Metric Induces Topology

Theorem
Let $M = \struct {A, d}$ be a metric space.

Then the topology $\tau$ induced by the metric $d$ is a topology on $M$.

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


 * $(1): \quad$ By Union of Open Sets of Metric Space is Open, the union of any collection of open sets of a metric space is open.


 * $(2): \quad$ By Finite Intersection of Open Sets of Metric Space is Open, a finite intersection of [Definition:Open Set of Metric Space|open sets]] of a metric space is open.


 * $(3): \quad$ By Open Sets in Metric Space, $\O \in \tau$ and $A \in \tau$.

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 motivation behind the definition of open sets in topology.