Closure of Open 1-Ball in Standard Discrete Metric Space

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

Let $x \in A$.

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

Then:


 * $\map \cl {\map {B_1} x} = \set x$

while:
 * $\set {y \in A: \map d {x, y} \le 1} = A$

Proof
By definition of the standard discrete metric:


 * $\map d {x, y} = \begin {cases}

0 & : x = y \\ 1 & : x \ne y \end {cases}$

That is:
 * $\forall \tuple {x, y} \in A: \map d {x, y} \le 1$

Thus:
 * $\set {y \in A: \map d {x, y} \le 1} = A$

From Open Ball in Standard Discrete Metric Space:
 * $\map {B_1} x = \set x$

Let $y \in A: y \ne x$.

By definition of closure:
 * $\map \cl {\map {B_1} x} = \paren {\map {B_1} x}^i \cup \paren {\map {B_1} x}'$

where:
 * $\paren {\map {B_1} x}^i$ denotes the set of isolated points of $\map {B_1} x$
 * $\paren {\map {B_1} x}'$ denotes the set of limit points of $\map {B_1} x$.

But:
 * from Point in Standard Discrete Metric Space is Isolated, all points in $\map {B_1} x$ are isolated:
 * $\paren {\map {B_1} x}^i = \map {B_1} x$


 * from Set in Standard Discrete Metric Space has no Limit Points, $\map {B_1} x$ has no limit points:
 * $\paren {\map {B_1} x}' = \O$

Hence: