Mixing
Mathematical meaning of repeated power set operations on $Acup B$
$begingroup$
Recognizing that $Acup B$ contains elements in $A$ or $B$, $mathscr{P}(Acup B)$ contains subsets of the union, and $mathscr{P}(mathscr{P}(Acup B))$ contains the ordered pair $(a,b)$ for some $ain A, bin B$, $mathscr{P}(mathscr{P}(mathscr{P}(Acup B)))$ contains the Cartesian product $Atimes B$ and also the function $f:Arightarrow B$, $mathscr{P}(mathscr{P}(mathscr{P}(mathscr{P}(Acup B))))$ 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 $mathscr{P}^5(Acup B)$, 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.
elementary-set-theory soft-question
asked Jan 23 at 7:58
MacrophageMacrophage
1,191115
$endgroup$
|
show 1 more comment
$begingroup$
Recognizing that $Acup B$ contains elements in $A$ or $B$, $mathscr{P}(Acup B)$ contains subsets of the union, and $mathscr{P}(mathscr{P}(Acup B))$ contains the ordered pair $(a,b)$ for some $ain A, bin B$, $mathscr{P}(mathscr{P}(mathscr{P}(Acup B)))$ contains the Cartesian product $Atimes B$ and also the function $f:Arightarrow B$, $mathscr{P}(mathscr{P}(mathscr{P}(mathscr{P}(Acup B))))$ 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 $mathscr{P}^5(Acup B)$, 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.
elementary-set-theory soft-question
asked Jan 23 at 7:58
MacrophageMacrophage
1,191115
$endgroup$
$begingroup$
I'm confused... $P^2(Acup B)$ consists of sets of subsets of $Acup B$, not ordered pairs $(a,b) in Atimes B$...
$endgroup$
– user7530
Jan 23 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 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 at 10:35
$begingroup$
@Macrophage $(a,a) = {{a},{a,a}} = {{a},{a}} = {{a}}$, :P.
$endgroup$
– Metric
Jan 24 at 23:53
$begingroup$
@Metric But $(a,a) in P(P(Acup B))$ so...? I'm not sure if I see what you are trying to imply.
$endgroup$
– Macrophage
Jan 25 at 4:36
|
show 1 more comment
$begingroup$
Recognizing that $Acup B$ contains elements in $A$ or $B$, $mathscr{P}(Acup B)$ contains subsets of the union, and $mathscr{P}(mathscr{P}(Acup B))$ contains the ordered pair $(a,b)$ for some $ain A, bin B$, $mathscr{P}(mathscr{P}(mathscr{P}(Acup B)))$ contains the Cartesian product $Atimes B$ and also the function $f:Arightarrow B$, $mathscr{P}(mathscr{P}(mathscr{P}(mathscr{P}(Acup B))))$ 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 $mathscr{P}^5(Acup B)$, 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.
elementary-set-theory soft-question
asked Jan 23 at 7:58
MacrophageMacrophage
1,191115
$endgroup$
Recognizing that $Acup B$ contains elements in $A$ or $B$, $mathscr{P}(Acup B)$ contains subsets of the union, and $mathscr{P}(mathscr{P}(Acup B))$ contains the ordered pair $(a,b)$ for some $ain A, bin B$, $mathscr{P}(mathscr{P}(mathscr{P}(Acup B)))$ contains the Cartesian product $Atimes B$ and also the function $f:Arightarrow B$, $mathscr{P}(mathscr{P}(mathscr{P}(mathscr{P}(Acup B))))$ 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 $mathscr{P}^5(Acup B)$, 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.
elementary-set-theory soft-question
elementary-set-theory soft-question
asked Jan 23 at 7:58
MacrophageMacrophage
1,191115
asked Jan 23 at 7:58
MacrophageMacrophage
1,191115
edited Jan 23 at 9:18
Macrophage
asked Jan 23 at 7:58
MacrophageMacrophage
1,191115
asked Jan 23 at 7:58
MacrophageMacrophage
1,191115
asked Jan 23 at 7:58
MacrophageMacrophage
1,191115
1,191115
$begingroup$
I'm confused... $P^2(Acup B)$ consists of sets of subsets of $Acup B$, not ordered pairs $(a,b) in Atimes B$...
$endgroup$
– user7530
Jan 23 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 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 at 10:35
$begingroup$
@Macrophage $(a,a) = {{a},{a,a}} = {{a},{a}} = {{a}}$, :P.
$endgroup$
– Metric
Jan 24 at 23:53
$begingroup$
@Metric But $(a,a) in P(P(Acup B))$ so...? I'm not sure if I see what you are trying to imply.
$endgroup$
– Macrophage
Jan 25 at 4:36
|
show 1 more comment
$begingroup$
I'm confused... $P^2(Acup B)$ consists of sets of subsets of $Acup B$, not ordered pairs $(a,b) in Atimes B$...
$endgroup$
– user7530
Jan 23 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 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 at 10:35
$begingroup$
@Macrophage $(a,a) = {{a},{a,a}} = {{a},{a}} = {{a}}$, :P.
$endgroup$
– Metric
Jan 24 at 23:53
$begingroup$
@Metric But $(a,a) in P(P(Acup B))$ so...? I'm not sure if I see what you are trying to imply.
$endgroup$
– Macrophage
Jan 25 at 4:36
$begingroup$
I'm confused... $P^2(Acup B)$ consists of sets of subsets of $Acup B$, not ordered pairs $(a,b) in Atimes B$...
$endgroup$
– user7530
Jan 23 at 9:58
$begingroup$
I'm confused... $P^2(Acup B)$ consists of sets of subsets of $Acup B$, not ordered pairs $(a,b) in Atimes B$...
$endgroup$
– user7530
Jan 23 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 at 10:30
$begingroup$
@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 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 at 10:35
$begingroup$
@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 at 10:35
$begingroup$
@Macrophage $(a,a) = {{a},{a,a}} = {{a},{a}} = {{a}}$, :P.
$endgroup$
– Metric
Jan 24 at 23:53
$begingroup$
@Macrophage $(a,a) = {{a},{a,a}} = {{a},{a}} = {{a}}$, :P.
$endgroup$
– Metric
Jan 24 at 23:53
$begingroup$
@Metric But $(a,a) in P(P(Acup B))$ so...? I'm not sure if I see what you are trying to imply.
$endgroup$
– Macrophage
Jan 25 at 4:36
$begingroup$
@Metric But $(a,a) in P(P(Acup B))$ so...? I'm not sure if I see what you are trying to imply.
$endgroup$
– Macrophage
Jan 25 at 4:36
|
show 1 more comment
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$begingroup$
I'm confused... $P^2(Acup B)$ consists of sets of subsets of $Acup B$, not ordered pairs $(a,b) in Atimes B$...
$endgroup$
– user7530
Jan 23 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 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 at 10:35
$begingroup$
@Macrophage $(a,a) = {{a},{a,a}} = {{a},{a}} = {{a}}$, :P.
$endgroup$
– Metric
Jan 24 at 23:53
$begingroup$
@Metric But $(a,a) in P(P(Acup B))$ so...? I'm not sure if I see what you are trying to imply.
$endgroup$
– Macrophage
Jan 25 at 4:36