Interior of Closure of Interior of Union of Adjacent Open Intervals

Theorem
Let $a, b, c \in R$ where $a < b < c$.

Let $A$ be the union of the two adjacent open intervals:
 * $A := \left({a \,.\,.\, b}\right) \cup \left({b \,.\,.\, c}\right)$

Then:
 * $A^{\circ - \circ} = A^{- \circ} = \left({a \,.\,.\, c}\right)$

where:
 * $A^\circ$ is the interior of $A$
 * $A^-$ is the closure of $A$.

Proof
From Interior of Union of Adjacent Open Intervals:
 * $A^\circ = A$

From Closure of Union of Adjacent Open Intervals:
 * $A^- = \left[{a \,.\,.\, c}\right]$

From Interior of Closed Real Interval is Open Real Interval:
 * $\left[{a \,.\,.\, c}\right]^\circ = \left({a \,.\,.\, c}\right)$

whence the result.