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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 counterexample 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 τ : N → N such that for all n ∈ N : y n = x τ ( n ) Let’s first prove the following lemma (L) by induction: ∀ strictly increasing τ : N → N we must have τ ( n ) ≥ n ∀ n ∈ N (i) initialization: for n = 1: τ ( n ) ≥ 1 as τ maps from N into N . (ii) induction hypothesis: ∀ n ≤ ¯ n : τ ( n ) ≥ n . (iii) induction step: ¯ n → ¯ n + 1 τ (¯ n + 1) > τ (¯ n ) (as τ () is strictly increasing) ≥ ¯ n (From (ii)). Since τ (¯ n + 1) ∈ N we therefore know that τ (¯ n + 1) ≥ ¯ n + 1 . Let’s 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 τ ( n ) = x ˆ n for some ˆ 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 τ ( 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 τ ( · ) denote the strictly increasing function from N to N that defines ( y n ) . From class we know that τ ( · ) increases without bound. Hence there exists M ∈ N , such that for all m > M , τ ( m ) > N epsilon1 so that y m = x τ ( 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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 Fall '07
 Simon
 Order theory, Necessary and sufficient condition, Monotonic function, Closed set, Boundary

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