1

CHAPTER 1: THE REAL NUMBER SYSTEM The real number system is the main frame in mathematical analysis. Thus it should be addressed in the first place as part of the introduction to analysis. The following diagram shows us the set of real numbers.

R Z

N

Figure 1.1: Set of real numbers, R

The natural numbers arise well and naturally. These are the numbers

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ………. Natural numbers are normally used for counting objects. Natural numbers are

denoted as N . The integers, Z are the numbers

-5, -4, -3,-2, -1, 0, 1, 2, 3, 4, 5, ….. , which extend indefinitely in either direction. The symbol Z is usually used to

denote the set of integers. Symbol Z to denote the set of positive integers. The symbol Z comes from the German word for number, Zahlen. When numbers are used for measurement, it is sometimes necessary to consider fractions of an integer. This leads to the concept of rational numbers. A rational

number is a number of the form q

p, where p and q are integers with 0q . The

set of all rational numbers is denoted as Q. In the event where the set of rational numbers is not big enough for the purpose of exact measurement, it is necessary to introduce a larger set of numbers known as real numbers and denoted by the symbol R.

2

Every rational number is a real number and the real number that are not rational numbers are called irrational numbers. Example of irrational number is

,.....5,3,2 .

1.1 FIELD In this section we discuss postulates / axioms that lead to the algebra of real numbers. We must study algebra in order to study analysis. Definition 1.1 : A field is a structure consisting of a set F and two binary operations, called addition and multiplication, satisfying the following axioms. 1. For every a and b in F ,

FbaandFba .

[ F is closed under addition and multiplication.]

2. For every a and b in F ,

abbaandabba .

[ Addition and multiplication are commutative. ]

3. For every ba , and c in F

cbacba ()

and,

)()( bcacab

[ Addition and multiplication are associative. ]

4. For every ba, and c in F

)()()( acabcba

[ Multiplication is distributive over addition. ] 5. There exists a unique element O of F such that

aOaFa ,

[ F has an additive identity.]

3

6. There exists a unique element 1 of F such that

01 and aaFa .1,

[ F has a multiplicative identity. ] 7. For every a in F there exists a unique elements a of F such that

0)( aa .

[ Each element of F has an additive inverse.]

8. For every 0a in F there exists a unique element

1a of F such that 11 aa .

[ Each non zero element of F has a multiplicative inverse. ] The most commonly encountered examples of fields are the rational number system and the real number system. 1.1.1 ORDERED FIELD A set S (not necessarily a field) is an ordered set if there is a binary relation < satisfying the following axioms.

9. For every ba, , c in S , ,ba ,cb then ca .

[ < is transitive ]

10. For every a and b in S exactly one of the following relations hold :

abbaba ,, .

[ < satisfies the trichotomy law ]

4

Definition 1.1.1 : A field F which is also an ordered set is an ordered field, provided the following axioms are satisfied.

11. For every ba , , c in F

.cbcaba

12. For every ba , ,c in F

.0 bcaccandba

Though only < is used in the axioms, we can easily define other inequality symbols in common use. We define ba

To mean ab

the symbol is frequently read as “ less than or equal to”, but it is best defined through the trichotomy law : )( ba means [not )( ba ]

Similarly )( ba means [not )( ba ].

1.2 ABSOLUTE VALUE, INTERVALS & INEQUALITIES Definition 1.2.1 : Given a real number x , the absolute value of x , denoted by

x , is defined by :

x , if 0x

x = 2x = x , if 0x

5

y

xy

x -3 -2 -1 1 2 3

Figure 1.2 : A graph of the absolute value function.

Theorem 1.2.1 : Let ba ,

i) 00;0 aaa ,

ii) aa , for every real number a .

iii) |||||| baab , for all real numbers .,ba

iv) baba (Triangle Inequality).

Proof : (i),(ii) & (iii) follows from the definition 1.2.1. We now prove the triangle inequality

22222)( babababa

22

2 bbaa

2)( ba

canceling out the square, we then obtained

baba . ■

Y

b

a

0 ba X

Figure 1.3 A geometric interpretation of this triangle inequality for 2

, ba

6

The common use of TI in analysis is in a slightly different form that the one stated in Theorem 1.2.1.

