201_syllectnt real الكتاب

8/6/2019 201_SylLectNt real

1/69

Math 201 (Introduction to Analysis) Fall 2004-2005

Instructor: Dr. Kin Y. Li

Office: Room 3471 Office Phone: 2358 7420 e-mail address: [email protected]

Office Hours: Tu. 3:30pm - 4:30pm (or by appointments)

Prerequisite: A-level Math or One Variable Calculus

Website for Lecture Notes and Practice Exercises: http://www.math.ust.hk/ makyli/UG.html

Grade Scheme: Homeworks (4%), Tutorial Presentations (6%), Midterm (40%), Final Examination (50%)

All records of grades will be put on the website http://grading.math.ust.hk/checkgrade/ as soon as they are

available. This course is essentially graded by curve with one exception, namely students who achieve 40% or

less of the overall grade will fail the course. Students who take the same midterm will be grouped under the

same curve.

Students should make copies of homeworks before submitting the originals. In case homeworks are not

received, students will be required to resubmit copies within a short period of time (may be less than a day).

For tutorial presentations, students willform groups (of 1 to 3 students) and present solutionsto assigned problems

in the tutorial sessions. All members of a group must attend the sessions to assist in answering possible questions from

presentations. Marks will be deducted for failure to present solutions or for absence in supporting his/her group.

Course Description: This is the first of two required courses on analysis for Math majors. It is to be followed by

Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, as appeared in one

variable calculus. Along the way to establish the proofs, many new concepts will be introduced. These include

countability, sequences and series of numbers, of functions, supremum/infimum, Cauchy condition, Riemann integrals

and improper integrals. Understanding them and their properties are important for the development of the present and

further courses.

Textbooks:

Kenneth A. Ross, Elementary Analysis : The Theory of Calculus, Springer-Verlag, 1980.

Robert Bartle and Donald Sherbert, Introduction to Real Analysis, 3rd ed., Wiley, 2000.

Ross is for students who are taking this course for the first time. Students who are repeating may use either Ross

or Bartle and Sherbert.

References:

1. J. A. Fridy, Introductory Analysis

2nd ed., Academic Press, 2000.

2. Manfred Stoll, Introduction to Real Analysis, Addison Wesley, 1997.

3. Walter Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, 1976.

4. Tom Apostol, Mathematical Analysis, 2nd ed., Addison-Wesley, 1974.

*5. Chinese Solution Manual to Tom Apostols Mathematical Analysis, 2nd ed.

Reminders: Students are highly encouraged to come to office hours for consultation. This is a difficult course for

many, but not all students. Although there are lecture notes, students should attend all lectures and tutorials as

lectures notes are only briefrecords of materials covered in class, which may contain typographical errors. Of course,

questions from students and answers from instructors or other digressions will not be recorded. Students are advised

to take your own notes. All materials presented in lectures and tutorials as well as proper class conduct are the

students responsibility. The instructor reserves the right to make any changes to the course throughout the

semester. The only way to succeed in this course is to do the work.

1


2/69

Objectives of the Course

The objectives of the course are to learn analysis and learn proofs.

Questions: What is analysis? How is it different from other branches of mathematics?

Analysis is the branch of mathematics that studies limit and concepts derived from limit, such as continuity,differentiation and integration, while geometry deals with figures, algebra deals with equations and inequalities

involving addition, subtraction, multiplication, division and number theory studies integers.

When we try to solve problems involving real or complex numbers, such as finding roots of polynomials or

solving differential equations, we may not get the right answers the first time. However, we can get approximations

and the limits of these approximations, we hope, will give us the right answers. At least, we can know solutions exist

even though we may not be able to write them explicitly.

Problems involving integers can be much harder since integers are discrete, i.e. there are minimum distance

separating distinct integers so that one cannot find integers arbitrarily close to another integer.

Try to see if there is a real solution to the equation 2x

x2

4x 4

1 987654321 Then try to see if there is an integer

solution. What are the differences in the way you solve these two problems?

Question: Why should we learn proofs?

A statement is true not because your teacher tells you it is true. A teacher can make mistakes! There were famous

mathematicians who made conjectures that were discovered to be wrong years later. How can we be certain the facts

we learned are true? How can we judge when more than one proposed solutions are given, which is correct?

Suppose we want to find limx 0

x 2 sin 1x

sinx

Since the numerator is between x 2 and x 2

the numerator has limit 0.

The denominator also has limit 0. So, by lHopitals rule, limx 0

x 2 sin 1x

sinx

limx 0

2x sin 1x

cos 1x

cosx

However, the new

numerator does not have a limit because cos 1x

has no limit as x

0

while the new denominator has limit 1. So the

limit of the original problem does not exist. Is this reasoning correct? No. Where is the mistake?

Sometimes we explain facts by examples or pictures. For instance, the statement that every odd degree polynomial

with real coefficients must have at least one root is often explained by some examples or some pictures. In our lifetime,

we can only do finitely many examples and draw finitely many pictures. Should we believe something is true by seeing

a few pictures or examples?

Draw the graphs of a few continuous functions on [0

1]

Do you think every continuous function on [0

1]

is differentiable in at least one point on

0

1

? Or do you think there exists a continuous function on [0

1] not

differentiable at any point of

0

1

?

Consider the function f n n2 n

41

Note f 1 41 f 2 43 f 3 47 f 4 53 f 5

61

f

6

71

f

7

83

f

8

97

f

9

113

f

10

131 are prime numbers. Should you believe that f

n

is a prime number for every positive integer n? What is the first n that f n is not prime?

In order to have confident, you have to be able to judge the facts you learned are absolutely correct. Almost

correct is not good enough in mathematics.

2


3/69

Chapter 1. Logic

To reason correctly, we have to follow some rules. These rules of reasoning are what we called logic. We will

only need a few of these rules, mainly to deal with taking opposite of statements and to handle conditional statements.

We will use the symbol (or

) to denote the word not. Also, we will use the symbol

to denote for all,

for any, for every. Similarly, the symbol will denote there is (at least one), there exists, there are (some)and usually followed by such that. The symbols

and

are called quantifiers.

Negation. Below we will look at rules of negation (i.e. taking opposite). They are needed when we do indirect

proofs (or proofs by contradiction). For any expression p

we have

p

p

Examples. (1)expression :

p

x

0 and

q

x

1

opposite expression : x

0 or x

1

rule :

p and q

p

or

q

(2) expression : x 0 or x 1

opposite expression : x 0 and x 1

rule :

p or q

p

and

q

(3)statement : For every x

0

x has a square root. (True)

quantified statement :

x 0 x has a square root

opposite statement : There exists x

0 such that x does not have a square root. (False)

quantified opposite statement :

x

0

x has a square root

(4) statement : For every x

0

there is y

0 such that y2 x

(True)

quantified statement :

x

0

y

0

y2 x

opposite statement : There exists x 0 such that for every y 0 y2 x (False)

quantified opposite statement :

x

0

y

0

y2 x

From examples (3) and (4), we see that the rule for negating statements with quantifiers is first switch every

to

and every

to

, then negate the remaining part of the statement.

If-then Statements. If-then statements occur frequently in mathematics. We will need to know some equivalent

ways of expressing an if-then statement to do proofs. The statement if p

then q may also be stated as p implies q,

p only ifq, p is sufficient for q, q is necessary for p and is commonly denoted by p

q. For example,

the statement if x 3 and y 4

then x 2

y2 25 may also be stated as x 3 and y 4 are sufficient for

x 2

y2 25 or x 2

y2 25 is necessary for x 3 and y 4.

Example. (5) statement : Ifx 0 then x x

(True)

opposite statement : x

0 and x x

(False)

rule :

p

q

p and

q

Remark. Notep q

p q

p and

q

p

or

q

p

or q

3


4/69

For the statement if p

then q

p

q

there are two related statements: the converse of the statement is if

q , then p (q p) and the contrapositive of the statement is if q then p ( q p).

Examples. (6) statement : Ifx 3 then x 2 9 (True)

converse : Ifx 2 9 then x 3 (False, as x may be 3.)

contrapositive : Ifx2

9

then x

3

(True)

(7) statement : x 3

2x 6 (True)

converse : 2x 6

x 3 (True)

contrapositive : 2x 6 x 3 (True)

(8) statement : If x 3 then x 3 (False, as x may be 3.)

converse : Ifx 3 then x 3 (True)

contrapositive : Ifx 3

then x 3

(False, as x may be 3.)

Remarks. Examples (6) and (7) showed that the converse of an if-then statement is not the same as the statement

nor the opposite of the statement in general. Examples (6), (7) and (8) showed that an if-then statement and itscontrapositive are either both true or both false. In fact, this is always the case because by the remark on the last page,

q

p

q

or

p

q or p

p

or q p

q

So an if-then statement and its contrapositive statement are equivalent.

