Getting the fundamental elements of a set of sets
$begingroup$
Let $A = {1, 2, 3}$, $B = {4, 5, 6}$, and $C = {A, B}$. Let's call the elements of $A cup B$ the fundamental elements.
It is easy enough to get the set of fundamental elements from $A$ and $B$ i.e. ${x ; | ; x in A lor x in B}$ or even $A cup B$.
It doesn't seem as straight forward if we want to get the fundamental elements from $C$ though. Perhaps ${ x ; | ; y in C land x in y }$?
This requires knowing how many subsets you need to go down before you get to the fundamental elements though. What about if you have an abstract set of subsets where the depth of the subsets until you get to the fundament elements is unknown/arbitrary?
elementary-set-theory
asked Jan 18 at 17:43
ojunkojunk
102
$endgroup$
|
show 2 more comments
$begingroup$
Let $A = {1, 2, 3}$, $B = {4, 5, 6}$, and $C = {A, B}$. Let's call the elements of $A cup B$ the fundamental elements.
It is easy enough to get the set of fundamental elements from $A$ and $B$ i.e. ${x ; | ; x in A lor x in B}$ or even $A cup B$.
It doesn't seem as straight forward if we want to get the fundamental elements from $C$ though. Perhaps ${ x ; | ; y in C land x in y }$?
This requires knowing how many subsets you need to go down before you get to the fundamental elements though. What about if you have an abstract set of subsets where the depth of the subsets until you get to the fundament elements is unknown/arbitrary?
elementary-set-theory
asked Jan 18 at 17:43
ojunkojunk
102
$endgroup$
$begingroup$
You may be interested in a Wikipedia article on urelements.
$endgroup$
– Somos
Jan 18 at 17:48
$begingroup$
If you work in ZF then, by foundation, you know there are only finitely many stages, so you can use the transitive closure.
$endgroup$
– MacRance
Jan 18 at 19:06
$begingroup$
In axiomatic set theory given by ZF (or by GBN), the set $Acup B$ is actualy derived from the set $C$. The Axiom of Pairing says that given $A$ and $B$, there is a set $C’$ that has $A$ and $B$ as elements (from which, the Axiom of separation allows you to construct the set $C$ whose elements are exactly $A$ and $B$). The Axiom of union says that given a set $x$, there is a set $y$ such that $zin y$ if and only if $exists win x(zin w)$. So that your set $Acup B$ is actually derived from your set $C$. That is, $C$ is more “basic” than $Acup B$ in ZF.
$endgroup$
– Arturo Magidin
Jan 18 at 23:15
$begingroup$
Thanks @Somos that is very interesting and urelements are definitely what I'm talking about but I couldn't find a reference of how to find them when the depth of subsets is unknown/arbitrary, can you suggest anything?
$endgroup$
– ojunk
Jan 19 at 11:26
$begingroup$
@MacRance to be perfectly honest I don't know if I'm in ZF. I did some set theory many years ago during my undergrad and never looked at it again until now when I think I might be able to define a solution space using set theory. It is important that create sets from subsets from subsets an arbitrary amount of times but it is also import that I can always retrieve the urelements at any point.
$endgroup$
– ojunk
Jan 19 at 11:55
|
show 2 more comments
$begingroup$
Let $A = {1, 2, 3}$, $B = {4, 5, 6}$, and $C = {A, B}$. Let's call the elements of $A cup B$ the fundamental elements.
It is easy enough to get the set of fundamental elements from $A$ and $B$ i.e. ${x ; | ; x in A lor x in B}$ or even $A cup B$.
It doesn't seem as straight forward if we want to get the fundamental elements from $C$ though. Perhaps ${ x ; | ; y in C land x in y }$?
