hw2sol - Math 131A - Section 2 Spring 2010 Homework 2...

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Math 131A - Section 2 Spring 2010 Homework 2 Graded problems: 4.8 (3 points), 4.11 (2 points), 4.12 (2 points), 4.14 (3 points), 5.6 (3 points), 7.4 (2 points) 4.4.(e). 0; (i). 0; (j). 2 / 3; (l). No inf; (v). - 1 / 2; (w). - 3 / 2 4.8.(a). Since T is nonempty, there exists t 0 T . Since s t 0 for all s S , we have S is bounded above. Similarly, since S is nonempty, there exists s 0 S . Since s 0 t for all t T , we have T is bounded below. Thus, sup S and inf T exists and are finite. (b). Fix t 0 T . Since s t 0 for all s S , we have that t 0 is an upper bound for S . Thus, sup S t 0 (since the supremum is the least upper bound). Our choice of t 0 T was arbitrary, we have that sup S t for all t T . Thus, sup S is a lower bound for T , and since the infimum is the greatest lower bound, we have that sup S inf T . (c). Take S = T = { 0 } . Then, S T = { 0 } 6 = , and the condition s t for all s S and t T holds (since 0 0). (d). Let S = (0 , 1) and T = (1 , 2). Then, S T = , but sup S = 1 = inf T . Notes: The most confusing part of this problem seemed to have been what to do when the book says “observe that. ..” in the beginning of a problem. In general, this should mean you prove the result they state, and your proof should be rather short and pretty much follow immediately from the assumptions in the problem. 4.11. Let
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This note was uploaded on 11/20/2010 for the course MATH 131A 131A taught by Professor Kim during the Spring '10 term at UCLA.

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hw2sol - Math 131A - Section 2 Spring 2010 Homework 2...

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