Closure of Interior of Closure of Union of Adjacent Open Intervals

Theorem
Let $A$ be the union of the two adjacent open intervals:
 * $A := \left({0 \,.\,.\, \dfrac 1 2}\right) \cup \left({\dfrac 1 2 \,.\,.\, 1}\right)$

Then:
 * $A^{- \circ -} = A^{\circ -} = A^- = \left[{0 \,.\,.\, 1}\right]$

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

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

Thus:

Hence the result.