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Here's Theorem 1.2 on page 6, Martin Andreas Väth's Nonstandard Analysis(See here on googlebooks)

The Dedekind completion $\overline{X}$ of a totally ordered field $X$ is a complete Archimedean field with $\Bbb{Q}_{\overline{X}}$ as the canonical copy of $\Bbb{Q}_{X}$.

$X$ has the Archimedean property. For each $x \in X$ there is some $n \in \Bbb{N}_{X}$ such that $n > x$. Each totally ordered field X contains a “canonical copy” of the set $\Bbb{N}_{X}$, namely $\{1_X, 1_X +1_X, 1_X +1_X +1_X, \ldots\}$.

${}^{\ast}\Bbb R$ is a totally ordered field without Archimedean property. Isn't it the case that its Dedekind completion doesn't have Archimedean property?

Metta World Peace
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1 Answers1

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The Dedekind-completion is an order completion, and if the field is non-Arcihmedean then its Dedekind-completion is not a field at all.

To see this simply note that in the completion, there is a point $t$ which is the realization of the cut $R=\{x\mid\exists n\in\Bbb N. x<n\}$. And $t-1$ cannot exist.

Asaf Karagila
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  • Thank you! So there's something wrong with the phrasing of the theorem. $X$ must be an Arcihmedean totally ordered field. Another question: Perhaps I misunderstand you, do you mean $t^{-1}$? – Metta World Peace Oct 06 '13 at 10:22
  • Yes, that does seem like a mistake in the phrasing. About the second question, either one, really. If $t-1$ exists then it has to be smaller than some $n\in\Bbb N$, and then so does $t$. Similar arguments can be made about $t^{-1}$ too. – Asaf Karagila Oct 06 '13 at 10:24