on propositions which corresponds to the word if... then...

in English. We will denote this operation . So AB can be read if A then B , or Aimplies B . A is called the antecedent, and B is called the consequent. The truth table forimplication is the following.

A B ABT T T T F F F T T F F T

So, as we see from the truth table which constitutes the denition of the operation im-plication, the implication is false only in the case in which the antecedent is true and theconsequent is false. It is true in the remaining cases.

Notice that our introduced implies differs from that used in the ordinary speech. Thereason for such a denition will become clear later. It proved to be useful in mathematics.Now we have to accept it. Thus in the mathematical meaning of implies the proposition

Snow is black implies grass is red

is true (since it corresponds to the last line of the truth table).The proposition AB can be also read as A is sufficient for B , B if A, A only if

B and B is necessary for A (the meaning of the latter will be claried later on).

3 January 28, 2006


8/111

1.1. PROPOSITIONAL CONNECTIVES

1.1.5 Equivalence

The last binary operation on propositions we introduce, is equivalence . Saying that A isequivalent to B we will mean that A is true whenever B is true, and vice versa. We denotethis by AB . So we stipulate that AB is true in the cases in which the truth values of A and B are the same. In the remaining cases it is false. This is given by the following truthtable.

A B ABT T T T F F F T F F F T

AB can be read as A is equivalent to B , or A if and only if B (this is usuallyshortened to A iff B ), or A is necessary and sufficient for B . The equivalence ABcan be also dened by means of conjunction and implication as follows

(AB )(BA).

Now let A be a proposition. Then A is also a proposition, so that we can construct itsnegation A. Now it is easy to understand that A is the same proposition as A as wehave only two truth values T and F .

A

A

The last equivalence is called the double negation law .

4 January 28, 2006


9/111


1.2 Logical laws

In our denitions in the previous section A, B , etc. stand for arbitrary propositions. Theythemselves may be built from simple propositions by means of the introduced operations.Logical laws or, in other words, logical tautologies are composite propositions built from

simple propositions A, B , etc. (operands) by means of the introduced operations, that aretrue , no matter what are the truth values of the operands A, B etc.

The truth value of a proposition constructed from A, B , etc. does not depend on thepropositions A, B , etc. themselves, but depends only on the truth value of A, B , etc. Hencein order to check whether a composite proposition is a law or not, one can plug in T or F instead of A, B , etc. in all possible combinations to determine the corresponding truth valuesof the proposition in question. If all the values are T then the proposition in question is alaw. If there is a substitution which gives the value F , then it is not a law.

Example 1.2.1. The proposition

(AB )(AB )

is a law.

The shortest way to justify this is to build the truth table for ( AB )(AB ).

A B AB AB (AB )(AB )T T T T T T F F T T F T F T T F F F F T

As we see the last column consists entirely of T s, which means that this is a law.

Below we list without proof some of the most important logical laws. We recommend thatyou verify them by constructing the truth tables of these laws.

Commutative law of disjunction(1.2.1) ( A

B )

(B

A)

Associative law of disjunction(1.2.2) [(AB )C ][A(BC )]

Commutative law of conjunction(1.2.3) ( AB )(BA)

Associative law of conjunction

(1.2.4) [(AB )C ][A(BC )]

5 January 28, 2006


10/111

1.2. LOGICAL LAWS

First distributive law

(1.2.5) [A(BC )][(AB )(AC )]

Second distributive law(1.2.6) [A(BC )][(AB )(AC )]

Idempotent laws(1.2.7) ( AA)A, (AA)A

Absorption laws(AT )A, (AF )F,(AT )T, (AF )A(1.2.8)

Syllogistic law(1.2.9) [(AB )(BC )](AC )

(1.2.10) ( AA)T

(1.2.11) ( AA)F

De Morgans laws

(AB )(AB )(1.2.12)(AB )(AB )(1.2.13)

Contrapositive law(1.2.14) ( AB )(B A)

(1.2.15) ( AB )(AB )

6 January 28, 2006


11/111


1.3 Sets

The notion of a set is one of the basic notions. We cannot, therefore, give it a precisemathematical denition. Roughly speaking:

A set is a collection of objects, to which one can assign a size.

It is obviously not a denition (what is collection? what is size?). A rigorous set theoryis constructed in the axiomatic way. We however conne ourselves to the naive set theory,introducing some axioms only in order to clarify the notion of a set. We accept as the basicnotions set, and the relation to be an element of a set. If A is a set we write aA toexpress that a is an element of the set A, or a belongs to A. If a is not an element of Awe write aA. So the following is true for any x and A:

(xA) (xA).One way to dene a set is just by listing its elements.

Example 1.3.1. (i) A = {0, 1}. This means that the set A consists of two elements, 0and 1.(ii) B = {0, {1}, {0, 1}}. The set B contains three elements: the number 0; the set {1}containing one element: the number 1; and the set containing two elements: the numbers

0 and 1.

The order in which the elements are listed is irrelevant. Thus

{0, 1

}=

{1, 0

}.

A set can be also specied by an elementhood test .

Example 1.3.2. (i) C = {xN | x is a prime number }. The set C contains all primes.We cannot list them for the reason that there are innitely many primes. (Here N isthe set of natural numbers 1 , 2, 3 . . . )

(ii) D = {xR | x2 x = 0}. The set D contains the roots of the equation x2 x = 0. Butthese are 0 and 1, so the set D contains the same elements as the set A in the previousexample. In this case we say that the sets A and D coincide, and write A = D . (HereR is the set of real numbers.)

Sets satisfy the following fundamental law (axiom)

If the sets A and B contain the same elements, then they coincide, (or are equal).

1.3.1 Subsets. The empty set

Denition 1.3.1. If each element of the set A is also an element of the set B we say that Ais a subset of B and write AB .

This denition can be expressed as a logical proposition as follows:

For every x [(xA)(xB )].

7 January 28, 2006


12/111

1.3. SETS

Note that by the denition

(A = B )[(AB )(BA)].This is the main way we will use to prove equalities for sets.

Since it happens that a set may contain no elements, the following denition is useful.

Denition 1.3.2. The empty set is the set which contains no elements.

Warning: there is only one empty set. It is denoted by .

This in particular means that the set of elephants taking this course of Analysis coincideswith the set of natural numbers solving the equation x2 2 = 0.

Let us illustrate the notion of a subset by a simple example.

Example 1.3.3. Let S = {0, 1, 2}. Then the subsets of S are:, {0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}, {0, 1, 2}.

Subsets of a set A which do not coincide with A are called proper subsets. In this exampleall subsets except for the last one are proper.

Denition 1.3.3. The set of all subsets of a set A is called the power set of A, and is denotedby P (A) (or 2A in some books).

Note that in the last example S contains 3 elements, whereas P (S ) has 8 elements (2 3 = 8).In general, if a set A has n elements then P (A) has 2n elements (provided n is nite). Proveit!

1.3.2 Operations on sets

Now we introduce operations on sets. The main operations are: union, intersection, differenceand symmetric difference.

Union of sets

Denition 1.3.4. The union of sets A and B is the set containing the elements of A and theelements of B , and no other elements.

We denote the union of A and B by AB .Note: existence of the union for arbitrary A and B is accepted as an axiom .

For arbitrary x and arbitrary A and B the following proposition is true .

(xAB )(xA)(xB ).

8 January 28, 2006


13/111


Intersection of sets

Denition 1.3.5. The intersection of sets A and B is the set containing the elements whichare elements of both A and B , and no other elements.

We denote the intersection of A and B by A B . Thus for arbitrary x and arbitrary Aand B the following proposition is true .(xA B )(xA)(xB ).

Difference of sets

Denition 1.3.6. The difference of sets A and B is the set containing the elements of Awhich do not belong to B .

We use the notation A B for the difference. The following is true for arbitrary x andarbitrary A and B :

(xA B )[(xA)(xB )].By the de Morgan law and by the double negation law

(x

A

B )

[

(x

A)

(x

B )].

Symmetric difference

Denition 1.3.7. The symmetric difference of the sets A and B is dened by

A B = ( A B )(B A).

Let us illustrate the introduced operations by a simple example.

Example 1.3.4. Let A = {0, 1, 2, 3, 4, 5}and B = {1, 3, 5, 7, 9}. ThenAB = {0, 1, 2, 3, 4, 5, 7, 9}.

A B = {1, 3, 5}.A B = {0, 2, 4}, B A = {7, 9}.

A B = {0, 2, 4, 7, 9}.Note that

AB = ( A B )(A B ).9 January 28, 2006


14/111

1.3. SETS

1.3.3 Laws for operations on sets

Commutative laws

AB = BA,(1.3.16)A B = B A.(1.3.17)

Associative lawsA(BC ) = ( AB )C,(1.3.18)A (B C ) = ( A B ) C.(1.3.19)

Distributive lawsA (BC ) = ( A B )(A C ),(1.3.20)A(B C ) = ( AB ) (AC ).(1.3.21)

Idempotent lawsAA = A, A A = A.(1.3.22)

The proofs of the above laws are based on the corresponding laws for conjunction anddisjunction.

Now let us formulate and prove some laws of the difference.