Letting ,zybandyxa the triangle inequality becomes

zyyxzx

Example 1.1 : 7|4||)47(||3| xxx .

Theorem 1.2.2 : (Schwarz Inequality).

Let n be a positive integer. If nn bbbandaaa ,....,,,....,, 2121 are real numbers and

0j

a for some Nj , then

)()()(1

2

1

22

1

n

kk

n

kkk

n

kk

baba .

Equality occurs if and only if there is a constant c such that kk

bca for all

integers ....,,2,1 nk

Proof : if ,01

2

n

k

ka then all of the 'ka s are 0 and both sides of the inequality are

0. So the above inequality is true. Suppose that .01

2

n

k

ka

Then for any constant ,c

n

kkk

n

kk

n

kkkk

n

k

bbacacbca1

2

11

222

1

2)(0

Hence

n

kkk

n

kk

n

kk

bbacac1

2

11

22 20

By choosing ,2

1

k

n

kkk

a

ba

c we then have

7

n

k

kn

k

k

n

k

n

k

kkkk

b

a

baba

1

2

1

2

1 1

22 )(2)(

0 ,

and,

n

k

k

k

n

k

kk

ba

ba

1

2

2

2

1

)(

0 .

Then we have )()()(1

2

1

22

1

n

kk

n

kkk

n

kk

abba .

The only way in which equality occur is if there is a value of c for which

,0kk

bca for ....,,2,1 nk Then both sides of the inequality is equal to

.)( 2

1

22

n

kk

ac

Conversely, suppose that the equality occurs, 2

11

2

1

2 )( k

n

k

k

n

k

k

n

k

k baba

since

,0j

a for some j , then .01

2

n

k

ka

Choosing ,

1

2

1

n

k

k

n

k

kk

a

ba

c yield

2

11

2 )(0 k

n

k

k

n

k

k bcaa

.

Since 0)(,0 2

11

2

k

n

kk

n

kk

bcathena for each k . Thus .kk bca ■

Theorem 1.2.3 ( Minkowski Inequality) : Let n

aaa ,...,,21

and n

bbb ,...,,21

be

real numbers. Then

2/1

1

2

2/1

1

2

2/1

1

2)(

n

kk

n

kk

n

kkk

baba .

8

Proof : From the Theorem 1.2.2, we have

2/1

1

22/1

1

2

11

)()()()(

n

kk

n

kkk

n

kkk

n

kk

bababa ,

then

n

kkk

ba1

2)( =

n

kk

a1

2+ 2

k

n

kkba

1

+

n

kk

b1

2

n

kk

a1

2 + 2

2/1

1

22/1

1

2 )()(

n

kk

n

kk

ba +

n

kk

b1

2.

= ( 2/1

1

2 )(

n

kk

a + 2/1

1

2 )(

n

kk

b )2

. ■

Definition 1.2.2 : A set S of real numbers is an interval if and only if S contains at

least two points and for any two points ,, Syx every real number between x and

y belongs to S .

Next, we define those possible familiar forms of interval. The symbol represent a quantity that is larger than any real number. Similarly the symbol

represents a quantity that is smaller than any number. Neither nor is a

real number. The real numbers together with these two symbols and are

known as the Extended Real Number System. Definition 1.2.3 : An interval has one of the following nine forms.

1. bxaxba :),( , a bounded, open interval.

2. bxaxba ), , a bounded, half open interval.

3. bxaxba :, , a bounded, half open interval.

4. bxaxba :, , a bounded, closed interval.

5. axxa :, , an unbounded open interval.

6. axxa :, , an unbounded closed interval.

7. bxxb :, , an unbounded, open interval.

8. bxxb :, , an unbounded, closed interval.

9

9. ,, an unbounded interval, both open & closed.

