Metric Induces Topology

Theorem
Let $\left({M, d}\right)$ 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 Subsets, the union of any collection of open subsets of a metric space is also open.


 * $(2): \quad$ By Intersection of Open Subsets, the intersection of a finite number of open subsets is open.


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

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 reason behind the definition of open sets in topology.