It is sufficient to show that a k sup s is an upper

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it is sufficient to show that (a) k sup S is an upper bound for kS and (b) if y is any upper bound for kS , then k sup S y . To show that k sup S is an upper bound for kS , suppose that x is in kS . Then there is an s in S so that x = ks . Then s sup S since sup S is an upper bound for S and so x = ks k sup S . Since x was arbitrary, k sup S is an upper bound for kS . To show (b), suppose that y is an upper bound for kS . Then ks y for every s in S , so s y/k for every s in S . Therefore, y/k is an upper bound for S and so sup S y/k . Multiplying by k we see that k sup S y as desired. To prove that inf( kS ) = k inf S there are two cases. If k = 0, then kS = 0 · S = { 0 } and so inf( kS ) = inf { 0 } = 0 = 0 · inf S = k inf S. Now suppose that k > 0. To show inf( kS ) = k inf S it is sufficient to show that (c) k inf S is a lower bound for kS and (d) if y is any lower bound for kS , then y k inf S . To show that k inf S is a lower bound for kS , suppose that x is in kS . Then there is an s in S so that x = ks . Then s inf S since inf S is a lower bound for S
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