# hw13solutions - ε = 1, and choosing n 1 so that x 1 >...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: February 24, 2010 Assignment 13 1. Let { x n } be a bounded sequence and let s := sup { x n : n N } . Show that if s 6∈ { x n : n N } , then there is a subsequence of { x n } that converges to s . We know that s is the supremum of the set if and only if for any given ε we can ﬁnd an element of the set y so that s y > s - ε . First, we show that since s 6∈ { x n } , we have that s is also the supremum of the set that is obtained after removing any ﬁnite number of elements. Indeed, by removing elements, the supremum can only decrease. On the other hand, after removing elements { y 1 ,...,y N } , we take ε = min {| y 1 - s | ,..., | y N - s |} in the above. Then there must be some element z of the original set so that s z > s - ε . Since s is not in the set, z cannot be s nor could that have been removed in the ﬁnite elements. Moreover, z cannot be any of the y i due to the choice of ε . As such, s remains the supremum of the set with a ﬁnite number of elements removed. We now proceed using, again, the characterization above. We start by setting
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Unformatted text preview: ε = 1, and choosing n 1 so that x 1 > s-1. We then deﬁne n k recursively. We know that s remains as the supremum of the set { x n }\{ x 1 ,...,x n k-1 } . We use ε = 1 /k and choose n k so that x n k > s-1 /k . It follows trivially that n k is an increasing sequence. Moreover, by construction, we have that | s-x n k | = s-x 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 n-x 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.

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