Closure of Half-Open Real Interval is Closed Real Interval

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Theorem

Let $\left({\R, \tau_d}\right)$ be the real number line under the usual (Euclidean) topology.

Let $H_1 = \left({a \,.\,.\, b}\right]$ and $H_2 = \left[{a \,.\,.\, b}\right)$ be half-open intervals of $\R$.


Then the closure of both $H_1$ and $H_2$ in $\R$ are the closed interval $\left[{a \,.\,.\, b}\right]$.


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 $\left({\R, \tau_d}\right)$ containing $H$ as a subset.

Let $A := \left({a \,.\,.\, b}\right)$ 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:

$\left({a \,.\,.\, b}\right)^- = \left[{a \,.\,.\, b}\right]$

Thus as $\left[{a \,.\,.\, b}\right]$ is the closure of $\left({a \,.\,.\, b}\right)$, it follows that:

$\left[{a \,.\,.\, b}\right]$ is the smallest closed set of $\left({\R, \tau_d}\right)$ containing $\left({a \,.\,.\, b}\right)$.

But $\left[{a \,.\,.\, b}\right]$ also contains as subsets both $H_1$ and $H_2$.

Thus we have:

$A \subseteq H_1 \subseteq \left[{a \,.\,.\, b}\right]$

and:

$A \subseteq H_2 \subseteq \left[{a \,.\,.\, b}\right]$

Thus it follows that $\left[{a \,.\,.\, b}\right]$ is the smallest closed set of $\left({\R, \tau_d}\right)$ containing $H_1$ and $H_2$.

Hence the result.

$\blacksquare$