Closure of Open Ball in Metric Space

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

Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball in $M$.

Let $y \in \map \cl {\map {B_\epsilon} x}$, where $\cl$ denotes the closure of $\map {B_\epsilon} x$.

Then $\map d {x, y} \le \epsilon$.

Proof
Suppose $\map d {x, y} > \epsilon$.

Then $\map {B_{\map d {x, y} - \epsilon} } y$ is an open set containing $y$ and not meeting $\map {B_\epsilon} x$.

Hence:
 * $y \notin \map \cl {\map {B_\epsilon} x}$