Subset of Standard Discrete Metric Space is Open

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

Let $S \subseteq A$ be a subset of $A$.

Then $S$ is an open set of $M$.

Proof
From the definition of standard discrete metric:
 * $\forall x, y \in A: \map d {x, y} = \begin {cases}

0 & : x = y \\ 1 & : x \ne y \end {cases}$

Let $\epsilon \in \R_{>0}$ be such that $0 < \epsilon \le 1$.

Let $x \in S$.

Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball of $x$.

Then by definition of $\epsilon$ and $d$:
 * $\map {B_\epsilon} x = \set x$

Thus:
 * $\forall x \in S: \map {B_\epsilon} x \subseteq S$

Hence the result by definition of open set.