Set in Standard Discrete Metric Space has no Limit Points

Theorem

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

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

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

Let $\alpha \in S$.

By definition, $\alpha$ is a limit point of $H$ if and only if every deleted $\epsilon$-neighborhood $\map {B_\epsilon} \alpha \setminus \set \alpha$ of $\alpha$ contains a point in $A$:

$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap H \ne \O$

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

Thus:

$\ds \map {B_1} \alpha$

$=$

$\ds \set {y \in S: \map d {\alpha, y} < 1}$

$\ds $

$=$

$\ds \set \alpha$

$\ds \leadsto \ \ $

$\ds \map {B_1} \alpha \setminus \set \alpha$

$=$

$\ds \O$

$\ds \leadsto \ \ $

$\ds \paren {\map {B_1} \alpha \setminus \set \alpha} \cap H$

$=$

$\ds \O$

So there exists a deleted $\epsilon$-neighborhood $\map {B_\epsilon} \alpha \setminus \set \alpha$ of $\alpha$ containing no points of $H$.

Hence $\alpha$ is not a limit point of $H$

As $\alpha$ is arbitrary, the result follows.

$\blacksquare$