This requires knowing how many subsets you need to go down before you get to the fundamental elements though. What about if you have an abstract set of subsets where the depth of the subsets until you get to the fundament elements is unknown/arbitrary?
elementary-set-theory
asked Jan 18 at 17:43
ojunkojunk
102
$endgroup$
Let $A = {1, 2, 3}$, $B = {4, 5, 6}$, and $C = {A, B}$. Let's call the elements of $A cup B$ the fundamental elements.
It is easy enough to get the set of fundamental elements from $A$ and $B$ i.e. ${x ; | ; x in A lor x in B}$ or even $A cup B$.
It doesn't seem as straight forward if we want to get the fundamental elements from $C$ though. Perhaps ${ x ; | ; y in C land x in y }$?
This requires knowing how many subsets you need to go down before you get to the fundamental elements though. What about if you have an abstract set of subsets where the depth of the subsets until you get to the fundament elements is unknown/arbitrary?
elementary-set-theory
elementary-set-theory
asked Jan 18 at 17:43
ojunkojunk
102
asked Jan 18 at 17:43
ojunkojunk
102
asked Jan 18 at 17:43
ojunkojunk
102
asked Jan 18 at 17:43
ojunkojunk
102
asked Jan 18 at 17:43
ojunkojunk
102
102
$begingroup$
You may be interested in a Wikipedia article on urelements.
$endgroup$
– Somos
Jan 18 at 17:48
$begingroup$
If you work in ZF then, by foundation, you know there are only finitely many stages, so you can use the transitive closure.
$endgroup$
– MacRance
Jan 18 at 19:06
$begingroup$
In axiomatic set theory given by ZF (or by GBN), the set $Acup B$ is actualy derived from the set $C$. The Axiom of Pairing says that given $A$ and $B$, there is a set $C’$ that has $A$ and $B$ as elements (from which, the Axiom of separation allows you to construct the set $C$ whose elements are exactly $A$ and $B$). The Axiom of union says that given a set $x$, there is a set $y$ such that $zin y$ if and only if $exists win x(zin w)$. So that your set $Acup B$ is actually derived from your set $C$. That is, $C$ is more “basic” than $Acup B$ in ZF.
$endgroup$
– Arturo Magidin
Jan 18 at 23:15
$begingroup$
Thanks @Somos that is very interesting and urelements are definitely what I'm talking about but I couldn't find a reference of how to find them when the depth of subsets is unknown/arbitrary, can you suggest anything?
$endgroup$
– ojunk
Jan 19 at 11:26
$begingroup$
@MacRance to be perfectly honest I don't know if I'm in ZF. I did some set theory many years ago during my undergrad and never looked at it again until now when I think I might be able to define a solution space using set theory. It is important that create sets from subsets from subsets an arbitrary amount of times but it is also import that I can always retrieve the urelements at any point.
$endgroup$
– ojunk
Jan 19 at 11:55
|
show 2 more comments
$begingroup$
You may be interested in a Wikipedia article on urelements.
$endgroup$
– Somos
Jan 18 at 17:48
$begingroup$
If you work in ZF then, by foundation, you know there are only finitely many stages, so you can use the transitive closure.
$endgroup$
– MacRance
Jan 18 at 19:06
$begingroup$
In axiomatic set theory given by ZF (or by GBN), the set $Acup B$ is actualy derived from the set $C$. The Axiom of Pairing says that given $A$ and $B$, there is a set $C’$ that has $A$ and $B$ as elements (from which, the Axiom of separation allows you to construct the set $C$ whose elements are exactly $A$ and $B$). The Axiom of union says that given a set $x$, there is a set $y$ such that $zin y$ if and only if $exists win x(zin w)$. So that your set $Acup B$ is actually derived from your set $C$. That is, $C$ is more “basic” than $Acup B$ in ZF.
$endgroup$
– Arturo Magidin
Jan 18 at 23:15
$begingroup$
Thanks @Somos that is very interesting and urelements are definitely what I'm talking about but I couldn't find a reference of how to find them when the depth of subsets is unknown/arbitrary, can you suggest anything?