Finally, we introduce the terminology p if and onlyifq to mean if p then q and ifq then p. The statement

p if and only ifq is the same as p is necessary and sufficient for q. We abbreviate p if and only ifq by p

q. So p

q means p q and q p. The phrase if and only if is often abbreviated as iff.

Caution! Note

and

but

For example, every student is assigned a

number is the same as

student,

number such that the student is assigned the number. This statement implies

different students may be assigned possibly different numbers. However, if we switch the order of the quantifiers, the

statement becomes

number such that

student, the student is assigned the number. This statement implies there

is a number and every student is assigned that same number!

4


5/69

Chapter 2. Sets

To read and write mathematical expressions accurately and concisely, we will introduce the language of sets. A

setis a collection of objects (usually numbers, ordered pairs, functions, etc.) If object x is in a set S, then we say x is

an element(or a member) of S and write x S

Ifx is not an element of S, then we write x S

A set having finitely

many elements is called a finite set, otherwise it is called an infinite set. The empty setis the set having no objects and

is denoted by

A set may be shown bylistingits elements enclosed in braces (eg. 1

2

3 is a set containing theobjects 1

2

3

the

positive integer 1

2

3

, the integer

2

1

0

1

2

, the empty set

) or by description

enclosed in braces (eg. the rational numbers mn

: m

n

the real numbers x : x is a real number

and the complex numbers x

iy : x

y

) In describing sets, the usual convention is to put the form of the

objects on the left side of the colon and to state the conditions on the objects on the right side of the colon. It is also

common to use a vertical bar in place of colon in set descriptions.

Examples. (i) The closed interval with endpoints a

b is [a

b] x : x

and a

x

b

(ii) The set of square numbers is1

4

9

16

25

n 2 : n

(iii) The set of all positive real numbers is

x : x and x 0 (If we want to emphasize this is a subset

of

we may stress x is real in the form of the objects and write

x

: x

0

If numbers are alwaystaken to mean real numbers, then we may write simply

x : x 0 )

(iv) The set of points (or ordered pairs) on the line m with equation y mx is

x

mx

: x

For sets A B we say A is a subset of B (or B contains A) iff every element of A is also an element of B In

that case, we write A B (For the case of the empty set, we have

S for every set S ) Two sets A and B are

equal if and only if they have the same elements (i.e. A B means A B and B A

) So A B if and only if

(x A

x B). If A B and A B

then we say A is a proper subset of B and write A B

(For example,

if A 1

2

B 1

2

3

C 1

1

2

3

then A B is true, but B C is false. In fact, B C

Repeated

elements are counted only one time so that C has 3 elements, not 4 elements.)

For a set S we can collect all its subsets. This is called the power set of S and is denoted by P S or 2 S For

examples, P

P

0

0

and P

0

1

0

1

0

1

For a set with n elements, its powerset will have 2n element. This is the reason for the alternative notation 2S for the power set of S Power set is one

operation of a set. There are a few other common operations of sets.

Definitions. For sets A1 A2 An

(i) their union is A1 A2 An

x : x A1 or x A2 or or x An

(ii) their intersection is A1 # A2 # # An

x : x A1 and x A2 and and x An

(iii) their Cartesian productis

A1 $ A2 $ $

An

x1 x2 xn : x1 A1 and x2 A2 and

and xn An

(iv) the complementof A2 in A1 is A1 ' A2

x : x

A1 and x

A2

Examples. (i)1

2

3

3

4

1

2

3

4

1

2

3

#

2

3

4

2

3

1

2

3

'

2

3

4

1

(ii) [ 2

4]#

1

2

3

4

[0

2]

[1

5]

[4

6] [0

6]

(iii)

[0

7]#

'

n2 : n

0

1

2

3

4

5

6

7 '

1

4

9

16

25

0

2

3

5

6

7

(iv)

$

$

x y z : x y z

$

'

a b : a is rational and b is irrational

Remarks. (i) For the case of the empty set, we have

A

A

A

A#

#

A

A$

$

A

A'

A and

'

A

5


6/69

(ii) The notions of union, intersection and Cartesian product may be extended to infinitely many sets similarly. The

union is the set of objects in at least one of the sets. The intersection is the set of objects in every one of the sets.

The Cartesian product is the set of ordered tuples such that the i-th coordinate must belong to the i-th set.

(iii) The set A1 A2 An may be written asn

k 1

Ak If for every positive integer k there is a set Ak then the

notation A1 A2 A3

may be abbreviated as

k 1

Ak or

k

Ak If for every x S there is a set Ax then

the union of all the sets Ax s for all x S is denoted by

x S

Ax Similar abbreviations exist for intersection and

Cartesian product.

Examples. (i)

[1

2]

[2

3]

[3

4]

[4

5]

#

[1

#

(ii)

n

0

1

1

n [0

2

#

0

11

2 #

0

11

3 #

0

11

4 #

[0

1]

(iii) For every k

let Ak

0

1

then

A1$

A2$

A3$

x1

x2

x3

: each xk is 0 or 1 for k

1

2

3

(iv) For each m

let m be theline withequation y mx on theplane, then

m

m

2'

0

y

: y

y 0

and

m

m

0

0

We shall say that sets are disjointiff their intersection is the empty set. Also, we say they are mutually disjointiff

the intersection of every pair of them is the empty set. A relation on a set E is any subset of E$

E The following is

an important concept that is needed in almost all branches of mathematics. It is a tool to divide (or partition) the set of

objects we like to study into mutually disjoint subsets.

Definition. An equivalence relation R on a set E is a subset R of E$

E such that

(a) (reflexive property) for every x E x x R

(b) (symmetric property) if

x

y

R

then

y

x

R

(c) (transitive property) if

x

y

y

z

R

then

x

z

R

We write x y if

x

y

R

For each x E

let [x] y : x y

This is called the equivalence class

containing x Note that every x [x] by (a) so that

x E

[x] E If x y then [x] [y] because by (b) and (c),

z [x]

z

x

z

y

z [y] If x y then [x]#

[y]

because assuming z [x]#

[y] will

lead to x z and z y which imply x y a contradiction. So every pair of equivalence classes are either the same

or disjoint. Therefore, R partitions the set E into mutually disjoint equivalence classes.

Examples. (1) (Geometry) For triangles T1 and T2 define T1

T2 if and only if T1 is similar to T2 This is an

equivalence relation on the set of all triangles as the three properties above are satisfied. For a triangle T [T] is the setof all triangles similar to T

(2) (Arithmetic) For integers m and n

define m n if and only ifm n is even. Again, properties (a), (b), (c) can

easily be verified. So this is also an equivalence relation on

There are exactly two equivalence classes, namely

[0]

4

2

0

2

4

(even integers) and [1]

5

3

1

1

3

5

(odd integers). Two integers

in the same equivalence class is said to be of the same parity.

(3) Some people think that properties (b) and (c) imply property (a) by using (b), then letting z x in (c) to conclude

x x R This is false as shown by the counterexample that E 0 1 and R 1 1 which satisfies properties

(b) and (c), but not property (a). R fails property (a) because 0 E but 0 0 R as 0 is not in any ordered pair in R

6


7/69

A function (or map or mapping) f from a set A to a set B (denoted by f : A

B) is a method of assigning to

every a A exactly one b B This b is denoted by f a and is called the value of f at a Thus, a function must be

well-definedin the sense that ifa a then f a f a The set A is called the domain of f (denoted by dom f)

and the set B is called the codomain of f (denoted by codom f). We say f is a B-valued function (eg. if B then

we say f is a real-valued function.) When the codomain B is not emphasized, then we may simply say f is a function

on A

The image or range of f (denoted by f

A

or im f or ran f) is the set f

x

: x A

(To emphasize this is a

subset of B we also write it as f x B : x A ) The set G

x f x : x A is called the graph of f Twofunctions are equal if and only if they have the same graphs. In particular, the domains of equal functions are the same

set.

Examples. The function f : given by f x x 2 has dom f codom f Also, ran f

0

1

4

9

16

This is different from the function g : given by g x x 2 because dom g dom f

Also, a function may have more than one parts in its definition, eg. the absolute value function h :

defined by

h

x

x if x

0 x if x

0

Be careful in defining functions. The following is bad: let x n

1

n and i

xn n

The

rule is not well-defined because x1

1 x3 but i x1 1 3 i

x3

Definitions. (i) The identity function on a set S is IS : S S given by IS x x for all x S

(ii) Let f : A

B

g : B

C be functions and f

A

B

The composition of g by f is the function

g f : A C defined by g f x

g f x for all x A

(iii) Let f : A

B be a function and C A

The function f C : C B defined by f C x f

x

for every x C

is called the restriction of f to C

(iv) A function f : A

B is surjective (or onto) iff f

A

B

(v) A function f : A B is injective (or one-to-one) iff f x f y implies x y

(vi) A function f : A

B is a bijection (or a one-to-one correspondence) iff it is injective and surjective.

