Point in Standard Discrete Metric Space is Isolated

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Theorem

Let $M = \struct {S, d}$ be the standard discrete metric space on a set $A$.

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

Let $\alpha \in H$.


The $\alpha$ is an isolated point of $H$.


Proof

By definition of the standard discrete metric:

$\map d {x, y} = \begin {cases} 0 & : x = y \\ 1 & : x \ne y \end {cases}$

Let $\alpha \in H$.

Let $\map {B_1} \alpha$ be the open $1$-ball of $\alpha$ in $M$.

Thus:

\(\ds \map {B_1} \alpha \cap H\) \(=\) \(\ds \set {y \in H: \map d {\alpha, y} < 1}\)
\(\ds \) \(=\) \(\ds \set \alpha\)

Hence the result by definition of isolated point.

$\blacksquare$