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Let $(k,+,.,0,1,<_k)$ be an ordered field, $| . |_k = x \mapsto \max(x,-x)$.

If $\lambda$ is an ordinal, you can define convergent $\lambda$-sequences: maps $f: \lambda \rightarrow k$ satisfying $\forall \varepsilon >_k 0(\exists \alpha \in \lambda(\forall \beta \ni \alpha(|f(\beta) - x|_k <_k \varepsilon)))$ for some $x \in k$. (you can also define Cauchy $\lambda$-sequences in a similar way)

If $cf(k)$ is the least order type of cofinal well ordered subsets of $k$ (or equivalently the least cardinal of cofinal subsets of $k$), then you get the nice property that $cf(k)$ is the least ordinal such that every bounded monotone $cf(k)$-sequence is a Cauchy $cf(k)$-sequence whenever cf(k) $> \aleph_0$. I can't find a way to do without this hypothesis but my method might not be the right one.

edit: This is actually false (see Eric Wofsey's answer).

This means that $cf(k)$ is the least ordinal such that $cf(k)$-sequentially closed sets are closed sets with respect to the order topology. Also, $k$ is complete (there is no dense embedding of $k$ into a proper linearly ordered field) iff its Cauchy $cf(k)$-sequences are convergent, but this result I can prove without assuming that $cf(k)$ is uncountable.

In the case when $cf(k) = \omega_0$, proving that bounded monotone sequences are Cauchy sequences is the same as proving that $k$ is archimedian. So my question is: are there non archimedian linearly ordered fields with countable cofinality? Do we even know the cofinality of some famous non archimedian fields such as fields of hyperreals? (without assuming CH)

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  • $\begingroup$ If, as is usually the case, one insists on $\omega_1$-saturation, the cofinality of ${^*\Bbb R}$ is uncountable; without this requirement there are models with countable cofinality, according to the answer to this MO question. $\endgroup$ – Brian M. Scott Jun 12 '15 at 23:48
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There are lots of non-archimedean ordered fields of countable cofinality. For instance, if $k_0$ is any archimedean field, you can take $k=k_0(x)$ where $x$ is greater than every element of $k_0$ (the powers of $x$ are cofinal).

I also believe your result is incorrect even for fields of uncountable cofinality. For instance, let $\lambda$ be an uncountable regular cardinal and let $k=\mathbb{Q}(x_\alpha)_{\alpha<\lambda+\lambda}$, ordered such that $x_\alpha$ is greater than every element of $\mathbb{Q}(x_\beta)_{\beta<\alpha}$. Then $k$ has cofinality $\lambda$ (the sequence of $x_{\lambda+\alpha}$ for $\alpha<\lambda$ is cofinal), but the sequence $(x_\alpha)_{\alpha<\lambda}$ is a bounded increasing sequence of length $\lambda$ that is not Cauchy.

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  • $\begingroup$ Thanks, there was indeed a supid error in my proof, and I realize now that my method was very bad too! $\endgroup$ – nombre Jun 13 '15 at 10:02

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