Let $X$ be a non-empty set. Let $f:X\times X\to \mathbb{R}$ satisfying the following properties,

  • $f(x,y)=0\iff x=y$ for all $x,y\in X$.

  • $f(x,y)=-f(y,x)$ for all $x,y\in X$.

  • $f(x,y)=f(x,z)+f(z,y)$ for all $x,y,z\in X$.

If such a function exists, call the function $f$ to be a pre-metric on $X$. Prove that,

  1. The function $d(x,y)=|f(x,y)|$ defined a metric on $X$ where $|\cdot|$ is the absolute value function of $\mathbb{R}$.

  2. From the previous result you can conclude that we can always get a metric from a pre-metric but is the converse always true?

  3. If the converse doesn't hold in general, what condition(s) on $d$ are needed to ensure that the converse also holds?

The first part of the problem is easy and I have proved it but for the second and third part I got nowhere. Can anyone help me?

  • $\begingroup$ What about the unit triangle? $\endgroup$ – Henricus V. Jan 31 '16 at 5:46
  • $\begingroup$ @HenryW: What is an unit triangle? $\endgroup$ – user 170039 Jan 31 '16 at 5:47
  • 1
    $\begingroup$ A metric space with only three elements. The distance between each pair of elements is $1$. $\endgroup$ – Henricus V. Jan 31 '16 at 5:47
  • 1
    $\begingroup$ "Set equipped with a pre-metric" seems very similar to "one-dimensional real affine space." $\endgroup$ – goblin Jan 31 '16 at 5:59
  • $\begingroup$ @goblin: Yes, you are right. When I asked my friend who gave me this exercise, he told me that the motivation for the problem came from "one-dimensional real affine space". $\endgroup$ – user 170039 Jan 31 '16 at 6:33

A necessary and sufficient condition on a metric space $(X,d)$ for $d$ to come from some pre-metric is that $(X,d)$ be isometric to a subspace of the metric space $\mathbb{R}$ with the ordinary distance.

We may assume $X \ne \varnothing$. (When $X = \varnothing$, both conditions are true.)

Let $f$ be a pre-metric on $X$, and fix an "origin" $a \in X$. Write $g(x) = f(a,x)$. For any $x,y \in X$, we have $f(x,y) = f(x,a) + f(a,y) = - f(a,x) + f(a,y) = g(y) - g(x)$. Moreover, if $g(x) = g(y)$, then by the foregoing, $f(x,y) = 0$, so $x = y$.

Thus $g$ is a bijection between $X$ and a subset of $\mathbb{R}$, and $f(x,y) = g(y) - g(x)$. The distance $d(x,y)=|f(x,y)|$ corresponds via the bijection $g$ to the ordinary distance on $\mathbb{R}$.

Conversely, if $g$ is an isometry of some metric space $X$ onto a subset of $\mathbb{R}$, the distance function on $X$ is obtained from the pre-metric $f(x,y) = g(y) - g(x)$.

  • $\begingroup$ Of course, there's nothing to stop one from trying to find characterizations of metric spaces isometric to subspaces of $\mathbb{R}$. $\endgroup$ – David Jan 31 '16 at 6:55
  • $\begingroup$ Interestingly enough, if we define a "pre-pseudometric" as a function $f:X\times X\to \mathbb{R}$ on a non-empty set $X$ such that, (1) $x=y\implies f(x,y)=0$, (2) $f(x,y)=-f(y,x)$ for all $x,y\in X$ and (3) $f(x,y)=f(x,z)+f(z,y)$ for all $x,y,z\in X$ then we can say that every pseudometric can be obtained from a pre-pseudometric and vice versa. $\endgroup$ – user 170039 Feb 3 '16 at 14:57

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.