Open Real Interval is not Closed Set

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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 $\R$.


Corollary

Let:

$I_a = \left({-\infty \,.\,.\, a}\right)$
$I_b = \left({b \,.\,.\, \infty}\right)$

be unbounded open real intervals.


Then neither $I_a$ nor $I_b$ are closed sets 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 $\R$.

$\blacksquare$