prelim1_solutions

prelim1_solutions - 1. For both of the following, assume...

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Unformatted text preview: 1. For both of the following, assume that A,B R and that neither are empty. (a) (10pts) Show that A B = sup A sup B . Solution. If B is not bounded above, then sup B = , and it is trivial. So let b := sup B < and a := sup A . Then b is an upper bound for B , so x A B implies x b , and hence b is an upper bound for A . Then a b by definition of a as a supremum. (b) (10pts) Show that sup( A B ) = max { sup A, sup B } . Solution. Show both inequalities. x A = x sup A = x { sup A, sup B } . Similarly, x B = x { sup A, sup B } . Since we have shown the inequality in either case, we have x ( A B ) , x max { sup A, sup B } = sup( A B ) max { sup A, sup B } , since max { sup A, sup B } is an upper bound for A B . A A B = sup A sup( A B ) by the first part. The same argument shows sup B sup( A B ). Therefore, the inequality holds for whichever is larger, i.e., max { sup A, sup B } sup( A B ). 2. (20pts) Let s 1 = 2 and s n +1 = p 2 + s n . Prove { s n } converges. If you use the fact that a < b a < b ( a,b > 0), please attach a copy of this proof. (Hint: show0), please attach a copy of this proof....
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prelim1_solutions - 1. For both of the following, assume...

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