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 n – A . The set  R n – A  is open because  A  is closed, so there is an open rectangle  W  such that  x Î W Í 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 1 ,  x 2 – a 2  ,  b 2 – x 2  , … ,  x n – a n  ,  b n – x n  } . Notice that  d > 0  and for all  y ,  y Î A  Þ  y Ï W  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 B  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 B ,  x  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 y – x ï + ï x – x k ï ³ ï y – x k ï ³ d ( x k )  but  ï x – x k ï £ d ( x k ) / 2 because  x Î U ( x k ) , therefore  ï y – x 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  x ,  U 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 n – A ) . Proof: First notice that  bd A = bd ( R n – A ) . So, we will only prove that  bd A =  Ç  . Now, using our result from (a):  Ç  = ( A È bd A ) Ç ( ( R n – A ) È bd ( R n – A ) ) = ( A È bd A ) Ç ( ( R n – A ) È bd A ) = ( A Ç ( R n – A ) ) 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 ³ … ³ 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 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 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  = k  Û  Û    Û  Û  f ( x )  is perpendicular to  f ¢ ( x ) .

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