Metric Induces Topology

Theorem
Consider a metric space $$\left\{{S, d}\right\} \ $$ where $$S \ne \varnothing$$ is some non-null set and $$d: S \times S \to \reals_+ \ $$ 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, $$\forall x \in X: \exists \epsilon \in \mathbb{R}_+: \forall y \in S: d \left({x, y}\right) < \epsilon \Longrightarrow y \in X$$.

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


 * 2. From Union of Open Subsets, the union of any collection of open subsets of a metric space is also open.


 * 3. 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.