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:
\(\ds \alpha\) | \(=\) | \(\ds \frac {a + b} 2\) | ||||||||||||
\(\ds \epsilon\) | \(=\) | \(\ds \frac {b - a} 2\) |
Then:
\(\ds a\) | \(=\) | \(\ds \alpha - \epsilon\) | ||||||||||||
\(\ds b\) | \(=\) | \(\ds \alpha + \epsilon\) |
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
$\blacksquare$