Metric Induced by a Pseudometric

Theorem
Let $$X$$ be a set on which there is a pseudo-metric $$d: X \times X \to \R$$.

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 Pseudo-Metric 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.