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 $\eqclass w \sim = \eqclass y \sim$
| \(\ds \map d {z, w}\) | \(\le\) | \(\ds \map d {z, x} + \map d {x, y} + \map d {y, w}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map d {x, y}\) | Definition of $\sim$ | |||||||||||
| \(\ds \map d {x, y}\) | \(\le\) | \(\ds \map d {x, z} + \map d {z, w} + \map d {w, y}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map d {z, w}\) | Definition of $\sim$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map d {x, y}\) | \(=\) | \(\ds \map d {z, w}\) |
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.
$\blacksquare$