Generalization of metric that can induce ordering

Question

I was wondering if there is some generalization of the concept metric to take positive and negative and zero values, such that it can induce an order on the metric space? If there already exists such a concept, what is its name?

For example on $\forall x,y \in \mathbb R$, we can use difference $x-y$ as such generalization of metric.

Thanks and regards!

Answer 1

If we have a function $\delta$ such that

1. $\delta(x,y)=0$ if and only if $x=y$
2. $\delta(x,y) = -\delta(y,x)$
3. If $\delta(x,y)\geq 0$ and $\delta(y,z)\geq 0$, then $0\leq \delta(x,z) \leq \delta(x,y)+\delta(y,z)$.

then $d(x,y) = |\delta(x,y)|$ clearly defines a metric. Furthermore $x\geq y$ if and only if $\delta(x,y)\geq 0$ defines a total order.

Conversely, if $d$ is a metric and $\geq$ a total order, $\delta(x,y) =\begin{cases} d(x,y) & x\geq y \\ -d(x,y) & x<y \end{cases}$ satisfies all the axioms above.

Answer 2

You will have to lose some of the other axioms of a metric space as well since the requirement that $d(x,y)\ge 0$ in a metric space is actually a consequence of the other axioms: $0=d(x,x)\le d(x,y)+d(y,x)=2\cdot d(x,y)$, thus $d(x,y)\ge 0$. This proof uses the requirements that $d(x,x)=0$, the triangle inequality, and symmetry.

There are notions of generalizations of metric spaces that weaken these axioms. I think the one closest to what you might be thinking about is partial metric spaces (where $d(x,x)=0$ is dropped).