analysis book bsc(hons) mathematics delhiuniversity

8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.

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Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444

P r o o f

1 . z = z + 0 ( E x i s t e n c e o f z e r o e l e m e n t )

= z + ( a + ( - a ) ) = ( z + a ) + ( - a )

= a + ( - a ) = 0 .

2 . I f u and b 0 are e lements in a n d u b = bTo S h o w t h a t - u = 1

C o n s i d e r u = u 1 (Exis tence of uni t )

= u (b ) ( E x i s t e n c e o f r e c i p r o c a l s )

= (u b) ( u s i n g a s s o c i a t i o n o f m u l t i p l i c a t i o n

= b ( ub= bis guven)

u = b = 1

u = 1

3 . I f a , t h e n a 0 = 0a + a 0 = a 1 + a 0 (Exis tence of uni t )

= a (1 + 0) = a 1 (Dis t r ibut iv i ty)

= a ( E x i s t e n c e o f u n i t )

a 0 = 0 (By (a) )

T h e o r e m

1 . I f 0 a

a r e s u c h t h a t a b = 1 b =

2 . I f a b = 0 then e i ther a = 0 or b = 0


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R a t i o n a l n u m b e r

A n y n u m b e r t h a t c a n b e w r i t t e n a s where a , b a n d b 0 ,i s c a l l e d a r a t i o n a l n u m b e r. T h e s e t of a l l r a t i o n a l n u m b e r s i s

denoted by .

I r r a t i o n a l n u m b e r

A n y n u m b e r w h i c h i s n o t r a t i o n a l i s c a l l e d i r r a t i o n a l .

OR

A number X is ca l led i r ra t ional i ff a and b(0) in

s u c h t h a t

X =

T h e o r e m T h e r e d o e s n o t e x i s t s a r a t i o n a l n u m b e rr s u c h t h a t r 2 = 2 .

P r o o f Let i f poss ib le ra t ional number

r = (q 0 , p , q ) :

= 2 .

N o t e - w i t h o u t t h e l o s s o f g e n e r a l i t y w e c a n a s s u m e t h a t

t h e r e i s n o c o mm o n F a c t o r b / w p a n d q , a n d p a n d q b o t h a r e

p o s i t i v e , N o w =2 p2 = 2 q p2 i s e v e n

(p i s odd p = 2n- 1 , n T h e n p 2 = ( 2 n - 1 ) 2

= 2 ( 2 n2

- 2 n + 1 ) - 1 , i s o d d )


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A l s o p a n d q d o n o t h a v e 2 a s a c o m m o n f a c t o r

q i s odd .. (1)

N o w a s p i s e v e n p = 2 m

Where m : 4m2 = 2 q 2 q2 = 2m 2

q2 i s even q i s even (2)

F r o m ( 1 ) a n d ( 2 ) w e ha v e a c o n t r a d i c t i o n .

A n d h e n c e t h e r e i s n o p a n d q i n

s u c h t h a t

( ) 2 = 2 any r : r2 = 2 .

O R D E R P R O P E RT Y O F -

T h e r e i s a n o n e m p t y s u b s e t P o f , c a l l e dThe s e t o f p o s i t i v e r e a l n u m b e r s ,

T h a t s a t i s f y t h e f o l l o w i n g pr o p e r t i e s .

1 . I f a , b b e l o n g t o P , then a + b P. 2 . I f a , b P then a b P 3 . I f a t h e n e x a c t l y o n e o f t h e f o l l o w i n g i s t r u e . .

a P , a = 0 , - a P .

3 r d property is cal led tr ichotomy propertyN e g a t i v e S e t

D e f i n e a s e t { - a : a P} , then th is se t i s ca l led se t of negat iven u m b e r s .

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N o t a t i o n s

1 . I f a P we wri te ,a > 02 . I f a P U {0} , we wri te a 0 3 . I f - a P we wri te a < 0 4 . I f - a P U {0} , we wri te a 0

a > 0 , a i s pos i t ive rea l number

a 0 , a i s non- n e g a t i v e r e a l n u m b e r

a < 0 , a i s n e g a t i v e r e a l n u m b e r

a 0 , a i s non- p o s i t i v e r e a l n u m b e r

D e f i n i t i o n - L e t a , b b e t h e e l e m e n t s i n ( a ) . I f a - b P, we wri te a > b or b < a

( b ) . I f a - b PU{0}, we wri te a b or b a

T h e o r e m - L e t a , b , c b e a n y e l e m e n t s o f ( a ) . I f a > b a n d b > c , t h e n a > c

( b ) . I f a > b t h e n a + c > b + c

( c ) . I f a > b a n d c > 0 , t h e n c a > c b

( d ) . I f a > b a n d c < 0 , t h e n c a < c b

N o t e - ( b ) i s c a l l e d m o n o t o n i c l a w o f a d d i t i o n .

( c ) i s c a l l e d m o n o t o n i c l a w o f m u l t i p l i c a t i o n .


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P r o o f -

( a ) . G i v e n - a > b a n d b > c

a b P a n d b c P

To S h o w t h a t - a > c i . e a c P

A s a b P and b c P

( B y p r o p e r t y 1 . o f P ) w e h a v e ( a - b ) + ( b - c) P

a + (- b + b ) - c P

a - c P a >c .

( b ) . I f a > b , t h e n a + c > b + c

P r o o f - G i v e n - a > b a- b P

N o w ( a + c ) - ( b + c ) = a + c - b - c

= a + c - c - b

= a - b P

( a + c ) - (b + c) P a + c > b + c .

( c ) . G i v e n - a > b a - b P

a n d c > 0 c P , by proper ty 2n d o f P

c ( a - b) P ca- cb P

ca > c b .

( d ) . G i v e n - a > b a - b P


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a 2 = ( - a ) ( - a) P (- a P (- a ) ( - a) P)

a2 P a2 > 0

( b ) . Now as 0 1

by (a) 12 > 0

1 = 12 > 0 1 > 0

(c) . We l l show us ing mathemat ica l induct ion

F o r n = 1 ; 1 > 0

Now assume resul t i s t rue for n = k k P

Now we l l show tha t resul t i s a l so t rue for n = k +1

Now k P and 1 P

k +1 P (us ing 1s t p r o p e r t y o f P )

resul t i s t rue for n = k +1 resul t i s t rue for a l ln n > 0 n . A b s o l u t e v a l u e

L e t a b e a n y r e a l n u m b e r t h e n a b s o l u t e v a l u e o f a i s

d e n o t e d b y | a | a n d i s d e f i n e d a s .

| a | =

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T h e o r e m

(a) . | ab | = | a | |b | a , b ( b ) . | a | 2 = a 2 a

(c) . i f c 0 , then |a | c i ff- c a c

( d ) . - | a | a |a | a P r o o f -

(a) . i f a = 0 or b = 0 | ab | = | a | |b |

So assume a 0 , b 0 .

C a s e 1 a > 0, b > 0 ab > 0 | ab | = ab

Now |ab | = ab = |a | |b |

( a > 0 | a | = a & b > 0 |b | = b )

C a s e 2 a < 0 , b < 0 ab > 0 | ab | = ab

N o w | a b | = a b = ( - a ) ( - b) = |a | |b | ( b < 0 |b | = - b, a < 0 | a | = - a )

C a s e 3 a < 0 , b > 0 ab < 0 | ab | =- a b

| ab | = (- a ) b = | a | | b |

( a < 0 | a | = - a , b > 0 |b | = b )

C a s e 4 a > 0, b < 0 ab < 0 | ab | =- a b

| a b | = a ( - b ) = | a | | b |

( a > 0 | a | = a , b < 0 |b | =- b )

in poss ib le cases we have

| a b | = | a | | b |

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(b) . a : a 2 0

a 2 = | a | 2 = |a a | = | a | | a | = |a | 2

| a | 2 = a 2

(c) . c 0 , then |a | c i ff- c a c

N o t e - To prove th is we l l prove the fo l lowing

M a x { a , - a } = | a |

When a = 0 the resul t i s t rue assume a 0 .

C a s e 1 a > 0 | a | = a .. (1)

a > 0 - a < 0 a > - a max {a ,- a } = a

max {a ,- a } = a = | a | ( B y ( 1 ) )

max {a ,- a } = | a |C a s e 2 a < 0 | a | > - a (2)

a < 0 - a > 0 - a > a max {a ,- a } = - a

max {a ,- a } = - a = | a | ( B y (2 ) )

max {a ,- a } = | a |

Now we l l show tha t | a | c - c a c

| a | c m a x { a , - a} c

a c & - a c

a c & a - c

- c a c

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(d) . Now |a | 0

D o y o u r s e l f -

Tr i a n g u l a r I n e q u a l i t y

I f a , b t h e n | a + b | | a | + | b | P r o o f -

C a s e - 1 a + b > 0

| a + b | = a + b . . (1)

and we know tha t a | a | , b |b |

a + b |a | + |b | . . (2)

U s i n g ( 2 ) i n ( 1 ) w e g e t | a + b | |a | + |b | .

C a s e - 2 a + b < 0

| a + b | = - ( a + b )

= - ( a ) + - (b) . (3)

N o w w e k n o w t h a t

- (a) | a | , - (b) |b |

- ( a ) + - (b) | a | + | b | (4)

Using (4) in (3) we get | a + b | | a | + |b | .

C a s e - 3 a + b = 0 a =- b a n d | a + b | = 0

| a | = |- b | = | b |

| a | + |b | = 2 |a |

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| a + b | = 0 |a | + |b |

| a + b | | a | + |b |

C o r o l l o r y - I f a , b

t h e n

( a ) . | | a | - | b | | | a - b | ,

( b ) . | a - b | | a | + |b | .

P r o o f - | a | = | a - b + b |

|a - b | + | b | ( u s i n g t r i a n g l e i n e q u a l i t y )

| a | - | b | | a - b | . . (1)

N o w | b | = | b a + a | |b - a | + | a |

|b | - | a | |b - a | = | a - b | . . (2)

max { |a | - | b | , | b | - | a | } | a - b |

max { |a | - | b | , - ( | a | - | b | ) } |a - b |

| | a | - | b | | | a - b | .

