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 is Contained in Exterior of Exterior:
 * $A^\circ \subseteq A^{ee}$

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

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

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

From Exterior of Exterior of Union of Adjacent Open Intervals:
 * $A^{ee} = \left({a \,.\,.\, c}\right)$

From Interior of Union of Adjacent Open Intervals:
 * $A^\circ := \left({a \,.\,.\, b}\right) \cup \left({b \,.\,.\, c}\right)$

Thus:
 * $b \in A^{ee}$

but:
 * $b \notin A^\circ$

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

Also see

 * Interior is Contained in Exterior of Exterior