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I'm now going through the completeness axiom. What is it used for? What can you do with it? Why is it called completeness? And beside that, how can you proof this theorem? Suppose that S is a nonempty subset of R and k is an upper bound of S. Then k is the least upper bound of S if and only if for each $\epsilon > 0$ there exists $s \in S$ such that $k - \epsilon < s$. I tried picking a random $\epsilon$, but then I come to the point $k-s<-\epsilon$..

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    You refer to it first as an axiom, then as a theorem. It can't be both, so which one is it?If it is an axiom, it is assumed , and needs no proof.2011-10-26
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    And you made a mistake in going from $k-e< s$ to $k-s< e$.2011-10-26
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    @gary: The axiom says that every bounded subset has a least upper bound. The second is a theorem to characterize this least upper bound.2011-10-26
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    The theorem was an exercise where you might use the completeness axiom. But I couldn't find the answer on it.2011-10-26
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    Verb: Prove (the thing you *do*). Noun: Proof (an often logical derivation describing why a theorem holds).2011-10-26
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    @Asaf: so which is the second one, i.e., what is the statement of the theorem, is it that every Cauchy sequence of reals converges?2011-10-26
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    @gary: The theorem is that if $k$ is *an* upper bound of $S$, then $k$ is the least upper bound if and only if for every $\epsilon>0$ there is some $s\in S$ such that $k-\epsilon. $$\begin{align}{}\\\end{align}$$ The axiom of completeness is that every bounded subset of $\mathbb R$ has a least upper bound.2011-10-26
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    @Asaf:Got it, thanks.2011-10-26
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    @Kevin: Just a note on how you phrased things. Picking a "random $\epsilon$" is not what what you mean to say. For instance, considering $0 < \epsilon$ we could pick a "random $\epsilon$" uncountably many times and find that it is true every single time. That doesn't make the statement $0 < \epsilon$ true for all $\epsilon$. A better phrasing is "I let an $\epsilon$ be given".2011-10-26
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    The completeness axiom formalizes the idea that any number you think should be there is actually there. The "existence of the $\sup$" is a precise and handy form of this principle.2011-10-26

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