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Unformatted text preview: a A . This shows that is an upper bound forA . Suppose  . If is to be the least upper bound forA , we must show that there exists an elementa A witha . 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 . Sincea 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.
 Spring '10
 MetCalfe
 Calculus

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