Subset of Standard Discrete Metric Space is Open

Theorem
Let $M = \left({A, d}\right)$ 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: d \left({x, y}\right) = \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 $B_\epsilon \left({x}\right)$ be the open $\epsilon$-ball of $x$.

Then by definition of $\epsilon$ and $d$:
 * $B_\epsilon \left({x}\right) = \left\{{x}\right\}$

Thus:
 * $\forall x \in S: B_\epsilon \left({x}\right) \subseteq S$

Hence the result by definition of open set.