analysis book bsc(hons) mathematics delhiuniversity
TRANSCRIPT
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
1/141
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
2/141
z
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
3/141
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
4/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 )
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
5/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 .
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
6/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 .
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
7/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
8/141
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
9/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 | =
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
10/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 |
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
11/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
(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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
12/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
(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 |
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
13/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
| 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 | < } .
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
14/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 .
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
15/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
16/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 .
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
17/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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.
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
18/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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.
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
19/141
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
20/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
]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 .
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
21/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 .
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
22/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
23/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
] 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 .
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
24/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
25/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 )
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
26/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 }
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
27/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
28/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
29/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
30/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
31/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 +
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
32/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
( 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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
33/141
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
34/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
( 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)
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
35/141
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
36/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 )
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
37/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
|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)
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
38/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 )
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
39/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 |
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
40/141
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
41/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 < u x : 0 < x < u x S1 a n d x < u , w h i c h i s a c o n t r a d i c t i o n a s u i s alower bound o f S 1
u > 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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
42/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
(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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
43/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
( 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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
44/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 )
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
45/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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)
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
46/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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)
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
47/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 or x = u x u
2 . Let K be any upper bound of S U {u} (T.S .T u K)
As K is upbd of S U {u} x K x S U {u} as uS U {u} u K
Sup {S U {u} } = u = Sup{ s* , u }
C a s e 2 I f s * > 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 }
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
48/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 }
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
49/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 * .
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
50/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 < u , then by condi t ion 2 s*
S : s* > 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 .
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
51/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 )
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
52/141
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
53/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 < u n )
( Q 4 ) L e t S b e a n y n o n e m p t y b o u n d e d s e t i n ;( a ) L e t a > 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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
54/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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)
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
55/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
56/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
57/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
58/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 )
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
59/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
60/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
61/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
62/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
63/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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.
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
64/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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.
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
65/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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)
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
66/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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}
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
67/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
68/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
69/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 >
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
70/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
71/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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)
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
72/141
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
73/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
i . e . I f < a - z 2
m> ( (2z+1) / (a- z 2 ) ) (a- z 2 > 0 )
S u c h a n m e x i s t s b y A . P.
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
74/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
75/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
C o n s i d e r ( u + ) 3 = u 3 + 1 / m 3 + 3u 2 / m + 3 u / m 2 < 2
< u 3 + 1 / m + 3 u 2 / m + 3 u 2 /m (u>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 )
S u c h a n m e x i s t s b y A . P.
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
76/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 ) < u 3 - 2
m > (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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
77/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
78/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
79/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
D I F F E R E N C E T H E O R E M
L e t X x , Y y t h e n X - Y x - y
( i . e i f l i m x n = x , l i m y n = y, t h e n l i m ( x n - y n = x - y )
P r o o f : L e t > 0 b e g i v e n ,( to f ind m : | ( x n - y n ) - ( X - Y ) | 0 be g iven ( to f ind m : | x n y n xy | < nm) C o n s i d e r | x n y n x y | = | x n y n x y n + xy n x y |
| xn y n x y | = | ( x n x ) y n + x ( y n y ) |
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
80/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
| 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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
81/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
82/141
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
83/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
P r o o f : Let >0 be g iven
A s l i m x n = l
m1 : l - < x n
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
84/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 | |
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
85/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
86/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
87/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
88/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 )
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
89/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
(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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
90/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 < b xn + 1 x n =
l im (xn + 1 - 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)
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
91/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
92/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
( 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)
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
93/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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 |
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
95/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
96/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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
-
8/11/2019 Analysis Book BSc(HONS) MATHEMATICS DELHIUNIVERSITY.
97/141
z
Notes prepared by sanjeev kumar shukla 9971245238 or 8750558444
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