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Least Upper Bound Property ⇒ Archimedean Principle
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[QUOTE="Someone2841, post: 6213820, member: 364392"] This post got moved to abstract algebra (I was thinking of ##\mathbb{F}## as a generalization of the real numbers), and so I realize now that I could have just proven that any linearly-ordered group ##\mathbb{G}## that is not Archimedean does not have the least upper bound property. It can be done similarly but more concisely: If ##G## is non-Archimedean linearly-ordered group, then there exists some ##g, h \in G## such that ##g^n < h## for all ##n \in \mathbb{N}##. This means that the set ##H=\{g^n : n \in \mathbb{N}\}## is both non-empty and bounded by ##h##. Now suppose ##H## has a least upperbound ##l \in G##. As a upper bound, ##g^{n+1}<l## for all ##n## (since ##n## is arbitrary). But this means that ##g^n < g^{-1}l < l##, and ##g^{-1}l## is a lower least upper bound for ##H##, which is a contradiction. Therefore, if ##G## is a non-Archimedean linearly-ordered group, it does not satisfy the least upper bound property. □ [/QUOTE]
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Least Upper Bound Property ⇒ Archimedean Principle
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