(1.3.23) A(B A) = AB.Proof.[xA(B A)] {(xA)[(xB )(xA)]}

[(xA)(xB )][(xA)(xA)][(xA)(xB )],

since [(xA)(xA)] is true.From the last formula it follows that the difference is not an inverse operation to the union

(which means that in general A(B A) = B ).

(1.3.24) A B = A (A B ).Proof.

[xA (A B )] {(xA)(xA B )}{(xA)[(xA)(xB )]{(xA)[(xA)(xB )]}{[(xA)(xA)][(xA)(xB )]}[(xA)(xB )]

(xA B )10 January 28, 2006


15/111


since [(xA)(xA)] is false.

De Morgans laws

A (B C ) = ( A B )(A C ),(1.3.25)A (BC ) = ( A B ) (A C ).(1.3.26)

The proof is based on de Morgans laws for propositions.

1.3.4 Universe. Complement

In many applications of the set theory one considers only sets which are contained in somexed set. (For example, in plane geometry we study only set containing points from the

plane.) This set is called a universe . We will denote it by U .Denition 1.3.8. Let U be the universe. The set U Ais called the complement to A. Itis denoted by Ac.It is easy to see that the following properties hold

(Ac)c = A,(1.3.27)(A B )c = AcB c,(1.3.28)(AB )

c = Ac B c.(1.3.29)Using the properties of the difference prove it!

11 January 28, 2006


16/111

1.4. PREDICATES AND QUANTIFIERS

1.4 Predicates and quantiers

In mathematics, along with propositions, one deals with statements that depend on one ormore variables , letters denoting elements of sets. In this case we speak of predicates. Forinstance, n is a prime number is a predicate. As we see, it may be true or false dependingon the value n. A predicate becomes a proposition after substituting the variable by axed element from the set of denition (the set to which the variable belongs). Generally, apredicate can be written as

A(x)(xS ),

where S is the set of denition (which we often omit when it is clear what S is).A subset of S containing all the elements of S which make A(x) true is called the truth

set for A(x).

For the truth set of A(x) we writeTruth set of A = {xS |A(x)}.

Example 1.4.1. Let A {xR |x2 x = 0}. (The notation R is used to denote the set of real numbers, which we will discuss in details later on.) Then

{xR |A(x)}= {0, 1}.We often want to say that some property holds for every element from S . In this case we

use the universal quantier . So for for all x S A(x) we write (x S ) A(x). Afterapplying the universal quantier to a predicate we obtain a proposition (which may be trueor false, as usual). The universal quantier substitutes for the words every, for every,any, for all.

Example 1.4.2. (i) The proposition Every real number has a non-negative square canbe written as

(xR ) [x2 0].

(ii) The proposition Every real number is non-negative can be written as

(xR ) [x 0].Evidently, (i) is true and (ii) is false.

Note that proposition [( xS ) A(x)] means that the truth set of A(x) is the whole setS . Thus if for an element x of S , A(x) is false, the proposition [( xS ) A(x)] is false, hencein order to state that it is false it is enough to nd one element in S for which A(x) is false.

To express that a property holds for some element of S , or in other words, there existsan element in S such that A(x) holds, we usethe existential quantier and write

(xS ) A(x).

substitutes for the words for some, there exists.

12 January 28, 2006


17/111


Note that in order to state that

[(xS ) A(x)] is trueit is enough to nd one element in S for which A(x) is true.

Example 1.4.3. The proposition Some real numbers are greater than their squares canbe written as

(xR ) [x2 < x ].

It is true, of course.

In propositions ( x) P (x) and (x) P (x) P (x) may itself contain quantiers.

Example 1.4.4. (i)(x

N)(yN) [y = x + x].

(ii)(xR )(yR ) [y < x ].

(N denotes the set of natural numbers.)

Quantiers negation laws

The following equivalences are laws

[(xS ) P (x)][(xS ) P (x)],

[(xS ) P (x)][(xS ) P (x)].

This means that we can obtain a proposition with initially placed quantier that is equiv-alent to the negation of a proposition with initially placed quantier by pushing the negationto the inside and changing to and to .

Example 1.4.5. Consider the following proposition

(xN) (y

N) (2y > x ).

The negation of it is the proposition

(xN) (yN) (2y x).Example 1.4.6.

[(xX )(yY )(zZ ) P (x,y,z )][(xX )(yY )(zZ ) P (x,y,z )].

13 January 28, 2006


18/111

1.5. ORDERED PAIRS. CARTESIAN PRODUCTS

1.5 Ordered pairs. Cartesian products

Let us talk about sets from a universe U . Recall that {a}denotes the set containing oneelement a. The set {a, b}contains two elements if a = b and one element otherwise. Obviously,{a, b}= {b, a}. In many problems in mathematics we need an object in which the order in apair is important. So, we want to dene an ordered pair .

You have already seen ordered pairs studying points in the xy plane. The use of x and ycoordinates to identify points in the plane works by assigning to each point in the plane anordered pair of real numbers x and y. The pair must be ordered because, for example, (2 , 5)and (5 , 2) correspond to different points.

How to dene an ordered pair formally? Whatever we dene the ordered pairs ( a, b) and(c, d) to be, it must come out that

(1.5.30) [(a, b) = ( c, d)][(a = c)(b = d)].Denition 1.5.1.

(a, b) = {{a}, {a, b}}.Let us prove that (1.5.30) is fullled.

Proof. 1)(if) [(a = c)(b = d)][(a, b) = ( c, d)].Suppose a = c and b = d. Then

(a, b) = {{a}, {a, b}}= {{c}, {c, d}}= ( c, d).

2)(only if) [( a, b) = ( c, d)][(a = c)(b = d)]. Suppose that ( a, b) = ( c, d), i.e.

{{a}, {a, b}}= {{c}, {c, d}}.There are two cases to consider.Case 1. a = b. In this case

{a, b}= {a}so {{a}, {a, b}}= {{a}}.Hence

{{a}}= {{c}, {c, d}}, so {c}= {a}and {c, d}= {c},i.e. a = b = c = d as required.

Case 2. a = b. In this case we have

[{c}= {a}][{c}= {a, b}].But [{c}= {a, b}] is false since a = b. Hence {c}= {a}, so a = c. Since a = b, it followsthat {a} = {a, b}. Therefore {c, d}= {a, b}and as a = c, {a, d}= {a, b}. Hence b = d, asrequired.

Thus the denition of ordered pair, however articial it may appear, gives to orderedpairs the crucial property that we require of them; and that is all we can reasonably requirefrom a mathematical denition.

Now we dene Cartesian product which is an essential tool for further development.

14 January 28, 2006


19/111


Denition 1.5.2. Let A and B are sets. The Cartesian product of A and B , denoted byA B , is the set of all ordered pairs ( a, b) in which aA and bB , i.e.

A B = {(a, b) |(aA)(bB )}.Thus

( pA B ) {(aA)(bB ) [ p = ( a, b)]}.Example 1.5.1. 1. If A = {red, green }and B = {1, 2, 3}then

A B = {(red , 1), (red , 2), (red , 3), (green , 1), (green , 2), (green , 3)}.2. R R = {(x, y ) |x and y are real numbers }. These are coordinates of all points in theplane. The notation R 2 is usually used for this set.

X X is called Cartesian square of X .The following theorem provides some basic properties of the Cartesian product.

Theorem 1.5.2. Let A,B,C,D be sets. Then

A (B C ) = ( A B ) (A C ),(1.5.31)A (BC ) = ( A B )(A C ),(1.5.32)

A = A = .(1.5.33)Proof of (1.5.31) . For arbitrary p

[ pA (B C )] {(aA)(xB C ) [ p = ( a, x )]}{(aA)(xB ) [ p = ( a, x )]}{(aA)(xC ) [ p = ( a, x )]}{ pA B}{ pA C }{ p(A B ) (A C )}

which proves (1.5.31).The proof of (1.5.32) and of (1.5.33) is left as an exercise.

15 January 28, 2006


20/111

1.6. RELATIONS

1.6 Relations

Denition 1.6.1. Let X, Y be sets. A set RX Y is called a relation from X to Y .If (x, y)R, we say that x is in the relation R to y. We will also write in this case xRy .Example 1.6.1. 1. Let A = {1, 2, 3}, B = {3, 4, 5}. The set R = {(1, 3), (1, 5), (3, 3)}isa relation from A to B since RA B .

2. G = {(x, y)R R |x > y }is a relation from R to R .Denition 1.6.2. Let R be a relation from X to Y . The domain of R is the set

D (R) = {xX |yY [(x, y)R]}.The range of R is the set

Ran (R) = {yY |xX [(x, y )R]}.The inverse of R is the relation R 1 from Y to X dened as followsR 1 = {(y, x )Y X | (x, y )R}.

Denition 1.6.3. Let R be a relation from X to Y , S be a relation from Y to Z . Thecomposition of S and R is a relation from X to Z dened as follows

S R = {(x, z )X Z |yY [(x, y )R][(y, z )S ]}.Theorem 1.6.2. Let R be a relation from X to Y , S be a relation from Y to Z , T be a relation from Z to V . Then

1. (R 1

) 1

= R.2. D (R 1) = Ran (R).