D e f i n i t i o n - Let a a n d > 0 . T h e n t h en e i g h b o r h o o d o f a i s t h e s e t v ( a ) d e f i n e d a sv (a) = { x

: | x a | < } .

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P r o o f - Since 0 |x a | < > 0

|x a | = 0 ( i f 0 a < > 0 then a = 0 )

x = a .

Q 1 : I f a , b

a n d b 0 s h o w t h a t

( a ) . | a | = ( b ) . | | =

S o l : ( a ) . C a s e 1 a > 0 | a | = a

| a | 2 = a 2

| a | = C a s e 2 a < 0 | a | = - a

| a | 2 = - ( a ) 2 = a 2

| a | =

C a s e - 3 a = 0 | a | = 0

| a | 2 = 0 2

| a | = = ( b ) . | | =

T h o e r e m - Let a . I f x b e l o n g s t o t h e n e i g h b o r h o o dv (a) > 0 , t h e n x = a .

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C a s e - 1 a > 0 , b > 0 > 0 = =

( a = | a | , b = |b | )

C a s e - 2 a < 0 , b < 0 > 0 = = = ( | a | = - a , | b | = - b | , a < 0 , b < 0 )

C a s e - 3 a < 0 , b > 0 < 0 = = = ( | a | = - a , | b | = b , a < 0 , b > 0 )

C a s e - 4 a > 0 , b > 0 . D o y o u r s e l f .

n e i g h b o r h o o d

Let a be any rea l number and le t > 0 be any number then

neighborhood of a i s def ined a s t h e s e t

v (a) = { x

: | x a | < } = (a , a + ) .

N e i g h b o r h o o d

L e t a b e a n y r e a l n u m b e r t h e n a n o n e m p t y s e t V i s c a l l e dneighborhood of a i ff > 0 :

v (a) V i . e (a , a + ) V.N o t e - Every neighborhood of a i s ne ighborhood of a

L e t V = v ( a )

We l l s h o w t h a t V i s a n b h d o f a .

A s v ( a ) i s an nbhd of a a lso v(a) v ( a ) = V

v(a) V V i s a nbhd of a


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O p e n S e t - A s e t G i s s a i d t o b e a n o p e n s e t i f f o r each x G

a nbhd V of x such tha t V G.

C l o s e d S e t A set F i s sa id to be c lose t se t i f i t s complement

F c = - F i s a n o p e n s e t .

T h e o r e m -

L e t A b e s u b s e t o f

t h e n A i s op e n i f f x A x > 0 :

]X - x , X+ x [ A

P r o o f - F i r s t a s s u m e A is open then x A nbhd o f Vof x : V A ..(1)

Now V is nbhd of x x > 0 : ] X - x , X+ x [ V (2)

F r o m ( 1 ) & ( 2 ) w e h a v e ] X - x , X+ x [ A

C o n v e r s e l y Assume tha t x A x > 0 : ] X - x , X+ x [ A

To s h o w t h a t - A i s o pe n I . e T o s h o w t ha t n b h d o f V o f x

x A

Let x A (be any e lement)

By our assumpt ion x > 0 : ] X - x , X+ x [ A

L e t V = ] X - x , X+ x [ A then V is a nbhd of x

V A A i s an open se t .


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N o t e - F r o m n o w o n w a r d s w h e n e v e r w e n e e d t o s h o w A i s

o p e n w e l l s h o w t h a t

x A x > 0 : ] X - x , X+ x [ A .

R e m a r k - G i s op e n i f f x G > 0 :

v (x) G G i s ne ighborhood of x

G i s open i ff i t i s ne ighborhood of each of i t s poin t s .

E x a m p l e E v e r y o p e n i n t e r v a l o f t h e t y p e ]a , b [ i s a n o p e n s e t .

P r o o f - Consider ]a , b[ Let x ]a , b[ be any poin t , we shal ls h o w ] a , b [ i s n b h d o f x .

Let x = m i n { x a , b - x } t h e n ] x - x , x + x [ ] a , b [

( x = m i n { x a , b - x } x x a , x b - x )

a x - x & x + x b )

]x - x , x + x [ ]a , b[

A n d h e n c e ] a , b [ i s o p e n s e t .

E x a m p l e , T h e S e t o f r e a l s i s an o p e n s e t .P r o o f - Let x

, t h e n

> 0

] x - x , x + x [ i s o p e n s e t .E x a m p l e S e t o f r a t i o n a l s Q i s n o t o p e n s e t .

P r o o f - Let x Q, then x Q and > 0 ] x - x , x + x [ Q( ]x - x , x + x [ c o n t a i n s i n f i n i t e n o. o f r a t i o n a l s w h i c h c a nb e i n Q ) Q cant be open.

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Q u e s t i o n - E m p t y s e t i s a n o p e n s e t .

S o l u t i o n - Since i s a se t wi th no e lements in i t t h e r ei s no e lement in o f which i t i s not a nbhd

i s nbhd of each of i t s poin ts

i s open se t .

Q u e s t i o n A n y f i n i t e s e t i s n o t o p e n .

S o l u t i o n - Let S be any f in i te se t we l l show S i s not nbhd ofa n y o f i t s points . Let x S be any

Then > 0

] x - x , x + x [ S ( ]x- x , x + x [ i s i n f i n i t es e t w h i l e s i s f i n i t e s e t )

S i s not open.

Q u e s t i o n E v e r y o p e n i n t e r v a l o f t h e t y p e ] - , a [ i s o p e n .

P r o o f

- Let x ]- ,a[ be any number x + < a

- < x + < a - < x - < x + < a

]x - , x + [ ]- ,a[

]- , a[ i s open.


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]x - , x + [ G i i = 1 , 2 , 3 .n

]x - , x + [

i s a n o p e n s e t .D e f i n i t i o n ( C l o s e d s e t ) -

A set G is c losed i ff i t s complement i s open.

P r o p e r t i e s o f C l o s e d s e t s

( a ) . A r b i t r a r y i n t e r s e c t i o n o f c l o se d s e t s i s a c lo s e d s e t .

( b ) . F i n i t e u n i o n o f c l o se d s e t s i s a c l o s e d s e t .

P r o o f - L e t { G } b e a f a m i l y o f c l o s e d s e t s .

To s h o w t h a t i s c l o s e d s e t .i . e To s h o w t h a t (

) c i s o p e n s e t

w e k n o w = . (1) Now as G i s c losed (G ) c i s open

i s open se t ( Arbi t rary union of opens e t s i s a n o p e n s e t ) .

By (1)

i s o p e n

c l o s e d s e t .( b ) . F i n i t e u n i o n o f c l o se d s e t s i s a c l o s e d s e t .

L e t G 1 , G2 , .. Gn b e f i n i t e n u m b e r o f c l o s e d s e t s .

T h e n w e h a v e t o s h o w t h a t

i s c l o s e d i . e To

s h o w t h a t

i s a n o p e n s e t .


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We k n o w t h a t = (1) N o w s i n c e G i i s c losed for i = 1 , 2 , . n

(Gi ) c i s open for i = 1 , 2 , . N i s open ( f in i te in tersec t ion of open se ts i s

o p e n )

N o t e - 1 A r b i t r a r y i n t e r s e c t i o n o f o p e n s e t s m a y n o t b e op e n .

E x a m p l e - L e t F = { ( - ) | n } a f a m i l y o f o p e n s e t s .Wel l show tha t

i s n o t a n o p e n s e tAs = { 0 } a n d { 0 } i s n o t a n o p e n s e t ,b e i n g a f i n i t e s e t .

N o t e - 2 A r b i t r a r y u n i o n o f c l o s e d s e t s m a y n o t b e c l o s e d .

E x a m p l e L e t F = { [ ] | n

} t h e n e a c h m e m b e r of F i s

c l o s e d s e t ( b e i n g c l o s e d i n t e r v a l s )

B u t *+ = ( 0 , 2 ] i s n o t c l o s e d .T h e o r e m E ve r y c l o s e d i n t e r v a l of t h e t y p e [ a , b ] i s a c l o s e d s e t .

P r o o f - L e t F = [ a , b ] T h e n F c = [ a , b ] c = ( - ,a)U(b, ) whichi s o p e n s e t ( b e i n g u n i o n o f o p e n s e t s )

F i s c losed [a , b] i s c losed se t .


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C l u s t e r p o i n t - ( l i m i t p o i n t )

A r e a l n u m b e r x i s sa id to be limi t poin t or c lus ter poin t of a se t A i f f e v e r y - nbhd of x conta ins a member of A other than x .i .e > 0 V (x) (A {x})

] x - , x + [ ( A {x}) > 0

E x a m p l e Consider the se t S = ]a , b[ then > 0 ] a - , a +[c o n t a i n s i n f i n i t e l y m a n y p o i n t s o f ] a , b [

a i s l i m i t p o i n t o f ] a , b [

Similar ly > 0 ] b - , b +[ c o n t a i n s i n f i n i t e l y m a n y p o i n t sof ]a ,b[ b i s l imi t poin t of ]a , b[

T h e o r e m - A set A is c losed i ff i s conta ins a l l of i t s l imi t poin ts .

P r o o f - F i r s t L e t u s a s s u m e t h a t A i s c l o s e d s e t .

L e t x b e a n y l i m i t p o i n t o f A

To show tha t x A.

Let x A x ~ ANow as A is c losed ~ A i s o p e n . ~ A i s ne ighborhood o f x > 0 : ] x - , x + [

~ A


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] x - , x + [ A =

x i s not a l imi t poin t of A which i s a cont radic t ion as xi s l i m i t p o i n t o f A

x A i s not poss ib le x A

A conta ins a l l i t s l imi t poin ts .

C o n v e r s e l y assume tha t A conta ins a l l i t s li m i t p o i n t s .