(vii) For an injective function f : A

B

the inverse function of f is the function f

1 : f

A

A defined by

f

1

y

x

f

x

y

Remarks. A function f : A

B is surjective means f

A

B

which is the same as saying every b B is an f

a

for at least one a A

In this sense, the values of f do not omit anything in B

We will loosely say f does not omit

any element of B for convenience. However, there may possibly be more than one a A that are assigned the same

b B

Hence, the range of f may repeatsome elements of B

If A and B are finite sets, then f surjective implies the

number of elements in A is greater than or equal to the number of elements in B

Next, a function f : A

B is injective means, in the contrapositive sense, that x y implies f

x

f

y

which we may loosely say f does not repeat any element of B

However, f may omit elements of B as there may

possibly be elements in B that are not in the range of f

So if A and B are finite sets, then f injective implies the

number of elements in A is less than or equal to the number of elements in B

Therefore, a bijection from A to B is a function whose values do not omit nor repeat any element of B If A and

B are finite sets, then f bijective implies the number of elements in A and B are the same.

Remarks (Exercises). (a) Let f : A

B be a function. We have f is a bijection if and only if there is a function

g : B

A such that g

f IA and f g IB (In fact, for f bijective, we have g

f

1 is bijective.)

(b) If f : A B and h : B C are bijections, then h

f : A C is a bijection.

(c) Let A

B be subsets of and f : A

B be a function. If for every b B

the horizontal line y b intersects the

graph of f exactly once, then f is a bijection.

To deal with the number of elements in a set, we introduce the following concept. For sets S1 and S2 we will

define S1

S2 and say they have the same cardinality (or the same cardinal number) if and only if there exists a

bijection from S1 to S2 This is easily checked to be an equivalence relation on the collection of all sets. For a set S the

equivalence class [S] is often called the cardinal numberof S and is denoted by card S or S This is a way to assign

a symbol for the number of elements in a set. It is common to denote, for a positive integer n card 1 2 n n

card

0 (read aleph-naught) and card c (often called the cardinality of the continuum).

7


8/69

Chapter 3. Countability

Often we compare two sets to see if they are different. In case both are infinite sets, then the concept of countable

sets may help to distinguish these infinite sets.

Definitions. A set S is countably infinite iff there exists a bijection f :

S (i.e. and S have the same cardinal

number 0 ) A set is countable iff it is a finite or countably infinite set. A set is uncountable iff it is not countable.

Remarks. Suppose f:

S is a bijection. Then f is injective means f

1

f

2

f

3

are all distinct and f

is surjective means f

1

f

2

f

3

S

So n

f

n

S is a one-to-one correspondence between

and S

Therefore, the elements of S can be listed in an orderly way (as f

1

f

2

f

3

) without repetition or

omission. Conversely, if the elements of S can be listed as s1 s2 without repetition or omission, then f : S

defined by f

n

sn will be a bijection as no repetition implies injectivity and no omission implies surjectivity.

Bijection Theorem. Let g: S

T be a bijection. S is countable if and only if T is countable.

(Reasons. The theorem is true because S countable implies there is a bijective function f :

S

which implies

h g

f :

T is bijective, i.e. T is countable. For the converse, h is bijective implies f g

1

h is bijective.)

Remarks. Similarly, taking contrapositive, S is uncountable if and only ifT is uncountable.

Basic Examples. (1) is countably infinite (because the identity function I

n

n is a bijection).

(2) is countably infinite because the following function is a bijection (one-to-one correspondence):

1

2

3

4

5

6

7

8

9

f

0

1

1

2

2

3

3

4

4

The function f :

is given by f

n

n2

ifn is even

n

1

2

ifn is oddand its inverse function g :

is given

by g

m

2m ifm 0

1

2m ifm 0

Just check g

f

I

and f

g

I

(3) $

m

n

: m

n

is countably infinite.

(Diagonal Counting Scheme) Using the diagram on the right, de-

fine f: $

by f 1 1 1 , f 2 2 1 , f 3 1 2 ,

f 4 3 1 , f 5 2 2 , f 6 1 3 , , then f is injective

because no ordered pair is repeated. Also, f is surjective because

m

n

f

m

n

2

k 0

k

n

f

m

n 2

m

n 1

2

n

1

1

1

2

1

3

1

4

2

1

2

2

2

3

3

1

3

2

4

1

(4) The open interval

0

1

x: x

and 0

x

1 is uncountable. Also, is uncountable.

f

1

0

a11a12a13a14

f

2

0

a21a22a23a24

f

3

0

a31a32a33a34

f

4

0

a41a42a43a44

Suppose

0

1

is countably infinite and f:

0

1

is a bijection as

shown on the left. Consider the number x whose decimal representation is

0 b1b2b3b4 , where bn

2 ifann 1

1 ifann

1. Then 0

x 1 and x f n

for all n because bn

ann . So f cannot be surjective, a contradiction. Next

is uncountable because tan x 1

2

provides a bijection from

0

1

onto

To determine the countability of more complicated sets, we will need the theorems below.

8


9/69

Countable Subset Theorem. Let A B

If B is countable, then A is countable. (Taking contrapositive, if A is

uncountable, then B is uncountable.)

Countable Union Theorem. If An is countable for every n then

n

An is countable. In general, if S is countable

(say f : S is a bijection) and As is countable for every s S then

s

S

As

n

Af n is countable. (Briefly,

countable union of countable sets is countable.)

Product Theorem. If A, B are countable, then A$

B a b : a A b B is countable. In fact, if A1 A2 Anare countable, then A1 $ A2 $ $ An is countable (by mathematical induction).

(Sketch of Reasons. For the countable subset theorem, if B is countable, then we can list the elements of B and to

count the elements of A, we can skip over those elements of B that are not in A

For the countable union theorem, if

we list the elements of A1 in the first row, the elements ofA2 in the second row, then we can count all the elements

by using the diagonal counting scheme. As for the product theorem, we can imitate the example of $

and also

use the diagonal counting scheme.)

Examples. (5)

n 1Sn where Sn

m

n: m

For every n

thefunction fn : Sn given by fn m

m

n

is a bijection (with f

1n

mn

m), so Sn is countable by the bijection theorem. Therefore, is countable by the

countable union theorem. (Then subsets of like '

0

0

#

0

1

are also countable.)

(6) '

is uncountable. (In fact, if A is uncountable and B is countable, then A'

B is uncountable as A'

B

countable implies

A#

B

A'

B

A countable by the countable union theorem, which is a contradiction).

(7) x

iy : x

y

contains and is uncountable, so by the countable subset theorem, is uncountable.

(8) Show that the set A r

m : m

r

0

1

is uncountable, but the set B r

m : m

r

#

0

1

is countable.

Solution. Taking m 1

we see that

0

1

A

Since

0

1

is uncountable, A is uncountable. Next we will

observe that B

m

Bm where Bm

r

m : r

#

0

1

r 0 1

r m for each m

Since

#

0 1 is countable and r m has 1 element for every r #

0 1 Bm is countable by the countable uniontheorem. Finally, since is countable and Bm is countable for every m B is countable by the countable

union theorem.

(9) Show that the set L of all lines with equation y mx

b

where m

b

is countable.

Solution. Note that for each pair m b of rational numbers, there is a unique line y mx

b in the set L So the

function f : $

L defined by letting f m b be the line y mx

b (with f 1 sending the line back to

m b ) is a bijection. Since $

is countable by the product theorem, so the set L is countable by the bijection

theorem.

(10) Show that ifAn

0

1 for every n then A1 $ A2 $ A3 $ is uncountable. (In particular, this shows that

the product theorem is not true for infintely many countable sets.)

Solution. Assume A1$

A2$

A3$

a1

a2

a3

: each ai

0 or 1

is countable and f :

A1 $ A2 $ A3$

is a bijection. Following example (4), we can change the n-th coordinate of f

n

(from 0 to

1 or from 1 to 0) to produce an element of A1 $ A2 $ A3 $

not equal to any f

n

which is a contradiction.

So it must be uncountable.

(11) Show that the power set P

of all subsets of is uncountable.

Solution. As in example (10), let An

0

1

for every n!

Define g : P

A1 $ A2 $ A3 $ by

g S a1 a2 a3 where am

1 ifm S

0 ifm S

(For example, g 1 3 5 1 0 1 0 1 ) Note

g has the inverse function g

1

a1 a2 a3

m : am 1

Hence g is a bijection. Since A1 $ A2 $ A3 $

is uncountable, so P

is uncountable by the bijection theorem.

