Consider the collection of all subsets \(A\) of the topological space \(X\). The operations of closure \(A \to \overline{A}\) and complementation \(A \to X - A\) are functions from this collection to itself.
Show that starting with a given set \(A\), one can form no more than 14 distinct sets by applying these two operations successively.
Find a subset \(A\) of \(\mathbb{R}\) (in its usual topology) for which the maximum of 14 is obtained.
Solution
Let \(k\) denote closure, \(i\) interior, and \(c\) complementation. Since \[ i=ckc, \] the monoid generated by \(k\) and \(c\) is the same as that generated by \(k,i,c\). The identities \[ k^2=k,\qquad i^2=i,\qquad c^2=1, \] and \[ ic=ck,\qquad kc=ci \] allow every complement symbol to be moved to the far left. Hence it suffices to bound the number of distinct words in \(k\) and \(i\).
After removing consecutive repetitions, such a word alternates. Moreover \[ kiki=ki,\qquad ikik=ik. \] For the first identity, the set \(U=iA\) is open, so \(U=iU\subseteq ikU\) and \[ U\subseteq ikU\subseteq kU, \] so taking closures gives \(kikU=kU\), hence \(kikiA=kiA\); the other identity follows by complementation. Thus every word in \(k,i\) reduces to one of \[ 1,\ k,\ i,\ ki,\ ik,\ kik,\ iki. \] There are at most seven such sets, and possibly their seven complements, for a total of at most \(14\).
The bound is sharp in \(\mathbb R\). Let \[ A=\{0\}\cup(1,2)\cup(2,3)\cup(\mathbb Q\cap(4,5)). \] The seven sets obtained from the seven reduced \(k,i\) words are \[ \begin{aligned} A&=\{0\}\cup(1,2)\cup(2,3)\cup(\mathbb Q\cap(4,5)),\\ iA&=(1,2)\cup(2,3),\\ kA&=\{0\}\cup[1,3]\cup[4,5],\\ kiA&=[1,3],\\ ikA&=(1,3)\cup(4,5),\\ kikA&=[1,3]\cup[4,5],\\ ikiA&=(1,3). \end{aligned} \] They are pairwise distinct. Each is bounded, whereas each complement is unbounded, so none of these seven equals the complement of another. Their complements are also pairwise distinct. Thus exactly \(14\) distinct sets occur.