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hw5 - Math 3110 Homework 5 Solutions Exercise 6.2.2 The...

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Math 3110 Homework 5 Solutions Exercise 6.2.2 The terms of a sequence { x n } take on only finitely many values a 1 , a 2 , . . . , a k . That is, for every n , x n = a i for some i (the index i will depend on n ). Claim. The sequence { x n } has a cluster point. Proof. We claim there exists an i ∈ { 1 , 2 , . . . , k } such that x n = a i for infinitely many n . Suppose to the contrary that for each i ∈ { 1 , 2 , . . . , k } , x n = a i for only finitely many n . So for every i ∈ { 1 , 2 , . . . , k } , there exists an N i such that x n = a i for all n N i . Let N = max { N 1 , N 2 , . . . , N k } . Then for all n N , x n / ∈ { a 1 , a 2 , . . . , a k } . But this is a contradiction. Therefore, there is an i ∈ { 1 , 2 , . . . , k } such that x n = a i for infinitely many n . So for all > 0, | x n - a i | = 0 < for infinitely many n . That is, a i is a cluster point of { x n } . Exercise 6.4.1 Claim. Every convergent sequence is a Cauchy sequence. Proof. Let { a n } be a convergent sequence, with lim n →∞ a n = L . Let > 0, so 2 > 0. Then there exists an N such that for all n N , | a n - L | < 2 . Then by the triangle inequality, for all m, n N we have | a n - a m | = | a n - a m + L - L | = | ( a n - L ) + ( L - a m ) | ≤ | a n - L | + | L - a m | = | a n - L | + | a m - L | < 2 + 2 = . That is, { a n } is a Cauchy sequence. 1
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Exercise 6.5.2 Let S be a non-empty bounded set of real numbers, and m = sup S . We prove that inf { m - x | x S } = 0 . First, notice that since m is an upper bound for S , then m - x 0 for all x S . So 0 is a lower bound for { m - x | x S } . Next, let K be a lower bound for { m - x | x S } . That is, m - x K for all x S . Then x m - K for all x S . So m - K is an upper bound for S . But since m is the least upper bound for S , then m - K m . That is, 0 K . This shows that 0 is the least upper bound for { m - x | x S } , completing the proof.
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