Open Real Interval is not Closed Set

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

Let $I = \left({a \,.\,.\, b}\right)$ be an open real interval.

Then $I$ is not a closed set of $\R$.

Proof
Consider the relative complement of $I$ in $\R$:
 * $J = \complement_\R \left({I}\right) = \R \setminus I = \left({-\infty \,.\,.\, a}\right] \cup \left[{b \,.\,.\, \infty}\right)$

Let $\epsilon \in \R_{>0}$.

Consider the open $\epsilon$-ball $B_\epsilon \left({a}\right)$.

Whatever the value of $\epsilon$ is, $a + \epsilon$ is not in $B_\epsilon \left({a}\right)$.

So, by definition, $J$ is not an open set of $\R$.

By Relative Complement of Relative Complement, $\complement_\R \left({J}\right) = I$.

By definition, it follows that $I$ is not a closed set of $\R$.