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Unformatted text preview: Section 1.4: Consequences of Completeness In this section, we deal with the consequences of completeness and why it is necessary for our study of analysis. Questions such as “Why are the rational numbers insufficient for analysis?” are addressed. Before we begin, a few lemmas are in order. Lemma: Let S be a nonempty subset of R that is bounded above and let α be an upper bound of S . Then sup S α = if and only if for every ε , there exists some s S ∈ such that s α ε α < ≤ . Proof The forward implication follows directly from the definition of least upper bound (see above). For if there exists some ε such that for every x S ∈ , x α ε ≤ , then α ε is an upper bound of S that is strictly less than α , which contradicts our hypothesis that sup S α = . Conversely, suppose that for every ε , there exists some s S ∈ such that s α ε α < ≤ . Suppose for contradiction that sup S α ≠ . Then there is an upper bound M with M α < . Let M ε α = . It follows that there is some s S ∈ such that M s α ε α = < ≤ which contradicts that M is an upper bound. Hence sup S α = . Of course a similar result holds for the greatest lower bound of a set. Lemma: Let S and T be subsets of R with T S ⊂ and S bounded above. Then sup sup T S ≤ . Proof Let sup S α = and lub T β = . Suppose for the sake of contradiction that α β < . Then by the lemma above, there is some t T ∈ with t α β < < . Since t α < and lub S α = , t is an upper bound of S . Note then for all s S ∈ , s t α ≤ < . It follows from this that t S ∉ . However, this contradicts our hypothesis that T S ⊂ . Thus β α ≤ . Now, let us begin discussing the consequences of completeness. Recall: Axiom of Completeness: Every nonempty subset of R that is bounded above has a least upper bound in R . Although we assume that R has the least upper bound property, we may prove that is has the corresponding “greatest lower bound” property....
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 Spring '08
 PLOTKIN
 upper bound, Archimedean Property, greatest lower bound

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