The 3 conditions for a metric space $(X,d)$ are that for all $x,y,z\in X, $ $$d(x,y)=d(y,x)$$ $$d(x,y)\geq0,d(x,y)=0 \iff x=y$$ $$d(x,y)+d(y,z)\geq d(x,z)$$ Are there any interesting results in altering the third condition so that we get$$d(x,y)+d(y,z)= d(x,z)$$ where the elements $x,y,z$ act in the way that vectors do, in so much as we get $xy→+yz→=xz→ $, we get in this "metric" that $d(x,y)+d(y,z)=d(x,z)$. ?
Thanks to @Arthur for the edit suggestion