# And so x ks k inf s since x was arbitrary k inf s is

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and so x = ks k inf S . Since x was arbitrary, k inf S is a lower bound for kS . To show (d), suppose that y is a lower bound for kS . Then ks y for every s in S , so s y/k for every s in S . Therefore, y/k is a lower bound for S and so inf S y/k . Multiplying by k we see that k inf S y as desired.

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3 12.10a Prove:If x and y are real numbers with x < y , then there are infinitely many rational numbers in the interval [ x, y ]. In class.
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