# Mathematical meaning of repeated power set operations on $A\cup B$

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.

• I'm confused... $P^2(A\cup B)$ consists of sets of subsets of $A\cup B$, not ordered pairs $(a,b) \in A\times B$... Jan 23 '19 at 9:58
• @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). Jan 23 '19 at 10:30
• @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? Jan 23 '19 at 10:35
• @Macrophage $(a,a) = \{\{a\},\{a,a\}\} = \{\{a\},\{a\}\} = \{\{a\}\}$, :P. Jan 24 '19 at 23:53
• @Metric But $(a,a) \in P(P(A\cup B))$ so...? I'm not sure if I see what you are trying to imply. Jan 25 '19 at 4:36