Let $R$ be a set of points and $\mathbb{D}$ be a totally ordered field. Further consider a function $\rho:R\times R \rightarrow \mathbb{D}$.

$\langle R,\mathbb D,\rho\rangle$ is a metric space if $\rho$ has the following properties:

1) $\rho(x,y) \geqslant 0$

2) $\rho(x,y) = 0$ iff $x = y$

3) $\rho(x,y) = \rho(y,x)$

4) $\rho(x,z) \leqslant \rho(x,y) + \rho(y, z)$

Does this less restricted definition properly induce a topology on $R$ in the way expected by a standard metric and are there spaces which metrizable under this concept of metric that are not metrizable under the usual restriction?

  • $\begingroup$ Every totally ordered field contains a copy of the rationals in them. And metric spaces that are typically defined with the reals can be redefined using the rationals because of their denseness. This kind of parallels how Cauchy completed the reals - he used the rationals as his totally ordered field. $\endgroup$ – Robert Wolfe Oct 26 '14 at 18:40
  • $\begingroup$ Interesting, but I'm not sure if this answers the question. Are you implying that using the collection of open epsilon balls under the above metric as the basis for a topology would not induce any topologies that cannot be induced using the standard collection of metrics? $\endgroup$ – Daniel Goldman Oct 26 '14 at 19:05

In general, these two concepts of metrisability are different. Consider an ultrapower $\mathbb{D}$ of $\mathbb{R}$ over a non-principal ultrafilter on $\mathbb{N}$. It is readily seen that $\mathbb{D}$ is non-archimedean: consider the element

$$[(1,2,3,4, \ldots )].$$

I claim that each basis of neighbourhoods of the origin induced from your non-standard metric is uncountable. If there were a countable basis of the origin, $\mathbb{D}$ would be archimedean. This proves that $\mathbb{D}$ with the corresponding topology is not first countable, hence non-metrisable.

  • $\begingroup$ So the class of "metrizable" topologies under the nonstandard definition would be larger, but would clearly include every topological space that is metrizable under normal conditions. $\endgroup$ – Daniel Goldman Oct 26 '14 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.