9


10/69

(12) Show that the set S of all polynomials with integer coefficients is countable.

Solution. Let S0

For n

the set ofSn of all polynomials of degree n with integer coefficients is countable

becausethe function f : Sn ' 0 $ $ $

defined by f

anxn an

1xn

1

a0

an an

1 a0

is a bijection and

'

0

$

$ $

is countable by the product theorem. So, S S0

n

Sn is countable

by the countable union theorem.

(13) Show that there exists a real number, which is not a root of any nonconstant polynomial with integer coefficients.

Solution. For each polynomial f with integer coefficients, let Rf denotes the set of roots of f Then Rf has

at most deg f elements, hence Rf is countable. Let S be the set of all nonconstant polynomials with integer

coefficients, which is the subset

n

Sn ofS in the last example. Then

f S

Rf is the set of all roots of nonconstant

polynomials with integer coefficients. It is countable by the countable union theorem. Since is uncountable,

'

f S

Rf is uncountable by the fact in example (6). So there exist uncountably many real numbers, which are

not roots of any nonconstant polynomial with integer coefficients.

Remarks. Any number which is a root of a nonconstant polynomial with integer coefficients is called an algebraic

number. A number which is not a rootof any nonconstant polynomialwith integer coefficients is called a transcendental

number. Are there any algebraic numbers? Are there any transcendental numbers? If so, are there finitely many orcountably many such numbers?

Since every rational number ab

is the root of the polynomial bx a every rational number is algebraic. There

are irrational numbers like

2

which are algebraic because they are the roots of x 2 2

Using the identity

cos 3

4 cos3

3 cos

the irrational number cos 20

is easily seen to be algebraic as it is a root of 8x 3 6x 1

Example (12) and the fact that every nonconstant polynomial has finitely many roots showed there are only countably

many algebraic numbers. Example (13) showed that there are uncountably many transcendental real numbers. It is

quite difficult to prove a particular number is transcendental. In a number theory course, it will be shown that

and e

are transcendental.

The following are additional useful facts concerning countability.

Theorem.

(1) (Injection Theorem) Let f : A B be injective. If B is countable, then A is countable. (Taking contrapositive,

if A is uncountable, then B is uncountable.)

(2) (Surjection Theorem) Let g : A

B be surjective. If A is countable, then B is countable. (Taking contrapositive,

if B is uncountable, then A is uncountable.)

(Reasons. For the first statement, observe that the function h : A

f

A

defined by h

x

f

x

is injective

(because f is injective) and surjective (because h A f A ). So h is a bijection. If B is countable, then f A is

countable by the countable subset theorem, which implies A is countable by the bijection theorem.

For the second statement, observe that B g

A

x A

g

x

If A is countable, then it is a countable union of

countable sets. By the countable union theorem, B is countable.)

A Famous Open Problem in Mathematics

For two sets A and B

it is common to define card A

card B if and only if there exists an injective function

f : A

B

This is a way to indicate B has at least as many elements as A

Continuum Hypothesis. If S is uncountable, then card card S (This means every uncountable set has at least as

many elements as the real numbers.)

In 1940, Kurt Godel showed that the opposite statement would not lead to any contradiction. In 1966, Paul Cohen

won the Fields Medal for showing the statement also would not lead to any contradiction. So proof by contradiction

may not be applied to every statement.

10


11/69

Chapter 4. Series

Definitions. A series is the summation of a countable set of numbers in a specific order. If there are finitely many

numbers, then the series is a finite series, otherwise it is an infinite series. The numbers are called terms. The sum of

the first n terms is called the n-th partial sum of the series.

An infinite series is of the form a1

1st term

a2

2nd term

a3

3rd term

or we may write it as

k 1

ak

The first partial sum is S1 a1. The second partial sum is S2

a1

a2. The nth partial sum is Sn

a1

a2

an .

Series are used frequently in science and engineering to solve problems or approximate solutions. (E.g. trigono-

metric or logarithm tables were computed using series in the old days.)

Examples.(1)1 1

2

1

4

1

8

1

16

? (Sn 1

1

2

1

4

1

2n 2 1

2n

1 1

2

1

4

1

8

1

16

limn

2 1

2n

2.)

We say the series converges to 2, which is called the sum of the series.

(2) 1

1

1

1

1

1

(Sn 1

1

1

n

n, limn

Sn

.) We say the series diverges (to

).

(3) 1 1

1 1

1 1

1 1

. (Sn

1 ifn is odd

0 ifn is even, lim

n

Sn doesnt exist.) We say the series diverges.

Definitions. A series

k 1

ak a1

a2

a3

converges to a number S iff limn

a1

a2

an lim

n

Sn S

In that case, we may write

k 1

ak S and say S is the sum of the series. A series diverges to

iff the partial sum Sn

tends to infinity as n tends to infinity. A series diverges iff it does not converge to any number.

Remarks. (1) For every series

k 1

ak, there is a sequence (of partial sums) Sn . Conversely, if the partial sum sequence

Sn is given, we can find the terms an as follows: a1 S1, a2

S2 S1, , ak

Sk Sk

1 for k 1. Then

a1

an S1

S2 S1

Sn Sn

1 Sn. So Sn is the partial sum sequence of

k 1

ak. Conceptually,

series and sequences are equivalent. So to study series, we can use facts about sequences.

(2) Let N be a positive integer.

k 1

ak converges to A if and only if

k N

ak converges to B A a1

aN

1

because

B limn

aN

an lim

n

a1

a2

an

a1

aN

1 A

a1

aN

1

So to see if a series converges, we may ignore finitely many terms.

Theorem. If

k 1

ak converges to A and

k 1

bk converges to B then

k 1

ak

bk A

B

k 1

ak

k 1

bk

k 1

ak bk

A B

k 1

ak

k 1

bk

k 1

cak c A c

k 1

ak

for any constant c

For simple series such as geometric or telescoping series, we can find their sums.

11


12/69


13/69

For a nonnegative series

k 1

ak (i.e. ak 0 for every k), we have S1 S2 S3 and limn

Sn must exist as a

number or equal to

. So either

k 1

ak converges to a number or

k 1

ak diverges to

. (In short, either

k 1

ak S

or

k 1

ak

.) For nonnegative series, we have the following tests.

Theorem (Integral Test). Let f : [1

decrease to 0 as x

. Then

k 1

f

k

converges if and only if

1

f

x

d x

. (Note in general,

k 1

f

k

1

f

x

d x

)

(Reason. Thisfollows from f

2

f

3

f

n

1

f

x

d x

f

1

f

2

f

n 1

as shown in the figures below.)

...

1 2 3 4 n...

(2)

(3)

1 2 3 4 ... n

...

(2)

(1)f

f

f

f

Examples. (1) Consider the convergence or divergence of

k 1

1

1

k2

As x

1

x 2

so1

1

x 2 0

Now

1

1

1

x 2d x arctan x

1

2

4

So

k

1

1

1

k2

converges.

(2) Consider the convergence or divergence of

k 2

1

kln kand

k 2

1

k ln k 2.

As x

x lnx and x lnx 2

so their reciprocals decrease to 0

Now

2

d x

x lnx ln

lnx

2

So

k 2

1

kln kdiverges. Next

2

d x

x

lnx

2

1

lnx

2

1

ln 2

So

k 2

1

k

ln k

2converges.

Theorem (p-test). For a real number p

p

k 1

1

kp 1

1

2p

1

3p

1

4p

converges if and only if p

1.

(Reason. For p 0 the terms are at least 1, so the series diverges by term test. For p 0 f x 1xp

decreases

to 0 as x

. Since

1

1

xpd x

x

p

1

p

1

1

1

p 1if p

1

1

1

xpd x

lnx

1

if p 1 and

1

1

xpd x

x

p

1

p

1

1

if p

1, the integral test gives the conclusion.)

Remarks. For even positive integer p

the value of

p

was computed by Euler back in 1736. He got

2

2

6

4

4

90

2n

1

n

1 2 2nB2n

2

2n !

13


14/69


15/69

(Reason. Since c1 c2 c3 0, we have 0 S2 S4 S6 S5 S3 S1 Since limn

Sn Sn

1

limn

cn 0, the distances between the partial sums decrease to 0 and so lim

n

Sn must exist.)

Examples. Both

k 2

1

k

kln kand

k 1

e k cos k converge by the alternating series test because as k

kln k

and ek

so 1 kln k

0 and e

k

0 and cos k 1 k

For series with arbitrary positive or negative term, we have the following tests.

Theorem (Absolute Convergence Test). If

k 1

ak converges, then

k 1

ak converges.

