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 $\map d {x, y} = 0$.

Let $\eqclass x \sim$ 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:
 * $\map {d^*} {\eqclass x \sim, \eqclass y \sim} = \map d {x, y}$

is a metric.

Hence $\struct {X^*, d^*}$ 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 $z \in \eqclass x \sim$ and $w \in \eqclass y \sim$.

From Equivalence Class Equivalent Statements:
 * $\eqclass z \sim = \eqclass x \sim$ and $\eqclass w \sim = \eqclass y \sim$

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 $\map {d^*} {a, b} > 0$ where $a \ne b$.

So let $\map {d^*} {a, b} = 0$ where $a = \eqclass x \sim, b = \eqclass y \sim$.

Then $\map d {x, y} = 0$ so $y \in \eqclass x \sim$ and so $a = b$.

Hence the result.