3. Ran (R 1) = D (R).

4. T (S R) = ( T S ) R.5. (S R) 1 = R 1 S 1.For the proof see [1], p.170.

Next we take a look at some particular types of relations. Let us consider relations fromX to X , i.e. subsets of X

X . In this case we talk about relations on X .

A simple example of such a relation is the identity relation on X which is dened as followsiX = {(x, y)X X |x = y}.

Denition 1.6.4. 1. A relation R on X is said to be reexive if

(xX ) (x, x )R.

2. R is said to be symmetric if

(xX )(yX ) {[(x, y)R][(y, x )R]}.3. R is said to be transitive if

(xX )(yX )(zX ){[((x, y )R)((y, z )R)][(x, z )R]}.16 January 28, 2006


21/111


Equivalence relations

A particularly important class of relations are equivalence relations.

Denition 1.6.5. A relation R on X is called equivalence relation if it is reexive, symmetricand transitive.

Example 1.6.3. 1. Let X be a set of students. A relation on X X : to be friends.It is reexive (I presume that everyone is a friend to himself/herself). It is symmetric.But its not transitive.

2. Let X = R , a be some positive number. Dene RX X asR = {(x, y) | |x y| a}.

R is reexive, symmetric, but not transitive.3. Let X = Z , mN . Dene the congruence mod m on X X as follows:

x y if (kZ |x y = km ).This is an equivalence relation on X .

Denition 1.6.6. Let R be an equivalence relation on X . Let xX . The equivalence classof x with respect to R is the set

[x]R = {yX | (y, x )R}.Let us take a look at several properties of classes of equivalence.

Proposition 1.6.1. Let R be an equivalence relation on X . Then 1.(xX ) x[x]R .2.(xX )(yX ) [(y[x]R )([y]R = [x]R )].

Proof. 1. Since R is reexive, (x, x )R, hence x[x]R .2.First, let y [x]R . (a) Suppose that z [y]R (an arbitrary element of [ y]R . Then

(x, y)R, (y, z )R and by transitivity ( x, z )R. Therefore z[x]R , which shows that

[y]R [x]R . (b) Suppose that z[x]

R . Similarly show that z[y]R .Therefore [ x]R = [y]R .

The implication ([ y]R = [x]R )(y[x]R ) is obvious.

From the above proposition it follows that the classes of equivalence are disjoint and everyelement of the set X belongs to a class of equivalence (the union of the classes equals to theset X ).

Remark 1. See more on this in [1].

17 January 28, 2006


22/111

1.7. FUNCTIONS

1.7 Functions

1.7.1 FunctionThe notion of a function is of fundamental importance in all branches of mathematics. Youmet functions in your previous study of mathematics, but without precise denition. Herewe give a denition and make connections to examples you learned before.

Denition 1.7.1. Let X and Y be sets. Let F be a relation from X to Y . Then F is calleda function if the following properties are satised

(i) (xX )(yY ) [(x, y)F ].In this case it is customary to write y = F (x).

(ii) (xX )(yY )(zY ){([(x, y )F ][(x, z )F ])(y = z)}.(In other words, for every xX there is only one yY such that ( x, y )F ).X is called the domain of F and Y is called codomain.

Let us consider several examples.

Example 1.7.1. (i) Let X = {1, 2, 3}, Y = {4, 5, 6}. Dene F X Y asF = {(1, 4), (2, 5), (3, 5)}.

Then F is a function.(In contrast to that dene GX Y as

G = {(1, 4), (1, 5), (2, 6), (3, 6)}.Then G is not a function.)

(ii)Let X = R and Y = R . Dene F X Y asF = {(x, y)R R |y = x2}.

Then F is a function from R to R .(In contrast to that dene GX Y as

G = {(x, y)R R |x2 + y2 = 1}.Then G is not a function.)

We often use the notationF : X Y

for a function F from X to Y .

Note that in order to dene a function F from X to Y we have to dene X , Y and asubset of X Y satisfying (i) and (ii) of the denition.

Let F : X Y be a function. If xX then we know that there is a unique y such that(x, y)F . For this y we write y = F (x). This y is called the image of x under F .Note: one can assign a value of y for each value xX by means of a rule or a formula.Though these notions is rather vague and cannot be used in a denition of a function.

18 January 28, 2006


23/111


Theorem 1.7.2. Let X, Y be sets, F, G be functions from X to Y . Then

[(

x

X )(F (x) = G(x))]

(F = G).

Proof. 1. (). Let x X . Then (yY )(y = F (x)). But G(x) = F (x), so (x, y ) G.Therefore F G. Analogously one sees that GF .2. (). Obvious since the sets F and G are equal.

The above theorem says that in order to establish that two functions are equal one has toestablish that they have the same domain and codomain and for every element of the domainthey have equal images. Note that equal function may be dened by different rules.

Example 1.7.3. Let f : R R , g : R R , h : R R + (By R + we denote the set of non-negative real numbers). Let xRf (x) = ( x + 1)

2, g(x) = x

2+ 2 x + 1 , h(x) = ( x + 1)

2.

Then f and g are equal, but f and h are not since they have different codomains.

Denition 1.7.2. Let f : X Y be a function. The setRan (f ) = {y | (xX )( f (x) = y)}

is called the range of f .

Example 1.7.4. Let f : R R , f (x) = 2xx2 +1 . Find Ran (f ).Solution . We have to nd a set of y such that the equation y = 2x

x2

+1has a solution x

R .The equation is equivalent to yx2 2x + y = 0. The existence of a real solution is equivalentto D = 4 4y2 0. Solving this inequality we obtain that Ran (f ) = [1, 1].

The denition of composition of relations can be also applied to functions. If f : X Y and g : Y Z theng f = {(x, z )X Z | (yY )[(x, y )f ][(y, z )g]}.

Theorem 1.7.5. Let f : X Y and g : Y Z . Then g f : X Z and (xX ) [(g f )(x) = g(f (x))].

Proof. We know that g f is a relation. So we must prove that for every xX there existsa unique element zZ such that ( x, z )g f .Existence: Let xX be arbitrary. Then yY such that y = f (x), or in other words,(x, y)f . Also zZ such that z = g(y), or in other words, ( y, z )g. By the denition itmeans that ( x, z )g f . Moreover, we see that

(g f )(x) = g(f (x)) .Uniqueness: Suppose that ( x, z1) g f and ( x, z2) g f . Then by the denition of composition (y1Y ) [(x, y1)f ][(y1, z1)g] and (y2Y ) [(x, y2)f ][(y2, z2)g].But f is a function. Therefore y1 = y2. g is also a function, hence z1 = z2.

19 January 28, 2006


24/111

1.7. FUNCTIONS

Example 1.7.6. Let f : R R , g : R R ,

f (x) =1

x2 + 2 , g(x) = 2 x 1.Find ( f g)(x) and ( g f )(x).Solution .

(f g)(x) = f (g(x)) =1

[g(x)]2 + 2=

1(2x 1)2 + 2

,

(g f )(x) = g(f (x)) = 2 f (x) 1 =2

x2 + 2 1.

Warning: As you clearly see from the above f

g = g

f .

Images and inverse images

Denition 1.7.3. Let f : X Y and AX . The image of A under f is the setf (A) = {f (x) |xA}.

Example 1.7.7. Let f : R R dened by f (x) = x2. Let A = {xR |0 x 2}. Thenf (A) = [0 , 4].

The following theorem (which we give without proof) establishes some properties of imagesof sets.

Theorem 1.7.8. Let f : X Y and AX , BX . Then (i) f (AB ) = f (A)f (B ),

(ii) f (A B )f (A) f (B ),(iii) (AB )[f (A)f (B )].

Remark 2. Note that in (ii) there is no equality in general. Consider the following example.f : R R dened by f (x) = x2. Let A = [1, 12 ] and B = [12 , 1]. Then f (A) = [0 , 1],f (B ) = [0 , 1], so that f (A) f (B ) = [0 , 1]. At the same time A B = [

12 ,

12 ], and hencef (A B ) = [0 , 14 ].

Denition 1.7.4. Let f : X Y and BY . The inverse image of B under f is the setf 1(B ) = {xX | f (x)B}.

Example 1.7.9. Let f : R R dened by f (x) = x2. Let B = [1, 4]. Then f 1(B ) =[2, 2].The following theorem (which we give without proof) establishes some properties of inverse

images of sets.

20 January 28, 2006


25/111


Theorem 1.7.10. Let f : X Y and AY , BY . Then (i) f 1(A

B ) = f 1(A)

f 1(B ),

(ii) f 1(A B ) = f 1(A) f 1(B ),(iii) (AB )[f

1(A)f 1(B )].

Remark 3. Note the difference with the previous theorem.

1.7.2 Injections and surjections. Bijections. Inverse functions

In the last section we saw that the composition of two functions is again a function. If f : X Y the is a relation from X to Y . One can dene the inverse relation f 1. Then thequestion comes : Is f 1 a function? In general the answer is no. In this section we willstudy particular cases of functions and nd out when the answer is yes.

Denition 1.7.5. Let f : X Y . Then f is called an injection if (x1X )(x2X )[(f (x1) = f (x2))(x1 = x2)].

The above denition means that f is one-to-one correspondence between X and Ran (f ).Using the contrapositive law one can rewrite the above denition as follows.

