Pseudometric Induces a Topology

Theorem

Consider a pseudometric space $\left({S, d}\right)$ where $S \ne \varnothing$ is some non-null set and $d: S \times S \to \R_+$ is a pseudometric.

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

Any topological space which is homeomorphic to such a $\left({S, \vartheta_{\left({S, d}\right)}}\right)$ is defined as pseudometrizable.

Proof

Let $\vartheta_{\left({S, d}\right)}$ be the set of all $X \subseteq S$ which are open in the sense that:

$\forall y \in X: \exists \epsilon \left({y}\right) > 0: N_{\epsilon \left({y}\right)} \left({y}\right) \subseteq X$

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

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

We need to show that $\vartheta_{\left({S, d}\right)}$ forms a topology on $S$.

We examine each of the criteria for being a topology separately.

$(1): \quad$ From Open Sets in Pseudometric Space $\varnothing \in \vartheta_{\left({S, d}\right)}$ and $S \in \vartheta_{\left({S, d}\right)}$.

$(2): \quad$ Let $\{U_i\}_{i\in I}$ be a family of open subsets, and let $V=\bigcup_{i\in I}U_i$ be the union of that collection of open subsets. Then for any $x\in V$, there is some $i\in I$ such that $x\in U_i$ and from the definition $\exists \epsilon \left({x}\right) > 0: N_{\epsilon \left({x}\right)} \left({x}\right) \subseteq U_i\subseteq V$. Hence $V$ is open by definition.

$(3): \quad$ Let $U$ and $V$ be two open sets, and $x\in U\cap V$. Then $\exists \epsilon_U \left({x}\right) > 0: N_{\epsilon_U \left({x}\right)} \left({x}\right) \subseteq U$ and $\exists \epsilon_V \left({x}\right) > 0: N_{\epsilon_V \left({x}\right)} \left({x}\right) \subseteq V$. Take $\epsilon=\min\{\epsilon_U,\epsilon_V\}$, then $N_{\epsilon \left({x}\right)} \left({x}\right) \subseteq U\cap V$. Hence $U\cap V$ is open by definition.

Hence the result.

$\blacksquare$

Note

Thus it can be seen that the concept of an open set as applied to a metric space is directly equivalent to that of an open set as applied to a topological space.

This is the reason behind the definition of open sets in topology.

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$: Uniformities