(Reason. From ak ak ak we get 0 ak

ak 2 ak Since

k 1

2 ak converges, so by the comparison test,

k 1

ak

ak converges. Then

k 1

ak

k 1

ak

ak

k 1

ak converges.)

Definition. We say

k 1

ak converges absolutely iff

k 1

ak converges. We say

k 1

ak converges conditionally iff

k 1

ak

converges, but

k 1

ak diverges.

Examples. Determine if the following series converge absolutely or conditionally

(a)

k 1

cos k

k3(b)

k 1

cos k

1

k.

Solutions. (a)

k 1

cos kk3

k 1

1k3

. Since

k 1

1k3

converges by p-test, it follows that

k 1

cos kk3

converges by the

comparison test. So

k 1

cos k

k3converges absolutely by the absolute convergence test.

(b)

k 1

cos k

1

k

k 1

1

1

kbecause cos k

1

k.

1

d x

1

x ln

1

x

1

k 1

1

1

kdiverges.

However,1

1

kdecreases to 0 as k

. So by the alternating series test,

k 1

cos k

1

k

k 1

1

k 1

1

k

converges. Therefore

k 1

cos k

1

kconverges conditionally.

Theorem (Ratio Test). If ak

0 for every k and limk

ak

1 ak exists, then

limk

ak

1

ak

1

k 1

ak converges absolutely

1

k 1

ak may converge e g

k 1

1

k2

or diverge

e

g

k 1

1

k

1

k 1

ak diverges

15


16/69


17/69

(Reason. Note a1 S1 and ak

Sk Sk

1 for k 1. So,

n

k 1

akbk S1b1

S2 S1 b2

Sn Sn

1 bn

Sn bn

S1 b2

b1

Sn

1 bn

bn

1

Example. Show that

k 1

sin k

kconverges.

Let ak sin k and bk

1

k

Using the identity sin m sin1

2

1

2

cos

m 1

2

cos

m

1

2

we have

Sk sin1

sin2

sin k cos 1

2 cos

k 1

2

2sin 12

This implies Sk 1sin

1

2

for every k

Applying summation by parts and noting that limn

Snn

0

we get

k 1

sin k

k lim

n

n

k 1

sin k

k lim

n

Sn

n

n

1

k 1

Sk 1

k

1

1

k

k 1

Sk 1

k

1

k

1

Now

k 1

Sk 1

k

1

k

1

1

sin

1

2

k 1

1

k

1

k

1

1

sin

1

2

by the telescoping series test. So by the

absolute convergence test,

k 1

sin k

k

k 1

Sk 1

k

1

k 1

converges.

Complex Series

Complex numbers S1 S2 S3 with Sn un

i

n are said to have limit limn

Sn u

i

iff limn

un u

and limn

n

A complex series is a series where the terms are complex numbers. The definitions of convergent,

absolutely convergent and conditional convergent are the same. The remarks and the basic properties following the

definitions of convergent and divergent series are also true for complex series.

The geometric series test, telescoping series test, term test, absolute convergence test, ratio test and root test are

also true for complex series. For zk xk

iyk we have

k 1

zk converges to z x

i y if and only if

k 1

xk converges

to x and

k 1

yk converges to y So complex series can be reduced to real series for study if necessary.

Examples. (1) Note limn

in 0 (otherwise 0 limn

in limn

1 is a contradiction). So

k 1

ik diverges by term test.

(2) If z

1

then

zk

k2

1

k2and

k 1

1

k2converges by p-test implies

k 1

zk

k2converges absolutely. However, if

z

1

then limk

zk

1

k

1

2

k2

zk

limk

k2

k

1

2z z

1 implies

k 1

zk

k2diverges by the ratio test.

17


18/69


19/69

(Sketch of reason. Let pk

ak ifak 0

0 ifak 0and qk

0 ifak 0

ak ifak 0

Then ak pk

qk and ak pk

qk

Now both

k 1

pk

k 1

qk must diverge to

(If both converges, then their sum

k 1

ak will be finite, a contradiction.

If one converges and the other diverges to

then

k 1

ak

k 1

pk

k 1

qk will diverges to

a contradiction

also.) Let un

n be sequences of real numbers having limits x and u n

n

1 0 Now let P1 P2 be the

nonnegative terms of

k 1

ak in the order they occur and Q1 Q2 be the absolute value of the negative terms in

the order they occur. Since

k 1

Pk

k 1

Qk differ from

k 1

pk

k 1

qk only by zero terms, they also diverges to

Let m1 k1 be the smallest integers such that P1

Pm1

1 and P1

Pm1 Q1

Qk1 u1

Let m2 k2 be the smallest integers such that P1

Pm1 Q1

Qk1

Pm1

1

Pm2

2 and

P1

Pm1 Q1

Qk1

Pm1

1

Pm2 Qk1

1

Qk2 u2 and continue this way. This is possible

since thesums ofPk and Qk are

Now ifsn tn are thepartial sumsof this series P1

Pm1 Q1

Qk1

whose last terms are Pmn Qkn respectively, then sn

n Pmn and tn un Qkn by the choices of mn kn Since

Pn Qn have limit 0, so sn tn must have limit x As all other partial sums are squeezed by sn and tn the series we

constructed must have limit x )

Dirichlets Rearrangement Theorem. If ak and

k 1

ak converges absolutely, then every rearrangement

k 1

a

k

converges to the same sum as

k 1

ak.

(Reason. Define pk qk as in the last proof. Since pk qk ak

k 1

pk

k 1

qk converge, say to p and q respectively.

Since a k

p k

q k we may view

k 1

p k as a rearrangement of the nonnegative terms of

k 1

ak and inserting

zeros where a k 0 For any positive integer m the partial sum sm

m

k 1

p k

k 1

pk p Since pk 0 the

partial sum sm is also increasing, hence

k 1

p

k

converges. Now, for every positiveinteger n

n

k 1

pk

k 1

p

k

p

As n

we get

k 1

p

k p

Similarly,

k 1

q

k q

Then

k 1

a

k p q

k 1

ak )

Example.

k 1

1

2

k

1

2

1

22

1

23

1

24

1

25

converges (absolutely) to

1

2

1

1

2

1

3.

1

2

1

22

1

24

1

23

2 terms

1

28

1

27

1

26

1

25

4 terms

1

216

1

215

1

214

1

213

1

212

1

211

1

210

1

29

8 terms

is a rearrangement of

k 1

1

2

k, so it also converges to 1

3.

Remarks. As a consequence of the rearrangement theorem, the sum of a nonnegative series is the same no matter how

the terms are rearranged.

19


20/69

Chapter 5. Real Numbers

Decimal representations and points on a line are possible ways of introducing real numbers, but they are not too

convenient for proving many theorems. Instead we will introduce real numbers by its important properties.

Axiomatic Formulation. There exists a set

(called real numbers) satisfying the following four axioms:

(1) (Field Axiom) is a field(i.e. has two operations

and

such that for any a, b, c

,

(i) a

b, a

b

, (ii) a

b b

a, a

b b

a, (iii)

a

b

c a

b

c

,

a

b

c a

b

c

,

(iv) there are unique elements 0, 1

with 1 0 such that a

0 a, a

1 a,

(v) for each x

, there is a unique element x

such that x

x

0; ifx 0

then there is a unique

element x

1 such that x

x

1

1

(vi) a

b

c a

b

a

c.)

(This axiom allows us to do algebra with equations. Define a b to mean a

b

ab to mean a

b ab

to

mean a

b

1

Also, define 2 1

1

3 2

1

)

(2) (Order Axiom) has an (ordering) relation

such that for any a, b

(i) exactly one of the following a b, a b, b a is true,

(ii) ifa b, b c, then a c,

(iii) ifa b, then a

c b

c,

(iv) ifa

b and 0

c, then ac

bc.

(This axiom allows us to work with inequalities. For example, using (ii) and (iii), we can see that ifa b

and c d then a

c b

d because a

c b

c b

d Also, we can get 0 1 (for otherwise 1 0

would imply by (iii) that 0 1

1

0

1

1

which implies by (iv) that 0

1

1

1

a contradiction). Now define a b to mean b a a b to mean a b or a b etc. Also, define

closed interval [a b] x : a x b

open interval a b x : a x b

etc. Part (i) of the

order axiom implies any two real numbers can be compared. We define max a1 an to be the maximum

of a1 an and similarly for minimum. Also, define x max

x x Then x x and x x

i.e. x x x Next x a if and only if x a and x a i.e. a x a Finally, adding

x

x

x and

y

y

y

we get

x

y

x

y

x

y

which is the triangle inequalityx

y x

y )

(3) (Well-ordering Axiom) 1

2

3

is a well-orderedsubset of (i.e. for any nonempty subset S of , there

is m S such that m

x for all x S. This m is the least element (or the minimum) of S).

