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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