Suppose that ACR is bounded below, and define Bb is a lower bound for A}. Prove that sup B=inf A.={b € R |Use (a) to explain why there is no need to assert that greatest upperbounds exist as part of the Axiom of Completeness.

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Answered: Suppose that ACR is bounded below, and…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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a)To prove this directly, you'd suppose that A is a subset of B. Then you'd need to show that sup(A)≤sup(B)sup(A)≤sup(B). Since A is bounded above, sup(A) is the least upper bound for A which, by definition of l.u.b., is less than or equal to all upper bounds for A. b)Note that, max{sup(A), sup(B)} is either sup(A) or sup(B). So you have a couple of natural cases to break things down into. c) I am not sure of
Let A be a nonempty subset of R that is both bounded above andbelow and let B be a nonempty subset of A. Prove or disprove eachof the following assertions.(a) inf(A) ≤ inf(B) ≤ sup(B) ≤ sup(A).(b) If inf(A) = sup(A), then A has exactly one element.(c) If inf(A) = inf(B) and sup(A) = sup(B), then A = B.
1. True or False? Prove your answer!Suppose A is a countably infinite set of real numbers that is bounded above, and let S = sup A. Then for every E > 0, the interval (S − E, S) contains a number in A and a number not in A.
whenever a E A and Let A and B be non-empty subsets of R with azb bEB. Prove that sup (A) and inf (1B) exist and that sup (A) < Could whenever inf B Sup (A) = inf (B) even though arb AEA and bEB
Suppose A C R and A 0 there exists a set G that is the union of finitely many disjoint bounded open intervals such that |A \ G| +|G\ A| < ɛ.

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Suppose that ACR is bounded below, and define B = {b € R | b is a lower bound for A}. Prove that sup B = inf A. Use (a) to explain why there is no need to assert that greatest upper bounds exist as part of the Axiom of Completeness.