Closed Real Interval is Closed Set

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

Let $\left[{a \,.\,.\, b}\right] \subset \R$ be a closed interval of $\R$.

Then $\left[{a \,.\,.\, b}\right]$ is a closed set of $\R$.

Proof
From the corollary to Open Real Interval is Open Set, both $\left({-\infty \,.\,.\, a}\right)$ and $\left({b \,.\,.\, \infty}\right)$ are open sets in $M$.

From Union of Open Subsets it follows that $\left({-\infty \,.\,.\, a}\right) \cup \left({b \,.\,.\, \infty}\right)$ is open in $\R$.

But $\left({-\infty \,.\,.\, a}\right) \cup \left({b \,.\,.\, \infty}\right)$ is the relative complement of $\left[{a \,.\,.\, b}\right]$ in $\R$.

The result follows by definition of closed set.