Pseudometric induces Topology

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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.


Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets

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.

$\Box$


Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets

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

Let $x \in U \cap V$.

Then:

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

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.

$\Box$


Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology

From Open Sets in Pseudometric Space:

$\O \in \tau_d$

and:

$S \in \tau_d$

$\Box$


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

$\blacksquare$



Also see


Sources