(This axiom allows us to formulate the principle of mathematical induction later.)

Definitions. For a nonempty subset S of , S is bounded above iff there is some M

such that x M for all

x S. Such an M is called an upper boundof S. The supremum or least upper bound(denoted by sup S or lub S)

of S is an upper bound

M of S such that

M M for all upper bounds M of S.

Examples. (1) For S 1n

: n 1 12

1

3

the upper bounds of S are all M 1 So sup S 1 S

(2) For S x

: x

0 , the upper bounds ofS are all M

0

So sup S 0 S

(4) (Completeness Axiom) Every nonempty subset of which is bounded above has a supremum in .

(This axiom allows us to prove results that have to do with the existence of certain numbers with specific

properties, as in the intermediate value theorem.)

Definitions. 1

2

3

4

is the natural numbers (or positive integers), 3 2 1 0 1 2 3

is the integers, mn

: m and n ! is the rational numbers and '

x ! : x is the irrational

numbers.

Remarks (Exercises). The first three axioms are also true if is replaced by

However, the completeness axiom is

false for

For example, S x : x

x 2

2

2

2

#

is bounded above in

but it does not have a

supremum in

20


21/69

As above, we define S to be bounded below if there is some m

such that m

x for all x S. Such an m is

called a lower boundof S. The infimum or greatest lower bound(denoted by infS or glb S) of S is a lower bound

m

of S such that m

m for all lower bounds m of S.

lower bounds

are here

upper bounds

are hereS

infS sup S

If S is bounded above and below, then S is bounded. (Note sup S, infS may or may not be in S. Also, if S is

bounded, then for all x S

x

M max

sup S

infS

)

Remarks (Exercises). (1) (Completeness Axiom for Infimum) Every nonempty subset of which is bounded below

has an infimum in . This follows from considering B x : x B (This is the reflection of B about 0.) If B

is bounded below, then B is bounded above and inf B sup B Similarly, if B is bounded above, then B is

bounded below and sup B inf B

( )-B B

0inf(-B) sup(-B) supBinfA supA

AinfB

(2) For a set B, if it is bounded above and c 0 then let cB cx : x B (This is the scaling of B by a factor of

c

) We have sup cB c sup B

If

A B, then infB

inf A whenever B is bounded below and sup A

sup B

whenever B is bounded above.

infB supB sup(c+B)inf(c+B)

c units

B c+B

(3) For c

let c

B c

x : x B

(This is a translation of B by c units.) It follows that B has a supremum

if and only ifc

B has a supremum, in which case sup

c

B

c

sup B

The infimum statement is similar, i.e.

inf

c

B

c

inf B

More generally, if A and B are bounded, then letting A

B x

y : x A

y B

we

have sup

A

B

sup A

sup B and inf

A

B

inf A

inf B

Simple Consequences of the Axioms.

Theorem (Infinitesimal Principle). For x, y , x y

for all

0 if and only if x y (Similarly, y

x

for all

0 if and only if y x.)

Proof. Ifx

y, then for all

0, x

y y

0

y

by (iv) of the field axiom and (iii) of the order axiom.

Conversely, if x

y

for all

0

then assuming x

y

we get x y

0 by (iii) of the order axiom. Let

0

x

y

then x

y

0 Since

0 0

we also have x

y

0 These contradict (i) of the order axiom. Sox

y

The other statement follows from the first statement since y

x is the same as y

x

Remarks. Taking y 0, we see that x

for all

0 if and only if x 0. This is used when it is difficult to

show two expressions a

b are equal, but it may be easier to show a b

for every

0

Theorem (Mathematical Induction). For every n , A n is a (true or false) statement such that A 1 is true and

for every k , A k is true implies A k

1 is also true. Then A n is true for all n .

Proof. Suppose A

n

is false for some n

. Then S n

: A

n

is false is a nonempty subset of . By the

well-ordering axiom, there is a least element m in S

Then A

m

is false. Also, if A

n

is false, then m

n. Taking

contrapositive, this means that ifn

m

then A

n

is true.

21


22/69

Now A

1

is true, so m 1 and m

imply m

2. So m 1

1

Let k m 1

then k m 1

m

implies A k is true. By hypothesis, A k

1

A m is true, a contradiction.

Theorem (Supremum Property). If a set S has a supremum in and

0

then there is x S such that

sup S

x

sup S.

Proof. Since sup S

sup S, sup S

is not an upper bound of S. Then there is x S such that sup S

x .

Since sup S is an upper bound of S, x sup S. Therefore sup S

x sup S.

Theorem (Infimum Property). If a set S has an infimum in and

0 then there is x S such thatinfS

x infS.

Proof. Since infS

infS, infS

is not a lower bound ofS. Then there is x S such that infS

x . Since

infS is a lower bound of S, x infS. Therefore infS

x infS.

Theorem (Archimedean Principle). For any x

, there is n

such that n

x .

Proof. Assume there exists x

such that for all n

we have n

x

Then n: n%

has an upper

bound x . By the completeness axiom, has a supremum in . By the supremum property, there is n

such that

sup 1 n, which yields the contradiction sup n

1

.

Question. How is

contained in

? How is

'

contained in

?

Below we will show that is dense in in the sense that between any two distinct real numbers x y, no matter

how close, there is a rational number. Similarly, '

is dense in

First we need a lemma.

Lemma. For every x there exists a least integer greater than or equal to x (In computer science, this is called

the ceiling of x and is denoted by

x ) Similarly, there exists a greatest integer less than or equal to x (This is denoted

by [x] In computer science, this is also called the floor of x and is denoted byx

)

Proof. By the Archimedean principle, there is n

such that n

x

Then n

x

n

By (iii) of the order axiom,

0

x

n

2n

The set S k

: k

x

n is a nonempty subset of because 2n S

By the well-ordering

axiom, there is a least positive integer m

x

n

Then m n is the least integer greater than or equal to x

So the

ceiling of every real number always exist.

Next, to find the floor of x

let k be the least integer greater than or equal to x

then k is the greatest integer

less than or equal to x

Theorem (Density of Rational Numbers). If x y, then there is mn

such that x m

n y.

Proof. By the Archimedean principle, there is n

such that n

1

y x

So ny nx

1 and hence nx

1

ny

Let m [nx ]

1

then m 1 [nx ]

nx

[nx]

1 m

So nx

m

nx

1

ny

i.e. x

mn

y

Theorem (Density of Irrational Numbers). If x

y, then there is

'

such that x

y.

Proof. Let 0 ' . By the density of rational numbers, there is

mn

such that x 0

mn

y

0

(If mn

0

then pick another rational number between 0 and y 0

So we may take mn

0

) Let

mn

0 then ' and

x

y

Examples. (1) Let S

3

4

5]

then S is not bounded below and so S has no infimum. On the other hand,

S is bounded above by 5 and every upper bound of S is greater than or equal to 5 S So sup S 5

(2) Let S

1n : n

1 12 13 14 In the examples followingthe definition of supremum, we saw sup S

1

Here we will show infS 0

(Note 0 S.) Since 1n

0 for all n

0 is a lower bound of S. So by the

completeness axiom for infimum, infS must exist. Assume S has a lower bound t

0

By the Archimedean

principle, there is n

such that n

1

t

Then t

1

n S

a contradiction to t being a lower bound of S

So

0 is the greatest lower bound of S

(3) Let S [2

6

#

Since 2

x

6 for every x S

S has 2 as a lower bound and 6 as an upper bound. We

will show infS 2 and sup S 6 (Note 2 S and 6 S.) Since 2 S so every lower bound t satisfy t 2

Therefore infS 2

For supremum, assume there is an upper bound u

6

Since 2 S

so 2

u

By the

density of rational numbers, there is a r

such that u

r

6

Then r [2

6

#

S

As u

r contradicts

u being an upper bound of S

so every upper bound u

6

Therefore, sup S 6

22


23/69

Chapter 6. Limits

Limit is the most important concept in analysis. We will first discuss limits of sequences, then limits of functions.

Definitions. An (infinite) sequence in a set S (e.g. S or S [0

1]) is a list x 1 x2 x3 of elements of S in a

specific order. Briefly it is denoted by xn (Mathematically it may be viewed as a function x: S with x n xn

for n ) We say the sequence xn is bounded above iff the set x1 x2 x3 is bounded above. (Bounded belowand bounded sequences are defined similarly.) We will also write sup x n for the supremum of the set x1 x2 x3

and inf xn for the infimum of the set x1 x2 x3

CAUTION: Since we seldom talk about a set with one element from now on, so notations like x n will denote

sequences unless explicitly stated otherwise.

