Example of a complete, non-archimedean ordered field

Question

I'm looking for a concrete example of a complete (in the sense that all Cauchy sequences converge) but non-archimedean ordered field, to see that these two properties are independent (an example of archimedean non-complete ordered field is obviously the rationals).

Thank you in advance.

Answer 1

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 $x^{-1}$.

Answer 2

The best example (in fact, provably the simplest answer to your question) is the field of formal Laurent series that Hurkyl mentioned. But actually, you can see that this field is complete by finding a metric that induces the order topology. (It's a surprise that this is even possible!)

The elements of this field look like $\sum_{i=n}^{\infty} a_i x^i$. Let's write $\textrm{ord}(f)$ for the index of the first nonzero term, or $\infty$ if all the terms are zero. Then the ordering is given by

$f = \sum_{i=n}^{\infty} a_i x^i \quad > \quad 0$

when

$a_{\textrm{ord}(f)} > 0$. (Notice that this induces an ordering on the entire field, where $f > g$ iff $f - g > 0$.)

Ok, now let's give this baby a metric! You can verify on your own that

$d(f, g) = 2^{-\textrm{ord}(f-g)}$

is a metric, and that it induces the same topology as the order on the field.

The last thing to check is that the space is complete under the given metric. This is easy once you have the right intuition. Think of it like this: each integer index is one of the display windows of a slot machine. A Cauchy sequence allows the values in each window to spin, but as you progress further down the sequence, each spinner (starting from the leftmost) eventually stops. Therefore the value to which such a sequence converges is simply the formal power series obtained by taking the coefficient of each wheel after it's already stopped.

~~~~~

Also note that Harry Altman is right in general -- "most" nonarchimedean fields aren't second-countable, and so sequences don't suffice to characterize their topology. In this case you'd need nets or filters instead; thankfully the above field is actually metrizable, so you don't have to worry about this. (There's a nice characterization of the nonarchimedean fields that are metrizable, by the way.)

If you're interested in different notions of completeness, you'll find several (such as Cantor completeness and spherical completeness), but when it comes to ordered fields, the "right" one is Hilbert completeness.

Answer 3

In this context, complete means "order complete" rather than complete in the sense of metric spaces. A non-archimedian ordered field will necessarily have infinitely small elements, so I suspect that if you try to impose a metric you'll get something pathological like the closure of 0 containing all of the infinitely small elements.

You can construct an order-complete example by taking the field $\mathbb{R}(x)$ of rational functions over $\mathbb{R}$. Order it such that $x$ is less than every positive number but bigger than 0, and take its order completion.

Answer 4

The set of functions $$\mathbb{R}((\mathbb{Q})):=\{f:\mathbb{Q}\to \mathbb{R}\ |\ supp(f)\mbox{ is well-ordered}\},$$ where $supp(f):=\{x\in \mathbb{Q}\ | \ f(x)\neq 0\}$, is a field under the addition and multiplication defined as follows: for every $f,g\in \mathbb{R}((\mathbb{Q}))$ and $x\in \mathbb{Q}$,

1. $(f+g)(x):=f(x)+g(x)$,
2. $fg(x):=\displaystyle\sum_{a+b=x}f(a)g(b)$

Consider $\lambda:\mathbb{R}((\mathbb{Q}))\setminus\{0\}\to \mathbb{Q}$, $\lambda(f)=\min\{supp(f)\}$. For $f,g\in \mathbb{R}((\mathbb{Q}))$ we define: $0<f\Leftrightarrow f\neq 0\mbox{ and }f(\lambda(f))>0.$ Then $f<g\Leftrightarrow g-f>0$. Then $(\mathbb{R}((\mathbb{Q})),\leq)$ is a Cauchy-complete non-Archimedean ordered field called Hann field.

If the map $|\;\;|:\mathbb{R}((\mathbb{Q}))\to \mathbb{Q}$ is defined by $$ |f|:=\begin{cases} e^{-\lambda(f)} &, f\neq 0 \\ 0 &, f=0 , \end{cases} $$
then $(\mathbb{R}((\mathbb{Q})),|\;\;|)$ is a Cauchy-complete non-Archimedean valued field.

It is important to say that the order topology and the valuation topology coincide.

Another good example is the Levi-Civita field. $$\mathcal{R}:=\{f:\mathbb{Q}\to \mathbb{R} \:\vert\: supp(f)\cap(-\infty,n]\mbox{ is finite for every }n\in\mathbb{Z}\}$$ is a subfield of $\mathbb{R}((\mathbb{Q}))$. When we restrict the order of $\mathbb{R}((\mathbb{Q}))$ to $\mathcal{R}$, the latter becomes a Cauchy-complete, non-Archimedean ordered field. When we restrict the valuation of $\mathbb{R}((\mathbb{Q}))$ to $\mathcal{R}$, the latter becomes a Cauchy-complete, non-Archimedean valued field.

Both fields are real-closed field extensions of $\mathbb{R}$ and their complexifications $\mathbb{C}((\mathbb{Q})):=\mathbb{R}((\mathbb{Q}))+i\mathbb{R}((\mathbb{Q}))$ and $\mathcal{C}:=\mathcal{R}+i\mathcal{R}$ are algebraically closed.