We know that every complete ordered field is isomorphic to $\mathbb R$, but are there examples of complete ordered fields different, not isomorphically different of course, from $\mathbb R$?
[Math] Examples of a complete ordered field
field-theoryreal-analysis
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Consider the ring of formal Laurent series $R((x))$ with the ordering where $x$ is a positive infinitesimal.
That is, a rational function is positive if and only if its Laurent series has a positive leading coefficient.
What ($\omega$-indexed) sequences converge to zero?
Well, for some $n$, we must have $s_m < x^2$ for all $m > n$. In particular, this means the leading term cannot be of the form $a x^j$ with $j < 2$, because such a thing would be greater than $x^2$. So for all $m > n$, the coefficient on $x_j$ is $0$ for all $j < 0$.
A similar argument can be used in each degree; so we have a simple characterization of sequences that converge to zero: they are the sequences bounded on the left for which the sequence of coefficients on each $x^i$ is eventually always zero. (Note that it's not enough for the coefficients to simply converge to zero!)
The bounded criterion rules out things like the sequence
$$ 1, x^{-1}, x^{-2}, x^{-3}, \cdots $$
which diverges to $+\infty$.
Correspondingly, there is a simple condition for Cauchy sequences: they are precisely the sequences which are bounded on the left and for which each the sequence of coefficients on each $x^i$ is eventually constant.
Therefore, $R((x))$ is Cauchy complete (in the sense that every Cauchy sequence converges), and it is non-Archimedean because it has positive infinite numbers, such as $x^{-1}$.
Any ordered field $F$ has characteristic $0$, so it contains a copy of $\mathbb{Z}$; by the universal property of the quotient field, the ring monomorphism $\mathbb{Z}\to F$ lifts to a monomorphism $\mathbb{Q}\to F$. We can identify $\mathbb{Q}$ with its image, so it's not restrictive to assume that $\mathbb{Q}\subseteq F$.
It's not really difficult: if $m/n\in\mathbb{Q}$, then we send it to $$ \frac{f(m)}{f(n)}\in F $$ where $f\colon \mathbb{Z}\to F$ is the (unique) monomorphism. Is this a field homomorphism? Just a check.
Now we come to the order. First of all, positive integers are positive in $(F,\prec)$: if $n>0$, then $$ n=\underbrace{1+1+\dots+1}_{\text{$n$ times}} $$ and therefore $0\prec n$. Conversely, if $n<0$, then $$ n=-(\,\underbrace{\,1+1\dots+1}_{\text{$-n$ times}}\,) $$ and so $n\prec0$.
Any element of $\mathbb{Q}$ can be represented as $m/n$ with $n>0$, because $a/b=(-a)/(-b)$, where $a,b\in F$, $b\ne0$. So, let $0\prec m/n$ in the ordering of $F$, with $n>0$. Then, by the properties of ordered fields, $$ 0\prec n\cdot\frac{m}{n}=m $$ and therefore $m>0$. So a rational which is positive in $(F,\prec)$ is also positive in the usual order. A rational which is negative in $(F,\prec)$ is the opposite of a positive rational (in both orders).
Best Answer
Any Dedekind-complete ordered field can be defined to be the reals, $R$, although it is sometimes useful to have some other relationships between the members of $R. $ Examples: Assume that we have "the" field $Q$ of rationals, we can define $R$ as the set of equivalence classes of Cauchy sequences in $Q$, or as the union of $Q$ with the set of its proper Dedekind cuts, or by the usual set of decimal representations....