Prob. 21, Sec. 17, in Munkres' TOPOLOGY, 2nd ed: Closure and complementation can result in at most 14 different sets

Question

Here's Prob. 21 in Sec. 17 of the book Topology by James R. Munkres, 2nd edition:

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.

(a) Show that, starting with a given set $A$, one can form no more than $14$ distinct sets by applying these two operations successively.

(b) Find a subset $A$ of $\mathbb{R}$ (in its usual topology) for which the maximum of $14$ is obtained.

My effort:

Based on the feedback I've had from the valued Mathematics Stack Exchange community, I would like to ammend my effort as follows:

Let $\mathcal{P} (X)$ denote the power set (i.e. the collection of all the subsets ) of $X$, and let the functions $f \colon \mathcal{P} (X) \to \mathcal{P} (X) $, $g \colon \mathcal{P} (X) \to \mathcal{P} (X) $, and $i \colon \mathcal{P} (X) \to \mathcal{P} (X) $ be defined as follows: $$f(A) \colon= \overline{A} \ \ \ \mbox{ for all } \ \ \ A \in \mathcal{P} (X),$$ $$g(A) \colon= X - A \ \ \ \mbox{ for all } \ \ \ A \in \mathcal{P} (X),$$ and $$i(A) \colon= \mathrm{Int} (A) \ \ \ \mbox{ for all } \ \ \ A \in \mathcal{P} (X).$$ Then we have the following couple of results $$f(f(A)) = f(A), \ \ \ i(i(A) ) = i(A), \ \ \ \mbox{ and } \ \ \ g(g(A)) = A \ \ \ \mbox{ for all } \ A \in \mathcal{P}(X).$$ Moreover, we also have the following two relations: $$f(g(A)) = g(i(A)) \ \ \ \mbox{ and } \ \ \ g(f(A)) = i(g(A)) \ \ \ \mbox{ for all } \ A \in \mathcal{P} (X).$$

How do we show that $$f(g(f(g(f(g(f(g(A)))))))) = f(g(f(g(A))))$$ and $$g(f(g(f(g(f(g(f(A)))))))) = g(f(g(f(A))))$$ for all $A \in \mathcal{P} (X)?$.

Please also refer to this link.

Answer 1

Note also that $$iA\subseteq A\subseteq fA$$ and $$A\subseteq B\implies fA\subseteq fB,\ iA\subseteq iB.$$

From $$fgA=giA,$$ it follows that $$gfgA=ggiA=iA$$ and $$fgfgA=fiA;$$ thus the equality $$fgfgfgfgA=fgfgA$$ can be rewritten in the form $$fifiA=fiA,$$ which I will now prove.

$\underline{fifiA\subseteq fi A}$: From $$ifiA\subseteq fiA$$ it follows that $$fifiA\subseteq ffiA=fiA.$$

$\underline{fiA\subseteq fifiA}$: From $$iA\subseteq fiA$$ it follows that $$iA=iiA\subseteq ifiA$$ and $$fiA\subseteq fifiA.$$

Substituting $gA$ for $A$ in the previous identity, we have $$fgfgfgfg(gA)=fgfg(gA),$$ i.e., $$fgfgfgfA=fgfA;$$ taking complements, $$gfgfgfgfA=gfgfA.$$