For x

y

the distance between x and y is commonly denoted by d

x

y

which equals x y

Below we will

need a quantitative measure of what it means to be close for a discussion of the concept of limit. For

0

the open

interval

c

c

is called the

-neighborhoodofc

Note x

c

c

if and only ifd

x

c

x c

i.e. every number in

c

c

has distance less than

from c

Limit of a sequence xn is often explained by saying it is the number the xns are closer and closer to as n gets

larger and larger. There are two bad points about this explanations.

(1) Being close or large is a feeling! It is not a fact. It cannot be proved by a logical argument.

(2) The effect of being close can accumulate to yield large separation! If two numbers having a distance less than or

equal to 1 are considered close, then 0 is close to 1 and 1 is close to 2 and 2 is close to 3,

99 is close to 100,

but 0 is quite far from 100.

So what is the meaning of close? How can limit be defined so it can be checked? Intuitively, a sequence x n gets

close to a number x if and only if the distance d xn x goes to 0. This happens if and only if for every positive

the

distance d xn x eventually becomes less than

The following example will try to make this more precise.

Example. As n gets large, intuitively we may thinkxn

2n2 1

n2

1gets close to 2. For

0

1

how soon (that is, for

what n) will the distance d

xn

2

be less than

? (What if

0

01? What if

0

001? What if

is an arbitrarypositive number?)

Solution. Consider d xn 2

2n2 1

n2

1 2

3

n2

1

Solving for n we get n2 3

1

If

0

1

then

n 29 So as soon as n 6 the distance between xn and 2 will be less than

0

1

(If

0

01

then n 299 So n 18 will do. If

0

001

then n 2999 So n 55 will do.

If 0

3

then n [ 3

1]

1 will do. If

3

then since 3n2

1

3

for every n

so

n 1 will do. So for every

0

there is a K so that as soon as n K the distance d xn 2 will be less

than

) Note the value of K depends on the value of

the smaller

is, the larger K will be. (Some people write K

to indicate K depends on

)

Definition. A sequence xn converges to a number x (or has limit x) iff for every

0, there is K

such that for

every n K it implies d xn x

xn

x

(which means x K x K

1 x K

2 x

x

Let us now do a few more examples to illustrate how to show a sequence converges by checking the definition.

Later, we will prove some theorems that will help in establishing convergence of sequences.

Examples. (1) Let

n c For every

0, let K 1, then n K implies

n c 0

. So

n converges to c

(2) Let n

c 1

n

For every

0, there exists an integer K

1

(by the Archimedean principle). Then n

K

implies n

c 1

n

1

K

. So n converges to c

23


24/69

(3) Let xn

n

cos n

n

Show that xn converges to 1 by checking the definition.

Solution. For every

0

there exists an integer K

1

1

by the Archimedian principle. Then n

K implies

n

cos n n

1

cos n

cos n n

1

n 1

1

K 1

So xn converges to 1

(4) Let zn n1 n

Show that zn converges to 1 by checking the definition.

Solution. (Let un zn

1 zn 1

By the binomial theorem,

n znn

1

un n

1

nu n

n

n 1

2u2n

unn n

n 1

2u2n

so that un

2

n 1

) For every

0

there exists integer K

1

2

2(by the Archimedean principle). Then n

K

implies zn 1 un

2

n 1

2

K 1

So zn converges to 1.

Remarks. (i) From the definition, we see that xn converges to x xn x converges to 0 and xn

x converges

to 0 are equivalent because in the definition, xn x is the same as xn

x 0 xn x 0

(ii) To show xn converges to x means for every

0

we have to finda K as in the definition or show such a K

exists. On the other hand, if we are given that xn converges to x , then for every

0

(which we can even

choose for our convenience,) there is a K as in the definition for us to use.

Theorem (Uniqueness of Limit). If xn converges to x and y, then x y (and so we may write lim

n

xn x).

Proof. For every

0, wewill show x y

. (By the infinitesimal principle,we will get x y.) Let

0

2

0

By the definition of convergence, there are K1, K2 such that n K1 xn x

0 and n K2 xn y

0.

Let K max

K1 K2 . By the triangleinequality, x y

x x K

x K y

x x K

x K y

0

0

.

Boundedness Theorem. If xn converges, then xn is bounded.

Proof. Let limn

xn x . For

1, there is K

such that n

K

xn x

1

xn

xn x

x

1

x .

Let M max

x1 , , x K

1 , 1

x

then for every n

xn M (i.e. xn [ M

M]).

Remarks. The converse is false. The sequence

1

n is bounded, but not convergent. In general, bounded sequences

may or may not converge.

Theorem (Computation Formulas for Limits). If xn converges to x and yn converges to y, then

(i) xn

yn converges to x

y, respectively, i.e. limn

xn

yn lim

n

xn

limn

yn,

(ii) xnyn converges to x y, i.e. limn

xnyn

limn

xn

limn

yn

,

(iii)

xn

yn

converges to x

y, provided yn

0 for all n and y

0.Proof. (i) For every

0, there are K1 K2 such that n K1 xn x

2 and n

K2 yn y

2.

Let K max

K1 K2 Then n K implies n K1 and n K2 So for these ns,

xn

yn

x

y

xn x

yn y

xn x

yn y

2

2

(ii) We prove a lemma first.

Lemma. If an is bounded and limn

bn 0, then lim

n

anbn 0.

Proof. Since an is bounded, there is M such that an M for all n. For every

0, since

M

0 and

bn converges to 0, there is K such that n K bn 0

M

an bn 0

M bn

.

24


25/69

To prove (ii), we write xnyn x y xnyn

xny

xny x y xn yn

y

y

xn x

. Since xn converges,

xn is bounded by the boundedness theorem. So by (i) and the lemma,

limn

xnyn lim

n

xnyn x y

limn

x y limn

xn yn y

limn

y

xn x

x y 0

0

x y x y

(iii) Note 1

2

y 0 Since yn

converges to y there is K0 such that n K0 implies yn

y 1

2

y By

the triangle inequality, y yn yn y

12

y

12

y

yn for n K0 Then for every n

yn m min

y1 yK0

1 1

2y

0

Next we will show limn

1

yn

1

y(then by (ii), lim

n

xn

yn

limn

xn1

yn

x1

y

x

y). For every

0

let

0 m y

0

Since limn

yn y 0, there is K

such that n

K

yn y

0 Then

n

K

1

yn

1

y

y yn

yn y

0

m y

Remarks. (1) As in the proofs of the uniqueness of limit and part (i) of the computation formulas, when we

have n

K1 an a

1 and n K2 bn b

2 we may as well take K max

K1 K2 to say

n

K

an

a

1 and bn

b

2 from now on.(2) By mathematical induction, we can show that the computation formulas also hold for finitely many sequences.

However, the number of sequences must stay constant as the following example shows

1 limn

1

n

1

n

n terms

limn

1

n

limn

1

n 0

0 0

Sandwich Theorem (or Squeeze Limit Theorem). If xn n yn for all n and limn

xn lim

n

yn z then

limn

n z.

Proof. For any

0, there is K such that n K xn z

and yn z

, i.e. xn , yn z

z

. Since

xn

n

yn

so

n

z

z

, i.e.

n

z

.

Example. Let n

[10n 2]

10n

for every n

(Note 1

1

4

2 1

41

3 1

414

4 1

4142

)

Then10n 2 1

10n n

10n 2

10n

2

Since limn

10n 2 1

10n

2

by the sandwich theorem, limn

n

2

Theorem (Limit Inequality). If a n 0 for all n and limn

an a, then a

0.

Proof. Assume a 0. Then for

a , there is K such that n K an a

a , which implies

a

an a

a

a

0, a contradiction.

Remarks. By the limit inequality above, if xn yn , limn

xn x , lim

n

yn y, then taking an

yn xn 0 we get

the limit y x

0

i.e. x

y

Also, ifa

xn b and limn

xn x

then limn

a a

limn

xn x

limn

b b

i.e. xn [a b] and limn

xn x imply x [a

b]

(This is false for open intervals as 1n

0

2

limn

1n

0

0

2

)

Supremum Limit Theorem. Let S be a nonempty set with an upper bound c

There is a sequence n in S converging

to c if and only if c sup S.

Proof. Ifc sup S

then for n

, by the supremum property, there is n S such that c

1

n sup S 1

n n

sup S c. Since limn

c 1

n

limn

c c, the sandwich theorem implies limn

n c sup S.

Conversely, if a sequence n in S converges to c then n sup S implies c

limn

n sup S Since c is an

upper bound of S

so sup S

c

Therefore c sup S

25


26/69

Infimum Limit Theorem. Let S be a nonempty set with a lower bound c

There is a sequence n in S converging

to c if and only if c inf S.

