Kuratowski's Closure-Complement Problem/Proof of Maximum
Theorem
Let $T = \struct {S, \tau}$ 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 $T = \struct {S, \tau}$.
To simplify the presentation:
- let $a$ be used to denote the operation of taking the complement of $A$ relative to $S$: $\map a A = S \setminus A$
- let $b$ be used to denote the operation of taking the closure of $A$ in $T$: $\map b A = A^-$
- let $I$ be used to denote the identity operation on $A$, that is: $\map I A = A$.
- let the parentheses and the reference to $A$ be removed, so as to present, for example:
- $\map a {\map b {\map a A} }$
- as:
- $a b a$
From Relative Complement of Relative Complement:
- $\map a {\map a A} = A$
or, using the compact notation defined above:
- $(1): \quad a a = I$
and from Closure of Topological Closure equals Closure:
- $\map b {\map b A} = \map b A = A^-$
or, using the compact notation defined above:
- $(2): \quad b b = b$
Let $s$ be a finite sequence of elements of $\set {a, b}$.
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:
- $b a b = b a b a \paren {b a b}$
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.
$\blacksquare$