What will be an example of an ordered field in which every Cauchy sequence is eventually constant?
I think an example exists. See alpha.math.uga.edu/~pete/Clark-Diepeveen_PLUS.pdf. This proves that all Cauchy sequences being eventually constant is equivalent to an ordered field being first countable in order topology. If there would have been no example, what would have been the use of this theorem?
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Take the surreal field. Since the surreals contains all transfinite ordinals, we have the topology is not first countable at any point, so no nonconstant convergent sequences.
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It is equivalent that the ordered field have uncountable cofinality, i.e. that any countable subset of said field be bounded (which is the same as the order topology not being first countable).
One simple example is the field $\mathbb{Q}((X_{\alpha})_{\alpha<\omega_1})$ generated by uncountably many indeterminates, where we impose $\mathbb{Q}[X_{\alpha}]<X_{\beta}$ for all $\alpha<\beta<\omega_1$.
This field also embeds in any other example, i.e. in any ordered field of uncountable cofinality (using AC).
If one wants an example that is a set, one can use the hyperreals instead of the surreals. I prefer to construct them as an ultrapower $\Bbb R^\omega/\mathscr{U}$ for some free ultrafilter on $\omega$. Here too the order topology is not first countable at any point.
To see that it’s not first countable at $[0]_\mathscr{U}$, the equivalence class of the constant $0$ sequence in $\Bbb R^\omega$, let $\left\{\left[x^{(n)}\right]_\mathscr{U}:n\in\omega\right\}$ be any countable set of positive hyperreals. If we set
$$x_n=\frac12\min\left\{x_n^{(k)}:k\le n\text{ and }x_n^{(k)}>0\right\}$$
for each $n\in\omega$, then
$$\left\{k\in\omega:0<x_k<x_k^{(n)}\right\}=\left\{k\in\omega:k\ge n\text{ and }x_k^{(n)}>0\right\}\in\mathscr{U}$$
and hence $[0]_\mathscr{U}<[x]_\mathscr{U}<\left[x^{(n)}\right]_\mathscr{U}$ for each $n\in\omega$. And of course if it’s not first countable at $[0]_\mathscr{U}$, it’s not first countable anywhere.