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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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 Spring '06
 Spielberg
 Limit, lim, Trigraph, sup

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