Open Real Interval is not Closed Set

Theorem

Let $\R$ be the real number line with the usual (Euclidean) metric.

Let $I = \openint a b$ be an open real interval.

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

Corollary

Let:

$I_a = \openint \gets a$

$I_b = \openint b \to$

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 $\R$:

$J = \relcomp \R I = \R \setminus I = \hointl \gets a \cup \hointr b \to$

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

Consider the open $\epsilon$-ball $\map {B_\epsilon} a$.

Whatever the value of $\epsilon$ is, $a + \epsilon$ is not in $\map {B_\epsilon} a$.

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

By Relative Complement of Relative Complement, $\relcomp \R J = I$.

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

$\blacksquare$