hw4solutions - a A . This shows that- is an upper bound...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: January 29, 2010 Assignment 4 1. Show that if a,b R and a 6 = b , then there exist ε -neighborhoods U of a and V of b such that U V = . Assume without loss of generality that b > a . Set ε = ( b - a ) / 2. Then V ε ( a ) = U and V = V ε ( b ) satisfy U V = . Indeed, suppose x U V . Then | x - a | < ε and | x - b | < ε . Then, by the triangle inequality, we would have b - a ≤ | x - a | + | x - b | < 2 ε = b - a which is a contradiction. 2. Let A be a nonempty set of real numbers which is bounded below. Let - A = {- x : x A } . Prove that inf A = - sup( - A ) . Let η = inf A . Then a η for all a A . And thus, - a ≤ - η for all
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Unformatted text preview: a A . This shows that- is an upper bound for-A . Suppose - . If- is to be the least upper bound for-A , we must show that there exists an element-a -A with-a . If - , then- . Since is the greatest lower bound for A , it follows that- is not a lower bound for A . Thus, there exists an a A with a - . I.e.,-a . Since-a -A when a A , this completes the proof. 1...
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.

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