Open Real Interval is Open Ball

Theorem
Let $\R$ be the real number line considered as a metric space under the usual metric.

Let $I := \left ({a \,.\,.\, b} \right) \subseteq \R$ be an open real interval.

Then $I$ is the open $\epsilon$-ball $B_\epsilon \left({\alpha}\right)$ of some $\alpha \in \R$.

Proof
Let:

Then:

Thus:
 * $\left ({a \,.\,.\, b} \right) = \left ({\alpha - \epsilon \,.\,.\, \alpha + \epsilon} \right)$

From Open Ball in Real Number Line is Open Interval:
 * $\left ({\alpha - \epsilon \,.\,.\, \alpha + \epsilon} \right) = B_\epsilon \left({\alpha}\right)$

where $B_\epsilon \left({\alpha}\right)$ is the open $\epsilon$-ball of $\alpha$ in $\R$.

Hence the result