Closure of Half-Open Real Interval is Closed Real Interval
Let $H_1 = \hointl a b$ and $H_2 = \hointr a b$ be half-open intervals of $\R$.
For an arbitrary $H \subseteq \R$, let $H^-$ denote the closure of $H$ in $\R$.
By the definition of closure:
Let $A := \openint a b$ be the open interval between $a$ and $b$.
By definition of $H_1$ and $H_2$:
- $A \subseteq H_1$
- $A \subseteq H_2$
- $\openint a b^- = \closedint a b$
Thus as $\closedint a b$ is the closure of $\openint a b$, it follows that:
But $\closedint a b$ also contains as subsets both $H_1$ and $H_2$.
Thus we have:
- $A \subseteq H_1 \subseteq \closedint a b$
- $A \subseteq H_2 \subseteq \closedint a b$
Hence the result.