To s h o w t h a t A i s c l o s e d f o r t h a t w e n e e d t o s ho w t h a t

~ A i s o p e n .Let x ~ A ( b e a n y p o i n t ) x A x i s not l imi t poin t of A

> 0 : V (x) A =

V(x)

~ A

~ A i s n b h d o f xS i n c e x w a s a n y a r b i t r a r y p o i n t o f ~ A t h i sf o l l o w s t h a t ~ A i s n b h d o f e a c h o f i t s p o i n t s

~ A i s o p e n

A i s c losed .


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E x a m p l e s o n C l u s t e r p o i n t .

( 1 ) E v e r y R e a l n u m b e r i s c l u s t e r p o i n t o f Q .

S o l u t i o n Let x be any rea l number then > 0]x - , x + [ c o n t a i n s i n f i n i t e l y m a n y r e a l n u m b e r s

w h i c h a r e r a t i o n a l s .

]x - , x + [ (Q {x})

x i s l imi t poin t of Q.

D e f i n i t i o n ( D e r i v e d S e t )

L e t S b e a n y s u b s e t o f r e a l n u m b e r s t h e n d e r i v e d s e t o f S i s

d e n o t e d b y S l a n d i s d e f i n e d a s

S l = { x : x i s l i m i t p o i n t o f S }

N o t e i n v i e w o f a b o v e r e s u l t a n d d e f i n i t i o n w e c a n s a y

t h a t Q l =

( 2 ) S e t o f i n t e g e r s h a s n o l i m i t p o i n t .

S o l u t i o n L e t denote the se t of in tegers , we l l show no rea lnumber x can b e l i m i t p o i n t o f

As x w e h a v e t w o c a s e s .(1 ) x ( 2 ) x

C a s e 1 x t h e n ] x - [ i s a n e i g h b o r h o o d o fw h i c h c o n t a i n s n o i n t e g e r s

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z


O t h e r t h a n x ( b e t w e e n a n y t w o i n t e g e r s t h e r e

i s d i s t a n c e o f a t l e a s t o n e u n i t )

x i s not l imi t poin t of

C a s e 2 x t h e n w e c a n c h o o s e t w o i n t e g e r s n a n dn + 1 s u c h t h a t n < x < n + 1

T h e n i f w e c h o o s e = m i n { x - n , n + 1 - x }

] x - , x+[ w i l l c o n t a i n n o i n t e g e r

x cant be l imi t poin t of A n d h e n c e n o r e a l n u m b e r c a n b e l i m i t p o i n t o f

l =

E x a m p l e : ( ] a , b [ ) l = [ a , b ]

S o l u t i o n :

i f x [a , b] then for every > 0 , ] x - , x + [c o n t a i n i n f i n i t e l y m a n y m e m b e r s o f ] a ,

b [

And one member of ]a , b [ , o ther than x

x i s a l imi t poin t of ]a ,b[

Now le t x [a , b]

C a s e 1 x < a t h e n c h o o s e ( 0 < < a - x )

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x + b , t h e n c h o o s e 0 < < x - b

x- > b

]x- , x+[ d o e s n o t c o n t a i n a n y m e m b e ro f ] a , b [

x i s not a l imi t poin t of ]a , b[

H e n c e ( ] a , b [ ) l = [ a , b ]

E x a m p l e S e t o f i n t e g e r s i s c l o s e d .

S o l u t i o n : As l = and Z l Z

c o n t a i n s a l l i t s l i m i t p o i n t s

i s c l o s e dE x a m p l e S e t o f n a t u r a l n u m b e r s h a s n o l i m i t p o i n t .

O R

(

) l =

A n d () l = i s c l o s e dE x a m p l e : L e t S = { }

T h e n S l = { 0 }

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z


S o l u t i o n : L e t > 0 b e a n y r e a l n u m b e r t h e n b y a r c h i m e d i a n

proper ty n

:

n > o r n <

Now S a n d ]0-, 0+ [ > 0 ] 0 - , 0 + [ c o n t a i n s a m e m b e r o f S 0 i s c l u s t e r p o i n t o f S

To s h o w - N o r e a l n u m b e r o t h e r t h a n 0 c a n b e l i m i t

p o i n t o f S

Let x 0 be any rea l number

C a s e 1 x < 0

T h e n c h o o s e 0 < 1

T h e n c h o o s e 0 < < x - 1

]x-, x +[ d o e s t n o t c o n t a i n a n y m e m b e r o fS ( x- >1 ]x-, x+ [ ]1 , [)

C a s e 3 x = 1 , t h e f o r = 1 / 3 ( x - , x+) = ( 2 / 3 , 4 / 3 )

a n d i t d o e s n o t c o n t a i n a n y p o i n t s o f S

O t h e r t h a n 1

1 i s not l imi t poin t of S


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C a s e 4 0

T h e n ( ) i s a n b h d o f x = 1 / n a n d i t

c o n t a i n s n o p o i n t o f S o t h e r t h a n x = 1 / n

x i s not l imi t poin t of S .

C a s e 5 Let 0

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C h a p t e r 11 . 1 ( G u i d e l i n e s Q u e s t i o n s )

( Q 1 ) I f x (0 , 1) and x = m i n { x , 1 - x } i f | u - x | < x , s h o wtha t u (0 , 1)

S o l u t i o n | u - x | < x x- x < u < x + x .. (1)

Now x = m i n { x , 1 - x } x x and x + x 1 . (2)

From (1) & (2) 0x- x < u < x + x 1

0

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3. [b ,)c = ( - ,b) which i s open se t [b , ) i s c losed( - , b ]c = (b , ) which i s open se t

(- , b] i s c losed se t .

( Q 3 ) S e e t h e P r o o f o f o p e n s e t p r o p e r t p a r t ( b ) o n

p a g e n o . ( 5 ) & ( 6 )

( Q 4 ) Show tha t (0 , 1] = then 0 < x 1 < 1 + 1 /n n

x (0 , 1 + 1 /n) n x (0 , 1]

Now we l l show tha t no o ther rea ls be long to

( 0 , 1 + 1 / n )i . e i f x < 0 then x ( 0 , 1 + 1 / n )and i f x > 1 then x ( 0 , 1 + 1 / n )

C a s e - 1 i f x < 0 , then x (0 ,1+1/n) n x

( 0 , 1 + 1 / n )

C a s e 2 i f x > 1 , t h e n x - 1 > 0 > 0

n : n > < x - 1 1+ < x

x (0 ,1+1/n) for some n

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x ( 0 , 1 + 1 / n )

( 0 , 1 + 1 / n ) = ( 0 , 1 ]

( Q 5 ) S e e o n p a g e ( 2 2 ) E x a m p l e

( Q 6 ) S e e p a g e ( 2 2 ) E x a m p l e , S = { } t h e n S l = { 0 }Here A = S U {0} Al = {0} and {0} A Al A

A i s c losed

C h a p t e r - 2 . 2 ( R . G B a r t l e ) E x e r c i s e s c o n t i n u e d f r o m p a g e

1 4 .

( Q 2 ) i f a , b , t h e n | a + b | = | a | + | b | ab 0S o l u t i o n : | a + b | = | a | + | b |

| a + b | 2 = ( | a | + | b | ) 2

( a + b ) 2 = | a | 2 + | b | 2 + 2 |a | |b | ( |x | 2 = x 2 )

a 2 + b 2 + 2 a b = a 2 + b 2 + 2 |ab | ( | x |2 = x 2 )

2 a b = 2 | a b | ( l e f t c a n c e l l a t i o n l a w o f

a d d i t i o n )

a b = | a b | ( l e f t c a n c e l l a t i o n l a w o f

a d d i t i o n )

ab 0

( Q 4 ) | x - a | < - < x - a < a - < x < a +

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( Q 3 ) i f x , y, z x z , s h o w t h a t x y z | x - y | + | y - z | =| x - z |

S o l u t i o n : f i r s t le t x y z (we l l show |x- y | + | y - z | = | x - z | )

As x y x- y 0 |x- y | = y - x .. (1)

y z y- z 0 |y- z | = z y . (2)

x z x- z 0 |x- z | = z - x . (3)

f r o m ( 1 ) a n d ( 2 )

| x - y | + | y - z | = z - x = | x - z | (by 3)

C o n v e r s e l y a s s u m e

| x - y | + | y - z | = z - x = | x - z | .( * )

( We l l show x y z )

As x z i t i s enough to show y x and y z

Let i f poss ib le y < x x y > 0 |x- y | = x - y

Also y < x z y < z |y- z | = z y

|x - y | + | y - z | x + z 2y |x - z |

we have a cont radic t ion to (*)

y x x y . (a)

Similar ly we can show tha t z y

y z . (b)

From (a) & (b) x y z


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( Q 7 ) | x + 1 | + | x - 2 | = 7

We h a v e f o l l ow i n g c a s e s ( 1 ) x < - 1 , ( 2 ) - 1 < x < 2, ( 3 ) x > 2

(1 ) x < - 1 x < 2

x + 1 < 0 and x- 2 < 0

|x +1 | =- ( x + 1 ) a n d | x - 2 | = 2 x

|x +1 | + |x- 2 | = - x 1 + 2 x = - 2 x + 1

- 2x + 1 = 7 x =- 3

( 2 ) - 1 < x < 2

|x + 1 | + x + 1 , |x- 2 | = 2 x

x + 1 + 2- x = 7 3 = 7 (an absurd)

(3) x > 2 x >-1 x + 1 + x- 2 = 7 2x = 8 x = 4

( Q 8 ) ( a ) | x - 1 | > |x + 1 | (x-1 ) 2 > (x+1) 2

(x- 1 ) 2 - ( x + 1 ) 2 > 0

( (x - 1 ) + ( x + 1 ) ) ( x 1 - x - 1 ) > 0

2x (- 2) > 0 - 4x > 0 x < 0

x (- , 0)


