Closure of Interior of Closure 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:
 * $A^{- \circ -} = A^{\circ -} = A^- = \left[{a \,.\,.\, c}\right]$

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

Proof
Then:

Hence the result.