UNIVERSITY OF TORONTO

DEPARTMENT OF MATHEMATICS

MAT 237 Y - MULTIVARIABLE CALCULUS

FALL-WINTER 2005-06

ASSIGNMENT #2. SOLUTIONS

 

    

1.  Question 1-21 from Spivak:

a) If  A  is closed and  x A , prove that there is a number  d > 0  such that

  for all  y A .

Proof:

x A x R n A . The set  R n A  is open because  A  is closed, so there is an open rectangle  W  such that  x W R n A .

Let  W = ( a 1 , b 1 ) ( a 2 , b 2 ) ( a n , b n )  and let  x = ( x 1 , x 2 , , x n ) .

We can take

d = min { x 1 a 1  ,  b 1 x 1x 2 a 2  ,  b 2 x 2  , ,  x n a n  ,  b n x n  } .

Notice that  d > 0  and for all  yy A    y    .

b) If  A  is closed,  B  is compact, and  A B = , prove that there is  d > 0  such that    for all  y A  and  x B .

Proof:

For every  x B  we can find  d ( x ) > 0  as in part (a).

Also, for every  x let  U ( x )  be the open sphere with centre  x  and radius  d ( x ) / 2 .

The collection of all these open sets  U ( x )  is an open cover of the compact set  B .

So, there exists a finite subcollection of such open sets  U ( x 1 ) , U ( x 2 ) , , U ( x m )  which also covers  B .

We can take  d = ( min { d ( x 1 ) , d ( x 2 ) , , d ( x m ) } ) / 2 .

Notice that  d > 0  and that for all  x Bx  belongs to at least one of the sets  U ( x 1 ) , U ( x 2 ) , , U ( x m ) , say  x U ( x k ) .

Now, for every  y A , we have:

y x + x x k y x k d ( x k )  but  x x k d ( x k ) / 2

because  x U ( x k ) , therefore  y x d ( x k ) / 2 d .

c) Give a counterexample in  R 2  if  A  and  B  are closed but neither is compact.

Solution:

Take  A = { ( n , 0 ) n Z }  and  B = { ( n , 1 / n ) n Z , n 0 } ,

then  ( n , 0 ) ( n , 1 / n ) = 1 / n  which approaches  0  as  n .

 

2. Define the  closure  of  A  as

 = { x For every open rectangle  U  containing  xU A } .

a) Show that   = A bd A .

Proof:

If  x A  or  x bd A  then for any open rectangle  U ,

( x U ) ( U A ) , therefore  x  . So,  ( A bd A )  .

Now, if  x   then  x ext A  ( the exterior of  A ) ,

therefore  x ( A bd A ) . So, we also have    ( A bd A ) .

That is:   = ( A bd A ) .

b) Prove  bd A =    = bd ( R n A ) .

Proof:

First notice that  bd A = bd ( R n A ) .

So, we will only prove that  bd A =    .

Now, using our result from (a):

   = ( A bd A ) ( ( R n A ) bd ( R n A ) )

= ( A bd A ) ( ( R n A ) bd A ) = ( A ( R n A ) ) bd A

= bd A = bd A .

 

3. Show that every bounded set in  R  has a least upper bound.

Proof:

First notice that the above property is sometimes used as the Completeness axiom to identify  R as a complete ordered field. So, in order to produce a proof for the given property, we will have to assume that the completeness of  R  has been given by another equivalent axiom. For example: in  R , every bounded and monotonic sequence is convergent.

Let  A  denote a bounded set in  R  and let  B = { y R y  is an upper bound of  A } .

If there exists  z  such that  z A  and  z B  then it immediately follows that  z  is the least upper bound of the set  A .

So, we have to prove the property only when  A B = .

Let  a A  and  b B .

We can now define two sequences  a n  and  b n  as follows:

a 1 = a  ,  b 1 = b  ,  a n + 1 = max ( { a n , ( a n + b n ) / 2 } A )  and

b n + 1 = min ( { b n , ( a n + b n ) / 2 } B ) .

So, for all  n ,  a n A  and  b n B .

Notice that  a 1 a 2 a n < b  and  b 1 b 2 b n > a .

Both sequences are monotonic and bounded.

From the completeness axiom we conclude that there exist two real numbers  a  and  b  such that  a n a  and  b n b  as  n .

Notice also that  b n a n = ( b a ) / 2 n 1 , so  ( b n a n ) 0  as  n , and  a  = b .

Finally, for any number  p < a , there exists  K  such that  p < a K < a ,

therefore  p  is not an upper bound of  A .

Also, for every number  q > a , there exists  L  such that  q > b L > a ,

therefore  q A .

That is,  a   is the least upper bound of  A .

 

4. Prove the following for a vector function  f : R R n :

 is constant if and only if  f ( x )  is perpendicular to  f ( x ) .

Proof:

Let  k  denote a constant, then

 =  

     

  f ( x )  is perpendicular to  f ( x ) . 

 

 

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