Closure of Open Ball may not equal Closed Ball of Same Radius

Theorem
Let $M = \struct {A, d}$ be a metric space.

Let $x \in A$.

Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball of $x$ of radius $\epsilon$ for some $\epsilon \in \R_{>0}$.

Then it is not necessarily the case that:
 * $\map \cl {\map {B_\epsilon} x} = \set {y \in A: \map d {x, y} \le \epsilon}$

where $\cl$ denotes closure.

Proof
Proof by Counterexample:

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

From Closure of Open $1$-Ball in Standard Discrete Metric Space we have that:
 * $\map \cl {\map {B_1} x} = \set x$

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

Hence the result.