Half-Open Real Interval is not Closed in Real Number Line

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

Let $\hointr a b \subset \R$ be a right half-open interval of $\R$.

Then $\hointr a b$ is not a closed set of $\R$.

Similarly, the left half-open interval $\hointl a b \subset \R$ is not a closed set of $\R$.

Proof
Consider:
 * $A := \R \setminus \hointr a b = \openint \gets a \cup \hointr b \to$

Let $A:= \openint \gets a$ and let $B := \hointr b \to$.

Let $\map {B_\epsilon} b$ be the open $\epsilon$-ball of $b$.

We have that $b - \epsilon < b$ and so $\map {B_\epsilon} b = \openint {b - \epsilon} {b + \epsilon}$ does not lie entirely in $\hointr b \to$.

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

However:
 * $\hointr b \to \cap \map {B_\epsilon} b \ne \O$

and so there are elements of $\map {B_\epsilon} b$ which are not in $\openint \gets a \cup \hointr b \to$.

So $\openint \gets a \cup \hointr b \to$ is not an open set of $\R$.

Thus, by definition, $\hointr a b$ is not a closed set of $\R$.

Hence the result.

Mutatis mutandis, the argument also shows that $\hointl a b \subset \R$ is neither an open set nor a closed set of $\R$.