solutionshw4-201141

solutionshw4-201141 - Solutions for HW 4 Problem 55: 7. For...

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Unformatted text preview: Solutions for HW 4 Problem 55: 7. For each n pick x n ∈ F n . We first show that ( x n ) is a Cauchy sequence. Let > 0. Then there exists N such that ρ ( F n ) < for all n ≥ N . Let n,m ≥ N and assume n < m . Then ρ ( x n ,x m ) ≤ ρ ( F n ) < . Hence ( x n ) is a Cauchy sequence. As X is complete there exists x such that x n → x . Now x m ∈ F n for all m ≥ n implies that x ∈ F n (as F n is closed) for all n ≥ 1. Hence x ∈ ∩ ∞ n =1 F n . Assume that also y ∈ ∩ ∞ n =1 F n , then ρ ( x ,y ) ≤ ρ ( F n ) → 0 implies that ρ ( x ,y ) = 0, i.e., x = y . Hence ∩ ∞ n =1 F n = { x } . Counterexamples: (1) Take X = Q , and F n = [ √ 2 , √ 2 + 1 n ] ∩ X . Then (ii) and (iii) hold, but ∩ ∞ n =1 F n = ∅ . (2) Take X = R , and F n = (0 , 1 n ). Then (i) and (iii) hold, but ∩ ∞ n =1 F n = ∅ . (3) Let X = R , and F n = [ n, ∞ ). Then (i) and (ii) hold, but ∩ ∞ n =1 F n = ∅ ....
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This note was uploaded on 02/05/2012 for the course MATH 703 taught by Professor Schep during the Fall '11 term at South Carolina.

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solutionshw4-201141 - Solutions for HW 4 Problem 55: 7. For...

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