Solution To 4 Advanced Calc Proofs

Content type

User Generated

Subject

Mathematics

Type

Homework

Uploaded By

creary0999

Pages

Rating

Question 4.63:

a) Firstly, since

is an increasing function on

 

,ab

and

12 n

a x x x b    

, by definition, we

have

( ) ( )

1kk

f x f x

+  −

for all

1,2, , 1kn=−

and

( ) ( )

,f a f x+  −

and

( ) ( )

f x f b+  −

Rearranging the terms in the sum on left hand side yields

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

1 1 2 2

1 1 2 1

k k n n

n n n

f x f x f x f x f x f x f x f x

f x f x f x f x f x f x

f x f x

−

+ − − = + − − + + − − + + − −

       

       

= − − + + − − + + + − − + +

   

   

 − − + +



( ) ( )

f b f a − − +

b) Denote

 

( ) ( )

S x a b f x f x



=  + − − 





From the results in part (a) and since

is bounded on

 

,ab

, we can deduce

( ) ( ) ( ) ( )

x S x S

f x f x f b f a



  + − −  −  







which is possible if

is finite. If

is a discontinuity of

, we will have

( ) ( )

f x f x−  +

, and hence

the set of discontinuities is given by

 

( ) ( )

 

, : 0

S x a b f x f x



=  + − − 

Since countable union of finite sets

is countable, it follows that the set of discontinuities of

countable.

c) Since the set of discontinuities of

is countable, it follows that the set of discontinuities of

has

measure

. But any open subinterval of

 

,ab

has non-zero measure. Hence, it must follow that

has

points of continuity in every open subinterval of

 

,ab

Question 7.32:

a) Each

is the finite union of closed intervals; therefore,

is a closed set for all

k

. Then,

CA=

is also a closed set. Moreover,

is bounded since

 

0,1C 

. By Heine-Borel theorem,

compact.

( )

:

Let

xC

. Denote

1 2 3

0.x a a a=

.Consider

. We have that

belongs to either







(in

which case let

0a =

) or







(in which case let

2a =

). Now, consider

, let

0a =

belongs

to the leftmost subinterval of the interval of

to which

is currently belong, and let

2a =

belongs to the rightmost subinterval of the interval of

to which

is currently belong. By continuing

this process, we end up with

 

0,2

a 

for all

( )

:

Denote





, where

 

0,2

a 

Locate

 

0,1

as the following: Choose the leftmost subinterval of

0a =

, or choose the

rightmost subinterval of

2a =

. The interval of

we have chosen will be subdivided into three

subintervals when we form

, and choose the leftmost (or rightmost) subinterval if

0a =

(or

2a =

Continuing this way, we can see that

xA

for all

as long as

 

0,2

a 

and hence

x A C



=

c) Proof by contradiction. Suppose

is countable, and we can list its elements

 

1 2 3

, , ,C x x x=

Consider the ternary expansion of those



. Let

1 11 12 13

2 21 22 23

1 2 3

n n n n

x a a a

where

 

0,2

a 

for all

,ij

Now, consider the number

1 2 3

0.y bb b=

where

b =

a =

and

b =

a =

We have

yx

since

i ii

ba

for all

1,2, , ,in=

. This implies

yC

, but this is a contradiction

since

 

0,2

b 

for all

i

implies

yC

Therefore,

is an uncountable set.

d) Note that

is also the characteristic function of

End of Preview - Want to read all 6 pages?

Access Now

Unformatted Attachment Preview

Question 4.63: a) Firstly, since f is an increasing function on  a, b and a  x1  x2  have f ( xk + )  f ( xk +1 − ) for all k = 1, 2,  xn  b , by definition, we , n − 1 and f ( a + )  f ( x1 − ) , and f ( xn + )  f ( b − ) Rearranging the terms in the sum on left hand side yields n   f ( x + ) − f ( x − ) =  f ( x + ) − f ( x − ) +  f ( x + ) − f ( x − ) k k =1 k 1 1 2 = − f ( x1 − ) +  f ( x1 + ) − f ( x2 − )  +  − f ( x1 − ) + f ( xn + ) 2 +  f ( xn + ) − f ( xn − )  +  f ( xn −1 + ) − f ( xn − )  + f ( xn + )  f (b −) − f ( a + )   b) Denote S n =  x   a, b  : f ( x + ) − f ( x − )  1  n From the results in part (a) and since f is bounded on  a, b , we can deduce 0 1  n    f ( x + ) − f ( x − )  f (b ) − f ( a )   xSn xSn which is possible if S n is finite. If x is a discontinuity of f , we will have f ( x − )  f ( x + ) , and hence the set of discontinuities is given by n S n =  x   a, b  : f ( x + ) − f ( x − )  0 Since countable union of finite sets Sn ' s is countable, it follows that the set of discontinuities of f is countable. c) Since the set of discontinuities of f is countable, it follows that the set of discontinuities of f has measure 0 . But any open subinterval of ? ...
Purchase document to see full attachment