Metric Induced by a Pseudometric

Theorem

Let $X$ be a set on which there is a pseudometric $d: X \times X \to \R$.

For any $x, y \in X$, let $x \sim y$ denote that $d \left({x, y}\right) = 0$.

Let $\left[\!\left[{x}\right]\!\right]$ denote the equivalence class of $x$ under $\sim$.

Let $X^*$ be the quotient of $X$ by $\sim$.

Then the mapping $d^*: X^* \times X^* \to \R$ defined by:

$d^* \left({\left[\!\left[{x}\right]\!\right], \left[\!\left[{y}\right]\!\right]}\right) = d \left({x, y}\right)$

is a metric.

Hence $\left({X^*, d^*}\right)$ is a metric space.

Proof

From Pseudometric Defines an Equivalence Relation we have that $\sim$ is indeed an equivalence relation.

First we verify that $d^*$ is well-defined.

Let $z \in \left[\!\left[{x}\right]\!\right]$ and $w \in \left[\!\left[{y}\right]\!\right]$.

Then we have:

$\left[\!\left[{z}\right]\!\right] = \left[\!\left[{x}\right]\!\right]$

and:

$\left[\!\left[{w}\right]\!\right] = \left[\!\left[{y}\right]\!\right]$

But:

• $d \left({z, w}\right) \le d \left({z, x}\right) + d \left({x, y}\right) + d \left({y, w}\right) \le d \left({x, y}\right)$

• $d \left({x, y}\right) \le d \left({x, z}\right) + d \left({z, w}\right) + d \left({w, y}\right) \le d \left({z, w}\right)$

So $d^*$ is independent of the elements chosen from the equivalence classes.

Now we need to show that $d^*$ is a metric.

To do that, all we need to do is show that $d^* \left({a, b}\right) > 0$ where $a \ne b$.

So let $d^* \left({a, b}\right) = 0$ where $a = \left[\!\left[{x}\right]\!\right], b = \left[\!\left[{y}\right]\!\right]$.

Then $d \left({x, y}\right) = 0$ so $y \in \left[\!\left[{x}\right]\!\right]$ and so $a = b$.

Hence the result.

$\blacksquare$