Recognizing that $A\cup B$ contains elements in $A$ or $B$, $\mathscr{P}(A\cup B)$ contains subsets of the union, and $\mathscr{P}(\mathscr{P}(A\cup B))$ contains the ordered pair $(a,b)$ for some $a\in A, b\in B$, $\mathscr{P}(\mathscr{P}(\mathscr{P}(A\cup B)))$ contains the Cartesian product $A\times B$ and also the function $f:A\rightarrow B$, $\mathscr{P}(\mathscr{P}(\mathscr{P}(\mathscr{P}(A\cup B))))$ contains the set of all functions from $A$ to $B$ denoted as $B^A$. Until now, everything makes sense mathematically. I wonder if we again construct the power set of the last set, which is $\mathscr{P}^5(A\cup B)$, is there some useful meaning that we can attach to it (along the line of primitive set-theoretic relationship of belonging)?

Sorry if the question is not well-stated...I will try to explain as best as I can.

Further clarification: of course, we can arbitrarily raise the power of the operation $\mathscr{P}$ if we consider ordered triples, quadruples, and so on from multiple sets and build more complex relationships on them. But I wonder if we can take it further with just two sets, as stated in the question - A and B.

  • $\begingroup$ I'm confused... $P^2(A\cup B)$ consists of sets of subsets of $A\cup B$, not ordered pairs $(a,b) \in A\times B$... $\endgroup$
    – user7530
    Jan 23 '19 at 9:58
  • $\begingroup$ @user7530 I'm not saying it consists of ordered pairs but just it contains an ordered pair, which is the same as saying an ordered paired belongs to P^2 (A U B). $\endgroup$
    – Macrophage
    Jan 23 '19 at 10:30
  • $\begingroup$ @user7530 To clarify, I'm trying to find the most complex yet meaningful object each set constructed by repeated power operation can contain. True, in P^2(A U B) we have "singleton" like {{a}} encased in brackets that is not an ordered pair, but we can simply ignore them in this case, can't we? $\endgroup$
    – Macrophage
    Jan 23 '19 at 10:35
  • $\begingroup$ @Macrophage $(a,a) = \{\{a\},\{a,a\}\} = \{\{a\},\{a\}\} = \{\{a\}\}$, :P. $\endgroup$
    – Metric
    Jan 24 '19 at 23:53
  • $\begingroup$ @Metric But $(a,a) \in P(P(A\cup B))$ so...? I'm not sure if I see what you are trying to imply. $\endgroup$
    – Macrophage
    Jan 25 '19 at 4:36

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