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Unformatted text preview: ε = 1, and choosing n 1 so that x 1 > s1. We then deﬁne n k recursively. We know that s remains as the supremum of the set { x n }\{ x 1 ,...,x n k1 } . We use ε = 1 /k and choose n k so that x n k > s1 /k . It follows trivially that n k is an increasing sequence. Moreover, by construction, we have that  sx n k  = sx n k < 1 /k , from which it is now standard to see that x n k → s . 2. Let { x n } be a Cauchy sequence such that x n is an integer for every n ∈ N . Show that { x n } is ultimately constant. That is, there is an M ∈ N and a c ∈ R so that x n = c for all n ≥ M . We apply the deﬁnition of Cauchy with ε = 1 / 2. Then, there is an M ∈ N so that  x nx m  < 1 / 2 , if n,m ≥ M. But for two integers, we can only have this if x n = x m . Thus, we see that x n ≡ c for all n ≥ M . 1...
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.
 Spring '10
 MetCalfe
 Calculus

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