Sufficient and necessary binary relation for given metric

Question

Problem:

Let $p$ be binary relation on the set $X$, for any $x,y$ $\in$ $X$ let:

$$ d(x,y) = \begin{cases} 3, & \text{for } \hspace{0.5cm}xpy \\ 0, & \text{for} \hspace{0.5cm} \neg xpy \end{cases} $$

Determine necessary and sufficient condition(characteristic of relation $p$) so that $d$ is metric on the set $X$.

Necessary conditions for metric:

$$d(x,y)=0 \Leftrightarrow x=y$$

$$d(x,y)= d(x,y)=0 $$

$$d(x,z)\leq d(x,y) + d(y,z)$$

Possible solution:

The binary relation must be symmetric, because of the second condition for metric, but I do not know whether that is necessary and sufficient. It seems to me that the relation $p$ is non-equality, is it possible? Any help is appreciated.

Answer 1

$d$ is a metric if and only if $p$ is equal to the relation $\ne$, or $$(x,y) \in p \iff x \ne y$$

Assume that $p$ is equal to $\ne$ and let's check that $d$ is a metric.

1. Clearly $d \ge 0$. For $x \in X$ we have $x = x$ so $(x,x) \notin p$ which implies $d(x,x) = 0$. Assume $d(x,y) = 0$ for some $x,y \in X$. Then $(x,y) \notin p$ so $x = y$.
2. We have $$d(x,y) = 3 \iff (x,y) \in p \iff x \ne y \iff y \ne x \iff (y,x) \in p\iff d(y,x) = 3$$
3. Since $d$ attains only the values $0,3$ we only have to check that it isn't possible that $d(x,z) = 3$ but $d(x,y) = d(y,z) = 0$. This would be equivalent to $x \ne z$ and $x = y = z$, which is a contradiction.

Conversely, if $p$ isn't equal to $\ne$, then either

1. $p$ doesn't contain $\ne$, so there exist $x,y \in X, x \ne y$ such that $(x,y) \notin p$. It follows $d(x,y) = 0$ but $x \ne y$ so $d$ isn't a metric.
2. $\ne$ doesn't contain $p$ so there exists $x \in X$ such that $(x,x) \in p$. This implies $d(x,x) = 3$ so $d$ isn't a metric.