Consider a discrete set of points $X$ and a distance function $d : X \times X \to \mathbb{R}_+$. $d(\cdot,\cdot)$ is said to be a metric over $X$ if the following three constraints are satisfied:
- $\forall x,y \in X,\ d(x,y) = 0 \Leftrightarrow x = y$.
- $\forall x,y \in X,\ d (x,y) = d (y,x)$ [symmetry].
- $\forall x,y,z \in X,\ d(x,y) + d(y,z) \ge d (x,z)$ [triangle inequality].
A $k$-metric space is $X$ equipped with a distance function that always satisfies the first two constraints, and $k$ of the $\binom{|X|}{2} (|\mathcal{X}| - 2)$ non-trivial$^*$ triangle inequality constraints.
What is the maximum $k$ such that there exists a $k$-metric space that is not a metric space?
$*$ A set of triangle inequalities is non-trivial if each is composed of distinct $x,y,z$ and inequalities remain distinct upon all possible symmetry transformations.
From what I understand, if we let $\text{d}(x,y)=1$ when $x\neq y$, except for the distance between $a$ to $b$ and $a$ to $c$ for some fixed $a,b, \text{ and } c \in X$; where we have $\text{d}(a,b)=0.1$ and $\text{d}(a,c)=0.1$, we will have $X$ as a $[\binom{|X|}{2}(|X|-2)-1]$-metric space.