$endgroup$
– ojunk
Jan 19 at 11:26
$begingroup$
@MacRance to be perfectly honest I don't know if I'm in ZF. I did some set theory many years ago during my undergrad and never looked at it again until now when I think I might be able to define a solution space using set theory. It is important that create sets from subsets from subsets an arbitrary amount of times but it is also import that I can always retrieve the urelements at any point.
$endgroup$
– ojunk
Jan 19 at 11:55
$begingroup$
You may be interested in a Wikipedia article on urelements.
$endgroup$
– Somos
Jan 18 at 17:48
$begingroup$
You may be interested in a Wikipedia article on urelements.
$endgroup$
– Somos
Jan 18 at 17:48
$begingroup$
If you work in ZF then, by foundation, you know there are only finitely many stages, so you can use the transitive closure.
$endgroup$
– MacRance
Jan 18 at 19:06
$begingroup$
If you work in ZF then, by foundation, you know there are only finitely many stages, so you can use the transitive closure.
$endgroup$
– MacRance
Jan 18 at 19:06
$begingroup$
In axiomatic set theory given by ZF (or by GBN), the set $Acup B$ is actualy derived from the set $C$. The Axiom of Pairing says that given $A$ and $B$, there is a set $C’$ that has $A$ and $B$ as elements (from which, the Axiom of separation allows you to construct the set $C$ whose elements are exactly $A$ and $B$). The Axiom of union says that given a set $x$, there is a set $y$ such that $zin y$ if and only if $exists win x(zin w)$. So that your set $Acup B$ is actually derived from your set $C$. That is, $C$ is more “basic” than $Acup B$ in ZF.
$endgroup$
– Arturo Magidin
Jan 18 at 23:15
$begingroup$
In axiomatic set theory given by ZF (or by GBN), the set $Acup B$ is actualy derived from the set $C$. The Axiom of Pairing says that given $A$ and $B$, there is a set $C’$ that has $A$ and $B$ as elements (from which, the Axiom of separation allows you to construct the set $C$ whose elements are exactly $A$ and $B$). The Axiom of union says that given a set $x$, there is a set $y$ such that $zin y$ if and only if $exists win x(zin w)$. So that your set $Acup B$ is actually derived from your set $C$. That is, $C$ is more “basic” than $Acup B$ in ZF.
$endgroup$
– Arturo Magidin
Jan 18 at 23:15
$begingroup$
Thanks @Somos that is very interesting and urelements are definitely what I'm talking about but I couldn't find a reference of how to find them when the depth of subsets is unknown/arbitrary, can you suggest anything?
$endgroup$
– ojunk
Jan 19 at 11:26
$begingroup$
Thanks @Somos that is very interesting and urelements are definitely what I'm talking about but I couldn't find a reference of how to find them when the depth of subsets is unknown/arbitrary, can you suggest anything?
$endgroup$
– ojunk
Jan 19 at 11:26
$begingroup$
@MacRance to be perfectly honest I don't know if I'm in ZF. I did some set theory many years ago during my undergrad and never looked at it again until now when I think I might be able to define a solution space using set theory. It is important that create sets from subsets from subsets an arbitrary amount of times but it is also import that I can always retrieve the urelements at any point.
$endgroup$
– ojunk
Jan 19 at 11:55
$begingroup$
@MacRance to be perfectly honest I don't know if I'm in ZF. I did some set theory many years ago during my undergrad and never looked at it again until now when I think I might be able to define a solution space using set theory. It is important that create sets from subsets from subsets an arbitrary amount of times but it is also import that I can always retrieve the urelements at any point.
$endgroup$
– ojunk
Jan 19 at 11:55
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The question assumes a set theory with urelements (or fundamental elements). Given a set $,S,$ we want to find $,{scr U}(S),,$ the set of all the urelements used in the construction of $,S.,$ The definition is recursive as follows. If $,S,$ is an urelement, then
$,{scr U}(S) = S,,$ or else,
$,{scr U}(S) = bigcup_{xin S} {scr U}(x).,$ With a recursive definition you don't need to know the depth of recursion in advance.
