WA 9 - Nicola Hodges David Walnut MATH 290-002 November 29,...

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David Walnut MATH 290-002 November 29, 2010 Writing Assignment 8 1. Prove that if x n L and y n M and r R, then x n y n LM Lemma 1: Suppose x n L. Then x n is bounded, hence there exists a Z such that for all n N, |x n | Z. Let e=1. Then for some N, N n implies that |x n -L|<1 so |x n | < | L|+1. Let L 1 =max(|x 1 |,|x 2 |,…,|x n-1 |). So for all n N, |x n | < max (L 1 , |L|+1 ). Hence, x n is bounded by Z. Let ε >0. Since x n converges to L, then it is bounded by Lemma 1, therefore |x n |<Z. Since x n L, then there exists N 1 N. Choose N 1 so that n N 1 . Therefore, |x n -L| < e / 2(|M| +1). Since y n M, then there exists N 2 N. Choose N 2 so that n N 2 . Therefore, |y n -M| < e / 2(|Z| +1). Let N 3 =max{N 1 ,N 2 }, and assume n>N 3 . Note, (x n y n )-(LM) = (x n y n –x n M +x n M – LM) = x n (y n -M) + M(x n -L) Therefore, |(x n y n )-(LM)| = |x n | |(y n -M)| + |M| |(x n -L)| < Z ( e / (2(|Z| +1))) + |M| (e/ (2(|M|+1)))

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This note was uploaded on 03/05/2011 for the course MATH 290 taught by Professor Alligood,k during the Fall '08 term at George Mason.

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WA 9 - Nicola Hodges David Walnut MATH 290-002 November 29,...

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