hw2solutions

# hw2solutions - Math 317 HW #2 Solutions 1. Exercise 1.3.3....

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Unformatted text preview: Math 317 HW #2 Solutions 1. Exercise 1.3.3. (a) Let A be bounded below, and define B = { b ∈ R : b is a lower bound for A } . Show that sup B = inf A . Proof. First, note that B is bounded above by every element of A , so, by the Axiom of Completeness, B has a supremum. Moreover, inf A ∈ B since inf A is a lower bound for A and B is the set of all lower bounds for A . Therefore, since sup B is an upper bound for B , it must be the case that sup B ≥ inf A . If it were the case that sup B > inf A , then := sup B- inf A > , so, by Lemma 1.3.7, there exists b ∈ B such that b > sup B- = sup B- (sup B- inf A ) = inf A. But b ∈ B means that b is a lower bound for A , so this means that inf A is not the greatest lower bound for A . This is plainly impossible, so it must not be the case that sup B > inf A . The only remaining possibility is that sup B = inf A , as desired. Note: By definition of the infimum, inf A ≥ b for all b ∈ B , since B is precisely the set of lower bounds of A . If you had Exercise 1.3.7 in hand, this would complete the proof (and, really, the bulk of the above proof is essentially the proof of Exercise 1.3.7). [Added 09.12.10] Here’s a better proof, borrowed (with minor modifications) from Zach: Proof. If b ∈ B , then b is a lower bound for A , meaning that b ≤ a for all a ∈ A . Therefore, every element of A is an upper bound for B , so B is bounded above and thus, by the Axiom of Completeness, has a least upper bound sup B . We want to see that sup B satisfies the two conditions for being the greatest lower bound of A . i. If sup B is not a lower bound for A , then there exists a ∈ A such that a < sup B ....
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## This note was uploaded on 09/08/2011 for the course MAT 472 taught by Professor Spielberg during the Spring '06 term at ASU.

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hw2solutions - Math 317 HW #2 Solutions 1. Exercise 1.3.3....

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