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How does one define the alephs without using proper classes
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I’ve been reading Dasgupta’s Set Theory book about how the aleph numbers are constructed. He defines them as follows: 
He uses transfinite recursion:
However, I don’t think in that form, transfinite recursion works because there is no set which has all the ordinal as numbers. I know that some books such as Devlin’s The Joy of Sets have a version of transfinite recursion using classes which then works but I want to avoid using classes and only use sets. How does one properly define the alephs using transfinite recursion for sets?
set-theory
asked Jul 16 at 22:20
Eigenfield
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I’ve been reading Dasgupta’s Set Theory book about how the aleph numbers are constructed. He defines them as follows:
He uses transfinite recursion:
However, I don’t think in that form, transfinite recursion works because there is no set which has all the ordinal as numbers. I know that some books such as Devlin’s The Joy of Sets have a version of transfinite recursion using classes which then works but I want to avoid using classes and only use sets. How does one properly define the alephs using transfinite recursion for sets?
set-theory
asked Jul 16 at 22:20
Eigenfield
598417
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I’ve been reading Dasgupta’s Set Theory book about how the aleph numbers are constructed. He defines them as follows:
He uses transfinite recursion:
However, I don’t think in that form, transfinite recursion works because there is no set which has all the ordinal as numbers. I know that some books such as Devlin’s The Joy of Sets have a version of transfinite recursion using classes which then works but I want to avoid using classes and only use sets. How does one properly define the alephs using transfinite recursion for sets?
set-theory
asked Jul 16 at 22:20
Eigenfield
598417
I’ve been reading Dasgupta’s Set Theory book about how the aleph numbers are constructed. He defines them as follows:
He uses transfinite recursion:
However, I don’t think in that form, transfinite recursion works because there is no set which has all the ordinal as numbers. I know that some books such as Devlin’s The Joy of Sets have a version of transfinite recursion using classes which then works but I want to avoid using classes and only use sets. How does one properly define the alephs using transfinite recursion for sets?
set-theory
asked Jul 16 at 22:20
Eigenfield
598417
asked Jul 16 at 22:20
Eigenfield
598417
asked Jul 16 at 22:20
Eigenfield
598417
asked Jul 16 at 22:20
Eigenfield
598417
598417
add a comment |Â
add a comment |Â
2 Answers
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You can define $omega_alpha$ for any given $alpha$ just using the theorem you cited here. Just apply it to $A=alpha+1$ and then we have $omega_alpha=F(alpha).$
If you want to define $(omega_alpha)_alphain Ord$ as a ordinal sequence on all ordinals, you need to realize that this is a class function, whereas recursion theorem you cite yields a set function $F:Ato V.$ You can't 'avoid classes and only use sets' here. It's true that you never need classes in a certain sense, but you don't replace them by sets, you replace them by metatheoretical reasoning about formulas. You want to prove the existence of a formula (that corresponds to the class function $omega_alpha:Ordto V$ that you want). If you insist on doing it without using any kind of class notation, you will just wind up proving an instance of the class version of transfinite recursion, only everything will be more verbose and difficult to parse. (And it's a good exercise for understanding how classes 'actually work'. You mention Devlin's book here and if I recall right, he has a pretty good discussion on this.)
answered Jul 17 at 2:04
spaceisdarkgreen
27.6k21547
Classes can be considered to be just abbreviations for formulas. For example when we write $xin$ On (the class of ordinals) it's an abbreviation for "$x$ is transitive and well-ordered by $in$."
– DanielWainfleet
Jul 17 at 3:07
@DanielWainfleet not sure if you're addressing me or OP with that, but I say as much (though perhaps not very clearly).
– spaceisdarkgreen
Jul 17 at 3:10
For the OP. I usually make that clear in a comment but this time I forgot to.
– DanielWainfleet
Jul 17 at 5:44
add a comment |Â
up vote
0
down vote
($Ord=$ collection of all ordinals)
($Ord_lim=$ collection of all limit ordinals)
(Class function= same as function but does not require the domain to be a set)
Here is how I saw it for the first time(note, this is not a definition, but a theorem):
Transfinite recursion 1: Suppose that $X$ is some class(collection of sets). Let $D$ be the class of functions $f:alphato X$ where $alphain Ord$, and suppose that $G:Dto X$ is some function. Then there is a (class) function $F:Ordto X$ such that for any ordinal $alpha$ we have $F(alpha)=G(Frestrictionalpha)$.
And we use this to define the alephs(or more accurate: to define the aleph function, $aleph:Ordto Card, aleph_alpha=aleph(alpha)$).