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z


C a s e - 2 - 2 < x < 1

4 < x + 2 + x- 1 < 5 4 < 2x +1 < 5 3 /2 < x < 2

x ( 3 / 2 , 2 ) U ( - 3 , - 5/ 2 )

( Q 1 2 ) ( a ) | x | = |y | x2 = y 2 y = x

( b ) | x | + | y | = 1

x 0 , y 0 x + y =1

x < 0 , y < 0 -x y = 1

x < 0 , y > 0 -x + y = 1

x > 0 , y < 0 x y = 1

( c ) | xy | = 2 xy = 2 ( d ) D o y o u r s e l f

( Q 1 3 ) D o y o u r s e l f

( Q 1 5 ) as a b | a- b | > 0 , L e t =

Then U = V ( a ) i s - n b h d o f a

V = V( b ) i s - n b h d o f b

(we l l show U V = )

Let i f poss ib le U V x U V

x U and x V

x V(a) , x V( b )

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z


|x - a | < , | x - b | < . (1)

N o w | a - b| = | a x + x - b |

| a- b | | x- a | + | x - b |

| a- b | < + | a- b | < 2 |a- b | < 2

(3 < 2) which i s a cont radic t i o n

U V i s not poss ib le U V =

( Q 1 6 ) ( a )

C a s e 1 a > b max {a , b } = a . (1)

Also a > b | a- b | = a - b

{a +b + |a- b | } = 1 / 2 { a + b + a - b} = a (2)

F r o m ( 1 ) & ( 2 ) m a x { a , b } = 1 / 2 { a + b + | a - b | }

C a s e - 2 a < b max {a , b} = b (3) and |a- b | = b a

A l s o { a + b + | a - b | } = { a + b + b - a} = b (4)

F r o m ( 3 ) & ( 4 ) m a x { a , b } = { a + b + | a - b | } ,

s i m i l a r l y w h e n a = b

max {a , b } = {a+b+|a- b |} a , b

A n d f o r m i n { a , b } = { a + b - | a - b | }

C o n s i d e r c a s e - 1 a > b min {a , b} = b (1)

Also a > b | a- b | = a - b

{a +b- | a - b | } = { a + b + b - a} = b (2)

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F r o m ( 1 ) & ( 2 ) m i n { a , b } = { a + b - | a - b | }

C a s e 2 a < b min {a , b} = a (3)

Also a < b | a- b | = b a

{a+b- | a - b | } = { a + b + a b } = a (4)

F r o m ( 3 ) a n d ( 4 ) m i n { a , b } = 1 / 2 { a + b - | a - b | }

C a s e 3 a = b min {a , b} = a (5) ( a = b)

{ a + b - | a - b | } = { a + a - | a - a |} ( a = b)

= {2a} = a (6)

F r o m ( 5 ) & ( 6 ) m i n { a , b } = { a + b - | a - b | }

( Q 1 4 ) f o r V (a) V ( a ) L e t r = m i n { , } ( then we l l showV(a) V ( a ) = V r ( a ) )

Let x V(a) V ( a )

x V(a) & x V ( a )

|x a | < A n d | x - a | < | x- a | < m i n { , } = r

| x- a | < r x Vr ( a )

V(a) V (a) Vr ( a ) (1)

L e t x V r ( a ) | x - a | < r = m i n { , }

|x a | < A n d | x - a | <

x V(a) , x V(a) x V(a) V ( a )

Vr (a) V(a) V(a) . (2)

F rom (1 ) & (2 ) V (a) V ( a ) = V r ( a )

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For V ( a ) U V ( a ) L e t r = m a x { , } ( then we l l showV( a ) U V ( a ) = V r ( a ) )

Let x V( a ) U V ( a ) x V(a) or x V ( a )

|x a | < A n d | x - a | < | x- a | < m a x { , } = r

x V r (a) V(a) V(a) Vr ( a ) . (3)

L e t x Vr ( a ) | x - a | < r = m a x { , } |x a |


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n m- 1 n

m- 1 i s a n u p p e r b o u n d o f

m m- 1 0 - 1 w h i c h i s n o t p o s s i b l e

nx : n x > x .

C h a p t e r ( 2 . 3 ) & ( 2 . 4 ) ( R . G B a r t l e ) ( G u i d e l i n e s Q u e s t i o n s )

( Q 1 ) S 1 = {x

: x 0}

1 . Clear ly 0 x x S1 0 i s a lower bound of S1 2 . Wel l show tha t 0 i s inf S1 , f o r t h a t w e o n l y n e e d t o

s h o w t h a t 0 i s g r e a t e s t l o w e r b o u n d o f S 1

L e t u b e a n y l o w e r b o u n d o f S 1

To s h o w t h a t - u 0 , le t i f poss ib le , u > 0 then u S1a l s o 0 0 i s not poss ib le u 0 Inf S1 = 0

3 . Wel l now show tha t S1 i s n o t b o u n d e d a b o v e .

L e t i f p o s s i b l e S 1 i s b o u n d e d a b o v e t h e n b y o r d e r

c o m p l e t e n e s s p r o p e r t y S 1 h a s s u p r e m u m . L e t

S u p S 1 = m , N o w c l e a r l y m > 0

m +1 > 0 m+1 S1

A s S u p S 1 = m m +1 < m

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(1< 0) (which i s a cont radic t ion)

S1 i s n o t b o u n d e d a b o v e .

( Q 2 ) D o y o u r s e l f .

( Q 3 ) S3 = {1/n : n }Sup S 3 = 1 ; c lear ly x 1 x S3

1 i s an upper bound ofS 3

N o w l e t u b e a n y u p p e r b o u n d o f S 3

Wel l show 1 u ; le t i f poss ib le 1 > u ;

Now 1 S3 a n d 1 > u i s a c o n t r a d i c t i o n A s u i s u p p e r b o u n d o fS 3

1 > u i s not poss ib le 1 u

Sup S 3 = 1

( Q 4 ) S4 = { 1 - ( - 1 ) n / n | n } f i n d s u p S 4 a n d I n f S 4S o l u t i o n S 4 = { 1 + 1 , 1 1 / 2 , 1 + 1 / 3 , 1 - 1/4.}

= {2, 0 .5 , 1 .333, 0 .75 , 1 .2 , . .}

Sup S 4 = 2 a n d I n f S 4 = 0 . 5


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( Q 5 ) L e t S be a n on - e m p t y s u b s e t o f

t h a t i s b o u n d e d b e l o w,

s h o w t h a t

I n f S = - S u p { - s : s S}

P r o o f L e t T = { - s : s S } a n d S u p T = m

To s h o w t h a t - I n f S = - m

C l a i m 1 m i s a l o w e r b o u n d o f s

L e t x S be any e lement then x T - x m ( m =S u p T )

x - m x S

- m i s a l o w e r b o u n d o f S

C l a i m 2 m i s g r e a t e s t l o w e r b o u n d o f S

L e t u b e a n y l o w e r b o u n d o f S

u x x S

- u - x - x T

- u i s a n u p p e r b o u n d o f T, A s m = s u p T

m -u - m u

- m i s g r e a t e s t l o w e r b o u n d o f S

inS =- m = - S u p T = - s u p { - s : s S}

N o t e S i n c e S i s b o u n d e d b e l o w k : k x x S


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Now y T ,- y S

- y k y - k y T

T i s bounded above and hence by orderc o m p l e t e n e s s p r o p e r t y T h a s a s u p r e m u m , w h i c h w e

c a l l e d

A s m , i n a b o v e P r o o f .

( Q 6 ) I f a se t S

c o n t a i n s o n e o f i t s u p p e r b o u n d s , s h o w

t h a t t h i s u p p e r b o u n d i s t h e s u p r e m u m o f S .

S o l u t i o n : L e t u b e u p b d o f S : u S (T.S .T Sup(S)= u)

A s u i s a l r e a d y a n u p p e r b o u n d

we only need to show u i s leas t upbd

L e t v b e a n y u p b d of S

A s V i s u p b d o f S (T.S .T u v)

And u S u v (By def in i t ion of upbd)

Sup(S) = u

( Q 7 ) ( ) Let u = upbd of S , Let t

a n d t > u ( T. S . T t s )

Let i f poss ib le t s as u = upbd of S t u which i sa cont radic t ion as t u ( t > u)

t s

( ) Let u b e :I f t

, t > u t h a n t s ( T. S . T u = u p b d o f S )


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Let s S (T.S .T s u)

Let i f poss ib le s u s > u now S

s S

s > u and s b y o u r h y p o t h e s i s s S , w h i c h i snot poss ib le as s S

s u i s not poss ib le

s u s S u = upbd of S .

