Kuratowski's Closure-Complement Problem/Proof of Maximum

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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:

$(1): \quad a a = I$

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:

$(2): \quad b b = b$


Let $s$ be a finite sequence of elements of $\left\{{a, b}\right\}$.

By successive applications of $(1)$ and $(2)$, it is possible to eliminate all multiple consecutive instances of $a$ and $b$ in $s$, and so reduce $s$ to one of the following forms:

$\text{a)}: \quad a b a b \ldots a$
$\text{b)}: \quad b a b a \ldots a$
$\text{c)}: \quad a b a b \ldots b$
$\text{d)}: \quad b a b a \ldots b$


From Closure of Complement of Closure is Regular Closed:

$b a b$ is regular closed.

By Interior equals Complement of Closure of Complement, the interior of $A$ is:

$a b a$

Recall the definition of regular closed:

a set $A$ is regular closed if and only if it equals the closure of its interior.

And so as $b a b$ is regular closed:

$b a b = b a b a \left({b a b}\right)$


So, adding an extra $b$ to either of $a b a b a b a$ or $b a b a b a$ will generate a string containing $b a b a b a b$ which can be reduced immediately to $b a b$.


It follows that the possible different subsets of $S$ that can be obtained from $A$ by applying $a$ and $b$ can be generated by none other than:

$I$
$a$
$a b$
$a b a$
$a b a b$
$a b a b a$
$a b a b a b$
$a b a b a b a$
$b$
$b a$
$b a b$
$b a b a$
$b a b a b$
$b a b a b a$

... a total of $14$.

Hence the result.

$\blacksquare$