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**Unformatted text preview: **(b) S is compact so S is closed and bounded. T is a closed subset of S . Since T ⊆ S and S is bounded, it follows that T is bounded. Therefore, T is closed and bounded so it is compact. 1 Section 16 6. (a) (Assume k negationslash = 0.) Let epsilon1 > 0. There exists a positive integer N such that N > | k | epsilon1 which implies that 1 N < epsilon1 | k | . Now, for all n > N vextendsingle vextendsingle vextendsingle vextendsingle k n- vextendsingle vextendsingle vextendsingle vextendsingle = | k | n = | k | 1 n < | k | 1 N < | k | epsilon1 | k | = epsilon1. Therefore, k/n → 0. (b) k > 0. Let epsilon1 > 0. There exists a positive integer N such that N > 1 k √ epsilon1 which implies that 1 N k < epsilon1 . Now, for all n > N vextendsingle vextendsingle vextendsingle vextendsingle 1 n k- vextendsingle vextendsingle vextendsingle vextendsingle = 1 n k < 1 N k < epsilon1. Therefore, 1 /n k → 0. (c) Let epsilon1 > 0. vextendsingle vextendsingle vextendsingle vextendsingle 3 n + 1 n + 2- 3 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle- 5 n + 2 vextendsingle vextendsingle vextendsingle vextendsingle = 5 1 n + 2 < 5 1 n . There exists a positive integer N such that N > 5 epsilon1 which implies that 1 N < epsilon1 5 . Now, for all n > N , vextendsingle vextendsingle vextendsingle vextendsingle 3 n + 1 n + 2- 3 vextendsingle vextendsingle vextendsingle vextendsingle < 5 1 n < 5 1 N < 5 epsilon1 5 = epsilon1. Therefore, 3 n + 1 n + 2 → 3. (e) First a calculation: If n ≥ 3, then n + 2 < ≤ 2 n and n 2- 3 ≥ n 2 / 2. Thus, for n ≥ 3 n + 2 n 2- 3 ≤ 2 n n 2 = 4 n . There exists a positive integer N such that N > 3 and N > 4 epsilon1 , which implies that 1 N < epsilon1 4 . Now, for all n > N , vextendsingle vextendsingle vextendsingle vextendsingle n + 2 n 2- 3- vextendsingle vextendsingle vextendsingle vextendsingle = n + 2 n 2- 3 ≤ 4 n < 4 1 N < 4 epsilon1 4 = epsilon1....

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- Fall '08
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- Math, Topology, 1 k, Metric space, 1 k, 1 5 2k