( Q 9 ) A , B a r e b o u n d e d S u b s e t s o f

, S h o w t h a t A U B i s

b o u n d e d a n d

S u p ( A U B ) = S u p { S u p A , S u p B }

S o l u t i o n A, B are bounded k , K a n d k l , Kl :K a K , kl b Kl a A, b B

Now x A U B x A or B

k x K or kl x Kl

min {k , kl } x max {K, Kl } x A U B

A U B is bounded

Now x A U B x A or x B x sup(A) or x S u p ( B )

x Sup {Sup(A) , Sup(B)} . . (1)

Now Let u be any upper bound of A U B(Wel lshow Sup{Sup(A) , Sup(B)} u )

a A as A A U B a A U B a u a A.(2)


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b B as B A U B b A U B b u b B.(3)

From (2) & (3) u i s and upbd of A and u i s anu p b d o f B

Sup(A) u and Sup(B) u

Sup {Sup(A) , Sup(B)} u (4)

F r o m ( 1 ) & ( 4 )

S u p ( A U B ) = S u p { S u p ( A ) , S u p ( B ) }

( Q 1 0 ) L e t S b e b o u n d e d s e t i n and S L e t S 0 S and S0 then

Inf (S) Inf (S0 ) Sup(S0 ) Sup(S)

S o l u t i o n - x S0 S

x S Inf (S) x x S0

I n f ( S ) i s a l o w e r b o u n d o f S 0

Inf (S) Inf (S0 ). . (1)

Again , x S0 , x S

x Sup(S) x S0

Sup(S) i s an upbd of S0

Sup(S0 ) Sup(S) . (2)

Now l e t s 0 S0 ( b e f i x e d t h e n )

I n f ( S 0 ) s0 S u p ( S 0 ) Inf (S0 ) Sup(S0 ) . (3)

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From (1) , (2) and (3) Inf (S) Inf (S0 ) Sup(S0 ) S u p ( S )

( Q 11 ) Let S a n d s u p p o s e s* = S u p ( S ) , s * S , I f u S ,

s h o w t h a t

S u p { S U { u } } = S u p { s * , u }

S o l u t i o n

C a s e 1 s * < u , S u p { s * , u} u

C l a i m S u p { S U { u } } = u

1 . Let x S U {u} x S or x = u x s* or x =u x u t h e n S u p { s * , u } = s *

Wel l c la im S u p { S U { u } } = s * = S u p { s * , u }

C l a i m 1 s*

i s a n u p p e r b o u n d o f S U { u }

A s s * = S u p S a n d s * > u

x s* x S and u < s*

x s* x S U {u}

s* i s a n u p p e r bo u n d of S U { u }

C l a i m 2 s * i s l e a s t u p p e r b o u n d o f S U { u }


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Let k be any upper bound of S U {u} , then x k xS U {u} . (1)

N o t e - w e a re g i v e n t h a t s * = SupS S

s* S S U {u}

U s e t h i s f a c t i n ( 1 ) w e g e t s * K s* i s l e a s tu p p e r b o u n d o f S U { u }

Sup{ S U {u}} = s* Sup{ S U {u}} = Sup{ s* , u } .

( Q 1 2 ) S h o w t h a t a n o n - e m p t y s u b s e t w h i c h i s f i n i t e , i n c o n t a i n s i t s s u p r e m u m .

P r o o f - Let S be any f in i te se t conta in ing n e lements we l lp r o v e t h e r e s u k t b y a p p l y i n g i n d u c t i o n o n

e l e m e n t s i n S .

w h e n s c o n t a i n s a s i n g l e e l e m e n t i . e n = 1

L e t S = { a 1 } t h e n c l e a r l y S u p { s } = a 1 S

N o w w h e n S c o n t a i n s t w o e l e m e n t s s a y a 1 & a 2

( W. L . O . G ) A s s u m e , a 1 > a 2 t h e n S u p { a 1 , a 2 } = a 1

S;

N o w l e t u s a s s u m e t h a t r e s u l t i s t r u e f o r a l l s e t s

conta in ing K e lements ; we l l show tha t resul t i s a l sot r u e f o r n = K + 1 i . e f o r a s e t c o n t a i n i n g ( K + 1 )

e l e m e n t s

L e t S = { a 1 , a 2 , . . ak , a k + 1 }

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T h e n S = S l U { a k + 1 } , w h e r e S l = { a 1 , a 2 , . . ak ,a k + 1 }

N o t e s ince O(S) = k + 1 ai aj i j

Now as S l i s a se t conta in ing k e lements by oura s s u m p t i o n S u p S l Sl , l e t S u p S l = s *

N o w b y t h e P r o o f o f a b o v e q u e s t i o n ( 2 )

We s e e S u p S = S u p { s * , a k + 1 }

SupS = s* o r S u p S = a k + 1

C a s e - 1 i f S u p S = s * Sl S

SupS S ,

C a s e 2 S u p S = a k + 1 S ( S = Sl U { a k + 1 } )

SupS S , in both of the cases .

SupS S

( Q 1 5 ) Wr i t e d o w n d e t a i l s i n P r o o f o f t h e f o l l o w i n g

t h e o r e m .

A number u i s the supremum of a non- e m p t y s u b s e t S o f

i f f u s a t i s f i e s c o n d i t i o n s

1. S u s S , 2 . I f v < u , then there exis ts s* Ss u c h t h a t v < s * .

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P r o o f F o r w a r d p a r t A s s u m e u = S u p S

S u s S , i f v < u, as u i s SupS u i s l .u .b andh e n c e v i s n o t a n u p p e r b o u n d

s* S : v < s*

C o n v e r s e l y a s s u m e t h a t 1 & 2 h o l d

Then 1 u i s an upper bound of S a lso i f v i s any upperb o u n d o f S

T.S.T u v , le t i f poss ib le v v , w h i c h i s a c o n t r a d i c t i o n a s v i s u p p e r b o u n do f S .

v < u i s not poss ib le u v

SupS = u

S u p r e m u m a n d I n f i m u m o f r e a l v a l u e d f u n c t i o n s d e f i n e d

o n a n y r e g i o n .

L e t D b e a n y r e g i o n i n s p a c e ( N o t e S p a c e s t a n d s f o r ,2 ,3 ,. .n )L e t f : D

, b e a n y f u n c t i o n t h e n S u p f , t h e s u p r e m u m o f

t h e s e t f ( D )

Defined as f (D) = {f (x) |xD}

i . e S u p f = S u p f ( D )

C o m p l e t e n e s s p r o p e r t y o f -

E v e r y n o n - e m p t y b o u n d e d a b o v e s u b s e t o f

h a s a

s u p r e m u m .

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S o l v e d Q u e s t i o n s E x e r c i s e 2 . 4

( Q 1 ) S h o w t h a t S u p { 1 - 1/n : n

} = 1

S o l u t i o n Let x {1- 1/n | n } b e a n y e l e m e n tt h e n c l e a r l y x = 1 - 1/n < 1 , n x < 1 x {1- 1/n | n } 1 i s an upper bound of {1- 1/ n | n }

N o w l e t u b e a n y u p p e r b o u n d o f { 1 - 1 / n | n

}

To s h o w t h a t - 1 u , Let i f poss ib le 1 > u 1- u > 0

> 0

By archimedian proper ty there exis ts n : n > 1- u> 1 / n

1- 1 / n > u

L e t x = 1 - 1/n; then x {1- 1/n | n } a n d x > u ; w h i c h i s ac o n t r a d i c t i o n a s u i s a n

U p p e r b o u n d o f { 1 - 1/n | n } our assumpt ion tha t 1 > u i swrong 1 u

1 = Sup{1- 1/n | n

}

( Q 2 ) I f S = { | m, n } , f i n d I n f S a n d S u p S .S o l u t i o n L e t S 1 = { | n } & S2 = { | m }

T h e n S = S 1 + S 2 and Sup(S) = Sup(S1 ) + Sup(S 2 ). . ( A )


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Now x S , x v < u- 1/n x < u- 1/n x S

u- 1 / n i s a n u p p e r b o u n d o f S , w h i c h i s a c o n t r a d i c t i o na s u - 1/n cant be an upper bound.

U > v i s not poss ib le u v

Sup(S) = u

C o n v e r s e l y L e t S u p ( S ) = u

x u x S x u < u +1/n x s n u +1/n i s an upper bound of S n

Also n , u - 1/n i s not an upper bound ( u i sl . u . b a n d u - 1/n 0 a n d l e t a S = { a s | s S} . Prove tha t inf (aS)

= a i n f ( S ) , S u p ( a S ) = a S u p ( S ) .(b ) Let b < 0 , and le t bS = {bS | s S} . Prove tha t

i n f ( b S ) = b S u p (S ) , S u p ( b S ) = b I n f ( s ) .

S o l u t i o n - ( a ) C l a i m 1 a Inf (S) i s a lower bound of aS

Let x aS be any e lement then x = as , s S .

Now Inf(S) s s S a Inf (S) as s S a Inf (S) x x aS

a Inf (S) i s a lower bound of aS.

C l a i m 2 i f V is any lower bound of aS then v aInf(S) .

As v i s a lower bound of aS, therefore v as ,

s S


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v /a s s S v /a i s a lower bound of S

v /a Inf (S) v a Inf (S)

f rom c la im 1 a n d c l a i m 2 , w e h a v e

I n f ( a S ) = a I n f ( S )

Now we l l prove tha t Sup(aS) = aSup(S)

C l a i m 1 a S u p ( S ) i s a n u p p e r b o un d o f a S .

Let x aS (be any) then x = as , s S (1)

Also s Sup(S) s S as aSup(S) , s S (2)

From (1) & (2) we have x aSup(S) .

aSup(S) i s an upper bound of aS.

C l a i m 2 i f V is any upper bound of aS, then aSup(S) v

A s v i s a n u p p e r b o u n d o f a S

v as s S

v /a s s S

v /a i s an upper bound of S

And hence Sup(S) v /a

aSup(S) v

A n d h e n c e S u p ( a S ) = a S u p ( S )

(b) we l l f i r s t prove Inf (bS) = bSup(S)

C l a i m 1 b S u p ( S ) i s a l o w e r b o u n d of b S .

Let x bS x = bs , s S . (1)

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Now s Sup(S) s S bs bSup(S) s S ( b < 0) (2)

From (1) & (2) x bSup(S) bSup(S) i s lower boundo f b S .

C l a i m 2 i f v i s any lower bound of bS, then bSup(S) v

A s v i s a l o we r b o u nd of b S

v bs s S v /b s s S

v /b i s an u p p e r b o u n d o f S

Sup(S) v /b bSup(S) v

F r o m c l a i m 1 & c l a i m - 2 , w e h a v e

I n f ( b S ) = b S u p ( S )

Now we l l show tha t Sup(bS) = bInf(s ) .

C l a i m 1 w e s h a l l s h o w t h a t

b I n f ( S ) i s a n u p p e r b o u n d o f b S

Let x bS be any e lement then we have tha t, x = b s f r o ms S (1)

N o w a s b y d e f i n i t i o n o f I n f ( S ) , w e h a v e

Inf (S) s s S

bInf (S) bs s S (2) ( b < 0)

F r o m ( 1 ) & ( 2 ) w e h a v e

X bInf(S) x bS

x bInf(S) x bS


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C l a i m 2 b I n f ( s ) i s l e a s t u p p e r b o u n d o f b S .

