4. Where in the proof of Theorem 2.3.3 did we use the following?(a) The least upper bound axiom(b) The assumption that {x } is bounded(c) The assumption that {x } is increasing5. Prove that a bounded decreasing sequence {xx } converges by9(a) using the result proved in the text for increasing sequences;(b) using the & – no definition of convergence and the set A = {xn | n EN}.Theorem 2.3.3. A bounded monotonic sequence converges.Proof. We prove the theorem for an increasing sequence; the decreasing sequencecase is left for the exercises (Problem 5).Assume that {In } is bounded and increasing. To show that { n } is convergent, weuse the & – no definition of convergence, and to use this definition, we need to knowthe limit of the sequence. We determine the limit using the set A = {xn | n EN},the set of points in R consisting of the terms of the sequence {x}. Because {x} isbounded, there is an M > 0 such that |X, | < M for all n, and this M is an upper boundfor A. Hence A is a bounded nonempty set of real numbers and so has a least upperbound. Let a =. lubA. This number a is the limit of {x}, which we now show.Take any & > 0. Since a = lubA, there is an element a E A greater than a – &;that is, there is an integer no such that Xno = a > a – E. But {x } is increasing, soXno < Xn for all n > no, and a an upper bound of A implies that In _ a for all n.Therefore, for n > no, a – & < xx < a, so |xn – a| <E.Remark Note that, from the above proof, it follows immediately that, if a is the limitof an increasing sequence {Xn}, then *n < a for all n E N. (Similarly, we have thatthe limit of a convergent decreasing sequence is a lower bound for the terms of thedecreasing sequence.)
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