Metric Induces Topology

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

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 metrizable.

Proof
Let $\vartheta_{\left({S, d}\right)}$ be the set of all $X \subseteq S$ which are open in the sense that $\forall y \in X: \exists \epsilon \left({y}\right) > 0: N_{\epsilon \left({y}\right)} \left({y}\right) \subseteq U$.

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

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 Metric Space $\varnothing \in \vartheta_{\left({S, d}\right)}$ and $S \in \vartheta_{\left({S, d}\right)}$.


 * $(2): \quad$ From Union of Open Subsets, the union of any collection of open subsets of a metric space is also open.


 * $(3): \quad$ From Intersection of Open Subsets, the intersection of a finite number of open subsets is open.

Hence the result.

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.