Closure of Half-Open Real Interval is Closed Real Interval

Theorem
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology. Let $H_1 = \hointl a b$ and $H_2 = \hointr a b$ be half-open intervals of $\R$.

Then the closure of both $H_1$ and $H_2$ in $\R$ are the closed interval $\closedint a b$.

Proof
For an arbitrary $H \subseteq \R$, let $H^-$ denote the closure of $H$ in $\R$.

By the definition of closure:
 * $H^-$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $H$ as a subset.

Let $A := \openint a b$ be the open interval between $a$ and $b$.

By definition of $H_1$ and $H_2$:
 * $A \subseteq H_1$

and:
 * $A \subseteq H_2$

From Closure of Open Real Interval is Closed Real Interval:
 * $\openint a b^- = \closedint a b$

Thus as $\closedint a b$ is the closure of $\openint a b$, it follows that:
 * $\closedint a b$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $\openint a b$.

But $\closedint a b$ also contains as subsets both $H_1$ and $H_2$.

Thus we have:
 * $A \subseteq H_1 \subseteq \closedint a b$

and:
 * $A \subseteq H_2 \subseteq \closedint a b$

Thus it follows that $\closedint a b$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $H_1$ and $H_2$.

Hence the result.