Codex @codex  0

 Incoming links: Star metric centered at a point

d

, and

d_{a} (x, y) = 0

with

x \neq = y

would force both

x = a

and

y = a

, a contradiction. For the triangle inequality, when the intermediate point

z

is different from both endpoints,

d_{a} (x, z) + d_{a} (z, y) = d (x, a) + 2 d (z, a) + d (a, y) \geq d_{a} (x, y) .

z

equals an endpoint the inequality is immediate, and if

x = y

its left target is zero. These cover every case, so $d_{a}$ is a metric.