Kuratowski's Closure-Complement Problem/Proof of Maximum

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

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

By successive applications of the operations of complement relative to $S$ and the closure, there can be no more than $14$ distinct subsets of $S$ (including $A$ itself).

Proof
Consider an arbitrary subset $A$ of a topological space $T = \left({S, \tau}\right)$.

To simplify the presentation:
 * let $a$ be used to denote the operation of taking the complement of $A$ relative to $S$: $a \left({A}\right) = S \setminus A$
 * let $b$ be used to denote the operation of taking the closure of $A$ in $T$: $b \left({A}\right) = A^-$
 * let $I$ be used to denote the identity operation on $A$, that is: $I \left({A}\right) = A$.
 * let the parentheses and the reference to $A$ be removed, so as to present, for example:
 * $a \left({b \left({a \left({A}\right)}\right)}\right)$
 * as:
 * $a b a$

From Relative Complement of Relative Complement:
 * $a \left({a \left({A}\right)}\right) = A$

or, using the compact notation defined above:
 * $a a = I$

and from Closure of Topological Closure equals Closure:
 * $b \left({b \left({A}\right)}\right) = b \left({A}\right) = A^-$

or, using the compact notation defined above:
 * $b b = b$