(x1X )(x2X )[(x1 = x2)(f (x1) = f (x2))].

Example 1.7.11. (i) Let X = {1, 2, 3}, Y = {4, 5, 6, 7}. Dene f : X Y asf = {(1, 5), (2, 6), (3, 7)}.

Then f is an injection.In contrast dene g : X Y as

g = {(1, 5), (2, 6), (3, 5)}.Then g is not an injection since g(1) = g(3).(ii)Let f : N N dened by f (n) = n2. Then f is an injection.In contrast let g : Z Z dened by g(n) = n2. Then g is not an injection since, for instance,g(1) = g(1).

Denition 1.7.6. Let f : X Y . Then f is called a surjection if (yY )(xX )[f (x) = y].

The above denition means that Ran (f ) = Y . For this reason surjections are sometimescalled onto .

21 January 28, 2006


26/111

1.7. FUNCTIONS

Example 1.7.12. (i) Let X = {1, 2, 3, 4}, Y = {5, 6, 7}. Dene f : X Y asf =

{(1, 5), (2, 6), (3, 7), (4, 6)

}.

Then f is an surjection.In contrast dene g : X Y as

g = {(1, 5), (2, 6), (3, 5), (4, 6)}.Then g is not an surjection since 7 is not in its range.(ii)Let f : Z Z dened by f (n) = n + 2. Then f is a surjection.In contrast let g : Z Z dened by g(n) = n2. Then g is not a surjection since, for instance,there is no integer such that its square is 2.

Denition 1.7.7. Let f : X Y . Then f is called a bijection if it is an injection an asurjection.

Example 1.7.13. (i) Let X = {1, 2, 3}and Y = {4, 5, 6}. Dene f : X Y byf = {(1, 4), (2, 5), (3, 6)}.

Then f is a bijection.(ii) Let X = Y = [0 , 1]. Dene g : X Y by g(x) = x2. Then g is a bijection.

Now we are ready to answer the question about the inverse to a function. First, recallthat if f : X

Y then f 1 is a relation from Y to X with the properties D (f 1) = Ran (f )

and Ran (f 1) = D (f ). So, if f 1 is a function from Y to X then D (f 1) = Y . Therefore weconclude that the condition Ran (f ) = Y is a necessary condition for f 1 to be a function.Which means, that f has to be surjective. Also, from the denition of the inverse relation f 1

it is clear that injectivity of f is also a necessary condition for f 1 to be a function. Hencebijectivity of f is a necessary condition for f 1 to be a function.

It turns out that this is also a sufficient condition.

Theorem 1.7.14. Let f : X Y . Then (f 1 : Y X )(f is a bijection ).

Proof. We have to show only that

(f is a bijection) (f 1 : Y X ).

Recall thatf 1 = {(y, x )Y X | (x, y )f }.

We have to verify two properties in the denition of a function.First, existence. We have to show that ( yY )(xX )[(y, x )f

1], or in other words,(yY )(xX )[(x, y )f ]. This follows from the surjectivity of f .

Second, Uniqueness. We have to show that ( x1 X )(x1 X ){[((y, x 1) f 1)((( y, x 2)f 1)](x1 = x2)}, or in other words (x1 X )(x1 X ){[f (x1) = f (x2)](x1 = x2)}

. But this follows from injectivity of f .

22 January 28, 2006


27/111


Denition 1.7.8. Let f : X Y . If f 1 is a function from Y to X we say that f isinvertible. In that case f 1 is called the inverse function.

Theorem 1.7.15. Let f : X Y . Let f 1 be a function from Y to X . Then f 1 f = iX and f f 1 = iY .

Proof. Let x X be arbitrary. Let y = f (x) Y . Then ( x, y) f so that ( y, x ) f 1.

Therefore f 1(y) = x. Thus

(f 1 f )(x) = f 1(f (x)) = f 1(y) = x.We have proved that

(

x

X )[f 1f (x) = x]

This is the same as to say that f 1 f = iX . The second statement is similar and is leftas an exercise.

Example 1.7.16. Let f : R R dened by f (x) = x+75 . Then f is a bijection.Indeed, Let x1, x2R be arbitrary and

x1 +75 =

x2 +75 . Then, of course, x1 + 7 = x2 + 7,

therefore x1 = x2, so f is an injection.Now let yR be arbitrary and y =

x+75 . Then, of course, x + 7 = 5 y and x = 5 y 7R .So we have proved that

(yR )(xR )[y = f (x)].

By denition this means that f is a surjection.So f is a bijection. Moreover f 1(y) = 5 y 7. This means that g(x) = 5 x 7 is the inversefunction to f .

23 January 28, 2006


28/111

1.8. SOME METHODS OF PROOF. PROOF BY INDUCTION

1.8 Some methods of proof. Proof by induction

1. First we discuss a couple of widely used methods of proof: contrapositive proof and proof by contradiction.

The idea of contrapositive proof is the following equivalence

(AB )(B A).So to prove that AB is true is the same as to prove that B A is true.

Example 1.8.1. For integers m and n, if mn is odd then so are m and n.

Proof. We have to prove that ( m, n N)

(mn is odd)[(m is odd)(n is odd)],

which is the same as to prove that

[(m is even)(n is even)](mn is even) .

The latter is evident.

The idea of proof by contradiction is the following equivalence

(AB )(AB ) (AB ).So to prove that AB is true is the same as to prove that AB is true or else that ABis false.Example 1.8.2. Let x, yR be positive. If x

2 + y2 = 25 and x = 3, then y = 4.

Proof. In order to prove by contradiction we assume that

[(x2 + y2 = 25)

(x = 3)]

(y = 4) .

Then x2 + y2 = x2 + 16 = 25. Hence ( x = 3) (x = 3) which is a contradiction.

2. In the remaining part of this short section we will discuss an important property of natural numbers. The set of natural numbers

N = {0, 1, 2, 3, 4, . . . }which we will always denote by N , is taken for granted.

We shall denote the set of positive natural numbers N+

= {1, 2, 3, 4, . . . }.24 January 28, 2006


29/111


The Principle of Mathematical Induction is often used when one needs to prove the state-ment of the form

(nN

+) P (n)

or similar types of statements.

Since there are innitely many natural numbers we cannot check one by one that they allhave property P . The idea of mathematical induction is that to list all natural numbers onehas to start from 1 and then repeatedly add 1. Thus one can show that 1 has property P and that whenever one adds 1 to a number that has property P , the resulting number alsohas property P .

Principle of Mathematical Induction. If for a statement P (n)

(i) P (1) is true,

(ii ) [P (n)P (n + 1)] is true,

then (nN+ ) P (n) is true.

Part (i) is called base case ; (ii) is called induction step .

Example 1.8.3. Prove that nN+ if n 1 then:

12 + 2 2 + 3 2 + + n2 =n(n + 1)(2 n + 1)

6.

Solution . Base case: n = 1. 1 2 = 1236 is true.Induction step: Suppose that the statement is true for n = k (k 1). We have to prove thatit is true for n = k + 1. So our assumption is

12 + 2 2 + 3 2 + + k2 =k(k + 1)(2 k + 1)

6.

Therefore we have

12 + 2 2 + 3 2 + + k2 + ( k + 1) 2 =k(k + 1)(2 k + 1)

6+ ( k + 1) 2

=(k + 1)( k + 2)(2 k + 3)

6,

which proves the statement for n = k + 1. By the principle of mathematical induction the

statement is true for all natural n .

Example 1.8.4. 1 + 3 + 5 + + (2 n 1) =?First we have to work out a conjecture. For this let us try several particular cases for n.

n = 1; 1 =1;

n = 2; 1 + 3 =4 = 2 2;

n = 3; 1 + 3 + 5 =9 = 3 2.

So a reasonable conjecture is that nN+

1 + 3 + 5 + + (2 n 1) = n2

.

25 January 28, 2006


30/111

1.8. SOME METHODS OF PROOF. PROOF BY INDUCTION

Proof. We need not make the base case, since it has been done working out the conjecture.Induction step:

Let 1 + 3 + 5 + + (2 k 1) = k2 (k 1).Then

1 + 3 + 5 + + (2 k 1) + (2 k + 1) = k2 + 2 k + 1 = ( k + 1) 2.This completes the proof by induction.

The base step can be started not necessarily from 1. It can start from any integer onwards.

Example 1.8.5. Prove that ( nN)(3|(n3 n)).Solution . Base case: n = 0. (3

|0) is true.

Induction step:Assume that (3 |(k3 k)) for k 0. Then

(k + 1) 3 (k + 1) = k3 + 3 k2 + 3 k + 1 k 1 = ( k3 k) divisible by 3

+ 3( k2 + k)

divisible by 3Example 1.8.6. Prove that ( nN)[(n 5)(2

n > n 2)].

Solution . Base case: n = 5. True.Induction step:Suppose that 2 k > k 2 (k 5). Then

2k+1 = 2 2k > 2k2.Now it is sufficient to prove that

2k2 > (k + 1) 2.

Consider the difference:

2k2 (k + 1) 2 = k2 2k 1 = ( k 1)2 2.Since k 5 we have that k 1 4 and ( k 1)2 2 14 > 0 which proves the aboveinequality.

