# Can an Ordered Ring be Considered as a Metric Space?

Consider an ordered ring $R$. One can define a function $$|\cdot|: R \rightarrow R$$ by |x| = \left\{ \begin{align} x &, x \geq 0 \\ -x&, x < 0 \end{align} \right. If we set $d(x,y) = |x - y|$, it can be shown that $d$ satisfies:

1. $d(x,y) \geq 0$ and $d(x,y) = 0$ iff $x=y$
2. $d(x,y) = d(y,x)$
3. $d(x,z) < d(x,y) + d(y,z)$

Here, $d$ looks very much like a metric except for the fact that $d$ actually produces an element of $R$, not an element $\mathbb{R}$. Otherwise, $(R, d)$ seems to have all of the features one would expect of a metric space. So, is there a way in which $(R, d)$ can be considered a metric space? Using this "metric", we can define Cauchy sequences and other items of interest but since the "metric" doesn't actually produce an element $\mathbb{R}$, anything that depends on the output of $d$ being a number would fail.

-
Problem: $|\cdot|: R \rightarrow R$. $d$ has $R \times R$ for domain. – Thomas Andrews Dec 7 '11 at 19:19
@ThomasAndrews Thanks for pointing that out; it's been fixed. – ItsNotObvious Dec 7 '11 at 19:21
While technically not a metric space, of course, since the distance isn't a real value, it has some of the properties of a metric space - specifically, a topology on $R.$ Is this topology the same as the order topology on $R$? Are the maps: $+:R\times R\rightarrow R$ and $\times:R\times R\rightarrow R$ continuous under this topology? – Thomas Andrews Dec 7 '11 at 19:29
@Jack: You can get such if and only if there is a $u\in R$ such that $0<u<1$. Then $0<ux<x$. – Thomas Andrews Dec 7 '11 at 20:18
(Whoops, the "only if" part of that last statement isn't true if $u$ is a zero divisor, which is apparently possible.) – Thomas Andrews Dec 7 '11 at 20:30

Added: The main results do not in fact depend on $R$ having no zero divisors, though I had to modify the arguments just a little in order to show this. I’ve made the necessary modifications below; significantly changed material is inset, like this note.

$R$ will be throughout a commutative ordered ring with identity, and $R^+\triangleq\{r\in R:r>0\}$. Define the $R$-norm $|\cdot|$ and $R$-metric $d$ as in the question. For $$x,r\in R let B(x,r)=\{y\in R:d(x,y)<r\}\,.$$

Prop. 1: The $R$-metric topology on $R$ coincides with the order topology.

Proof: If $a<x<b$, let $r=\min\{x-a,b-x\}>0$; clearly $x\in B(x,r)\subseteq(a,b)$. Conversely, $B(x,r)=(x-r,x+r)$. $\dashv$

Henceforth I assume that $R$ bears this topology.

Prop. 2: $R$ is discrete iff $R^+$ has a minimum element, and in that case $R$ is of course metrizable. $\dashv$

Assume henceforth that $R$ is not discrete.

Prop. 3: For each $x\in R^+$ there is a $y\in R^+$ such that $2y\le x$.

Proof: Let $x\in R^+$. Since $R$ is not discrete, there is a positive $u<x$, and there is a positive $v<x-u$. Let $y=\min\{u,v\}$; then $2y\le u+v=x$. $\dashv$ (This argument no longer depends on multiplication.)

Theorem 4: If there is a strictly decreasing sequence $\langle r_n:n\in\omega\rangle$ in $R^+$ that is co-initial in $R^+$, then $R$ is metrizable.

Proof: For $n\in\omega$ let $\mathscr{B}_n=\{B(x,r_n):x\in R\}$. Suppose that $V$ is an open nbhd of some $x\in R$. By Prop. 1 there is some $r\in R^+$ such that $B(x,r)\subseteq V$. By Prop. 3 there is an $n\in\omega$ such that $2r_n\le r$, whence $\operatorname{st}(x,\mathscr{B}_n)\subseteq V$. Thus, $\{\mathscr{B}_n:n\in\omega\}$ is a development for $R$. It’s well known that every linearly ordered topological space is collectionwise normal, so $R$ is metrizable by the Moore metrization theorem (also sometimes called the Bing metrization theorem). $\dashv$

Added: It takes a bit more work, but one can show that in this case $R$ has a metric $d$ such that the $R$-Cauchy sequences are precisely the ordinary Cauchy sequences in $\langle R,d\rangle$. (This is trivially true in the discrete case.)

