104_Sp08_MT1_sol - Midterm 1 for MATH 104, Spring 2008...

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Midterm 1 for MATH 104 , Spring 2008 Solutions Problem 1 [15P] (a) State the least upper bound principle for R precisely. (b) For the following sets of real numbers, determine whether they are non-empty and bounded from above. If so, determine the least upper bound. Justify your answer. (i) { n - 1 n : n N } ; Solution. Clearly, the set is non-empty, since it contains for example 0. Assume the set is bounded from above by M R . Then by the Archimedean principle there exists an N Nat such that N > M . This implies N + 1 > M . But since M is an upper bound, we must have N = N + 1 - 1 < N + 1 - 1 N + 1 M < N , contradiction. ± (ii) T n = 1 (1 - 1 n , 1 + 1 n ). Solution. We claim that T n = 1 (1 - 1 n , 1 + 1 n ) = { 1 } . Clearly, 1 is an element of every interval (1 - 1 n , 1 + 1 n ), so 1 is an element of the intersection. On the other hand, suppose x , 1. Assume x > 1. Since 1 + 1 n 1, there exists an N such that 1 + 1 N < x , and hence x is not an element of the intersection. The proof for
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104_Sp08_MT1_sol - Midterm 1 for MATH 104, Spring 2008...

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