system
Given the following post, respond as if you're commenting on Reddit:
human
Are there metrics with triangle inequality defined stricter than necessary?

A metric must meet the following criteria, copied from Wikipedia:


* *d*(*x*, *y*) ≥ 0     (non-negativity, or separation axiom)
* *d*(*x*, *y*) = 0   if and only if   x = y     (identity of indiscernibles, or coincidence axiom)
* *d*(*x*, *y*) = *d*(*y*, *x*)     (symmetry)
* *d*(*x*, *z*) ≤ *d*(*x*, *y*) + *d*(*y*, *z*)    (subadditivity / triangle inequality).

Are there metrics in which last criterion is stricter:

*d*(*x*, *z*) < *d*(*x*, *y*) + *d*(*y*, *z*)

Does such change give any significant/interesting implications?
gpt
There can't be. Let *x* = *y* = *z* in the triangle inequality. Because of property two, you are then left with 0 < 0. You will need also the criterion that strict inequality only applies when *x*, *y* and *z* are all different.

You can create metric spaces with this property by restricting the Euclidean metric to a suitable subset. For example, let us define the following metric on the real numbers.

*d*(*x*, *y*) = | (*x*, *x*^(2)) – (*y*, *y*^(2)) |

Where | · | is the two-dimensional Euclidean norm. Essentially, this measures the distance between the two points on the parabola *f*(*x*) = *x*^(2) which correspond to two real numbers. The triangle inequality in a Euclidean space is only non-strict when the three points are collinear, but there are no three collinear points on a parabola.

As for your last question, allow me to answer in terms of embedded metric spaces in Euclidean spaces. (Maybe someone else knows more about more exotic things.)

Due to the collinearity requirement, such spaces cannot be linear, not even in a subset. They have to be everywhere curved, although that doesn't mean they can't be *locally* Euclidean (like a manifold—my example is one). They also have to be curved everywhere in the same direction, so if all the principal curvatures are never zero and of the same sign everywhere, that should work.

But don't quote me on that.