P.3. Problem Set (Medium-Hard)

P.3. Problem Set (Medium-Hard) - #1 Recall the Axiom of...

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Recall the Axiom of Completeness , which states that every nonempty subset of R that is bounded above has a least upper bound in R. In this problem, we consider some of the consequences of the completeness of R . Theorem (Archimedean Property): For every real number x , there exists a positive integer n such that . Theorem (Greatest Lower Bound Property): Every nonempty subset of R that is bounded below has a greatest lower bound. Exercise: Define what it means for to be the least upper bound of a set S . Prove that if is the least upper bound of S , then for every , there is some x in S such that . Proof (Archimedean Property) Suppose not for contradiction. Then there is some x in R such that for all positive integers n . It follows that N is bounded above. Put , which we know exists by the axiom of completeness. Then is not an upper bound of N . Exercise: Use this and the exercise above to find a contradiction. Conclude that N is not bounded above. This completes the proof of the first theorem. Proof (Greatest Lower Bound Property): Let S be a nonempty subset of real numbers that is bounded above, and suppose . Define the set T by Exercise: Show that is a lower bound for the set T . Exercise:
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P.3. Problem Set (Medium-Hard) - #1 Recall the Axiom of...

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