Definition 1.2.4 : The extended real number system consists of the set R and the two symbols + and . Both addition and multiplication together with the order on R are being treated like usual. In fact, for any real number a , we have

(1) a + = , ;)()(,, a

(2) If ,0a then )(a and ;)( a

(3) If ,0a then )(a and ;)( a

(4) ;))((),)(()()(

(5) );/(0/ aa

(6) . a

Take notes that the operation and 0 are not well defined. The importance of defining the Extended Real Number System is to introduce the concept of supremum and infimum. 1.3 THE COMPLETENESS AXIOM Definition 1.3.1 : Let S be a nonempty set of real numbers.

(i) The set S is bounded above if there is a number M such that Mx for

all Sx . The number M is called an upper bound of S .

(ii) The set S is bounded below if there is a number m such that mx for

all Sx , the number m is called a lower bound of .S

(iii) The set S is bounded if there is a number M such that Mx for all

.Sx The number M is called a bound for .S

Examples 1.2:

(i) The set Z is bounded below : any negative number is a lower bound.

(ii) The set

0&:1

xQxx

xS is bounded : any number 1M is

an upper bound of .S

(iii) The set 0: xRxA is bounded above, but not bounded.

10

NOTE : A set S is unbounded if for each number M there is a point ,Sx such

that .Mx

Definition 1.3.2 : Let S be a non empty set of real numbers. A number b is said

to be the supremum of S , denoted as ,bSSup if

(i) b is an upper bound of S and.

(ii) bb 'for any upper bound .' Sofb

NOTE : Supremum S is also known as least upper bound of .S

Definition 1.3.3 : Let S be a non empty set of real numbers. A number is the

infimum of S denoted as ,inf S if

(i) is a lower bound of ,S and

(ii) ,' for any lower bound .' Sof

NOTE : Infimum S is also known as greatest lower bound of .S

The following are direct consequences from the definition of supremum and infimum: 1. Let be an upper bound of a set A. Asup if and only if for each 0 , there

exist Ax such that x .

2. Let be a lower bound of a set .A Ainf if and only if for each ,0 there

exist Ax such that x .

Examples 1.3 :

(i) 10: xxS interval 1,0

0inf&1sup SS

(ii) ,11: xxS

Ssup does not exist

1inf S

(iii) ,........,3,2,1NS

sup S does not exist since S has no upper bound.

11

(iv)

....,1

....,3

1,

2

1,1:

1

nNn

nS

.0inf&1sup SS

Completeness Axiom for R: Each nonempty set of real numbers that is bounded above has a supremum. NOTE : The completeness axiom guarantees the existence of the supremum of a nonempty set of real numbers that is bounded above. It does not indicate how the supremum is found, but only states that such a number exists. The following Corolary is the consequences of the completeness axiom. Corolary 1.3.1 : Each nonempty set of real numbers that is bounded below has an infimum. Proof : Suppose that S is a nonempty set of real numbers that is bounded below

and define a set T as .: SSxxT Since S is bounded below, then

there exist a number b with ,xb for each .Sx Then ,bx for each

bSx . is the upper bound of set .T Using the completeness axiom guarantees

the existence of supremum ,T say .sup T We are now about to show that

.inf S

(i) as an upper bound of T means that , x for each ,Sx and

this indicate that x then is a lower bound of .S

(ii) Now take any lower bound, b of set .S Since ,xb for each ,Sx

then bx implying that b is an upper bound of T . This shows

that ,b since .supT Therefore yields ,b for any ,b

a lower bound of S . Then .inf S

Note : From what we can derive from this proof is :

)sup(supinf STS .

12

Theorem 1.3.1 : For each real number ,x there exist a unique integer denoted

as x that satisfies

xxx 1 .

Proof : Define set .: xnNnS S is bounded from above then using the

completeness axiom exist supremum of S , say .supSx Next we show that x

is an integer.

Since x is the supremum of S and using the fact that ,1 xx there exist

Sk , such that kx 1 .

][x m

1x Sk x

If there is an km , then xkm 1 and from the definition of supremum of

set S , this indicates that Sm . In other words if Sm , then km so k is an

upper bound of S , and kx but Sk , therefore xk . Now we have

(i) kx

kx an integer

(ii) xk

Since Skx , implies that xx , and Sx 1 implying that xx 1 , and

this gives 1 xx . Hence

xxx 1 .

