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Unformatted text preview: 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 6 = 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 Exercise 6.5.2 Let S be a nonempty 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....
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This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell University (Engineering School).
 '08
 RAMAKRISHNA
 Math

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