26 January 28, 2006


31/111


32/111

2.1. VARIOUS SORTS OF NUMBERS

2.1.2 Rational Numbers

Let a

Z , b

Z . The equation

(2.1.1) ax = b

need not have a solution x Z . In order to enable one to solve (2.1.1) (for a = 0) wehave to widen our system of numbers again so that it included fractions b/a (existence of multiplicative inverse in Z {0}. This motivates the followingDenition 2.1.1. Rational numbers, or rationals, is the set

{r =pq | pZ , qN}.

The set of rational numbers will be denoted by Q . When writing p/q for a rational wealways assume that the numbers p and q have no common factor greater than 1.

All the arithmetical operations in Q are straightforward. Let us introduce a relation of order for rationals.

Denition 2.1.2. Let bN , dN , with both b, d > 0,. Then

ab

>cd

ad > bc .

The following theorem provides a very important property of rationals.

Theorem 2.1.1. Between any two rational numbers there is another (and, hence, innitely many others).

Proof. Let bN dN, both b, d > 0, and

ab

>cd

.

Notice that(mN) [

ab

>a + mcb + md

>cd

] .

Indeed, since b, d and m are positive we have

[a(b + md) > b (a + mc)]

[mad > mbc ]

(ad > bc ),

and[d(a + mc) > c (b + md)](ad > bc ).

2.1.3 Irrational Numbers

Suppose that aQ and consider the equation

(2.1.2) x2 = a.

In general (2.1.2) does not have rational solutions. For example the following theorem holds.

Theorem 2.1.2. No rational number has square 2.

28 January 28, 2006


33/111

CHAPTER 2. NUMBERS

Proof. Suppose for a contradiction that pq with p Z , qN , q = 0 and p, q have nocommon factors greater than 1, is such that ( pq )

2 = 2. Then p2 = 2 q2. Hence p2 is even, and

so is p. Hence, (kZ ) [ p = 2 k ]. This implies that

2k2 = q2

and therefore q is also even. The last statement contradicts our assumption that p and q haveno common factor.

Theorem 2.1.2 provides an example of a number which is not rational, or irrational. Hereare some other examples of irrational numbers.

Theorem 2.1.3. No rational x satises the equation x3 = x + 7 .

Proof. First we show that there are no integers satisfying the equation x3 = x + 7. For acontradiction suppose that there is. Then x(x + 1)( x 1) = 7 from which it follows that xdivides 7. Hence x can be only 1, 7. Direct verication shows that these numbers do notsatisfy the equation.

Second, show that there are no fractions satisfying the equation x3 = x + 7. For acontradiction suppose that there is. Let mn with m Z , n N , n = 0, and m, n have nocommon factors greater than 1, is such that ( mn )

3 = mn + 7. Multiplying this equality by n2

we obtain m3

n = mn2 + 7 n2, which is impossible since the right-hand side is an integer and

the left-hand side is not.

Example 2.1.4. No rational satises the equation x5 = x + 4.

Prove it!

Algebraic numbers A correspond to all real solutions of polynomial equations with integercoefficients. So x is algebraic if there exists nN, a0, a 1, . . . , a n Z such that

a0 + a1x + + an xn = 0 .

2.1.4 Cuts of the Rationals

In this subsection we discuss how a particular irrational number, say 2 is tted in amongthe rationals.

In trying to isolate a number whose square is 2, we rst observe from Theorem 2.1.2 thatthe positive rational numbers fall into two classes, those whose squares are greater than 2and those whose squares are less than 2. Call these classes the left-hand class L and theright-hand class R, corresponding to their relative positions when represented graphically ona horizontal line. Examples of numbers lL are 7/ 5 and 1 41, and of numbers r R are17/ 12 and 1 42. At this stage we shall have to convince ourselves that for elements of theclasses L and R the following properties hold.

(i) (lL)(r R)[r > l ];

(ii) (l1L)(l2L)[l2 > l 1];

(iii) (r1R)(r

2R)[r

2 < r 1].

29 January 28, 2006


34/111

2.1. VARIOUS SORTS OF NUMBERS

The last three statements become more concrete if we use the arithmetical rule for squareroot to nd, to as many decimal places as we please, a set of numbers lL

1, 1 4, 1 41, 1 414, . . . ,each of which is greater than the preceding (or equal to it if the last digit is 0) and eachhaving its square less than 2. Moreover, the numbers got by adding 1 to the last digit of thesenumbers lL for a set of numbers rR

2, 1 5, 1 42, 1 415, . . . ,each having its square greater than 2 and each less than (or equal to) the preceding.

If we are now given a particular number a whose square is less than 2, we shall by goingfar enough along the set of numbers 1 , 1

4, 1

41, 1

414, . . . come to one which is greater

than a.If, then, we are building a number-system starting with integers and then including the

rational numbers, we see that an irrational number (such as 2) corresponds to and can bedened by a cutting of the rationals into two classes L, R of which L has no greatest memberand R no least member. This is Dedekinds denition of irrationals by the cut .

There are cuts which correspond to non-algebraic numbers : the transcendental numbers,denoted by T . For example, the cut corresponding to 2 x = 3 is transcendental. (Check thatno rational x satises 2 x = 3). The proof that xA is more complicated. Summarising:

N

Z

Q

A

T

A=

R.

30 January 28, 2006


35/111

CHAPTER 2. NUMBERS

2.2 The Field of Real Numbers

In the previous sections, starting from integers, we have sketched the building-up of the setof real numbers which is the union of the sets of rationals and irrationals and denoted by R .Now we are making the list of the basic properties which the real numbers satisfy. Since

we provide no proof of these statements we present them in form of axioms. Naturally theyfall into two groups which concern algebraic operations with real numbers and the relation of order for them.

Those of you familiar with basic concepts of algebra will nd that the rst group of axiomsdenes R to be a eld.

A.1. (aR )(bR ) [(a + b)R ] .

A.2. (aR )(b

R ) [a + b = b + a] .A.3. (aR )(bR )(cR ) [(a + b) + c = a + ( b + c)] .

A.4. (0R )(aR ) [0 + a = a] .

A.5. (aR )(!xR ) [a + x = 0] . We write x = a.

Notation : We use the symbol ! in order to say that there exists a unique... So ( !xR )has to be red : there exists a unique real number x.The axioms A.6-A.10 that follow are analogues of A.1-A.5 for the operation of multipli-

cation.

A.6. (aR )(b

R ) [abR ].

A.7. (aR )(bR ) [ab = ba].

A.8. (aR )(bR )(cR ) [(ab)c = a(bc)].

A.9. (1R )(aR ) [1 a = a].A.10. (aR {0})(!yR ) [ay = 1] . We write y = 1 /a .

The last axiom links the operations of summation and multiplication.

A.11. (aR )(bR )(cR ) [(a + b)c = ac + bc] .

From the axioms above the familiar rules of manipulation of real numbers can be deduced.As an illustration we present

Example 2.2.1. (aR )[0 a = 0]. Indeed, by A.11 and A.4 we have1 a + 0 a = (1 + 0) a = 1 a.

Then by A.2 and A.4 0 a = 0.

Let us observe that integers do not form a eld since they do not satisfy axiom A.10.

Now we are adding the relevant axioms of order for real numbers.

31 January 28, 2006


36/111


37/111


38/111

2.2. THE FIELD OF REAL NUMBERS

Example 2.2.6. Prove that

(aR + )(b

R + )(cR + )(d

R + )a + b + c + d

4 4

abcd .Proof. By Example 2.2.4 and by (O.2) we have

a + b + c + d4

2ab + 2 cd4

4abcd.The equality holds iff a = b = c = d.

3. The third idea is to use the transitivity property

(

a

R+

)(

b

R+

)(

c

R+

)[(a

b)

(b

c)

(a

c)].

In general this means that when proving that a c we have to nd b such that a b andb c.Example 2.2.7. Let n 2 be a natural number. Prove that

1n + 1

+1

n + 2+ +

12n

>12

.

Proof.1

n + 1+

1n + 2

+ +1

2n>

12n

+1

2n+ +

12n

n= n

12n

=12

.

4. Inequalities involving integer may be proved by induction as we discussed before. Herewe give an inequality which is of particular importance in analysis.

Theorem 2.2.8. (Bernoullis inequality)

(nN)( > 1)[(1 + )n 1 + n ].Proof. For n = 0 , 1 it is true.Suppose that it is true for n = k (k 1), i.e. (1 + )k 1 + k . We have to prove that it istrue for n = k + 1, i.e.

(1 + )k+1 1 + ( k + 1) .Indeed,

(1 + )k+1 = (1 + )k (1 + ) (1 + k )(1 + )= 1 + ( k + 1) + k 2 1 + ( k + 1) .

34 January 28, 2006


39/111

CHAPTER 2. NUMBERS

2.3 Bounded sets of numbers

Simplest examples of bounded sets of numbers is nite sets of numbers. In case of a nite setof real numbers we can always nd the maximal number in the set and the minimal number.In general we can dene these numbers for any set of real number by the following

Denition 2.3.1. Let S R . Then

(x = max S ) {(xS )[(yS )(y x)]},(x = min S ) {(xS )[(yS )(y x)]}.

