# Open Real Interval is Open Ball

## Theorem

Let $\R$ denote the real number line with the usual (Euclidean) metric.

Let $I := \openint a b \subseteq \R$ be an open real interval.

Then $I$ is the open $\epsilon$-ball $\map {B_\epsilon} \alpha$ 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:

$\openint a b = \openint {\alpha - \epsilon} {\alpha + \epsilon}$
$\openint {\alpha - \epsilon} {\alpha + \epsilon} = \map {B_\epsilon} \alpha$

where $\map {B_\epsilon} \alpha$ is the open $\epsilon$-ball of $\alpha$ in $\R$.

Hence the result

$\blacksquare$