L e t v b e a n y u p p e r b o u n d o f b S

To s h o w t h a t (bInf(S) v)

As v i s upper bound of bS v bs s S

v /b s s S v /b i s a lower bound of S

v /b Inf (s ) v bInf(S)

H e n c e p r o v e d .

G u i d e l i n e Q u e s t i o n s f r o m C h a p t e r 2 . 4 ( B a r t l e )

( Q 6 ) L e t A a n d B b e b o u n d e d n o n - e m p t y s e t s i n a n dA + B = {a + b : a A, b B}.

S h o w t h a t

( a ) S u p ( A + B ) = s u p ( A ) + S u p ( B )

( b ) i n f ( A + B ) = I n f ( A ) + I n f ( B )

S o l u t i o n L e t S u p A = m 1 & S u p B = m 2 .

( a ) To s h o w t h a t S u p ( A + B ) = m 1 + m 2

C l a i m 1 m 1 + m 2 i s a n u p p e r b o u n d o f A + B

Let x A + B (be any e lement)

x = a1 + b 1 , a 1 A, b1 B

Now a 1 m1 , b 1 m2

x m1 + m 2 m1 + m 2 i s a n u p b d o f A + B


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C l a i m 2 m 1 + m 2 i s l e a s t u p p e r b o u n d

L e t v b e a n y u p b d o f A + B ( T. S . T m 1 + m 2 v )

a + b v a A, b B

b v a b B , a A

v a i s na upbd of B a A

m2 v a a A

a v- m 2 a A v- m 2 i s a n u p b d o f A

m1 v - m 2

m1 + m 2 v

m1 + m 2 i s l . u . b o f A + B

Sup(A + B) = m1 + m 2 = s u p ( A ) + S u p ( B )

(b ) L e t i n f A = m 1 a n d i n f B = m 2 ( T. S . T i n f ( A + B ) = m 1 +

m 2 )

( 1 ) t h e n x A + B , x = a + b , a A , b B

x m1 + m 2 ( a m1 , b m2 )

m1 + m 2 i s a l o w e r b o u nd o f A + B

( 2 ) L e t v b e a n y l o w e r b o u n d o f A + B ( m 1 + m 2 v)

Then a + b v a A, b B a v b a A, b B

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m1 v b b B b v- m 1 b B v- m 1 i slower bound of B m2 v - m 1

m2 + m 1 v inf (A+B) = m1 + m 2

Q1 2 Given any x , s h o w t h a t t h e r e e x i s t s a u n i q u en s u c h t h a tn 1 x < n .

Q1 3 I f y > 0 , show tha t there exis ts n

s u c h t h a t

< y

Q 1 8 I f u > 0 , i s a n y r e a l n u m b e r a n y x < y, s h o w t h a t

t h e r e e x i s t s a r a t i o n a l n u m b e r r s u c h t h a t

x < r u < y.

S o l u t i o n 1 2 x

, Def ine E = {m

: m > x} ( then by A .P

E )

Also E , E By W.O.P , E has a leas t e lement( u n i q u e )

Say n E n > x (I )

A l so n - 1 E ( n i s leas t e lement in E )

n 1 x . ( I I )

F r o m ( I ) & ( I I ) n 1 x < n

S o l u t i o n 1 3 As y > 0 1 /y > 0 By A.P, n :n > 1 / y

n

: 1 /n < y ( I )

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Now we know n , n < 2 n 1 /2 n < 1/n (I I ) F rom ( I ) & ( I I ) 1 /2 n < 1/n < y 1 /2n < y , f o rsome n

S o l u t i o n 1 8 x < y and u > 0 1 /u > 0 x /u < y /u

By dens i ty theorem r Q :

x /u < r < y /u x < ru < y

Q1 4 S h o w t h a t t h e r e e x i s t s a + v e r e a l n u m b e r y : y 2 = 3

Q15 I f a > 0 , t h e n s h o w t h a t t h e r e e x i s t s a + v e r e a l

n u m b e r z : z 2 = a

Q 1 6 S h o w t h a t t h e r e e x i s t s a + v e r e a l n u m b e r u s u c h t h a t

u 3 = 2

S o l u t i o n 1 4 D o a s q u e s t i o n 1 5 b e l o w ( r e p l a c e , z b y y a n d ab y 3 )

S o l u t i o n 1 5 . Define S = {S : s 0 a n d s 2 < a }C a s e 1 0 < a < 1 C h o o s e 0 < s < a , t h e n s 2 < a 2 < a( a < 1)

s2

< a a l s o s > 0

s S s

Now we l l show tha t S i s bounded above Let s S (bea n y ) t h e n s 2 < a < 1 ( a < 1)

S i s bounded above by 1

C a s e 2 a > 1 , s S then s2 < a a n d s > 0

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As a > 1 12 = 1 < a a lso 1 > 0 1 S S

And we l l show S i s bounded above by a .L e t i f p o s s i b le s S : s > a s2 > a 2 > a ( a > 1)

s2 > a (I)

But s S s2 < a . ( I I )

( I ) & (I I ) c o n t r a d i c t e a c h ot h e r S i s b d d a b o v e b y a

in each case S i s non empty bdd above se t Byc o m p l e t e n e s s p r o p e r t y S h a s s u p r e m u m

L e t S u p ( S ) = Z , C l a i m - Z 2 = a

C l a i m 1 Z 2 a , Let i f poss ib le Z2 < a

(we l l f ind m

: z + 1 /m S i .e (Z+1/m)2 < a , w h i c hi s c e r t a i n l y a c o n t r a d i c t i o n a s S u p ( S ) = Z )

C o n s i d e r - ( Z + 1 / m ) 2 = Z 2 + 1 / m 2 + 2 z / m

= Z 2 + ( 2 z + 1) (1/m2 < 1 / m )

N o w ( Z + 1 / m ) 2 < a i f Z 2 + ( 2 z + 1 ) < a

i f (2z + 1 ) < a - Z 2

m > ( ) ( a - Z 2 > 0 )A n d s u c h a n m e x i s t s b y A . P

m : ( Z + 1 / m ) 2 < a z + 1 /m SAs Sup(S) = Z z + 1 /m < Z

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1 /m 0

(1 0) not poss ib le Z2 aC l a i m - 2 Z 2 a , Let i f poss ib le Z2 > a

(we l l f ind m : z - 1/m S i .e (Z- 1 / m ) 2 < a )C o n s i d e r - ( Z - 1 / m ) 2 = Z 2 + 1 / m 2 - 2 z / m > Z 2 -

2 z / m

N o w ( Z - 1 / m ) 2 > a i f Z 2 - 2 z / m > a i . e i f m > ( )

( Z2 > a )

A n d s u c h a n m e x i s t s b y A . P

m : ( Z - 1 / m ) 2 > a Now Let y S (be any )T h e n y 2 < a < ( Z - 1 / m ) 2 y < Z - 1/m y S

which i s not poss ib le as Sup(S) = Z i s leas t up . Bd of S Z2 a a nd hence Z = a

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S o l . 1 6

S = { s

: s > 0 , s 3 < 2 }

(1 ) Clear ly S ( 1 >0 and 13 =12 s3 > 8 . . . . . . . . . . . . ( i )

Now sSs3

< 2 . . . . . . . . . . . . ( i i )

( i ) a n d ( i i ) c o n t r a d i c t e a c h o t h e r

s i s bounded above by 2 .

By comple teness proper ty S has a Supremum say u i .eS u p ( s ) = u

C l a i m u 3 = 2

C l a i m - 1 u 3 u )

= u 3 +

Now < 2 I f u 3 + ( 1 + 6 u 2 ) < 2

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m> ( 2-u 3 > 0 )

S u c h a n m e x i s t s b y A . P.

m : 2

(Wel l c la im m : > 2 )

C o n s i d e r = u 3 +

> u 3

> u 3

Now m 3 >m < >

> u3

> u 3 ( 1 + 3 u 2 )

> 2 I f u 3 - ( 1+3u 2 ) > 2

m > ( u3 - 20)

a n d s u c h a n m e x i s t s b y A . P.

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m : > 2Let s S ( be any )

t h e n s 3 < 2 < s < u- s S

u - i s a n u p p e r b ou n d o f S

As Sup(s) = u u u-

0 - 1 w h i c h i s n o t p o s s i b l e

u3 2

a n d h e n c e b y c l a i m 1 a n d c l a i m 2 u 3 = 2

S e c t i o n 2 . 4

S o l 6 . I f A a n d B a r e b o u n d e d s u b s e t s o f T h e n ( 1 ) S u p ( A + B ) : S u p A + S u p B

( 2 ) I n f ( A + B ) : I n f A + I n f B

Where A+B : {a+b/aA,bB}

P r o o f . L e t S u p A = m 1 a n d S u p B = m 2

To s h o w t h a t S u p ( A + B ) : m 1 +m 2

C l a i m 1 : m 1 + m 2 i s a n u p p e r b o u n d o f A + B

Note : Q.E.D stand for quod erat demonstration ,

which is Latin for which was to be demonstrated.


