Open Real Interval is Open Ball

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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:

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


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

$\blacksquare$


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