While thinking about properties of real closed fields, I came across the following contradiction:

Let $(k,+,.,0,1,\leq)$ be a non archimedian totally ordered field. Let us assume that $(k,\leq)$ is complete.

Let $\mathbb{N}(k)$ be the set of scalars of the form $n.1 = \sum \limits_{i=1}^n 1, n \in \mathbb{N}$.

Since $k$ is not archimedian, there exists $(x,y) \in ({k_+}^*)^2$ such that $\forall n \in \mathbb{N}, n.x \leq_k y$.

Therefore $\forall n \in \mathbb{N}, n.1 \leq_k \frac{y}{x}$.

$\mathbb{N}(k)$ is bounded so it has a supremum $\alpha$.

$\alpha - 1$ isn't an upper bound of $\mathbb{N}(k)$, which yields the existence of $n \in \mathbb{N}$ such that $n.1 >_k \alpha - 1$. Then $(n+1).1 >_k \alpha$, which contradicts the definition of $\alpha$, and more precisely the fact that it is an upper bound of $\mathbb{N}(k)$.

Therefore, there can be no totally orderd non archimedian complete field.

However, given a totally ordered non archimedian field $k$ (such as an hyperreal field), we may consider the field $\widetilde{k}$ of Dedekind cuts over $k$, with arithmetic operations defined in a similar way as they are defined in the Dedekind cuts definition of $\mathbb{R}$, and the order being $\subset$.

$\widetilde{k}$ is not archimedian. This is because the dense embedding $\varphi: k \ni x \mapsto ]-\infty;x[$ maps $\mathbb{N}(k)$ onto $\mathbb{N}(\widetilde{k})$. And $\widetilde{k}$ is complete.

Does anyone see where I'm wrong?


Correct: You showed is that a non-archimedean ordered field is not complete.

Completion by cuts defined "in a similar way as $\mathbb R$"? This is not possible. Some "cuts" $\xi$ will have $\xi+1 = \xi$, so you fail to get a field at all.

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  • $\begingroup$ Oh yes, I shouldn't have overlooked the details of the proof that there are opposites in the set of Dedekind cuts. Thank you! $\endgroup$ – nombre Mar 25 '15 at 19:05

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