HW15solns

# Mathematical Thinking: Problem-Solving and Proofs (2nd Edition)

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Unformatted text preview: 1. By the definition of a minimum, x for every x S, so is a lower bound for S. If is an another lower bound for S, then, in particular, since S. Hence must be the greatest lower bound of S. In other words, = inf S. 2. a is a lower bound for (a, b) simply by the definition of (a, b). Now suppose x > a. We will show x is not a lower bound of S. If x b, then (a + b)/2 is a member of S that is less than x. If x < b, then (a + x)/2 is a member of S that is less than x. In either case, we see that x is not a lower bound for S, as claimed. Thus a = inf(a, b). 3. False. Let S = {0, 1}. 4. Let S = {-f (x) : x I} , T = {f (x) : x I} , = sup S, and = - inf T . We want to show = . Since is an upper bound for S, -f (x) for every x I. Thus - f (x) for every x I, which implies - is a lower bound for T . Since - = inf T , - -, which implies that . Conversely, since - is a lower bound for T , - f (x) for every x I. Thus f (x) for every x I, so is an upper bound for S. But this implies that . It now follows that = , as desired. 5. (a) We have 0 0 and 1 - 1/n 1. (b) We can write S as the union of the sets S1 = {1/n : n N and n is odd} and S2 = {3/n : n N and n is even} . Then 1 is a maximum of S1 and 3/2 is a maximum of S2 , so sup S = 3/2. Since 3/2 S, we have the constant sequence 3/2 3/2 = sup S. Further, we see that 0 is a lower bound for both S1 and S2 and thus for S as well. By the Archimedean Principle, we must have inf S = 0. Then the sequence < 1/2n > in S2 S converges to 0 = inf S. 6. One possible set of examples are as follows: (a) Let f (x) = sin(x), g(x) = sin(x) + 3. (b) Let f (x) = sin(x), g(x) = sin(x) + 2. (c) Let f (x) = sin(x), g(x) = sin(x) + 1. 1 ...
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## This document was uploaded on 03/25/2011.

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