ans6 - MAA 4200: Homework 6 Solutions July 13, 2009 [3.35]...

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Unformatted text preview: MAA 4200: Homework 6 Solutions July 13, 2009 [3.35] Proof: Suppose E compact and nonempty. By definition of compact, E is bounded and so sup E and inf E exist. Suppose sup E is not in E . The proof for inf E is similar. Let Q be any neighborhood of sup E , then Q contains an interval (sup E- , sup E + ) for some > 0. By definition of sup E , the number sup E- is not an upper bound for E , so there exists x E such that sup E- < x < sup E . Therefore, x Q E . We have shown that every neighborhood of sup E contains a point of E , so sup E is an accumulation point of E by the Lemma at the bottom of p.39. Since E is compact, it is closed, and hence E contains its accumulation points, so sup E is in E . [3.44] Proof: Let g ( x ) = f ( x )- x . Then g : [ a,b ] R is continuous. Since the range is [ a,b ] we have a f ( x ) b for all x [ a,b ], i.e. f ( a ) a and f ( b ) b ....
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This note was uploaded on 08/13/2009 for the course MAT 371 taught by Professor Thieme during the Spring '07 term at ASU.

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ans6 - MAA 4200: Homework 6 Solutions July 13, 2009 [3.35]...

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