# 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:

$A \subseteq H_2$
$\openint a b^- = \closedint a b$

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.

$\blacksquare$