Of course, you need some kind of axiom of foundation to ensure that the recursion "bottoms out". It is needed to eliminate the possibility of having "sets" such as
$$S_0 := {S_1},, S_1 := {S_2},, dots,, S_n := {S_{n+1}},, dots. tag{1}$$ Note the similarity to
$$S_1 := {S_0},, S_2 := {S_1},, dots,, S_{n+1} := {S_n},, dots, tag{2}$$ but they are very different cases when used with $,{scr U}(S_n).,$
answered Jan 19 at 14:23
SomosSomos
15.1k11437
$endgroup$
$begingroup$
It probably ought to be noted that even if your set theory has urelements, it is by no means given that the numbers $1$, $2$, $3$, etc., will be urelements.
$endgroup$
– Henning Makholm
Jan 19 at 15:00
$begingroup$
In the context of the question, it is clear that $1,2,3,4,5,6$ are assumed to be urelements.
$endgroup$
– Somos
Jan 19 at 15:04
$begingroup$
Thanks @Somos I only know the basics of set theory but is an axiom of foundation totally necessary? Could you just say ${scr U }(S) = cup_{x in S} {scr U}(x)$ (until ${scr U } (x) = x$)? Extending this idea, how about: Let $Gamma$ be the set of all urelements and let $$Phi(S) := {x ; | ; x in S}$$ then $${scr U } (S) := {Phi^n(S) ; | ; exists n ge 0 ; : ; x in Gamma ; forall x in Phi^n(S) }?$$
$endgroup$
– ojunk
Jan 19 at 18:34
$begingroup$
Sorry if thisis a really silly thing to ask but can I just say "Let ${scr U}(S)$ be an operator that returns all the urelements that makeup $S$"?
$endgroup$
– ojunk
Jan 19 at 20:23
$begingroup$
@ojunk Well, actually you just did write it and so, of course, you can write it. Human natural languages are very expressive. Mathematical languages are deliberately more restricted and precise. There are several related complex issues involved. For example, here, what does "makeup" mean? Lipstick?
$endgroup$
– Somos
Jan 19 at 21:19
add a comment |
$begingroup$
$cup$C = $cup${A,B} = A $cup$ B
immediately gives the fundamental elements.
answered Jan 18 at 23:00
William ElliotWilliam Elliot
9,2142820
$endgroup$
$begingroup$
And what about if $C$ has sets of subsets of subsets... of an unknown/arbitrary depth of subsets until the fundamental elements are found?
$endgroup$
– ojunk
Jan 19 at 11:24
add a comment |
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2 Answers
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2 Answers
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active
oldest
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active
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active
oldest
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$begingroup$
The question assumes a set theory with urelements (or fundamental elements). Given a set $,S,$ we want to find $,{scr U}(S),,$ the set of all the urelements used in the construction of $,S.,$ The definition is recursive as follows. If $,S,$ is an urelement, then
$,{scr U}(S) = S,,$ or else,
$,{scr U}(S) = bigcup_{xin S} {scr U}(x).,$ With a recursive definition you don't need to know the depth of recursion in advance.
Of course, you need some kind of axiom of foundation to ensure that the recursion "bottoms out". It is needed to eliminate the possibility of having "sets" such as
$$S_0 := {S_1},, S_1 := {S_2},, dots,, S_n := {S_{n+1}},, dots. tag{1}$$ Note the similarity to
$$S_1 := {S_0},, S_2 := {S_1},, dots,, S_{n+1} := {S_n},, dots, tag{2}$$ but they are very different cases when used with $,{scr U}(S_n).,$
answered Jan 19 at 14:23
SomosSomos
15.1k11437
$endgroup$
$begingroup$
It probably ought to be noted that even if your set theory has urelements, it is by no means given that the numbers $1$, $2$, $3$, etc., will be urelements.
