The discrete metric $d_0$ can take two values $0$ and $1$. Can a metric function $d_X$ on a set $X$ attain exactly three distinct values?
Going through the route of actually finding a metric instead of disproving one exists, I tried finding a metric which has values $0$, $a$ and $b$. But by trying to define it in a similar way as the discrete metric, I had a hard time finding one that could satisfy the three requirements to be a metric.
Disproving it on the other hand sounds pretty easy tho, by just taking $x, y, z \in X$, I set the two values to random distances like $d_X(x, y) = a = d_X(y,x)$ , $d_X(x, z) = b = d_X(z, x)$ and finally $d_X(y, z) = c = d_X(z, y)$, with $c$ being either $a$ or $b$. Applying this to the triangle inequality gives us.
$d_X(x, z) \leq d_X(x, y) + d_X(y,z)$
$d_X(x, y) \leq d_X(x, z) + d_X(z,y)$
$d_X(y, z) \leq d_X(y, x) + d_X(x,z)$
Which gives us the linear inequalities:
$b \leq a + c$
$a \leq b + c$
$c \leq a + b$
This cannot hold if $c$ is either $a$ or $b$. Is this proof correct? And more importantly, isn't there a better proof, because this sounds like a very inefficient proof.