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Let xA+B (be any e lement)

xa1 +b 1 ; a 1A , b1B .(1)

A l s o , M 1 = Sup(A) am1 aA

M 2 = Sup(B) bm2 bB

a1 m1 a n d b 1 m2

a1 + b 1 M1 + M 2 .(2)

F r o m ( 1 ) a n d ( 2 ) :

a1 + b 1 M1 + M 2 xA+B

xM1 + M 2 xA+B

M 1 +M 2 a r e u p p e r b o u n d o f A + B

C l a i m 2 : I f v i s a n y u p p e r b o u n d o s A + B t h e n M 1 + M 2 v A s v = u p p e r b o u n d o f A + B

a_bv aA and bB

bv- a aA and bB

v- a i s a n u p p e r bound of B aA

M2 v a aA(sup(B)


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Sup(A+B) : M1 + M 2 = S u p ( A ) + S u p ( B )

S o l 7 . f ,g :X

To p r o v e :

( i ) Sup{f(x) +g(x) : xX} Sup{f(x) :xX} +Sup{g(x) :xX}

( i i ) I n f { f ( x ) + g ( x ) : xX} Inf{f(x) :xX} + Inf{g(x) :xX}

S o l ( i ) : Let Sup{f(x) : xX} = M1 and Sup{g(x) :nX}=M2

C l a i m : M 1 + M 2 i s an upper bound of {f (x)+g(x) : xX}

As : M 1 = Sup{f(x) :xX}

M 2 =Sup{g(x) :xX} g (x)M2xX

f (x) +g(x) M1 +M 2 x M 1 +M 2 i s a n d upper bound of f (x)+g(x) :xX}

Sup{(f (x)+g(x)}:xX}

M1 + M 2

Sup{f(x) :xX} + Sup{g(x) :x}( i i ) : Let Inf{f (x) : xX} = M1 and Inf{g(x) :nX}=M2

C l a i m : M 1 + M 2 i s an lower bound of {f (x)+g(x) : xX}

As : M 1 = Inf{f(x) :xX} f (x)M1xX

M 2 =Inf{g(x) :xX} g (x)M2xX

f (x) +g(x) M1 +M 2 x M 1 +M 2 i s and lower bound of f (x)+g(x) :xX}


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Inf{( f (x)+g(x)}:xX}

M1 + M 2

Inf{f (x) :xX} + Inf{g(x) :x}S o l 8 : X = ( 0 , 1 )

Y = ( 0 , 1 )

h:XXY

h ( x , y ) = 2 x + y

a ) xX,f ind f (x)=Sup{h(x ,y) :yY} t h e n f i n d Inf{f (x) :xX}

b ) yY,f ind g(y)=Inf{h(x ,y) :xX} then f ind Sup{g(y) :yY}

a) f (x) = Sup{h(x ,y) :yY}=Sup{2x+y|yY} =2x+Sup{y:yY}

= 2 x + S u p ( ( 0 , 1 ) ) = 2 x + 1

f (x)=2x+1 .(1)

Inf{f (x) :xX}=Inf{2x+1 |xX}=Inf{2x+1:o


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g(y)=y

Now Sup{g(y) :yY}=Sup{y:y (0 ,1)}

= S u p ( 0 , 1 ) = 1

S o l 9 : X = ( 0 , 1 ) = Y

h:XXY h ( x , y ) =

{

for any xX

a ) f (x)=Sup{h(x ,y) :yY}

=Sup({h(x ,y) :x x }( then by A.P. E )

Also, E, E B y W. O . P , E h a s a l e a s t e l e m e n t o f E


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n>x(1)

N o w , n - 1E ( n i s leas t e lement of E)

n- 1x.. (2)

F r o m ( 1 ) a n d ( 2 )

n- 1x 0

By A.Ap n n >

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To s h o w t h a t z>0 :

z 2 = a , a > 0

D e f i n e :

S={s: s 0 a n d s 2 < a }C a s e 1 : 0 < a < 1

C h o o s e ; 0 < s < a

T h e n , s 2 < a 2

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s2 < sS

s < sS

z- i s a n d U . B . of S

zz- = ( Sup(s) = z)

0 0- 1 01

which i s an cont radic t ion

z2>a i s not poss ib le

z2a

B y c l a i m 1 a n d c l a i m 2 w e h a v e

za and z2a

z2 =a

C a s e 2 : S={S| S > 0 , S 2 a} I f a > 1 , t h e n , 1 > 0

And 1 2 = 1 a

1S

s

Now we l l show S i s bounded above by a

Let id poss ib le spme sS:

s>a s2 >a 2 .(1)


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i . e . I f < a - z 2

m> ( (2z+1) / (a- z 2 ) ) (a- z 2 > 0 )


m:( z + 2 a

Wel l f ind m

( z + 2 >a

CONS. :

( z + 2 = z 2 + 1 / m 2 2 z / m > z 2 -

N o w, ( z + 2 > a I f z 2 - >a

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i . e . I f < z 2 - a m > (2z/z2 - a) (z2 - 1 )

a n d s u c h a n m e x i s t s b y A . P a

m : ( z+ 2 > aS o l 1 6 : S={s: s > 0 ; s 3 < 2 }

(1 ) Clear ly, s ( 1>0 and 13 = 1 < 2

1S)

A l s o , S i s b o u n d e d a b o v e b y 2

I f S i s n o t b o u n d e d a b o v e b y 2

sSLs>2 s3 >8... . (1)

Now, sS s3 2 s2 sS

S i s bounded above

By comple teness proper ty, S has a supremum

L e t S u p ( S ) = u

C l a i m u 3 = 2

C l a i m 1 : U 32

L e t i f p o s s i b l e U 3

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C o n s i d e r ( u + ) 3 = u 3 + 1 / m 3 + 3u 2 / m + 3 u / m 2 < 2

1 u2 > u )

= u 3 + 1 / m ( 1 + 3 u 2 + 3 u 2 )

= u 3 + 1 / m ( 1 + 6 u 2 )

N o w, ( u + ) 3 (1+6u2 ) / ( 2 - u 3 ) (2- u 3 > 0 )


m:(u+ ) 3 < z

(u+ ) S

u+ u

u+ u

< 0 10 which i s a cont radic t ion

u3 < 2 i s n o t p o s s i b l e

C l a i m 2 : u 22

L e t i f p o s s i b l e u 3 >2

Wel l c la im m

( u - ) 3 2

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C O N S . ( u - ) 3 = u 3 - 1 / m 3 - 3u 2 / m + 3 u / m 2

> u 3 - 1 / m 3 - 3u 2 / m + 3 u / m 2

(u- ) 3 > u 3 1 / m 3 (3u 2 / m )

N o w, m 3 > m

1 /m3 < 1 / m

- 1 / m

3

> - 1 / m

(u- ) 3 > u 3 1 / m - ( 3 u 2 / m )

= u 3 1 /m (1+3u 2 )

N o w, ( u - ) 3 > 2

I f u 3 - ( 1 + 3 u 2 ) > 2

( 1 + 3 u 2 ) (1+3u2 ) / ( u 3 - 2 )

S u n c h a n m e x i s t s :

B y A . P. : m

( u - ) 3 > 2

Let sS (be any)

s3

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u - i s a n U B o f S

uu- 0- 1 01 (which i s not poss ib le )

S o l 1 8 : G i v e n : x < y

A n d u > 0 >0

<

By dens i ty theorem : r

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T h e o r e m : ( S u m T h e o r e m 0

L e t X x a n d Y y ( i . e l i m x n = x , l i m y n = y )

T h e n X + Y x + y ( i . e l i m ( x n + y n ) = x+y)

P r o o f : Let >0 be g iven

( to f ind m

L | ( x n + y n ) = (x+y) | 0 m1 : | x n - x | < nm1 ..(2) A l s o l i m y n = y fo r > 0 m2

: | y n - y | < nm2 ..(3)

L e t m = m a x { m 1 , m 2 } then nm, |xn - x | < ; | y n - y | < ..(5)

U s e ( 5 ) i n ( 1 ) w e g e t

| ( x n + y n ) = - ( x + y ) | < + nm

| (xn +y n ) (x+y) | < nm l im (xn + y n ) = x + y

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| xn y n xy | |yn | | x n x | + | x | | y n y | .(1)

N o w ( y n ) i s cg t . (yn ) i s bounded M>0 : |yn | M n. . ( 2 )U s e ( 1 ) i n ( 2 )

|xn y n xy | M|xn x | + | x | | y n y |..(3)

L e t M * = m a x { M , | x | } t h e n | x n y n xy |M* ( |xn x | + | y n y | )..(4)

N o w a s l i m x n = x for >0 m1 : | x n - x | < nm1 .. . . . (5)

A l s o l i m y n =y for >0 m2 : | y n - y | < nm2 .(6)

L e t m = m a x { m 1 , m 2 | then nm we have

| x n - x | < , | y n - y | ..(7)

U s e ( 7 ) i n ( 4 ) w e o b t a i n

| x n y n x y | < M * ( + ) nm

| xn y n xy | 0 and m1:| y n |>k nm1


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z


P r o o f : As y0 |y |>0 choose 00 m1

:

| y n - y |k n>m1

Q U O T I E N T T H E O R E M

L e t X x a n d Y y wi th Y0 and y0, T h e n

( i . e I f l i m x n = x , l i m y n = y, y n 0 n a n d y 0t h e n l i m ( x n / y n ) = x / y

P r o o f : Wel l show l im (1 /yn ) = 1 / y , t h e n b y p r o d u c t

t h e o r e m


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z


P r o o f : Let >0 be g iven

A s l i m x n = l

m1 : l - < x n


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l im xn = 0

C a s e 2 : I f x0

Let >0 be g iven ( to f id m

: | x n - x | x 1 / (xn + x) 1 /x(2)

F r o m ( 1 ) a n d ( 2 ) w e h a v e

|x n - x | | x n - x |(3) L i m x n =x for x >0 m:|x n - x| x nm |xn - x| nm l im xn = x

M O D U L U S T H E O R E M

L e t l i m x n = x t h e n l i m | x n | = | x |

P r o o f : Let >0 be g iven ( to f ind m

: | | x n | - | x | |

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z


Q4 . L e t x 1 = 1 , x n + 1 = n S h o w t h a t ( x n ) i s convergent f ind i t s l imi t

S o l n : x 1 = 1 , x 2 = = 1 . 7 3 2x 3 = = = = .

