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: $$f(A) \colon= \overline{A} \ \ \ \mbox{ for all } \ \ \ A \in \mathcal{P} (A),$$ $$g(A) \colon= X - A \ \ \ \mbox{ for all } \ \ \ A \in \mathcal{P} (A),$$ 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.

  • $\begingroup$ I believe you wrong in second step. It is $X-\overline{A}$ $\endgroup$ – sinbadh Jan 25 '16 at 7:42
  • $\begingroup$ @MichaelAlbanese using the terminology of the first answer to the question on the page whose link you've sent me, how do I prove that $$fgfgfgfg S = fgfg S?$$ $\endgroup$ – Saaqib Mahmood Jan 25 '16 at 9:02
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    $\begingroup$ @SaaqibMahmuud The edit I can see is that you added "Please also refer to this link" at the end of your post, linking to the question indicated as a duplicate. How does this edit makes the question more specific? Did you even read Michael Albanese's answer there before asking for reopening here? If you did, which parts of your question are not fully answered there? To me, instantly asking for reopening without any substantial modification of the question smells like a clear abuse of the site, so please, convince me it is not. $\endgroup$ – Did Jan 25 '16 at 19:35
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    $\begingroup$ Users Jon Mark Perry, Travis, user 1, Yiorgos S. Smyrlis, bof: Why the vote to reopen? $\endgroup$ – Did Jan 26 '16 at 14:45
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    $\begingroup$ @SaaqibMahmuud If your question is about the general problem stated at the beginning, then it is definitely a duplicate. (So I do not think it should have been reopened.) If your main questions is about the two equalities stated at the end of the current version of your post (namely $f(g(f(g(f(g(f(g(A)))))))) = f(g(f(g(A))))$ and $g(f(g(f(g(f(g(f(A)))))))) = g(f(g(f(A))))$) then perhaps it is not a duplicate. However, if this is the case, you should make this clearer in your post. $\endgroup$ – Martin Sleziak Jan 26 '16 at 14:58

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, $$gfgfgfgfA=gfgfA.$$

  • $\begingroup$ Nice and elementary proof! +1 $\endgroup$ – ZFR Sep 19 '19 at 21:46

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