# Inf n 1 n n n min n 1 n n n 0 126a let s be a non

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inf n + ( - 1) n | n N = min n + ( - 1) n | n N = 0 12.6a Let S be a non-empty bounded subset of R . Prove that sup S is unique. Suppose that x and y are both suprema for S . Then (1) x is an upper bound for S . (2) If z is any upper bound for S , then x z . (3) y is an upper bound for S . (4) If z is any upper bound for S , then y z . By (1), x is an upper bound for S and so by (4), y x . By (3), y is an upper bound for S and so by (2), x y . Therefore x y and y x , so x = y by Trichotomy. 12.7a Let S be a non-empty bounded subset of R and let k be in R . Define kS = { ks | s S } . Prove the following: If k 0, then sup( kS ) = k sup S and inf( kS ) = k inf S . To prove that sup( kS ) = k sup S there are two cases. If k = 0, then kS = 0 · S = { 0 } and so sup( kS ) = sup { 0 } = 0 = 0 · sup S = k sup S. Now suppose that k > 0. The number sup( kS ) is the unique real number with the properties (a) sup( kS ) is an upper bound for kS and (b) if y is any upper bound for kS , then sup( kS ) y Therefore, because sup( kS ) is the only number for which properties (a) and (b) are true, to show sup( kS ) = k sup S

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• Fall '08
• staff
• Math, Supremum, Order theory, upper bound, Ks, sup

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