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nphw6 - β As(rearranging z-β< α there exists a ∈ A...

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Numbers and Polynomials Homework 6 Let the nonempty subsets A and B of R be bounded above. Let S = { a + b : a A, b B } , P = { a b : a A, b B } . (i) Verify that S has a least upper bound. (ii) Compare lub S to lub A + lub B . (iii) What about P ? Solution Write α = lub A and β = lub B for convenience. (i) S is nonempty, of course. If a A and b B then a + b α + β so S is bounded above, too. LUB applies. (ii) We have just shown that α + β is an upper bound for S ; we claim that it is least (so the two numbers to be compared are equal). For this, let z < α
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Unformatted text preview: + β . As (rearranging) z-β < α , there exists a ∈ A such that z-β < a (recall that α = lub A ). As (rearranging) z-a < β , there exists b ∈ B such that z-a < b (recall that β = lub B ). The resulting inequality z < a + b shows that z is not an upper bound of S . (iii) Though nonempty, P need not be bounded above: indeed, A and B may contain arbitrarily large negative numbers. 1...
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