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.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $2$
• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.11$