Pseudometric induces Topology

Theorem
Let $S \ne \O$ be a non-empty set.

Consider a pseudometric space $\struct {S, d}$ where $d: S \times S \to \R_{\ge 0}$ is a pseudometric.

Then $\struct {S, d}$ gives rise to a topological space $\struct {S, \tau_d}$ whose topology $\tau_d$ is defined (or induced) by $d$.

Proof
Let $\tau_d$ be the set of all $X \subseteq S$ which are open in the sense that:
 * $\forall y \in X: \exists \epsilon > 0: \map {B_\epsilon} y \subseteq X$

where $\map {B_\epsilon} y$ is the open $\epsilon$-ball of $y$.

Equivalently:
 * $\forall x \in X: \exists \epsilon \in \R_{>0}: \forall y \in S: \map d {x, y} < \epsilon \implies y \in X$

We need to show that $\tau_d$ forms a topology on $S$.

We examine each of the open set axioms in turn.

Let $\family {U_i}_{i \mathop \in I}$ be an indexed family of open sets of $S$.

Let $\ds V = \bigcup_{i \mathop \in I} U_i$ be the union of $\family {U_i}_{i \mathop \in I}$.

Then by the definition of union:
 * $\forall x \in V: \exists i \in I: x \in U_i$

and so by the definition of open set:
 * $\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U_i \subseteq V$

Hence $V$ is open by definition.

Let $U$ and $V$ be open sets of $S$.

Let $x \in U \cap V$.

Then:
 * $\exists \epsilon_U \in \R_{>0}: \map {B_{\epsilon_U} } x \subseteq U$
 * $\exists \epsilon_V \in \R_{>0}: \map {B_{\epsilon_V} } x \subseteq V$

Let $\epsilon := \min \set {\epsilon_U, \epsilon_V}$.

Then:
 * $\map {B_\epsilon} x \subseteq U \cap V$

Hence $U \cap V$ is open by definition.

From Open Sets in Pseudometric Space:
 * $\O \in \tau_d$

and:
 * $S \in \tau_d$

All the open set axioms are fulfilled, and the result follows.

Also see

 * Pseudometrizable: any topological space which is homeomorphic to such a $\struct {S, \tau_d}$.