Point in Standard Discrete Metric Space is Isolated
Jump to navigation
Jump to search
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.
$\blacksquare$