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Unformatted text preview: SOLUTIONS FOR HOMEWORK 3 4.11. Suppose, for the sake of contradiction, that the set S = Q ∩ ( a,b ) is finite. Enumerate the members of this set in the increasing order: that is, let x 1 = min S , x 2 = min( S \{ x 1 } ), x 3 = min( S \{ x 1 ,x 2 } ), etc.. Then the open interval ( a,x 1 ) contains no rational points, a contradic tion. 4.14. (a) Let S = A + B = { a + b  a ∈ A,b ∈ B } , s = sup S , a = sup A , and b = sup B . Then a + b ≥ a + b whenever a ∈ A and b ∈ B , hence a + b is an upper bound for S . Thus, s ≤ a + b . On the other hand, for every n ∈ N there exists a ∈ A s.t. a > a 1 /n . Indeed, otherwise, a 1 /n is an upper estimate for A , which means that a is not the least upper bound. Similarly, for every n ∈ N there exists b ∈ B s.t. b > b 1 /n . Therefore, for any n ∈ N there exists s ∈ S satisfying s > a + b 2 /n (just add a and b as above). Suppose, for the sake of contradiction, that s < a + b . By the Archimedian property of R , there exists n ∈ N s.t. n = 2 / (( a + b ) s ), or, equivalently, 2 /n < ( a + b ) s . By the above, there exists s ∈ S satisfying s > a + b 2 /n > s , a contradiction ( s turns out not to be an upper bound). Thus, s = a + b . (b) Use part (a) and the fact that, for any set S , inf S = sup( S ) (this is the essence of the proof of Corollary 4.5). 4.15. Suppose, for the sake of contradiction, that c = a b > 0. By the Archimedian Property of R , there exists n ∈ N such that cn > 1....
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This note was uploaded on 05/08/2010 for the course MATH 140a taught by Professor Staff during the Winter '08 term at UC Irvine.
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