Consider a pseudometric on $\Bbb N$ by $d(p_k,p_n)=0;$ else $d(x,y)=|x-y|.$ Here $p_k,p_n$ are distinct primes.
Edit 5/14/2020: From the comment below the definition I gave does not satisfy the triangle inequality. So I think this will work? $\frac{1}{z-x}d(x,z) \le d(x,y)+d(y,z).$
In this model, distinct primes have zero distance between them and the primes are seen as a degenerate case of the natural numbers.
So it's weird because for example $d(2,137)=0=d(2,3).$
I read this blurb from "Encyclopedia of Mathematics:"
A topology on $X$ is determined by a pseudo-metric $d$ on $X$ as follows: A point $x$ belongs to the closure of a set $A⊆X$ if $d(x,A)=0,$ where $d(x,A)=\inf_{a∈A}d(x,a) .$ This topology is completely regular but is not necessarily Hausdorff: singleton sets can be non-closed. Every completely-regular topology can be given by a collection of pseudo-metrics as the lattice union of the corresponding pseudo-metric topologies. Analogously, families of pseudo-metrics can be used in defining, describing and investigating uniform structures.
So apparently $X=\Bbb N$ in this case. How do you define a geometry based on this metric?
The inequality indeed holds for $x\ne z$, but it doesn’t make the function $d$ a pseudometric. On the other hand, we can define a pseudometric $d’$ on $\Bbb N$ putting for each natural $m$ and $n$, $d(m,n)$ equal to $0$, if $m$ or $n$ is prime and equal to $1$, otherwise. Although in this pseudometric the prime numbers constitute a clique of zero diameter, I think that it is
politicallymathematically incorrect to say that the primes are degenerated natural numbers. All points of a pseudometric space are equal, although as natural numbers they can have special properties. Finally, I don’t know what means to define a geometry based on a (pseudo)metric, but a topology defined by $d’$ is rather boring. The set of prime numbers is endowed with the antidiscrete topology and all other points of the space are isolated.