Open Real Interval is Open Set

Theorem
Let $\R$ be the real number line considered as an Euclidean space.

Let $\left({a \,.\,.\, b}\right) \subset \R$ be an open interval of $\R$.

Then $\left({a \,.\,.\, b}\right)$ is an open set of $\R$.

Proof
Let $c \in \R$ such that $a < c < b$.

Let $\epsilon < \min \left\{{b - c, c - a}\right\}$.

From the definition of positive it follows that $\epsilon \in \R_{>0}$.

Let $B_\epsilon \left({c}\right) = \left({c - \epsilon \,.\,.\, c + \epsilon}\right)$ be the open $\epsilon$-ball of $c$.

We have that $c + \epsilon < b$ and $a < c - \epsilon$.

Thus:


 * $B_\epsilon \left({c}\right) \subseteq \left({a \,.\,.\, b}\right)$

It follows that, by definition, $\left({a \,.\,.\, b}\right)$ is a neighborhood of $c$.

The result follows by definition of open set.