$endgroup$
– Henning Makholm
Jan 19 at 15:00
$begingroup$
In the context of the question, it is clear that $1,2,3,4,5,6$ are assumed to be urelements.
$endgroup$
– Somos
Jan 19 at 15:04
$begingroup$
Thanks @Somos I only know the basics of set theory but is an axiom of foundation totally necessary? Could you just say ${scr U }(S) = cup_{x in S} {scr U}(x)$ (until ${scr U } (x) = x$)? Extending this idea, how about: Let $Gamma$ be the set of all urelements and let $$Phi(S) := {x ; | ; x in S}$$ then $${scr U } (S) := {Phi^n(S) ; | ; exists n ge 0 ; : ; x in Gamma ; forall x in Phi^n(S) }?$$
$endgroup$
– ojunk
Jan 19 at 18:34
$begingroup$
Sorry if thisis a really silly thing to ask but can I just say "Let ${scr U}(S)$ be an operator that returns all the urelements that makeup $S$"?
$endgroup$
– ojunk
Jan 19 at 20:23
$begingroup$
@ojunk Well, actually you just did write it and so, of course, you can write it. Human natural languages are very expressive. Mathematical languages are deliberately more restricted and precise. There are several related complex issues involved. For example, here, what does "makeup" mean? Lipstick?
$endgroup$
– Somos
Jan 19 at 21:19
add a comment |
$begingroup$
The question assumes a set theory with urelements (or fundamental elements). Given a set $,S,$ we want to find $,{scr U}(S),,$ the set of all the urelements used in the construction of $,S.,$ The definition is recursive as follows. If $,S,$ is an urelement, then
$,{scr U}(S) = S,,$ or else,
$,{scr U}(S) = bigcup_{xin S} {scr U}(x).,$ With a recursive definition you don't need to know the depth of recursion in advance.
Of course, you need some kind of axiom of foundation to ensure that the recursion "bottoms out". It is needed to eliminate the possibility of having "sets" such as
$$S_0 := {S_1},, S_1 := {S_2},, dots,, S_n := {S_{n+1}},, dots. tag{1}$$ Note the similarity to
$$S_1 := {S_0},, S_2 := {S_1},, dots,, S_{n+1} := {S_n},, dots, tag{2}$$ but they are very different cases when used with $,{scr U}(S_n).,$
answered Jan 19 at 14:23
SomosSomos
15.1k11437
$endgroup$
$begingroup$
It probably ought to be noted that even if your set theory has urelements, it is by no means given that the numbers $1$, $2$, $3$, etc., will be urelements.
$endgroup$
– Henning Makholm
Jan 19 at 15:00
$begingroup$
In the context of the question, it is clear that $1,2,3,4,5,6$ are assumed to be urelements.
$endgroup$
– Somos
Jan 19 at 15:04
$begingroup$
Thanks @Somos I only know the basics of set theory but is an axiom of foundation totally necessary? Could you just say ${scr U }(S) = cup_{x in S} {scr U}(x)$ (until ${scr U } (x) = x$)? Extending this idea, how about: Let $Gamma$ be the set of all urelements and let $$Phi(S) := {x ; | ; x in S}$$ then $${scr U } (S) := {Phi^n(S) ; | ; exists n ge 0 ; : ; x in Gamma ; forall x in Phi^n(S) }?$$
$endgroup$
– ojunk
Jan 19 at 18:34
$begingroup$
Sorry if thisis a really silly thing to ask but can I just say "Let ${scr U}(S)$ be an operator that returns all the urelements that makeup $S$"?
$endgroup$
– ojunk
Jan 19 at 20:23
$begingroup$
@ojunk Well, actually you just did write it and so, of course, you can write it. Human natural languages are very expressive. Mathematical languages are deliberately more restricted and precise. There are several related complex issues involved. For example, here, what does "makeup" mean? Lipstick?