To proof that x is unique, we assume that there exist an integer a that satisfies

xax 1 .

a x

1x x

13

111 xxxxax

1ax

Since both x and a are integers, and this only happen when 0ax

ax .

Examples 1.4:

(i) 02.0

(ii) 45.4

(iii) 65.5

(iv) 4]4[

NOTE : x is also called the greatest integer less than or equal to .x

Theorem 1.3.2 : (Archimedean Principle)

If Rba , and 0a , there exists a positive integer n such that bna .

Proof : We assume that 0b , if not, we can choose 1n and ba . If 0b ,

then 01 a

b . Now define

a

bn 1 .

According to Theorem 1.3.1, we have

a

bn

a

b

a

b

a

b

a

b

a

b

1,01

1111

bnaa

bn ■

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Corolary 1.3.1 : For each positive real number ,a there exists a positive integer

n such that an

1.

Proof : Choose 1b and apply Theorem 1.3.2, gives us n such that 1na and

an

1.

Corolary 1.3.1 is very useful in showing the infimum of the set

Nnn

S :1

is

equal to 0 . It is obvious that 0 is a lower bound of the set S . Then to show 0 is

the greatest lower bound, we argue as follows : If there exist a number 0a ,

then by corolary 1.3.1, there exists (as a lower bound of S ) Nn such that an

1

, which contradicts the definition of lower bound. Theorem 1.3.3 (Density of rational numbers): Between any two distinct real numbers there is a rational number. Proof : Suppose x and y are distinct real numbers with yx . By the

Archimedean principle of the real numbers, there exists a positive integer n such

that

1 xyn

then

nxny 1 .

Define integer nxm 1 . Then using Theorem 1.3.1 we have

nxnxnx 1111 .

Then we have

nymnx

Since 0n , then yn

mx . Hence between any two numbers yx , there is a

rational number n

m.

15

NOTE : In terms of decimal expansions, there is a simple distinction between rational and irrational numbers. Rational numbers has a repeating pattern of

decimal expansion. For example .....851485148514.0101

15 , whereas an

irrational number has no repeating pattern. For example the irrational number

2 = 1.41421356237…, its decimal expansion has no repeating pattern.

Theorem 1.3.4 : There is no rational number p such that 22 p .

Proof : Assume now that there exist such a rational number p which satisfies the

equation, then p can be written as

0,,; nZnmn

mp , where the greatest common divisor (gcd) of nm & is 1.

22

2

2,2 nmn

m

.

This indicates that 2m is an even number which implies that m also is even.

Then m can be written as

Zqqm ,2 .

Thus we have

,2,42 22222 qnqmn where n is also an even number.

This contradicts the fact that the greatest common divisor of nm, is 1.

Hence there no such rational number p such that .22 p ■

Even though the set of all rational numbers, Q are dense in R , the set Q does

not satisfy the completeness axiom. That is, there exist a set of rational numbers

that is bounded above but do not have the supremum in Q . The following

example supports this statement.

Example 1.5 : Show that the set }20;{ pQpS does not have a

supremum in Q .

Solution : Clearly the set S is a nonempty set and bounded above in Q . Now

we show that there do not exists a rational number as the supremum of S .

16

We argue using contradiction. Assume that there is a rational number Qq

such that Sq sup . If 20 q , then theorem 1.3.3 implies that there exist a

rational number p such that 2 pq . The number p is in S , and this

contradicts the assumption that q is an upperbound of S . Then q must satisfies

q2 . Then by theorem 1.3.3 implies that there exists a rational number r

such that .2 qr Since q is the supremum, then this is also impossible.

Therefore there do not exists a rational number as the supremum of S . 1.4 SET AND FUNCTION 1.4.1 SET A set is a collection of objects. The objects in a set usually have some features in common, such as the set of real numbers or the set of continuous functions, but a set can also be any random collection of objects. The objects in a set are called elements or points in the set.

x belongs to a set S is usually written as Sx .

Set A is a subset of a set B then this is written as BABA , .

