psAnalysis3Ans - Fall 2007 ARE211 Problem Set #03- Answer...

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Unformatted text preview: Fall 2007 ARE211 Problem Set #03- Answer key Third Analysis Problem Set Problem 1 For the following problem, consider an arbitrary universe X and an arbitrary metric d defined on X X . State whether the following statements are true or false. If they are true, give a proof. If they are wrong, give a counter-example a) If a sequence x n converges then every subsequence of x n must also converge. b) If every subsequence of the sequence x n converges, then x n must also converge. c) If every subsequence of the sequence x n converges, then they all converge to the same point. d) If one subsequence of the sequence x n converges, then x n must also converge. Ans: Recall the definitions (C) Convergence : A sequence x n in X converges to an element x X in the metric d if epsilon1 > N N such that n > N we have d ( x n , x ) < epsilon1 (S) Subsequence : A sequence { y 1 ,y 2 ,...,y n ,... } is a subsequence of another sequence if there exists a strictly increasing function : N N such that for all n N : y n = x ( n ) Lets first prove the following lemma (L) by induction: strictly increasing : N N we must have ( n ) n n N (i) initialization: for n = 1: ( n ) 1 as maps from N into N . (ii) induction hypothesis: n n : ( n ) n . (iii) induction step: n n + 1 ( n + 1) > ( n ) (as () is strictly increasing) n (From (ii)). Since ( n + 1) N we therefore know that ( n + 1) n + 1 . Lets consider the problems: a) True: If a sequence x n converges then so does every subsequence of x n . (1) You are given that a sequence x n converges, i.e., x X s.t. epsilon1 > N N such that n > N we have d ( x n , x ) < epsilon1 (2) Using lemma (L) in the definition of a subsequence (S) we know that n : y n = x ( n ) = x n for some n n . (3) But (2) and (1) imply that epsilon1 > N N such that n > N we have d ( y n , x ) < epsilon1 . Hence y n converges to x . b) True: If every subsequence of the sequence x n converges then so does x n . This is trivial: Since every subsequence convergence, pick the subsequence ( n ) = n which is simply the sequence itself. Hence the sequence converges. c) True: from b), ( x n ) itself converges. Let x be the point to which it converges. Fix epsilon1 > ; N epsilon1 N such that n > N epsilon1 , d ( x n ,x ) < epsilon1 . Now consider an arbitrary subsequence ( y n ) of ( x n ) and let ( ) denote the strictly increasing function from N to N that defines ( y n ) . From class we know that ( ) increases without bound. Hence there exists M N , such that for all m > M , ( m ) > N epsilon1 so that y m = x ( m ) B ( x,epsilon1 ) . Hence ( y n ) converges to x . 2 Problem 2 For the following problem, consider an arbitrary universe X and an arbitrary metric d defined on...
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This note was uploaded on 08/01/2008 for the course ARE 211 taught by Professor Simon during the Fall '07 term at University of California, Berkeley.

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psAnalysis3Ans - Fall 2007 ARE211 Problem Set #03- Answer...

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