$endgroup$
– Somos
Jan 19 at 21:19
add a comment |
$begingroup$
The question assumes a set theory with urelements (or fundamental elements). Given a set $,S,$ we want to find $,{scr U}(S),,$ the set of all the urelements used in the construction of $,S.,$ The definition is recursive as follows. If $,S,$ is an urelement, then
$,{scr U}(S) = S,,$ or else,
$,{scr U}(S) = bigcup_{xin S} {scr U}(x).,$ With a recursive definition you don't need to know the depth of recursion in advance.
Of course, you need some kind of axiom of foundation to ensure that the recursion "bottoms out". It is needed to eliminate the possibility of having "sets" such as
$$S_0 := {S_1},, S_1 := {S_2},, dots,, S_n := {S_{n+1}},, dots. tag{1}$$ Note the similarity to
$$S_1 := {S_0},, S_2 := {S_1},, dots,, S_{n+1} := {S_n},, dots, tag{2}$$ but they are very different cases when used with $,{scr U}(S_n).,$
answered Jan 19 at 14:23
SomosSomos
15.1k11437
$endgroup$
The question assumes a set theory with urelements (or fundamental elements). Given a set $,S,$ we want to find $,{scr U}(S),,$ the set of all the urelements used in the construction of $,S.,$ The definition is recursive as follows. If $,S,$ is an urelement, then
$,{scr U}(S) = S,,$ or else,
$,{scr U}(S) = bigcup_{xin S} {scr U}(x).,$ With a recursive definition you don't need to know the depth of recursion in advance.
Of course, you need some kind of axiom of foundation to ensure that the recursion "bottoms out". It is needed to eliminate the possibility of having "sets" such as
$$S_0 := {S_1},, S_1 := {S_2},, dots,, S_n := {S_{n+1}},, dots. tag{1}$$ Note the similarity to
$$S_1 := {S_0},, S_2 := {S_1},, dots,, S_{n+1} := {S_n},, dots, tag{2}$$ but they are very different cases when used with $,{scr U}(S_n).,$
answered Jan 19 at 14:23
SomosSomos
15.1k11437
edited Jan 19 at 21:30
answered Jan 19 at 14:23
SomosSomos
15.1k11437
answered Jan 19 at 14:23
SomosSomos
15.1k11437
answered Jan 19 at 14:23
SomosSomos
15.1k11437
15.1k11437
$begingroup$
It probably ought to be noted that even if your set theory has urelements, it is by no means given that the numbers $1$, $2$, $3$, etc., will be urelements.
$endgroup$
– Henning Makholm
Jan 19 at 15:00
$begingroup$
In the context of the question, it is clear that $1,2,3,4,5,6$ are assumed to be urelements.
$endgroup$
– Somos
Jan 19 at 15:04
$begingroup$
Thanks @Somos I only know the basics of set theory but is an axiom of foundation totally necessary? Could you just say ${scr U }(S) = cup_{x in S} {scr U}(x)$ (until ${scr U } (x) = x$)? Extending this idea, how about: Let $Gamma$ be the set of all urelements and let $$Phi(S) := {x ; | ; x in S}$$ then $${scr U } (S) := {Phi^n(S) ; | ; exists n ge 0 ; : ; x in Gamma ; forall x in Phi^n(S) }?$$
$endgroup$
– ojunk
Jan 19 at 18:34
$begingroup$
Sorry if thisis a really silly thing to ask but can I just say "Let ${scr U}(S)$ be an operator that returns all the urelements that makeup $S$"?
$endgroup$
– ojunk
Jan 19 at 20:23
$begingroup$
@ojunk Well, actually you just did write it and so, of course, you can write it. Human natural languages are very expressive. Mathematical languages are deliberately more restricted and precise. There are several related complex issues involved. For example, here, what does "makeup" mean? Lipstick?
$endgroup$
– Somos
Jan 19 at 21:19
add a comment |
$begingroup$
It probably ought to be noted that even if your set theory has urelements, it is by no means given that the numbers $1$, $2$, $3$, etc., will be urelements.
