Interior may not equal Exterior of Exterior

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $A \subseteq S$ be a subset of the underlying set $S$ of $T$.

Let $A^e$ be the exterior of $A$.

Let $A^\circ$ be the interior of $A$.

Then it is not necessarily the case that:
 * $A^{ee} = A^\circ$

Proof
We have from Interior Contained in Exterior of Exterior:
 * $A^\circ \subseteq A^{ee}$

It remains to be shown that there exist $A \subseteq S$ such that:
 * $A^\circ \ne A^{ee}$

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

From Exterior of Exterior of Union of Half-Unit Open Intervals:
 * $A^{ee} = \left({0 \,.\,.\, 1}\right)$

From Interior of Union of Half-Unit Open Intervals:
 * $A^\circ := \left({0 \,.\,.\, \dfrac 1 2}\right) \cup \left({\dfrac 1 2 \,.\,.\, 1}\right)$

Thus:
 * $\dfrac 1 2 \in A^{ee}$

but:
 * $\dfrac 1 2 \notin A^{ee}$

and so:
 * $A^\circ \subsetneq A^{ee}$

Also see

 * Interior Contained in Exterior of Exterior