Example 2

On any set $X$ the pseudometric $d$, defined by the relations $d(x, x) = 0$ and $d(x, y) = 1$ if $x \neq y$, is a metric. The uniformity it defines on $X$ is the discrete uniformity.

We have a definition equivalent to Definition 1 if we say that a metric is a finite pseudometric such that the uniformity defined by this pseudometric is Hausdorff; a finite pseudometric which is equivalent to a metric is therefore a metric.

Uniform spaces defined by a single pseudometric (which we may assume to be finite) can be reduced to metric spaces when the pseudometric is not a metric. Let $f$ be such a pseudometric on a set $X$, and let $\mathscr{U}$ be the uniformity defined by $f$; $\mathscr{U}$ is not Hausdorff, and the intersection of the entourages of $\mathscr{U}$ is the subset of $X \times X$ defined by the equivalence relation $f(x, y) = 0$. Let $R$ denote this relation. If $x \equiv x' \pmod{R}$, then by the triangle inequality we have

and similarly $f(x', y) \leq f(x, y)$, so that $f(x, y) = f(x', y)$; in other words, $f$ is a function compatible (in $x$ and $y$) with the equivalence relation $R$ (Set Theory, R, § 5, no. 7). Let $\overline{f}$ be the function induced by $f$ on the quotient set; $\overline{f}$ is defined on $(X/R) \times (X/R)$, and if $x$ and $y$ are any two points of $X$ and if $\dot{x}$ and $\dot{y}$ denote the equivalence classes (mod $R$) of $x$ and $y$ respectively, then we have $\overline{f}(\dot{x}, \dot{y}) = f(x, y)$. It follows immediately that $\overline{f}$ is a metric on $X/R$; it is called the metric associated with the pseudometric $f$; furthermore, the uniformity it defines on $X/R$ is precisely the Hausdorff uniformity associated with $\mathscr{U}$ by the definition of this uniformity (Chapter II, § 3, no. 8, Remark). Thus, by passing to a suitable quotient space, the uniform structure defined by a single pseudometric can be reduced to the structure of a metric space.

Proposition 1 of § 1, no. 3 determines the structure of the completion of a metric space: