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Unformatted text preview: Math 104  Homework 2 Solutions Lectures 2 and 4, Fall 2011 1. Find the supremums of the following subsets of R and prove that your answers are correct. ( a ) (0 , 2) ( R \ Q ) ( b ) n n + 1 n N ( c ) 3 n 2 n 2 + n 1 n N and n 2 (Recall that R \ Q = { x R  x / Q } is the set of irrational numbers.) Suggestion: For each, use thecharacterization of supremums. Solution. (a) First, since any number in this set is in particular in (0 , 2), any such number is less than 2, so 2 is an upper bound of the given set. We claim that 2 is the supremum. To see this, let > 0. By the denseness of the irrationals in R , there is some x R \ Q such that 2 < x < 2 . This x is then an element of the given set larger than 2 , showing that nothing smaller than 2 can be any upper bound of the given set. We conclude that the supremum is 2 as claimed. (b) Since n < n +1 or every n N , dividing through by n +1 shows that anything in the given set is smaller than 1, so 1 is an upper bound. Now, let > 0 and choose N N such that 1 N < , which exists by the Archimedean Property of R . Since 1 N +1 < 1 N , we also have 1 N +1 < so 1 < 1 1 N + 1 = N N + 1 . In other words, N/ ( N + 1) is an element of the given set which is larger than 1 , so nothing smaller than 1 can be an upper bound of the given set and thus its supremum is 1. (c) Since 3 n 2 < 3 n 2 + 3 n 3 = 3( n 2 + n 1) for every n 2, we see that 3 is larger than all elements of this set so 3 is an upper bound. To show that 3 is the supremum, let > 0 and pick N N such that 3 > 1 N . Then > 3 N = 3 N N 2 > 3( N 1) N 2 + N 1 , where the last inequality is true since the numerator in the final fraction is smaller than that in the previous fraction and its denominator is larger. Thus 3 < 3 3( N 1) N 2 + N 1 = 3 N 2 N 2 + N 1 , showing that 3 is not an upper bound of the given set for any > 0, so 3 is the supremum as claimed. Note that in all of these, the final presented solutions do not hint at the scratch work that actually went into finding the N s that would work. As Ive said, this is common in most proofs you will see, where what is given is the final product but actual (hard) work goes into figuring out...
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 Fall '08
 RIEMAN
 Sets

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