$endgroup$
– Henning Makholm
Jan 19 at 15:00
$begingroup$
In the context of the question, it is clear that $1,2,3,4,5,6$ are assumed to be urelements.
$endgroup$
– Somos
Jan 19 at 15:04
$begingroup$
Thanks @Somos I only know the basics of set theory but is an axiom of foundation totally necessary? Could you just say ${scr U }(S) = cup_{x in S} {scr U}(x)$ (until ${scr U } (x) = x$)? Extending this idea, how about: Let $Gamma$ be the set of all urelements and let $$Phi(S) := {x ; | ; x in S}$$ then $${scr U } (S) := {Phi^n(S) ; | ; exists n ge 0 ; : ; x in Gamma ; forall x in Phi^n(S) }?$$
$endgroup$
– ojunk
Jan 19 at 18:34
$begingroup$
Sorry if thisis a really silly thing to ask but can I just say "Let ${scr U}(S)$ be an operator that returns all the urelements that makeup $S$"?
$endgroup$
– ojunk
Jan 19 at 20:23
$begingroup$
@ojunk Well, actually you just did write it and so, of course, you can write it. Human natural languages are very expressive. Mathematical languages are deliberately more restricted and precise. There are several related complex issues involved. For example, here, what does "makeup" mean? Lipstick?
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– Somos
Jan 19 at 21:19
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It probably ought to be noted that even if your set theory has urelements, it is by no means given that the numbers $1$, $2$, $3$, etc., will be urelements.
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– Henning Makholm
Jan 19 at 15:00
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It probably ought to be noted that even if your set theory has urelements, it is by no means given that the numbers $1$, $2$, $3$, etc., will be urelements.
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– Henning Makholm
Jan 19 at 15:00
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In the context of the question, it is clear that $1,2,3,4,5,6$ are assumed to be urelements.
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– Somos
Jan 19 at 15:04
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In the context of the question, it is clear that $1,2,3,4,5,6$ are assumed to be urelements.
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– Somos
Jan 19 at 15:04
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Thanks @Somos I only know the basics of set theory but is an axiom of foundation totally necessary? Could you just say ${scr U }(S) = cup_{x in S} {scr U}(x)$ (until ${scr U } (x) = x$)? Extending this idea, how about: Let $Gamma$ be the set of all urelements and let $$Phi(S) := {x ; | ; x in S}$$ then $${scr U } (S) := {Phi^n(S) ; | ; exists n ge 0 ; : ; x in Gamma ; forall x in Phi^n(S) }?$$
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– ojunk
Jan 19 at 18:34
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Thanks @Somos I only know the basics of set theory but is an axiom of foundation totally necessary? Could you just say ${scr U }(S) = cup_{x in S} {scr U}(x)$ (until ${scr U } (x) = x$)? Extending this idea, how about: Let $Gamma$ be the set of all urelements and let $$Phi(S) := {x ; | ; x in S}$$ then $${scr U } (S) := {Phi^n(S) ; | ; exists n ge 0 ; : ; x in Gamma ; forall x in Phi^n(S) }?$$
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– ojunk
Jan 19 at 18:34
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Sorry if thisis a really silly thing to ask but can I just say "Let ${scr U}(S)$ be an operator that returns all the urelements that makeup $S$"?
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– ojunk
Jan 19 at 20:23
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Sorry if thisis a really silly thing to ask but can I just say "Let ${scr U}(S)$ be an operator that returns all the urelements that makeup $S$"?
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– ojunk
Jan 19 at 20:23
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@ojunk Well, actually you just did write it and so, of course, you can write it. Human natural languages are very expressive. Mathematical languages are deliberately more restricted and precise. There are several related complex issues involved. For example, here, what does "makeup" mean? Lipstick?