Examples. (1) Let S 1n

: n

1

12

13

Since 0

1n

for all n

0 is a lower bound of the set S

Now

the sequence 1n

in S converges to 0

By the infimum limit theorem, inf S 0

(2) Let S

x 1

y : x ! # 0 1] y [1 2] Since1

2 x

1

y for all x # 0 1] and y [1 2] 1

2 is alower bound of S

Now the sequence 1n

1

2 in S converges to 1

2

By the infimum limit theorem, inf S 12

(3) Let A and B be bounded in Prove that ifA 2B a 2b : a A b B then sup A 2B sup A 2inf B

Solution. Since A and B are bounded, sup A and infB exist by the completeness axiom. For x A 2B

we have

x a 2b for some a A and b B

So x a 2b

sup A 2infB

Hence sup A 2 sup B is an upper bound

for A 2B

By the supremum limit theorem, there is a sequence an A such that an converges to sup A By the

infimum limit theorem, there is a sequence bn B such that bn converges to inf B Then an 2bn is a sequence in

A 2B and an 2bn converges to sup A

2infB by the computation formulas for limits. By the supremum limit

theorem, therefore sup

A 2B

sup A 2infB

Definition. A subsequence of xn is a sequence xnj , where nj and n1 n2 n3 .

Examples. For the sequence x1 x2 x3 x4 x5 x6 if we set nj j 2

we get the subsequence x1 x4 x9 x16

If nj 2j

1

then we get the subsequence x3 x5 x7 x9 If nj is the j -th prime number, then we get the

subsequence x2 x3 x5 x7

Remarks. (1) Taking nj j

we see that every sequence is a subsequence of itself. A subsequence can also be

thought of as obtained from the original sequence by throwing away possibly some terms. Also, a subsequence of a

subsequence of xn is a subsequence of xn

(2) By mathematical induction, we have nj j for all j because n1 1 and nj

1 nj j implies nj

1 j

1

Subsequence Theorem. If limn

xn x, then lim

j

xnj x for every subsequence xnj of xn . (The converse is

trivially true because every sequence is a subsequence of itself.)

Proof. For every

0, there is K

such that n

K

xn

x

. Then j

K

nj

K

xnj

x

.

Question. How can we tell if a sequence converges without knowing the limit (especially if the sequence is given by

a recurrence relation)?

For certain types of sequences, the question has an easy answer.

Definitions. xn is

increasing

decreasing

strictly increasing

strictly decreasing

iff

x1 x2 x3

x1 x2 x3

x1 x2 x3

x1 x2 x3

respectively. xn is

monotone

strictly monotone

iff

xn is

increasing or decreasing

strictly increasing or decreasing

respectively.

Monotone Sequence Theorem. If xn is increasing and bounded above, then limn

xn

sup xn . (Similarly, if xn

is decreasing and bounded below, then limn

xn inf xn .)

Proof. Let M sup xn , which exists by the completeness axiom. By the supremum property, for any

0, there is

x K such that M

x K M. Then j K M

x K xj M xj M M xj

Remark. Note the completeness axiom was used to show the limit of xn exists (without giving the value).

Examples. (1) Let 0

c

1 and xn c1 n

Then xn 1 and cn

1

cn

xn c1 n

c1 n

1

xn

1 So by the

monotone sequence theorem, xn has a limit x Now x22n

c1 2n

2 c1 n xn Taking limits and using the

subsequence theorem, we get x 2 x

So x 0 or 1. Since 0

c x1 x the limit x is 1. Similarly, ifc 1

then cn will decrease to the limit 1.

26


27/69

(2) Does

2

2

2

represent a real number?

Here we have a nested radical defined by x1

2 and xn

1

2

xn The question is whether xn

converges to a real number x . (Computing a few terms, we suspect that xn is increasing. To find an upper bound,

observe that if limn

xn x then x 2

x implies x 2 ) Now by mathematical induction, we can show that

xn

xn

1 2 (Ifxn

xn

1 2 then 2

xn

2

xn

1 4 so taking square roots, we get xn

1 xn

2 2 )

By the monotone sequence theorem, xn has a limit x We have x2 lim

n

x 2n

1 lim

n

2

xn 2

x Then

x 1 or 2. Since 2 x1 x so x 2

Another common type of sequences is obtained by mixing a decreasing sequence and an increasing sequence into

one of the form a1 b1 a2 b2 a3 b3 In the next example, we will have such a situation and we need two theorems

to handle these kind of sequences.

Nested Interval Theorem. If In [an bn ] is such that I1 I2 I3 , then

#

n 1In

[a b] where

a limn

an limn

bn b If lim

n

bn an

0, then

#

n 1In contains exactly one number.

[ ][ [ ] ]a1 a2 a3 b3 b2 b1...

Proof. I1 I2 I3 implies an is increasing and bounded above by b1 and bn is decreasing and bounded

below by a1. By the monotone sequence theorem, an converges to a sup an and bn converges to b

inf bn .

Since an bn for every n , taking limits, we have an a b bn. Consequently, x [an bn ] (i.e. an x bn)

for all n if and only if limn

an a x b lim

n

bn So

#

n 1In

[a b]. If 0 limn

bn an

b a then a b

and

#

n 1In

a

Remarks. Note in the proof, the monotone sequence theorem was used. So the nested interval theorem also implicitly

depended on the completeness axiom.

Intertwining Sequence Theorem. If x2m

and x2m

1 converge to x then xn

also converges to x

Proof. For every

0

since x2m converges to x there is K0 such that m K0 x2m x

Since x2m

1

also converges to x

there is K1 such that m K1 x2m

1 x

Now ifn

K max

2K0 2K1 1

then either n 2m

2K0 xn x x2m

x

or n 2m 1

2K1 1

xn x x2m

1 x

Example. Does1

1

1

1

represent a number?

Here we have a continued fraction defined by x1 1 and xn

1 1

1

xn We have x1 1

x2 1

2

x3

2

3

x4 3

5

Plotting these on the real line suggests 1

2

x2n x2n

2 x2n

1 x2n

1 1 for all n

This can be easily established by mathematical induction. (If 1

2

x2n x2n

2 x2n

1 x2n

1 1 then

1

x2n 1

x2n

2 1

x2n

1 1

x2n

1 Taking reciprocal and applying the recurrence relation, we have

x2n

1 x2n

3 x2n

2 x2n Repeating these steps once more, we get x2n

2 x2n

4 x2n

3 x2n

1 )

Let In [x2n x2n

1] then I1 I2 I3 Now

xm xm

1

1

1

xm

1

1

1

xm

xm

1 xm

1

xm

1 1

xm

xm

1 xm

1 1

2

1 1

2

4

9xm

1 xm

Using this, we get x2n

1 x2n

4

9x2n

2 x2n

1 4

9

4

9x2n

3 x2n

2

4

9

4

9

2n

2

x1 x2

4

9

2n

2 1

2

By the sandwich theorem, limn

x2n

1 x2n

0

So by the nested interval theorem,

#

n 1In

x for some

27


28/69

x and limn

x2n x lim

n

x2n

1 By the intertwining sequence theorem, limn

xn x

So x limn

xn

1

limn

1

1

xn 1

1

x

Then x

1

5

2

Since x I1 x

1

5

2

(Instead of estimating the lengths of the In s and squeezing them to 0 to see their intersection is a single point,

we can also do the following. Let In [x2n x2n

1] be as above so that I1 I2 I3

By the nested interval

theorem, #

n 1 In

[a

b]

where a

limn

x2n and b

limn

x2n 1

Taking limits on both sides ofx2n 1

1

1

x2n and

x2n

1

1

x2n

1

we get b 1

1

aand a

1

1

b

Then b

1

a

1 a

1

b

which yields b

ab a

ab

so a b

Hence

#

n 1In is a single point.)

Back to answering the question above in general, French mathematician Augustine Cauchy (17891857) intro-

duced the following condition.

Definition. xn is a Cauchy sequence iff for every

0, there is K

such that m, n K implies xm xn

.

Remark. Roughly, the condition says that the terms of these sequences are getting closer and closer to each other.

Example. Let xn

1

n2

(Note that if m

n

K

say m

n

then we have xm xn

1

n2

1

m2

1

n2

1

K2

)

For every

0

we can take an integer K

1

(by the Archimedean principle). Then m

n

K implies

xm xn

1

K2

So xn is a Cauchy sequence.

Theorem. If xn converges, then xn is a Cauchy sequence.

Proof. For every

0

since limn

xn x

there is K

such that j

K

xj x

2. For m, n

K, we

have xm xn xm

x

x xn

2

2

. So xn is a Cauchy sequence.

The converse of the previous theorem is true, but it takes some work to prove that. The difficulty lies prim

201_syllectnt real الكتاب

Documents