# how do you call a function that satisfies the metric axioms except for the coincidence axiom?

Hello everyone.

I am studying a bivariate function $d$ on a set $\mathcal{X}\times \mathcal{X}$ which is symmetric and satisfies all triangle inequalities but does not agree with the coincidence axiom (which says that $d(x,y)=0\Leftrightarrow x=y$).

Pseudodistances are such that only $d(x,y)=0 \Leftarrow x=y$.

In the case I am concerned with, the function is actually such that $d(x,y)>0$ for all $x,y$ (including $d(x,x)>0$) and so only the implication $\Rightarrow$ would be valid, trivially because $d(x,y)$ is never 0.

I have been looking for a proper generalization of distance functions that would take into account that case but I have not found any. Do such functions bear a name and have they ever been considered in the context of generalized metrics?

One example would be: $d(x,y) := \delta(x,y)+\delta(x,a)+\delta(y,b)$, where $\delta(x,y)$ is a usual metric, and $a \neq b$ are two arbitrary points in space (to have an example slightly less trivial than $d(x,y) := \delta(x,y) + \epsilon$... –  Suvrit Apr 28 '13 at 18:36
Thanks for all your comments! I gave it some extra thought. I see two links: negative definite kernels (in the sense of Berg Christensen Ressel. They consider negative definite kernels $\psi$ (p.82) that may not be such that $\psi(x,x)=0$. Negative definite kernels and distances are different, but they are somewhat related (the bigger, the more different). The other thing that's easy to check is that $1_{x\ne y} d(x,y)$ is itself a distance. It's not continuous.. but still a distance! –  mcuturi Apr 29 '13 at 4:49