Note that the denition says nothing about existence of max S and min S . And indeedthey not always exist.

Denition 2.3.2. (i) A set S R is called bounded above if

(K R )(xS )(x K ).

The number K is called an upper bound of S .

(ii) A set S R is called bounded below if

(kR )(xS )(x k).The number k is called a lower bound of S .

(iii) A set S R is called bounded if it is bounded above and below.

Note that if S is bounded above with upper bound K then any real number greater than

K is also an upper bound of S . Similarly, if S is bounded below with lower bound k then anyreal number smaller than k is also a lower bound of S .

Example 2.3.1. Let a, bR and a < b .

(i) A = {x R |a x b}= [a, b] - closed interval. It is bounded above since ( x A)(x b + 1) and bounded below since ( xA)(x a 1).Therefore max A = b, min A = a.

(ii) B = {xR |a < x < b }= ( a, b) - open interval. It is bounded above and below for thesame reasons as A.Therefore B has no maximal or minimal number .

Note that if AR is a bounded set and BA then B is also bounded. Prove it!Example 2.3.2.

A =n

n + 1 |nN .A is bounded. Indeed, ( nN)

nn +1 0) , so A is bounded below. Moreover,

(nN)n

n + 1< 1)

(Why? Prove this inequality) so A is bounded above, and therefore is bounded.Therefore min A = 12 , max A does not exist.

35 January 28, 2006


40/111

2.4. SUPREMUM AND INFIMUM

2.4 Supremum and inmum

If a set S R is bounded above one tries to nd the sharp, or in other words, the least upperbound (existence of which is non-trivial and will be proved below). The least upper bound of

a set is called supremum of this set. Let us give it a precise denition.

Denition 2.4.1. Let S R be bounded above. Then aR is called supremum of S (theleast upper bound of S ) if

(i) a is an upper bound of S , i.e.(xS )(x a);

(ii) there is no upper bound of S less than a, i.e.

( > 0)(bS )(b > a ).

Analogously one denes the greatest lower bound for a set bounded below.

Denition 2.4.2. Let S R be bounded below. Then a R is called inmum of S (thegreatest lower bound of S ) if

(i) a is a lower bound of S , i.e.(xS )(x a);

(ii) there is no lower bound of S greater than a, i.e.

( > 0)(bS )(b < a + ).

The next theorem is one of the corner-stones of analysis.

Theorem 2.4.1. Let S R be non-empty and bounded above. Then sup S exists.

Proof. Divide R into two classes L and R as follows.

(xL)(sS )(s > x ),(x

R)

(

s

S )(s

x).

Then

1. L R = , and L = , R = .Indeed, since S = , it follows that sS . Then x = s 1L. Let K be an upperbound of S . Then K R.2.

(lL)(rR) (l < r ).

Indeed, let lL and r R. The by the denition of L there exists s S such that

s > l . By the denition of R we have that r s, hence l < r .36 January 28, 2006


41/111

CHAPTER 2. NUMBERS

From this it follows that the classes L and R satisfy Dedekinds axiom. Therefore

(

R )(

> 0) [(

L)

( +

R)].

From Dedekinds axiom one cannot conclude whether L or R.We claim that in our circumstances R.Let us prove this claim by contradiction.For a contradiction suppose that L. Then by the denition of L there exists sS suchthat s > . Fix this s.Take = 12 (s + ). Then < < s . Hence R since > . Therefore by the denition of R we conclude that s which is a contradiction. This proves that R.So we obtain that satises

(1) (

s

S ) (s

);

(2) ( > 0)(sS ) (s > ).By the denition of the supremum we conclude that = sup S , which proves the theorem.

We use the following notation

S = {xR | xS }.Theorem 2.4.2. Let S R be non-empty and bounded below. Then inf S exists and is equal to sup(S ).Proof. Since S is bounded below, it follows that

k

R such that (

x

S )(x

k). Hence

(xS )(x k) which means that S is bounded above. By Theorem 2.4.1 there exists = sup( S ). We will show that = inf S . First, we have that ( x S )(x ), sothat x . Hence is a lower bound of S . Next, let is another lower bound of S , i.e.(xS )(x ). Then (xS )(x ). Hence is an upper bound of S , and by thedenition of supremum sup(S ) = . Hence which proves the required.

Archimedian Principle and its consequences.

Theorem 2.4.3. (The Archimedian Principle) Let x > 0 and yR . Then there existsnZ such that y < nx .

Proof. Suppose for a contradiction that

(nZ ) (nx y).Then the set A := {nx |nZ}is bounded above and y its upper bound. Then by Theorem2.4.1 there exists sup A. The number sup Ax < sup A is not an upper bound of A, so thereexists mZ such that mx > sup A x, or otherwise(2.4.3) ( m + 1) x > sup A.

But m +1 Z , hence ( m +1) xA, and (2.4.3) contradicts to the denition of supremum.

Corollary 2.4.1. N is unbounded.

37 January 28, 2006


42/111

2.4. SUPREMUM AND INFIMUM

Corollary 2.4.2.(x > 0)(y > 0)(nN) (

yn

< x ).

Proof. By the Archimedian Principle (Theorem 2.4.3)

(nZ ) (nx > y ).

Since y > 0 it follows that nN . Thereforeyn < x .

Corollary 2.4.3.(y > 0) inf

yn |nN , n = 0 = 0 .

Proof. Since all the elements of the setyn |nN are positive, we conclude that 0 is its

lower bound. By Corollary 2.4.2 any positive number x is not its lower bound, which provesthe assertion.

Example 2.4.4.

inf 1n |nN , n = 0 = 0 .

Example 2.4.5. Let

A :=n 1

2n |nN , n = 0 .Then sup A = 12 and inf A = 0.

Proof. All the element of A are positive, except for the rst one which is 0. Thereforemin A = 0, and hence inf A = 0.Now notice that

(nN) n = 0 n 1

2n=

12

12n

0)(nN)n 1

2n>

12 .

Notice thatn 12n >

12

12

12n

> 12

1

2n< [n (2) > 1].

By the Archimedian Principle there exists such nN .

Note that for any set AR if max A exists then sup A = max A. Also if minA existsthen inf A = min A.

Theorem 2.4.6. For any interval (a, b) there exists a rational r(a, b). In other words,

(aR )(b

R )[(b > a )(rQ )(a < r < b )].

38 January 28, 2006


43/111

CHAPTER 2. NUMBERS

Proof. Let h = ba > 0. Then by the Archimedian Principle

(nN)

1n < h .

Fix this n. By the Archimedian Principle ( m, m N) mn

< a a and (m N)mn

> a .Now the set of integers between m and m is nite. Let m be the least element of thatset such that

mn

> a.

Set r = mn . Then we have

a < r =mn

=m 1

n+

1n a +

1n

< a + h = b.

The next theorem is a multiplicative analogue of the Archimedian Principle.

Theorem 2.4.7. Let x > 1, y > 0. Then

(nN) (xn > y ).

Prove this theorem repeating the argument from the proof of Theorem 2.4.3.

Theorem 2.4.8. (

!x > 0) (x2 = 2) .

Proof. Dene a setA := {y > 0 |y2 < 2}.

A = since 1A. A is bounded above since ( yA) (y < 2). Then by Theorem 2.4.1there exists sup A = x. We will prove that x2 = 2.1) First let us suppose that x2 > 2. Let = x

2 22x . Then > 0 and

(x )2 = x2 2x + 2 > x 2 2x = x2 2xx2 2

2x= 2 .

Hence x is another upper bound for A, so that x is not the least upper bound for A. Thisis a contradiction.2) Now suppose that x2 < 2. Let = 2 x

2

2x+1 . Then by assumption 0 < < 1 so that

(x + )2 = x2 + 2 x + 2 < x 2 + 2 x + =

= x2 + (2x + 1) = x2 +2 x22x + 1

(2x + 1) = 2 .

Hence x + is also in A, in which case x cannot be an upper bound for A, which is acontradiction.

In the same way one can prove the following theorem.

Theorem 2.4.9. (

a > 0)(

n

N)[n = 0

(

!x > 0) (xn = a)].

39 January 28, 2006


44/111

Chapter 3

Sequences and Limits

3.1 Sequences

In general any function f : N {0} X with domain N+ and arbitrary codomain X iscalled a sequence. This means that a sequence ( an )nN + is a subset of X each element of which has its number, i.e. an = f (n). (We do not call this a series - this term is used for another concept. )

In this course we conne ourselves to sequences of numbers. In particular, in this chapterwe always assume that X = R . So we adopt the following denition.

Denition 3.1.1. A sequence of real numbers is a function f : N+ R . We use thefollowing notation for the general term an = f (n) and for the whole sequence ( an )nN + .

Example 3.1.1. 1. an = 1n . The sequence is 1 ,12 ,

13 ,

14 , . . . .

2. an = n2. The sequence is 1 , 4, 9, 16, 25, . . . .

3. an = ( 1)n

n . The sequence is 1, 12 , 13 , 14 , . . . .

3.2 Null sequences

In the example 1. in the previous section nth member of the sequence becomes smaller as n

becomes larger. Making computations we see that an tends to zero. The same refers to theexample 3. Such sequences are called null sequences.

We will give the precise mathematical denition to the notion of a null sequence.

