Pseudometric induces Topology

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

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

Then $\left({S, d}\right)$ gives rise to a topological space $\left({S, \tau_d}\right)$ 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:
 * $\forall y \in X: \exists \epsilon > 0: B_\epsilon \left({y}\right) \subseteq X$

where $B_\epsilon \left({y}\right)$ is the open $\epsilon$-ball of $y$.

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

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

We examine each of the open set axioms in turn.

$(O1)$: Union of Open Sets
Let $\left \langle{U_i}\right \rangle_{i \in I}$ be an indexed family of open sets of $S$.

Let $\displaystyle V = \bigcup_{i \mathop \in I} U_i$ be the union of $\left \langle{U_i}\right \rangle_{i \in I}$.

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:
 * $\exists \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \subseteq U_i \subseteq V$

Hence $V$ is open by definition.

$(O2)$: Intersection of Open Sets
Let $U$ and $V$ be open sets of $S$.

Let $x \in U \cap V$.

Then:
 * $\exists \epsilon_U \in \R_{>0}: B_{\epsilon_U} \left({x}\right) \subseteq U$
 * $\exists \epsilon_V \in \R_{>0}: B_{\epsilon_V} \left({x}\right) \subseteq V$

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

Then:
 * $B_\epsilon \left({x}\right) \subseteq U \cap V$

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

$(O3)$: Empty Set and Set Itself
From Open Sets in Pseudometric Space:
 * $\varnothing \in \tau_d$

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 $\left({S, \tau_d}\right)$.