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solutionshw4-201141

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

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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 0 such that x n x 0 . Now x m F n for all m n implies that x 0 F n (as F n is closed) for all n 1. Hence x 0 ∈ ∩ n =1 F n . Assume that also y 0 ∈ ∩ n =1 F n , then ρ ( x 0 , y 0 ) ρ ( F n ) 0 implies that ρ ( x 0 , y 0 ) = 0, i.e., x 0 = y 0 . Hence n =1 F n = { x 0 } . 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 = . Problem 55: 13. (1) Ω( x ) - x > 0, so Ω has no fix point. (2) One can see using calculus that Ω 0 ( x ) = 1 - e x (1+ e x ) 2 . We claim that 0 < Ω 0 ( x ) < 1 for all x . The right hand side of this inequality is obvious. For the left hand side we need that e x

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