Two sets A and B are equal then we write ABBA &

Set A is a proper subset of B if BABA ,

A set that has no elements is called the empty set denoted by the symbol .

A set can be described in various ways. For example, each of the following represents the set of all odd positive integers.

(i) ...,9,7,5,3,1

(ii) :Zn 2 does not divide n

(iii) Znn :12

The following definitions are very useful. Definition 1.4.1 : Let A and B be two sets.

(i) The intersection of A and B , denoted ,BA is the set of all elements

that belongs to both A and .B

17

Y

X

1 2

1

2

3

0

Y

X

1 2

1

2

3

0 3

A x B B x A

AxxBA : and Bx

(ii) The union of A and ,B denoted as ,BA is the set of all elements that

belong to either A or B or to both A and .B

AxxBA : or .Bx

(iii) The complement of B relative to ,A denoted as ,\ BA is the set of all

elements that belong to A but do not belong to .B

AxBA \ and Bx .

(iv) The Cartesian product of A and ,B denoted as A x ,B is the set of all

ordered pairs ba, where Aa and .Bb

A AabaB :, and Bb

Example 1.6 : Let 2,1A and 3,21,0 B

Then

(i) 3,0 BA

(ii) 2,1 BA ( A set consists of 1 and 2 (iii) 2,121:\ xxBA

(iv) 3,21,0\ AB

Figure 1.1 cartesian product of A and B

18

Note : A x BB x A

Theorem 1.4.1 : Let BA, and C be sets. Then

(i) ;CABACBA

(ii) ;CABACBA

(iii) ;\\\ CABACBA

(iv) .\\\ CABACBA

Proof :

(i) Assume AxCBAx and AxCBx and Bx or

BAxCx or .CABAxCAx

(ii) Need to show that CABACBA and

CBACABA . Assume that .CBAx

Then Ax or .CBx If Ax then BAx and ,CAx hence

.CABAx If CBx then, Bx and ,Cx hence

again BAx and .CAx Then

CABACBACABAx . .

Conversely. If ,CABAx then BAx and .CAx

Now if Ax , then CBAx . If ,Ax then ,CBx so again

. .x A B C A B A C A B C

(iii) AxCBAx \ and CBx

Ax and Bx and .Cx

BAx \ and CAx \

CABAx \\

The set operations shown above involve two sets, but it is easy to extend them to any finite number of sets. However, it is sometimes necessary to consider infinite collections of sets. Definition 1.4.2 : Let A be a nonempty set and suppose that for each element

,Aa there is a set aX . Then the set A is known as an index set.

Note : Set A could have infinite or finite elements.

De Morgan’s Laws.

19

Definition 1.4.3 :

(i) The symbol a

Aa

X

represent the union of the collection AaX a : .

aa

Aa

XxxX

: for some Aa

(ii) The symbol a

Aa

X

represent the intersection of the collection

AaX a :: .

aa

Aa

XxxX

: for each Aa

Example 1.7:

(i) Suppose ,3,2,1A and for each ,Aa assume that aX is a closed interval

3, aaXa

. Then

321

3

1

XXXXXa

aa

Aa

6,16,35,24,1 .

321

3

1

XXXXXa

aa

Aa

4,36,35,24,1 .

ii) Let NA (infinite set) and for ,An suppose we put .,...,3,2,1 nX n Then

.....3,2,12,111

NXXn

nn

An

.11

n

nn

An

XX

Theorem 1.4.2 : Let Y be a set and A be a collection of subsets of .Y Then

(i) a

Aa

a

Aa

XYXY

(ii) a

Aa

a

Aa

XYXY

(iii) a

Aa

a

Aa

XYXY \\

(iv) a

a

a

Aa

XYXY \\

De Morgan’s Laws

20

Proof :

a

bAaba

XXYXY \

a

bAab

XYXY\

a

cbAacb

XYXYXY ,\

.... dcb XYXYXY

#.a

Aaa

XYXY

Note : If the set Y is a universe set, then we can write the last two relations as

Aa

C

a

C

Aaa

XX

and

Aa

C

a

C

Aaa

XX

.

