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 Ï A , prove that there is a number  d > 0  such that

for all  y Î A .

Proof:

x Ï A Û x Î R nA . The set  R nA  is open because  A  is closed, so there is an open rectangle  W  such that  x Î W Í R nA .

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 1a 1  ,  b 1x 1x 2a 2  ,  b 2x 2  , … ,  x na n  ,  b nx n  } .

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

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

Proof:

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

Also, for every  x Î 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 Î Bx  belongs to at least one of the sets  U ( x 1 ) , U ( x 2 ) , … , U ( x m ) , say  x Î U ( x k ) .

Now, for every  y Î A , we have:

ï yx ï + ï xx k ï ³ ï yx k ï ³ d ( x k )  but  ï xx k ï £ d ( x k ) / 2

because  x Î U ( x k ) , therefore  ï yx ï ³ d ( x k ) / 2 ³ d .

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

Solution:

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

then  ½( n , 0 ) – ( n , 1 / n )½ = 1 / n  which approaches  0  as  n ® ¥ .

2. Define the  closure  of  A  as

= { x ½ For every open rectangle  U  containing  xU Ç A ¹ Æ } .

a) Show that   = A È bd A .

Proof:

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

( x Î U ) Þ ( U Ç A ¹ Æ ) , therefore  x Î  . So,  ( A È bd A ) Í  .

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

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

That is:   = ( A È bd A ) .

b) Prove  bd A =  Ç  = bd ( R nA ) .

Proof:

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

So, we will only prove that  bd A =  Ç  .

Now, using our result from (a):

Ç  = ( A È bd A ) Ç ( ( R nA ) È bd ( R nA ) )

= ( A È bd A ) Ç ( ( R nA ) È bd A ) = ( A Ç ( R nA ) ) È bd A

= Æ È 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 Î R ½ y  is an upper bound of  A } .

If there exists  z  such that  z Î A  and  z Î 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 Ç B = Æ .

Let  a Î A  and  b Î 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 } Ç A )  and

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

So, for all  n ,  a n Î A  and  b n Î B .

Notice that  a 1 £ a 2 ££ a n £ … < b  and  b 1 ³ b 2 ³³ b n ³ … > 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 na n = ( ba ) / 2 n – 1 , so  ( b na 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 Ï 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

= Û

Û    Û

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

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