Which topological spaces admit a nonstandard metric?

Question

My question is about the concept of nonstandard metric space that would arise from a use of the nonstandard reals R* in place of the usual R-valued metric.

That is, let us define that a topological space X is a nonstandard metric space, if there is a distance function, not into the reals R, but into some nonstandard R* in the sense of nonstandard analysis. That is, there should be a distance function d from X² into R*, such that d(x,y)=0 iff x=y, d(x,y)=d(y,x) and d(x,z) <= d(x,y)+d(y,z). Such a nonstandard metric would give rise to the nonstandard open balls, which would generate a metric-like topology on X.

There are numerous examples of such spaces, beginning with R* itself. Indeed, every metric space Y has a nonstandard analogue Y*, which is a nonstandard metric space. In addition, there are nonstandard metric spaces that do not arise as Y* for any metric space Y. Most of these examples will not be metrizable, since we may assume that R* has uncountable cofinality (every countable set is bounded), and this will usually prevent the existence of a countable local basis. That is, the nested sequence of balls around a given point will include the balls of infinitesimal radius, and the intersection of any countably many will still be bounded away from 0. For example, R* itself will not be metrizable. The space R* is not connected, since it is the union of the infinitesimal neighborhoods of each point. In fact, one can show it is totally disconnected.

Nevertheless, it appears to me that these nonstandard metric spaces are as useful in many ways as their standard counterparts. After all, one can still reason about open balls, with the triangle inequality and whatnot. It's just that the distances might be nonstandard. What's more, the nonstandard reals offer some opportunities for new topological constructions: in a nonstandard metric space, one has the standard-part operation, which would be a kind of open-closure of a set---For any set Y, let Y+ be all points infinitesimally close to a point in Y. This is something like the closure of Y, except that Y+ is open! But we have Y subset Y+, and Y++=Y+, and + respects unions, etc.

In classical topology, we have various metrization theorems, such as Urysohn's theorem that any second-countable regular Hausdorff space is metrizable.

Question. Which topological spaces admit a nonstandard metric? Do any of the classical metrization theorems admit an analogue for nonstandard metrizability? How can we tell if a given topological space admits a nonstandard metric?

I would also be keen to learn of other interesting aspects of nonstandard metrizability, or to hear of interesting applications.

I have many related questions, such as when is a nonstandard metric space actually metrizable? Is there any version of R* itself which is metrizable? (e.g. dropping the uncountable cofinality hypothesis)

Answer 1

The uniformity defined by a *R-valued metric is of a special kind.

Let $(n_i)_{i<\kappa}$ be a cofinal sequence of positive elements in *R. We may assume that $i < j$ implies that $n_i/n_j$ is infinitesimal.

Given a *R-valued metric space $(X,d)$ we can define a family $(d_i)_{i<\kappa}$ of pseudometrics by $d_i(x,y) = st(b(n_id(x,y)))$, where $st$ is the standard part function and $b(z) = z/(1+z)$ to make the metrics bounded by $1$. The topology on $X$ is defined by the family of pseudometrics $(d_i)_{i<\kappa}$. Note that these pseudometrics have a the following special property:

(+) If $i < j < \kappa$ and $d_j(x,y) < 1$ then $d_i(x,y) = 0$.

Every uniform space can be defined by a family of pseudometrics, but it is rather unusual for the family to have property (+) when $\kappa > \omega$.

On the other hand, given a family of pseudometrics $(d_i)_{i<\kappa}$ bounded by $1$ with property (+), then we can recover a *R-valued metric by defining $d(x,y) = d_i(x,y)/n_i$, where $i$ is minimal such that $d_i(x,y) > 0$.

Answer 2

These are $\omega_\mu$-metrizable spaces, where $\omega_\mu$ is the cofinality of ${}^*\mathbb{R}$. Take a decreasing sequence $\langle x_\alpha:\alpha<\omega_\mu\rangle$ of positive elements that converges to $0$; because $\omega_\mu$ is regular uncountable one can ensure that $x_{\alpha+1}/x_\alpha$ is always infinitesimal. Any `metric' $d:X\times X\to{}^*\mathbb{R}$ can be converted to an ultrametric $\rho$ with values in $\{x_\alpha:\alpha<\omega_\mu\}$; it satisfies the strong triangle inequality: $\rho(x,z)\le\max\{\rho(x,y),\rho(y,z)\}$. These spaces are also called linearly uniformizable: the sets $A_\alpha=\lbrace (x,y):\rho(x,y) < x_\alpha \rbrace $ form a (linearly ordered) base for a uniformity that generates the same topology as $\rho$. Also, because of the uncountable cofinality, these spaces are $P$-spaces: $G_\delta$-sets are open. The only metrizable such spaces must be discrete. Here is one of the earliest papers that I am aware of to treat this kind of thing systematically.

Answer 3

I don' t know how relevant this is to your question, but some topologists have investigated quite extensively topological spaces with metrics in a lattice-ordered group, ie a group with a lattice order, compatible with the group operation. Your R*-metric spaces seem to be a chapter of this general theory (and similarly the theory of spaces with metrics in ON, which maybe are also of interest to you), so perhaps a few questions you have touched upon in your post can be answered from a broader perspective.

Here is an interesting paper by several authors, one of whom is Ralph Kopperman:

http://www.wku.edu/~tom.richmond/Papers/l-Groups.pdf

PS Kopperman has two desiderable features from your geolocation's standpoint, the first is working in the NYC area (CCNY if I am not mistaken), and the second one is being a friend and collaborator of Prabhud Misra (and thus, via the small world law, very reachable). If I remember well, Ralph is an expert in these type of things, and so you may ask him for further infos, if needs be.

Answer 4

I once considered the following situation: $[0;1]^X$ is metrizable (with values in $\mathbb{R}$), if $X$ is countable. So I wondered whether $[0;1]^\mathbb{R}$ is nonstandard-metrizable for some totally ordered field $F$.

However it turns out that it does not admit a totally ordered local basis, so it can't be nonstandard-metrizable. I could give more details, if somebody is interested in them.