Half-Open Real Interval is neither Open nor Closed

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

Let $\left[{a \,.\,.\, b}\right) \subset \R$ be a half-open interval of $\R$.

Then $\left[{a \,.\,.\, b}\right)$ is neither an open set nor a closed set of $\R$.

Similarly, the half-open interval $\left({a \,.\,.\, b}\right] \subset \R$ is neither an open set nor a closed set of $\R$.

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

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

We have that $a - \epsilon < a$ and so $B_\epsilon \left({a}\right) = \left({a - \epsilon \,.\,.\, a + \epsilon}\right)$ does not lie entirely in $\left[{a \,.\,.\, b}\right)$.

Thus $\left[{a \,.\,.\, b}\right)$ is not a neighborhood $a$.

It follows that $\left[{a \,.\,.\, b}\right)$ is not an open set of $\R$.

Now consider:
 * $A := \R \setminus \left[{a \,.\,.\, b}\right) = \left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$

Let $A:= \left({-\infty \,.\,.\, a}\right)$ and let $B := \left[{b \,.\,.\, \infty}\right)$.

Let $B_\epsilon \left({b}\right)$ be the open $\epsilon$-ball of $b$.

We have that $b - \epsilon < b$ and so $B_\epsilon \left({b}\right) = \left({b - \epsilon \,.\,.\, b + \epsilon}\right)$ does not lie entirely in $\left[{b \,.\,.\, \infty}\right)$.

Now $b - \epsilon$ may itself lie in $A$.

However:
 * $\left[{a \,.\,.\, b}\right) \cap B_\epsilon \left({b}\right) \ne \varnothing$

and so there are elements of $B_\epsilon \left({b}\right)$ which are not in $\left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$.

So $\left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$ is not an open set of $\R$.

Thus, by definition, $\left[{a \,.\,.\, b}\right)$ is not a closed set of $\R$.

Hence the result.

Mutatis mutandis, the argument also shows that $\left({a \,.\,.\, b}\right] \subset \R$ is neither an open set nor a closed set of $\R$.