Unformatted text preview: β + 1 n ∈ T. This contradicts the assumption that β is the supremum of T . Case 2 : Suppose β 2 > b . Let ± = β 2b . Note that ( β1 n ) 2 = β 22 β n + 1 n 2 ≥ β 22 β n . Let n be chosen so that 2 β n ≤ ±. Then ( β1 n ) 2 > b. So β1 n is an upper bound of T contradicting the assertion that β is the least upper bound. So β 2 = b. Q.E.D. 1...
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 Spring '08
 DANG
 upper bound, Douglas Weber

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