Standard Discrete Metric induces Discrete Topology

Theorem
Let $M = \left({A, d}\right)$ be the discrete metric space on $A$.

Then $d$ induces the discrete topology on $A$.

Thus the discrete topology is metrizable.

Proof
A set $U \subseteq A$ is open in $M$ iff:


 * $\forall x \in U: \exists \epsilon \in \R_+: \forall y \in A: d \left({x, y}\right) < \epsilon \implies y \in A$

Let $a \in A$ and consider the set $\left\{{a}\right\}$.

Let $\epsilon = 1$.

From the definition of the discrete metric on $A$, we have that:


 * $\forall x, y \in A: d \left({x, y}\right) < 1 \implies x = y$.

That is:
 * $\forall x \in U: \forall y \in \left\{{a}\right\}: d \left({x, y}\right) < 1 \implies y = a$

So $\left\{{a}\right\}$ is open in $M$.

This holds for all $a \in A$.

Note that any $\epsilon \le 1$ works here. $1$ is used because it is obvious.

From Metric Induces a Topology it follows that $\left\{{a}\right\}$ is an open set in $\left({A, \vartheta_{A, d}}\right)$.

Hence the result from Basis for Discrete Topology.