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– Somos
Jan 19 at 21:19
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@ojunk Well, actually you just did write it and so, of course, you can write it. Human natural languages are very expressive. Mathematical languages are deliberately more restricted and precise. There are several related complex issues involved. For example, here, what does "makeup" mean? Lipstick?
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– Somos
Jan 19 at 21:19
add a comment |
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$cup$C = $cup${A,B} = A $cup$ B
immediately gives the fundamental elements.
answered Jan 18 at 23:00
William ElliotWilliam Elliot
9,2142820
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And what about if $C$ has sets of subsets of subsets... of an unknown/arbitrary depth of subsets until the fundamental elements are found?
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– ojunk
Jan 19 at 11:24
add a comment |
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$cup$C = $cup${A,B} = A $cup$ B
immediately gives the fundamental elements.
answered Jan 18 at 23:00
William ElliotWilliam Elliot
9,2142820
$endgroup$
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And what about if $C$ has sets of subsets of subsets... of an unknown/arbitrary depth of subsets until the fundamental elements are found?
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– ojunk
Jan 19 at 11:24
add a comment |
$begingroup$
$cup$C = $cup${A,B} = A $cup$ B
immediately gives the fundamental elements.
answered Jan 18 at 23:00
William ElliotWilliam Elliot
9,2142820
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$cup$C = $cup${A,B} = A $cup$ B
immediately gives the fundamental elements.
answered Jan 18 at 23:00
William ElliotWilliam Elliot
9,2142820
answered Jan 18 at 23:00
William ElliotWilliam Elliot
9,2142820
answered Jan 18 at 23:00
William ElliotWilliam Elliot
9,2142820
answered Jan 18 at 23:00
William ElliotWilliam Elliot
9,2142820
9,2142820
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And what about if $C$ has sets of subsets of subsets... of an unknown/arbitrary depth of subsets until the fundamental elements are found?
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– ojunk
Jan 19 at 11:24
add a comment |
$begingroup$
And what about if $C$ has sets of subsets of subsets... of an unknown/arbitrary depth of subsets until the fundamental elements are found?
$endgroup$
– ojunk
Jan 19 at 11:24
$begingroup$
And what about if $C$ has sets of subsets of subsets... of an unknown/arbitrary depth of subsets until the fundamental elements are found?
$endgroup$
– ojunk
Jan 19 at 11:24
$begingroup$
And what about if $C$ has sets of subsets of subsets... of an unknown/arbitrary depth of subsets until the fundamental elements are found?
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– ojunk
Jan 19 at 11:24
add a comment |
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You may be interested in a Wikipedia article on urelements.
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– Somos
Jan 18 at 17:48
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If you work in ZF then, by foundation, you know there are only finitely many stages, so you can use the transitive closure.
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– MacRance
Jan 18 at 19:06
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In axiomatic set theory given by ZF (or by GBN), the set $Acup B$ is actualy derived from the set $C$. The Axiom of Pairing says that given $A$ and $B$, there is a set $C’$ that has $A$ and $B$ as elements (from which, the Axiom of separation allows you to construct the set $C$ whose elements are exactly $A$ and $B$). The Axiom of union says that given a set $x$, there is a set $y$ such that $zin y$ if and only if $exists win x(zin w)$. So that your set $Acup B$ is actually derived from your set $C$. That is, $C$ is more “basic” than $Acup B$ in ZF.
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– Arturo Magidin
Jan 18 at 23:15
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Thanks @Somos that is very interesting and urelements are definitely what I'm talking about but I couldn't find a reference of how to find them when the depth of subsets is unknown/arbitrary, can you suggest anything?
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– ojunk
Jan 19 at 11:26
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@MacRance to be perfectly honest I don't know if I'm in ZF. I did some set theory many years ago during my undergrad and never looked at it again until now when I think I might be able to define a solution space using set theory. It is important that create sets from subsets from subsets an arbitrary amount of times but it is also import that I can always retrieve the urelements at any point.
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– ojunk
Jan 19 at 11:55