Denition 3.2.1. (an )nN + is a null sequence if

( > 0)(N N+ )(nN

+ ) [(n N )(|an | < )].Now we are ready to verify rigorously this denition for some examples.

Example 3.2.1. (an )nN + with an =1

nis a null sequence.

40


45/111

CHAPTER 3. SEQUENCES AND LIMITS

Proof. We have to prove that

( > 0)(N N

+)(n

N+

) (n > N )

1n < .

Otherwise we need n to satisfy the inequality n > 1 . By AP (the Archimedian Principle)

(N N+ ) (N > 1 ). (In particular one can take N =

1

+ 1.) Then [( n > N )(N >1 )](n >

1 ) which proves the statement.

The next theorem follows straight from the denition of a null sequence and the simplefact that ( aR ) (|a| = ||a||).Theorem 3.2.2. A sequence (an )nN + is a null sequence if and only if the sequence (|an |)nN +is a null sequence.

Example 3.2.3. (an )nN + with an =(1)n

nis a null sequence.

Denition 3.2.2. Let xR and > 0. The set

B (x, ) := ( x , x + ) = {yR | |y x| < }is called the -neighbourhood of the point x.

Using this notion one can say that a sequence ( an )nN + is a null sequence if an only if for any > 0 starting from some number all the elements of the sequence belong to the-neighbourhood of zero.

3.3 Sequence converging to a limit

A null sequence is one whose terms approach zero. This a particular case of a sequencetending to a limit.

Example 3.3.1. Let an = nn +1 . Then an tends to 1 as n .

Denition 3.3.1. A sequence ( an )nN + converges to the limit aR if

( > 0)(N N+ )(nN

+ ) [(n > N )(|an a| < )].We write in this case that

limn

an = a

or abbreviating:lim

nan = a

or, again

an a as n .41 January 28, 2006


46/111

3.3. SEQUENCE CONVERGING TO A LIMIT

Remark 4. 1. From the above denition it is clear that

limn

an

= a

((an

a)n

is a null sequence) .

2. limn an = a if an only if for any > 0 starting from some number all the elementsof the sequence belong to the -neighbourhood of a.

3. Note that in Denition 3.3.1 N depends on .

Let us prove now the statement of Example 3.3.1.

Proof. Let > 0 be given. We have to nd N N+ such that for n > N |an a| < . Thelast inequality for our example reads as follows

nn + 1 1 < .

An equivalent form of this inequality is

1n + 1

< n >1 1 .

Chose n to be any integer greater than or equal to 1 1 (which exists by AP). Then, if n > N ,we haven > N >

1 1.

Therefore1

n + 1<

nn + 1 1 < .

If a sequence ( an )n does not converge, we say that it diverges.

Denition 3.3.2. 1. The sequence ( an )nN + diverges to if (M R )(N N

+ )(nN+ ) [(n > N )(an > M )].

2. The sequence ( an )nN + diverges to if (M R )(N N

+ )(nN+ ) [(n > N )(an < M )].

Example 3.3.2. 1. The sequence an = n2 diverges to .2. The sequence an = n2 diverges to .3. The sequence an = ( 1)n diverges.

Theorem 3.3.3. A sequence (an )nN + can have at most one limit.

42 January 28, 2006


47/111


Proof. We have to prove that the following implication is true for all a, bR

limn an = a limn an = b (a = b).

By the denition

( > 0)(N 1N+ )(nN

+ ) [(n > N )(|an a| < )],( > 0)(N 2N

+ )(nN+ ) [(n > N )(|an b| < )].

Suppose for a contradiction that the statement is not true. Then its negation

limn

an = a limn an = b (a = b)

is true.Fix = 13 |a b| > 0 and take N = max {N 1, N 2}. Then, if n > N , we have

|a b| = |(a an ) + ( an b)| |an a|+ |an b| < 2 =23|a b|,

which is a contradiction.

Theorem 3.3.4. Any convergent sequence is bounded.

Proof. We have to prove that the following implication is true

limn

an = a [(m, M R )(nN+ ) (m an M )].

Take = 1 in the denition of the limit. Then N N+ such that for n > N we have

(|an a| < 1)(a 1 < a n < a + 1) .Set M = max {a + 1 , a 1, . . . , a N }and m = min {a 1, a 1, . . . , a N }. Then

(nN+ ) (m an M ).

Note that Theorem 3.3.4 can be expressed by the contrapositive law in the following way

If a sequence is unbounded then it diverges.So boundedness of a sequence is a necessary condition for its convergence!

Theorem 3.3.5. If limn

an = a = 0 then

(N N+ )(nN

+ ) (n > N )(|an | > |a|2

) .

Moreover, if a > 0 then for the above n a n >a2 , and if if a < 0 then for the above n a n N

|a|2 > |a an | |a| |an |,which implies |an | > |a| |a |2 = |a |2 , and the rst assertion is proved. On the other hand

( |a|2

> |a an |) a |a|2

< a n < a + |a|2 .Thus if a > 0 then for n > N

an > a |a|2

=a2

.

Also if a < 0 then for n > N

an < a + |a|2 = a2 .

Theorem 3.3.6. Let a, bR , (an )n , (bn )n be real sequences. Then

limn

an = a limn bn = b (nN+ ) (an bn ) (a b).

Proof. For a contradiction assume that b < a . Fix < a b2 so that b + < a and chooseN 1 and N 2 such that the following is true[(

n > N 1)( an > a

)]

[(

n > N 2)(bn < b + )].

If N max{N 1, N 2}thenbn < b + < a < a n ,

which contradicts to the condition ( nN+ ) (an bn ).

Theorem 3.3.7. Sandwich rule . Let aR and (an )n , (bn )n , (cn )n be real sequences. Then

{[(nN+ ) (an bn cn )]( limn an = limn cn = a)} ( limn bn = a).Proof. Let > 0. Then one can nd N 1, N 2N

+ such that

[(n > N 1)(a < a n )][(n > N 2)(cn < a + )].Then choosing N = max {N 1, N 2}we have

(a < a n bn cn < a + )(|bn a| < ) .

The next theorem is a useful tool in computing limits.

Theorem 3.3.8. Let limn

an = a and limn bn = b. Then

(i) limn (an + bn ) = a + b;

44 January 28, 2006


49/111


(ii) limn

(an bn ) = ab.

(iii) If in addition b = 0 and (nN

+

) (bn = 0) then lim

n

anbn

=ab

.

Proof. (i) Let > 0. Choose N 1 such that for all n > N 1 |an a| < / 2. Choose N 2 suchthat for all n > N 2 |bn b| < / 2. Then for all n > max{N 1, N 2}|an + bn (a + b)| |an a|+ |bn b| < .

(ii) Since (bn )n is convergent, it is bounded by Theorem 3.3.4. Let K be such that(nN

+ ) ( |bn | K ) and |a| K . We have

|an bn ab| = |an bn abn + abn an bn | |an a||bn |+ |a||bn b| K |an a|+ K |bn b|.

Choose N 1 such that for n > N 1 K |an a| / 2.Choose N 2 such that for n > N 2 K |bn b| / 2.Then for n > N = max {N 1, N 2}|an bn ab| < .

(iii) We have

an

bn

a

b=

an babnbn b

|an a||b|+ |a||bn b||bn b|

.

Choose N 1 such that for n > N 1

|bn | >12|b| (possible by Theorem 3.3.5) .

Choose N 2 such that for n > N 2

|an a| N 3

|a||bn b| N := max {N 1, N 2, N 3}anbn

ab

< .

45 January 28, 2006


50/111

3.4. MONOTONE SEQUENCES

3.4 Monotone sequences

In this section we consider a particular class of sequences which often occur in applications.Denition 3.4.1. (i) A sequence of real numbers ( an )nN + is called increasing if

(nN+ ) (an an+1 ).

(ii) A sequence of real numbers ( an )nN + is called decreasing if

(nN+ ) (an an+1 ).

(iii) A sequence of real numbers ( an )nN + is called monotone if it is either increasing ordecreasing.

Example 3.4.1. (i) an = n2 is increasing.

(ii) an =1n

is decreasing.

(iii) an = ( 1)n1n

is not monotone.

The next theorem is one of the main theorems in the theory of converging sequences.

Theorem 3.4.2. If a sequence of real numbers is bounded above and increasing then it isconvergent.

Proof. Let ( an )nN + be a bounded above increasing sequence of real numbers. Since the set

{an |nN+ }is bounded above, there exists supremuma := sup

n {an}.By the denition of supremum

( > 0)(N N+ ) (aN > a ).

Fix this N . Since (an )n is increasing, (

n > N ) (an > a N ). Therefore

(nN+ ) [(n > N )(an > a ).

Also by the denition of supremum

( > 0)(nN+ ) (an < a + ).

Therefore for n > N

(a < a n < a + )(|an a| < ).

In complete analogy to the previous theorem one proves the following one.

46 January 28, 2006


51/111


Theorem 3.4.3. If a sequence of real numbers is bounded below and decreasing then it isconvergent.

In many cases before computing the limit of a sequence one has to prove that the limitexists, i.e. the sequence converges. In particular, this is the case when a sequence is denedby a recurrent formula, i.e. the formula for the nth member of the sequence via previousmembers with numbers n 1 and less.