1.4.2 Function. Definition 1.4.4 : Let A and B be two non empty sets. A function BAf : is a

rule of correspondence that assigns to each element of the set A exactly one element of the set B (unique).

If f assigns the element b of B to the element a of A , then we write baf .

We say that the image of a is b and a is a pre-image of ,b sometime denoted by

.1 abf The set A is known as the domain of f and the set B is known as the

codomain of f .

The set Aaaf : which is a subset of B is known as the range of f and

sometimes denoted by Af .

Example 1.8 : Let 3,2,1A and dcbaB ,,, . Define functions BAf : and

BAg : by

dgaf 1;1

bgbf 2;2

cgcf 3;3

21

The range of f is cba ,, and the range of g is dcb ,, .

Definition 1.4.5 : Let XAYXf ,: and YB .

(i) Image of set A under ,f denoted by Af is a set given by the

following notation :

AxxfAf : .

(ii) Pre-image of B under ,f denoted by )(1 Bf is a set given as follows.

BxfXxBf :1 .

X Y

A Af

f

X Y

Bf 1 B

f

Figure 1.2: Image and pre-image.

Example 1.9 : Let RRf : and 2xxf . Find the image of 2,0 and the pre-

image of 4,0 under f .

Solutions :

2,0:2,0:2,0 2 xxxxff

4,0 .

22

40:4,0:4,0 21 xxxfxf

2,2 .

Theorem 1.4.3 : Let XBAYXf ,,: and ., YVU Then

(i) ,BfAf if ;BA

VfUf 11 , if ;VU

(ii) ;BfAfBAf

;111 VfUfVUf

(iii) ;BAfBAf

;111 VfUfVUf

Proof :

(i) Assume ,Afy then there exists Ax such that .yxf Since ,BA

then .Bx Therefore Bfyxf .

Thus we have BfAf .

Assume that .1 Ufx From the definition of preimage, we then have

Uyxf and since ,VU then VUxf and .Vxf Therefore

.1 Vfx

.11 VfUf

Definition 1.4.6 : Let BAf :

(i) We say that f is one to one (injective) function if 21 afaf implies

21 aa that is, each element in the range of f has a unique pre-image.

(ii) We say that f is an onto function (surjective) if B is the range of .f

Note : According to this definition, function f is one to one, denoted by ,11 if

21 aa implies .21 afaf The function f is onto if each ,Bb there exist

Aa such that .baf

23

Example 1.10:

(i) The function RRf : defined by xxf 2 is one to one but not onto.

(ii) The function RRg : defined by xxxg 3 is onto but not one to

one.

(iii) The function RRh : defined by 32 xxh is both one to one and

onto.

(iv) The function RRq : defined by 2xxq is neither one to one and

onto. Definition 1.4.7: Let BAf : define a one to one function and let BC be the

range of the function. Then we define the inverse function of ,f denoted by 1f

as

ACf :1

with xyf 1 for Cy and Ax if and only if .yxf

Loosely speaking, we find the inverse function by solving for x in the equation

,xfy if it is possible to do so.

Example 1.11:

(i) The function 2xxf with domain ,0 has the inverse xxf 1 ,

for ,0x .

(ii) The function ,1

x

xxf for 1\ Rx , has the inverse

.1

1

x

xxf

We put y

yxyxxyxxy

x

xy

11

1.

Note : When the inverse exists, the two equations xfy and yfx 1 yield

(i) xxff 1 for all x in the domain f .

(ii) yyff 1 for all y in the range of f .

Definition 1.4.8 : Let CBf : and BAg : be functions. Let

.: BxgAxD A composite function gf is defined by

,agfagf for each .Da

24

A B C

D xg

g

f

gof

Figure 1.3 : Composite function .gf

The composite function gf has a domain .: BxgAxD In general

fggf .

Example 1.12 : Let xxgRg cos,,: and Rf ,0: defined

as xxf 1 . Find the domain and the range of gf and .fg

Solution :

The domain of gf is the set

,0:, xgxD 2

,2

0cos:, xx

,cos1cos xxfxgfxgf then the range of ,gf for

2,1Dx

Using a similar method, the domain of fg is

.,:,0 xfxD

}1:,0 xx

1,0 .