We c a n s e e x 1 < x 2 < x 3

S o w e o b s e r v e t h a t ( x n ) i s ( increas ing)

Now we l l prove (xn ) i s i . e . x n xn + 1 n. ( 1 )T h e r e s u l t i s t r u e f o r n = 1 x2 = 1 . 7 3 2 > x 1

N o w l e t u s a s s u m e t h a t r e s u l t i s t r u e f o r n = k

xk xk + 1 (2)

Wel l show tha t resul t i s t rue for n=k+1 ( i . e . x k + 1 xk + 2 )

A s x k xk + 1 b y ( 2 )

2+xk 2+xk + 1 xk + 1 xk + 2

By P.M.I xn xn + 1 n

xn i s ..(*)

Now we l l show tha t (xn ) i s b o u n d e d a b o v e b y 2

( i . e x n 2 n)F o r n = 1 , x 1 =1


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z


A s s u m e t h a t r e s u l t i s t r u e f o r k

xk 2(3)

Wel l show resul t i s t r u e f o r n = k + 1 ( i . e x k + 1 2)

A s x k 2 2+xk 4

= 2 2 xk + 1 2

resul t i s t rue n

B y P. M . I

xn 2 n ( x n ) i s b o u n d e dB y M . C . T ( x n ) i s c o n v e rg e n t

L e t l i m i t = x a s x n + 1 = l im xn + 1 =

x = x2 = 2+x x2 - x - 2 = 0

x2 2 x + x - 2 = 0

(x- 2 ) ( x + 1 ) = 0

x=2 , x=-1B u t x n 1 n l im xn 1 x=- 1 i s n o t p o s s i b l e

x=2 l im xn =2

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z


Q5: L e t y 1 = , p > 0 , y n + 1 = , n.S h o w t h a t ( y n ) i s c o n v e rg e n t . f i n d i t s l i m i t .

S o l u t i o n :

y 1 = , y 2 = = > = y 1 y1 = Wel l show tha t (yn ) i s ( i .e yn + 1 yn n

)

F o r n = 1 , a s y 2 y1 resul t i s t rue

A s s u m e t h a t r e s u l t i s t r u e f o r n = k

yk + 1 yk .(1)

As y k + 1 yk

p+ yk + 1 p+ yk yk + 2 yk + 1 resul t i s t rue foe n+k+1 too and hence by P.M.I yn + 1 yn n

(yn ) i s

Now we l l shoe (yn ) i s b o u n d e d

A s y n + 1 yn yn (p+yn ) yn 2

yn 2 y n p 0

(yn - ) ( y n - ) < 0

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z


A s y n y1 = yn > >0 y n ( < 0 ) yn < n ( y n ) i s bounded (yn ) i s

c o n v e rg e n t

N o w l e t l i m y n = y

As y n + 1 = l im yn + 1 = y= y2 - y - p = 0

y = As yn > 0 n

l i m y n 0 y 0

Now y = , y = < 0 ( > 1 1- 0 , L e t z n + 1 =

n

S h o w t h a t ( z n ) c o n v e rg e s , f i n d i t s l i m i t

S o l : z n + 1 = zn + 1 2 = a + z n

zn 2 = a + z n - 1

(zn + 1 ) 2 - z n 2 = ( a + z n ) - ( a + z n - 1 )

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(zn + 1 ) 2 - z n 2 = z n z n - 1 (1)

I f z n > z n - 1 (zn + 1 ) 2 z n 2 > 0 zn + 1 > z n

And z n < z n - 1 zn + 1 < z n

i . e z n + 1 > z n > z n - 1 n a n d n > 1or z n + 1 < z n 1 (zn ) i s or (zn ) i s zn i s m o n o t o n e

Now z n + 1 > z n o r z n + 1 > z n

> z n o r < z n a+zn > z n 2 o r a + z n < z n 2

zn 2 z n a < 0 o r z n 2 z n a > 0

(zn - ) ( z n - ) < 0 o r i t s > 0B u t ( z n - ) 0 ( < 0 a n d z n > 0 ) (zn - ) < 0 o r ( z n - ) > 0 zn < n

o r z n > n

i . e . I f z n + 1 > z n n t h e n z n < n (zn ) i s bounded and sequence (zn ) i s c g t

Now i f z n + 1 < z n t h e n z n > n (zn ) i s monotone and bounded (zn ) i s c g t

90/141

z


L e t l i m z n = z , a s z n + 1 = l im zn + 1 =

z = z2 = a + z z2 - z - a = 0 z = ( zn 0 n l i m z n 0 z0 )Q 7 : x 1 = a > 0 , x n + 1 = x n + for n;

D e t e r m i n e I f ( x n ) c o n v e rg e s o r d i v e rg e s .

S o l : C l e a r l y x n >0 n x n + 1 x n = >0 n xn + 1 x n >0 n x n + 1 > x n (xn ) i s , I f (xn ) i s c g t t h e n i t h a s t o b o u n d e d

b>0 : xn b n

Now (x n ) i s cg t l im (xn + 1 x n ) =0..(1)

A l s o i n t h a t c a s e x n 0(2)

( 1 ) a n d ( 2 ) a r e c o n t r a r y t o e a c h o t h e r (xn ) c a n b ec o n v e rg e n t ( x n ) i s d i v e rg e n t

Q11 : L e t y n = + + .+ n S o l : y n + 1 y n = + - >0 n

yn + 1 y n > 0 n

( y n ) i s (1)

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A l s o y n < + + + yn < 1 n( y n ) i sbounded.. (2)

B y ( 1 ) a n d ( 2 ) a n d u s i n g M . C . T ( y n ) i s b g t

Q12 : x n = + + + + n C l e a r l y x n + 1 x n = > 0 n (xn ) i s ..(1)

N o w o b s e r v e t h a t

k2

- k2

xn 1+ (1- ) + ( - ) + ( - ) + .. +( - )

xn 2- < 2 n

xn i s b o u n d e d xn i s c g t ( b y M . C . T )

Q 1 3 :

( a ) l i m = l im ( )

= l i m l i m

= e - 1 = e

( b ) l i m = l i m

= = e 2


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z


( c ) l i m = = = e

( d ) l i m =

= = e - 1 - =

E x a p m l e : L e t a > 0 , L e t S 1 > 0 , a n d S n + 1 = ( S n + ) f o r

n

D e t e r m i n e i f ( s n ) i s c o n v e rg e n t

S o l n : A s S n + 1 = (S n + )

2Sn + 1 = (S n + )

2Sn + 1 S n = S n 2 + a

Sn 2 - 2S n + 1 S n + a = 0

N o w t h i s e q n m u s t h a v e r e a l r o o t s

Discreminent 0

4 S2 n + 1 4 a > 0 S2 n + 1 > a S n + 1 >

n

Sn > m i n { , S1 } n (Sn ) i s bounded below..(1)

Now we l l show i t s decreas ing

S n S n + 1 = S n - (S n + ) = 0 2 (sn ) i s decreas ing.(2)

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z


F r o m ( 1 ) a n d ( 2 ) ( S n ) i s c o n v e rg e n t ( B y M . C . T ) .

L e t l i m S n =a

S= S= + S

S= S2 = a s+ l im Sn =

D e f i n i t i o n : ( S U B S E Q U E N C E )

L e t X = ( x n ) b e a s e q u e n c e o f r e a l n u m b e r s a n d l e t n 1 < n 2 0 be g iven, to f ind m

:

| | < Km

A s l i m x n = x m:| x n - x |


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z


Q8 : G i v e n : ( 1 ) ( a n ) i s (bn ) i s sequence of rea l

( 2 ) a n bn n

S h o w t h a t : l i m a n l im bn a n d I f l i m a n = and l im bn = t h e m a n bn

S o l : a n b1 n (an ) i s and bounded above by b1

(an ) i s c g t . ( B y M . C . T )

A l s o l i m ( a n ) = S u p { a n : n|L e t l i m a n = (1)

Now (b n ) a lso a1 bn n (bn ) i s m o n o t o n e a n d b o u n d e d

(bn ) i s c o n v e rg e n t , a n d l i m b n = I n f { b n L n}L e t l i m b n = .(2)

We h a v e t o s h o w : .

As =Sup{an :n

} a n n

. ( 3 )

= Inf{bn : n| b n .(4) Now a n bn n a n - b n 0 l im (an - b n )0

=0 ..(5)

D r o m ( 3 ) , ( 4 ) , a n d ( 5 ) w e g e t


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An bn n

Q9 : A i s i n f i n i t e s u n s e t o f , t h a t i s b o u n d e d a b o v eLet u = Sup(A) (T.S .T (xn ) w i th x nA and l im xn = u )

S o l : A s S u p ( A ) = u u- i s not an upper bound of A n

n

x nA : xn > u- (1)

A l s o x nA xn u n (S u p ( A ) = u )u- xn u n , b y s q u e e z e t he o r e m l i m x n = u

B l o z a n o w e i e r s t r a s s t h e o r e m

A b o u n d e d s e q u e n c e o f r e a l n u m b e r s h a s a c o n v e rg e n t

s u b s e q u e n c e

P r o o f : B y m o n o t o n e s u b s e q u e n c e t h e o r e m i f ( x n ) i s a

s e q o f r e a l s t h e n ( x n ) h a s a m o n o t o n e s u b s e q u e n c e

s a y X 1 a l s o ( x n ) i s b o u n d e d

X1 i s b o u o n d e d

As X1 i s m o n o t o n e b y M . C . T X 1 i s c o n v e rg e n t


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z


T h e o r e m I f X = x n i s a s e q u e n c e o f r e a l n u m b e r s t h e n X x i f f e a c h o f

i t s s u b s e q u e n c e s c o n v e rg e s t o x .

P r o o f : i f ( x n ) x t h e n w e h a v e p r o v e d e v e r y s u b s e q u e n c e

of (x n ) a l s o c o n v e rg e s t o z .

N o w l e t e v e r y s u b s e q u e n c e o f X c o n v e rg e s t o x t h e n X

b e i n g a s u b s e q o f i t s e l f c o n v e rg e s t o x

E x r e s i c e 3 . 4

Q 1 : L e t ( x n ) b e a s e q d e f i n e d a s

x 2 n - 1 = 2 n - 1 a n d x 2 n =

(xn ) = (1 , , 3 , , 5 , .) C l e a r l y ( x n

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