# 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:

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

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.

$\blacksquare$