)1cos()( xxfg and the range of fg ]1cos,1[ .

25

Theorem 1.4.4 : Let BAg : and CBf :

(i) If f and g are one to one, then gf is one to one.

(ii) If f and g are onto, then gf is onto.

Proof :

(i) Let bgfagf , then we have bgfagf , since f is

one to one, this implies that bgag and since g is one to one, we

have .ba Therefore gf is one to one.

(ii) Here we need to show that the range of gf is C . Since

,: CAgf then the range of .Cgf Now we need to show

that C range of gf .

For this we assume that .Cc Since f is onto then, there exists Bb

Such that .cbf The function g is also onto, then there exists Aa

Such that .bag Then the element a is mapped to the element c ,

that is .cbfagfagf Hence c is in the range of

.gf

the set C is the range of .gf

1.5 COUNTABLE AND UNCOUNTABLE SET Definition 1.5.1 : Let A and B be sets. A and B are equivalent sets, written as

BA ~ if there exists a one to one function from A onto .B

Example 1.13 :

(i) Let naaaaA ,....,,, 321 and nB ,....,3,2,1 .

Define a function BAf : as nkkaf k ,....,3,2,1,

Easily seen that f is a one to one function from A onto .B Therefore

.~ BA

(ii) Let .,...,8,6,4,22

nZ The function nZZf 2: defined as nnf 2 is

a one to one and onto function.

Then

nZZ 2~ .

26

Z

nZ2

f

11

Figure 1.4: function f maps nZZ 2

(iii) Let A be a closed interval ba, and dcB , with ba and .dc

Define BAf : by

caxab

cdxf

.

As we can see the above graph of f is one to one from ba, onto dc, .

Hence dcba ,~, .

1

2

3

4

.

.

.

.

.

.

2

4

6

8

.

.

.

.

.

.

&

onto

o

Figure 1.5: Graph of function caxab

cdxf

Y

X

a b

c

d

27

Theorem 1.5.1 : Let U be a universe set and .: UAAX The relation ~ is

an equivalent relation on XX , that is

(i) AA ~ for each XA (reflexsive)

(ii) ,~ BA then ,~ AB (symmetric)

(iii) BA ~ and ,~ CB then CA ~ (transitive).

Now according to this theorem 1.5.1 and definition 1.5.1, we can say that if

,~ AB with A has n elements, then B also has n element. Further more we say

AAB ~ and B have the same number of elements.

Definition 1.5.2 : Let A be an arbitrary set.

(i) The set A is finite if A or

,.....,,2,1~ nA for some Nn

(ii) The set A is infinite if it is not finite.

(iii) The set A is countable if A is finite or .~ NA

(iv) The set A is uncountable if A is not countable. A set can be classified into finite and infinite. The distinction between finite sets and infinite sets is generally easy to grasp. A finite set is eventually exhausted when you start listing out its elements, where as an infinite set is not. For example, the number of license plate using four single-digit numbers is finite. There are many of them but in theory if you start writing down all the possibilities, eventually the list would end. The set of positive integers is infinite because a list of positive integers never ends. We then can categorized a set A as follows :

A

finite

countable

infinite

countable uncountable

Figure 1.6 classification of A

28

A function ANf : is called a sequence in .A The value of the sequence f at

an integer n is normally denoted as nanf . From the definition 1.5.2, if A is

countably infinite, that is ,~ NA then A is the range of a sequence. We then can

write A as an infinite sequence as follows :-

......,....,,, 321 naaaaA

with ji aa for .ji

Let .,...,,4,3,2,1 NnnZn The function AZfn: is called a finite

sequence in A . If

nZA ~ then A can be written as naaaaA ,...,,, 321 .

Theorem 1.5.2 : A nonempty set A is countable if and only if A can be listed as a finite or infinite sequence.

Proof : Assume A is a countable set. If A is finite, then

nZA ~ for some Nn .

For that, there exists a function AZf n : which is one to one and onto. That is

A is a range of a sequence and we then can list A as a finite sequence. If A is countably infinite, then NA ~ and there exist ANf : which is one to one and

onto. Therefore A is the range of an infinite sequence.

