# 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:

 $\displaystyle \alpha$ $=$ $\displaystyle \frac {a + b} 2$ $\displaystyle \epsilon$ $=$ $\displaystyle \frac {b - a} 2$

Then:

 $\displaystyle a$ $=$ $\displaystyle \alpha - \epsilon$ $\displaystyle b$ $=$ $\displaystyle \alpha + \epsilon$

Thus:

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

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

Hence the result

$\blacksquare$