4,5,7,8 - SOLUTIONS FOR HOMEWORK 3 4.11 Suppose for the...

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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 0 = sup S , a 0 = sup A , and b 0 = sup B . Then a 0 + b 0 a + b whenever a A and b B , hence a 0 + b 0 is an upper bound for S . Thus, s 0 a 0 + b 0 . On the other hand, for every n N there exists a A s.t. a > a 0 - 1 /n . Indeed, otherwise, a 0 - 1 /n is an upper estimate for A , which means that a 0 is not the least upper bound. Similarly, for every n N there exists b B s.t. b > b 0 - 1 /n . Therefore, for any n N there exists s S satisfying s > a 0 + b 0 - 2 /n (just add a and b as above). Suppose, for the sake of contradiction, that s 0 < a 0 + b 0 . By the Archimedian property of R , there exists n N s.t. n = 2 / (( a 0 + b 0 ) - s 0 ), or, equivalently, 2 /n < ( a 0 + b 0 ) - s 0 . By the above, there exists s S satisfying s > a 0 + b 0 - 2 /n > s 0 , a contradiction ( s 0 turns out not to be an upper bound). Thus, s 0 = a 0 + b 0 . (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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