Let us illustrate this point by an example.

Example 3.4.4. Let ( an )n be a sequence dened by

(3.4.1) an =

an 1 + 2 , a1 = 2.

1. (an )n is bounded.Indeed, we prove that

(nN+ ) (0 < a n 2).

Proof. Positivity is obvious.For n = 1 a1 = 2 2, and the statement is true.Suppose that it is true for n = k (k 1), i.e. ak 2.For n = k + 1 we have

ak+1 = ak + 2 2 + 2 = 2 ,and by the principle of induction the statement is proved.

2. (an )n is increasing.We have to prove that

(nN+ ) (an +1 an ).

Proof. This is equivalent to prove

[(nN+ ) (an + 2 an )][(nN+ ) (an + 2 a2)].

Indeedan + 2 a2n = (2 an )(an + 1) 0.

By Theorem 3.4.2 we conclude that ( an ) is convergent. Let lim n an = x. Then, of course,limn an 1 = x. Passing to the limit in (3.4.1) we obtain that

x = x + 2 ,from which we easily nd that x = 2.

Now we are ready to prove a very important property of real numbers.

Theorem 3.4.5. Let ([an , bn ])nN + be a sequence of closed intervals on the real axis such that:

(i) (nN ) ([an +1 , bn +1 ][a

n , bn ]);

47 January 28, 2006


52/111

3.5. CAUCHY SEQUENCES

(ii) limn

(bn an ) = 0(i.e. the length of the intervals tends to zero).

Then there is a unique point which belongs to all the intervals.

Proof. The sequence ( an )n is increasing and bounded above, and the sequence ( bn )n is de-creasing and bounded below. Therefore ( an )n , (bn )n are convergent.Let a := lim n an ; b := lim n bn .Then

a bby Theorem 3.3.6. Moreover,

ba = limn (bn an ) = 0 .Hence

a = bnN +

[an , bn ].

3.5 Cauchy sequences

In this short section we are going to derive the intrinsic criterion for convergence of sequences(i.e. the criterion which does not use the value of the limit and is expressed in terms of acondition on the members of the sequence).

First we introduce an important notion of a Cauchy sequence.

Denition 3.5.1. A sequence of real numbers ( an )nN + is called a Cauchy sequence if

( > o )(N N) {[(n > N )(m > N )](|an am | < )}.

Proposition 3.5.1. If a sequence (an )nN + is convergent then it is a Cauchy sequence.

Proof. Let a := lim n an . Let > 0. Then there exists N

N such that for all n > N

|an a| < / 2.Let m, n > N . Then by the triangle inequality

|an am | |an a|+ |am a| < .

It turns out that the inverse to Proposition 3.5.1 is also true. This is expressed in thenext theorem that delivers the Cauchy criterion for convergence of sequences.

Theorem 3.5.1. If (an )nN + is a Cauchy sequence of real numbers then it is convergent.

48 January 28, 2006


53/111


Proof. Let (an )nN + be a Cauchy sequence. Take = 1. There exists N N such that for all n,m > N

|an

am

|< 1.

Fix m > N . Then

|an | |am | |an am | < 1.Therefore

(n > N )(|an | < |am |+ 1) .It follows that ( an )nN + is bounded (Why?)For any n

N dene n := inf

k>nak , n := sup

k>nak .

Then we have(n

N) ([ n +1 , n +1 ][ n , n ]).

(Justify the above)The length of the interval [ n , n ] converges to zero as since from the Cauchy condition itfollows that for any > 0 there exists N N+ such that for all k > N

aN < a k < a N + ,from which we conclude that

aN N N aN + ,or else

(nN+ ) [(n > N )(2 N N n n )].

By Theorem 3.4.5 there exists an

N +

[ n , n ].

Fix > 0. Find N such that | N N | < . Clearly (k > N ) (ak [ N , n ]). It follows that(k > N ) (|ak a| < ).

Therefore limn

an = a.

49 January 28, 2006


54/111

3.6. SERIES

3.6 Series

In this section we discuss innite series. Let us start from an example which is likely to befamiliar from high school. How to make sense of the innite sum

12

+14

+18

+1

16. . .

This is the known sum of a geometric progression. One proceeds as follows. Dene the sumof the rst n term

sn =12

+14

+18

+ +1

2n= 1

12n

and then dene the innite sum s as limn sn , so that s = 1.This idea is used to dene formally an arbitrary series,

Denition 3.6.1. Let ( an )nN + be a sequence of real (complex) numbers. Let

sn = a1 + a2 + + an =n

k=1

ak .

We say that the series

k=1

ak = a1 + a2 + . . .

is convergent if the sequence of the partial sums ( sn )nN + is convergent. The limit of thissequence is called the sum of the series.

If the series is not convergent we say that it is divergent (or diverges).

Theorem 3.6.1. If the series

n =1an

is convergent then limn

an = 0 .

Proof. Let sn = nk=1 ak . Then by the denition the limit lim n sn exists. Denote it by s.Then of course lim

ns

n 1= s. Note that a

n= s

n s

n 1for n

2. Hence

limn

an = limn sn limn sn 1 = s s = 0 .The same idea can be used to prove the following theorem.

Theorem 3.6.2. If the series

n =1an

is convergent then

s2n sn = an +1 + an +2 + + a2n 0 as n .50 January 28, 2006


55/111


The above theorem expresses the simplest necessary condition for the convergence of aseries.

For example, each of the following series is divergent1 + 1 + 1 + 1 + . . . ,

1 1 + 1 1 + . . .since the necessary condition is not satised.

Let us take a look at several important examples.

Example 3.6.3. The series

k=0

xk = 1 + x + x2 + . . .

is convergent if and only if 1 < x < 1.Indeed,

sn =1

1 x xn

1 x if x = 1 ,n if x = 1 .

Example 3.6.4. Harmonic series The series

1 +12

+13

+14

+ +1n

+ . . .

diverges.

Indeed,s2n sn =

1n + 1

+1

n + 2+ +

12n

>12

.

Example 3.6.5. Generalised harmonic series The series

1 +12 p

+13 p

+14 p

+ +1

n p+ . . .

converges if p > 1 and diverges if p 1.

3.6.1 Series of positive terms

In this subsection we conne ourselves to considering series which terms are non-negativenumbers, i.e.

k=1

ak with ak 0 for every kN+ .

The main feature of such a case that the sequence of the partial sums sn = nk=1 ak isincreasing .

Indeed, sn sn 1 = an 0.This allows us to formulate a simple criterion of convergence for such a series.

51 January 28, 2006


56/111

3.6. SERIES

Theorem 3.6.6. A series of positive terms is convergent if and only if the sequence of partial sums is bounded above.

Proof. Indeed, if a series is convergent then the sequence of partial sums is convergent by thedenition, and therefore it is bounded.

Conversely, if the sequence of partial sums is bounded above then it is convergent byTheorem 3.4.2, and therefore the series is convergent.

Comparison tests

Here we establish tests which make possible to infer the convergence or divergence of a seriesby means of comparing it with another series convergence or divergence of which is known.

Theorem 3.6.7. Let

k=1

ak ,

k=1

bk be two series of positive terms. Assume that

(3.6.2) (nN+ ) (an bn ).

Then

(i) If

k=1

bk is convergent then

k=1

ak is convergent.

(ii) If

k=1 ak is divergent then

k=1 bk is divergent.

Proof. Let sn = nk=1 ak , sn =nk=1 bk . Then

(nN+ ) (sn sn ).

So if the theorem follows now from Theorem 3.6.6.

Remark 5. Theorem 3.6.7 holds if instead of (3.6.2) one assumes

(K > 0)(nN+ ) (an Kbn ).

Now since for p 1(nN

+ )1

n p 1n

,

we obtain one of the assertions of Example 3.6.5 from the above theorem.

The following observation (which follows immediately from the denition of convergenceof a series) is important for the next theorem.

The convergence or divergence of a series is unaffected if a nite number of terms are

inserted, or suppressed, or altered.

52 January 28, 2006


57/111


Theorem 3.6.8. Let (an )n , (bn )n be two sequences of positive numbers. Assume that

limn anbn = L > 0. (the limit is positive and nite ).

Then the series

k=1

ak is convergent if and only if the series

k=1

bk is convergent.

Proof. By the denition of the limit

(N N)(n > N )12

L N )(

nan < r )].Or otherwise an < r n . The convergence follows now from the comparison with the convergentseries n =1 r

n .Next suppose that l > 1. Then

(N N)(nN+ )[(n > N )(

nan > 1)].

Or otherwise an > 1. Hence the series diverges.

53 January 28, 2006


58/111

3.6. SERIES

Theorem 3.6.11. (DAlemberts Test or The Ratio Test) Let (an )nN + be a sequence of positive numbers. Suppose that

limn an +1an = l.

Then if l < 1, the series

n =1an converges; if l > 1, the series

n =1an diverges. If l = 1 , no

conclusion can be drawn.

Proof. Suppose that l < 1. Choose r such that l < r < 1. Then

(N N)(nN+ )[(n > N )(

an +1an

< r )].

Therefore

an =an

an 1 an 1an

Documents

Intro to Analysis Notes