Conversely, assume that naaaA ...,,, 21 or ...,...,,, 21 naaaA with

., ijaa ji Now we define a function f from A to

nZ or N as .kaf k Then

f is one to one and onto, so A is finite or countably infinite.

Theorem 1.5.3: Each subset of a countable set is countable. Proof : Assume A as a countable set and AB . If B finite, then B is countable.

Let B be infinite. Since A is countable, then ,...,, 321 aaaA .

Now let 1

n , be the smallest integer such that ,1

Ban and 2n an integer such that

12 nn and Ban 2

. Repeat this process for each Nk with kn as the smallest

integer such that 1 kk nn and Bakn . Then we have

,......,....,,,321 knnnn aaaaB

that is, B as a sequence. According to theorem 1.5.2, B is countable.

29

Theorem 1.5.4: A finite union of finite sets is again a finite set.

Note : This theorem can also be described as: if NnAA k

n

k

,1

and kA is a

finite set for each k , then A is also finite.

Theorem 1.5.5 : A countable union of countable sets is again a countable set. Proof : Similarly we can write the set as

n

k

kAA1

or

1k

kAA ,

where Nn and each kA is countable. For the earlier case we are supported by

theorem 1.5.4. For the later case (countably infinite), since each kA is countably

infinite, then we can write .....,....,2,1, nkAk

as follows :-

,...,, 14,1312111 aaaaA

,...,, 24,2322212 aaaaA

,...,, 34,3332313 aaaaA

,...,, 44,4342414 aaaaA

.

.

.

,...,, 4,321 nnnnn aaaaA

.

.

.

then

1k

kAA ,....,,, 13,2231,122111 aaaaaa .

Since A can be written in sequence, then by theorem 1.5.2, A is countable. Theorem 1.5.6 : The set of rational numbers is countable. Proof :

0,,: nZnmn

mQ

30

we can write Q as

,1

k

kAQ where

,....3

,3

,2

,2

,1

,1

,0kkkkkk

Ak

Now define a function ,: kk ANf

as nfk

n

k

n

nk

n

,2/1

,2/

then each kf is one to one and onto, that is kA is countable for each

,........3,2,1k According to Theorem 1.5.5, Q is countable.

1.5.1 Uncountable Set R As we have mentioned earlier that real numbers can be represented by decimal expansions. A decimal representation of non-negative real number x is an

expression of the sum

1 10i

i

ia

x ,

where ...., ,21 aa are integers 90 ia . This can also be written as

......101010

.....3

3

2

2103210

aaaaaaaax

This turns out that 0a is the integer part of x , not necessarily between 0 and 9.

,......,, 321 aaa are the fractional part of x .

Example 1.14 : Decimal expansion .......143143143143.0 represents the sum

........10

3

10

4

10

1

10

3

10

4

10

165432

even

odd.

31

Theorem 1.5.7 : The set 1,0 is uncountable.

Proof : Assume 1,0 as a countable set. Then 1,0 must be countable,

according to theorem 1.5.3. Thus 1,0 can be written as the following sequence

,....,,1,0 321 aaa .

Each 1,0n

a has a decimal representation for each na , we write

......0 321 nnnn aaaa

with 9.....,,2,1,0nka . Now define a real number ......0 321 bbbb by

,1 if 1kka

kb

,2 if 1kka

For each ,, nnn abn then ban . But 1,0b , and this contradict the assumption

that every element in 1,0 is also the element in the set ...,,, 321 aaa . Therefore

1,0 is uncountable and so as ]1,0[ .

Corollary 1.5.1: The set R is uncountable. This corollary resulted from Theorem 1.5.7 and Theorem 1.5.3. Corollary 1.5.2 : The set of irrational numbers is uncountable.

Proof : This will be another proof by contradiction. Let cQ be the set of irrational

numbers and suppose that cQ is countable. Since the rational numbers Q are

countable and

cQQR ,

It follows from theorem 1.5.5 that R is countable, and this contradicts corollary 1.5.1.