Open Real Interval is Open Ball

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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:

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


From Open Ball in Real Number Line is Open Interval:

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

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


Hence the result

$\blacksquare$


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