Up to here, then, $R$-metrics don’t really give us anything new. However, there are non-metrizable ordered rings. The easiest example that comes to mind is a countable ultrapower of $\mathbb{Q}$.

Example 5: Let $X=\mathbb{Q}^\omega$, with addition and multiplication defined coordinatewise, and let $\mathscr{U}$ be a free ultrafilter on $\omega$. Let $I=\big\{x\in X:\{n\in\omega:x(n)=0\}\in\mathscr{U}\big\}$; $I$ is a maximal ideal in $X$, so $R=X/I$ is a field. For $[x],[y]\in R$ define $[x]\le[y]$ iff $\{n\in\omega:x(n)\le y(n)\}\in\mathscr{U}$; the fact that $\mathscr{U}$ is an ultrafilter ensures that this is a linear order on $R$, and it’s easily seen to be compatible with the ring structure on $R$.

Now suppose that $\{x_n:n\in\omega\}\subseteq R^+$.

Now suppose that $\{[x_n]:n\in\omega\}\subseteq R^+$; I’ll show that there is an $x\in R^+$ such that $x<x_n$ for every $n\in\omega$, thereby showing that $R$ is not first countable (and hence certainly not metrizable).

Let $U_0=\{k\in\omega:x_0(k)>0\}$. Given $U_n$ for some $n\in\omega$, let $$U_{n+1}=U_n\cap\{k>n:x_{n+1}(k)>0\}.\,\tag{1}$$ Since $[x_n]>[0]$ for each $n\in\omega$, each $U_n\in\mathscr{U}$. Clearly $U_0\supseteq U_1\supseteq\dots$, and $\bigcap\limits_{n\in\omega}U_n=\varnothing$. Now define $x\in X$ as follows:

$$x(k)=\begin{cases} 0,&\text{if }k\in\omega\setminus U_0\\\\ \frac12\min\{x_m(k):m\le n\},&\text{if }k\in U_n\setminus U_{n+1}\;. \end{cases}$$

It follows from $(1)$ that if $k\in U_n\setminus U_{n+1}$, $x_m(k)>0\,$ for every $m\le n$ and hence that $x(k)>0$. Thus, $x(k)>0$ for every $k\in U_0\in\mathscr{U}$, and hence $[x]>[0]$. On the other hand, for each $n\in\omega$ we have $x(k)<x_n(k)$ for every $k\in U_n\in\mathscr{U}$, so $[x]<[x_n]$. $\dashv$

Definition 6: A sequence $\langle x_n:n\in\omega\rangle$ is $R$-Cauchy iff for each $r\in R^+$ there is an $n_r\in\omega$ such that $d(x_n,x_m)<r$ whenever $n,m\ge n_r$.

Obviously if $R$ is discrete, $R$-Cauchy sequences are eventually constant, while in the non-discrete metrizable case they behave like ordinary Cauchy sequences in metric spaces.

Prop. 7: If $R$ is not first countable, a sequence in $R$ is $R$-Cauchy iff it is eventually constant.

Proof: Clearly $R$ has a nested base at $0$, and this base cannot be countable: if it were, $R$ would be first countable. Let $\langle x_n:n\in\omega\rangle$ be an $R$-Cauchy sequence in $R$, and let $$D=\{d(x_n,x_m):n,m\in\omega\}\cap R^+,$$ the set of non-zero distances between terms of the sequence. $D$ is countable, so there is an $r\in R^+$ such that $r<d\,$ for every $d\in D$. Fix $n_r\in\omega$ such that $d(x_n,x_m)<r$ whenever $n,m\ge n_r$; clearly $d(x_n,x_m)=0$ and $x_n=x_m$ whenever $n,m\ge n_r$, i.e., the sequence is eventually constant. $\dashv$

Cor. 8: If $R$ is not discrete, $R$ is metrizable iff it has an $R$-Cauchy sequence that is not eventually constant. $\dashv$

-
Thank you for the very nice and comprehensive answer, though it will take me a bit to fully digest it. – ItsNotObvious Dec 9 '11 at 15:24
So, in summary... yes. – Pete L. Clark Dec 9 '11 at 19:56