But there is another version I saw that may be more along the lines what you searched:
Transfinite recursion 2: given $X$ a class, $G_1$ set and $G_2,3$ are (class) functions then there exists an unique (class) function $F:Ordto X$ such that:
1: $F(0)=G_1$
2: $forallalphain Ord(F(alpha+1)=G_2(F(alpha)))$
3: $forall betain Ord_lim(F(beta)=G_3(Frestriction beta))$
Using the second theorem it may be easier to understand as it is a lot more similar to classic recursion and to induction.
answered Jul 17 at 0:13
Holo
4,2512629
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You can define $omega_alpha$ for any given $alpha$ just using the theorem you cited here. Just apply it to $A=alpha+1$ and then we have $omega_alpha=F(alpha).$
If you want to define $(omega_alpha)_alphain Ord$ as a ordinal sequence on all ordinals, you need to realize that this is a class function, whereas recursion theorem you cite yields a set function $F:Ato V.$ You can't 'avoid classes and only use sets' here. It's true that you never need classes in a certain sense, but you don't replace them by sets, you replace them by metatheoretical reasoning about formulas. You want to prove the existence of a formula (that corresponds to the class function $omega_alpha:Ordto V$ that you want). If you insist on doing it without using any kind of class notation, you will just wind up proving an instance of the class version of transfinite recursion, only everything will be more verbose and difficult to parse. (And it's a good exercise for understanding how classes 'actually work'. You mention Devlin's book here and if I recall right, he has a pretty good discussion on this.)
answered Jul 17 at 2:04
spaceisdarkgreen
27.6k21547
Classes can be considered to be just abbreviations for formulas. For example when we write $xin$ On (the class of ordinals) it's an abbreviation for "$x$ is transitive and well-ordered by $in$."
– DanielWainfleet
Jul 17 at 3:07
@DanielWainfleet not sure if you're addressing me or OP with that, but I say as much (though perhaps not very clearly).
– spaceisdarkgreen
Jul 17 at 3:10
For the OP. I usually make that clear in a comment but this time I forgot to.
– DanielWainfleet
Jul 17 at 5:44
add a comment |Â
up vote
3
down vote
accepted
You can define $omega_alpha$ for any given $alpha$ just using the theorem you cited here. Just apply it to $A=alpha+1$ and then we have $omega_alpha=F(alpha).$
If you want to define $(omega_alpha)_alphain Ord$ as a ordinal sequence on all ordinals, you need to realize that this is a class function, whereas recursion theorem you cite yields a set function $F:Ato V.$ You can't 'avoid classes and only use sets' here. It's true that you never need classes in a certain sense, but you don't replace them by sets, you replace them by metatheoretical reasoning about formulas. You want to prove the existence of a formula (that corresponds to the class function $omega_alpha:Ordto V$ that you want). If you insist on doing it without using any kind of class notation, you will just wind up proving an instance of the class version of transfinite recursion, only everything will be more verbose and difficult to parse. (And it's a good exercise for understanding how classes 'actually work'. You mention Devlin's book here and if I recall right, he has a pretty good discussion on this.)
answered Jul 17 at 2:04
spaceisdarkgreen
27.6k21547
Classes can be considered to be just abbreviations for formulas. For example when we write $xin$ On (the class of ordinals) it's an abbreviation for "$x$ is transitive and well-ordered by $in$."
– DanielWainfleet
Jul 17 at 3:07
@DanielWainfleet not sure if you're addressing me or OP with that, but I say as much (though perhaps not very clearly).
– spaceisdarkgreen
Jul 17 at 3:10
For the OP. I usually make that clear in a comment but this time I forgot to.
– DanielWainfleet
Jul 17 at 5:44
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You can define $omega_alpha$ for any given $alpha$ just using the theorem you cited here. Just apply it to $A=alpha+1$ and then we have $omega_alpha=F(alpha).$
If you want to define $(omega_alpha)_alphain Ord$ as a ordinal sequence on all ordinals, you need to realize that this is a class function, whereas recursion theorem you cite yields a set function $F:Ato V.$ You can't 'avoid classes and only use sets' here. It's true that you never need classes in a certain sense, but you don't replace them by sets, you replace them by metatheoretical reasoning about formulas. You want to prove the existence of a formula (that corresponds to the class function $omega_alpha:Ordto V$ that you want). If you insist on doing it without using any kind of class notation, you will just wind up proving an instance of the class version of transfinite recursion, only everything will be more verbose and difficult to parse. (And it's a good exercise for understanding how classes 'actually work'. You mention Devlin's book here and if I recall right, he has a pretty good discussion on this.)
answered Jul 17 at 2:04
spaceisdarkgreen
27.6k21547
You can define $omega_alpha$ for any given $alpha$ just using the theorem you cited here. Just apply it to $A=alpha+1$ and then we have $omega_alpha=F(alpha).$
If you want to define $(omega_alpha)_alphain Ord$ as a ordinal sequence on all ordinals, you need to realize that this is a class function, whereas recursion theorem you cite yields a set function $F:Ato V.$ You can't 'avoid classes and only use sets' here. It's true that you never need classes in a certain sense, but you don't replace them by sets, you replace them by metatheoretical reasoning about formulas. You want to prove the existence of a formula (that corresponds to the class function $omega_alpha:Ordto V$ that you want). If you insist on doing it without using any kind of class notation, you will just wind up proving an instance of the class version of transfinite recursion, only everything will be more verbose and difficult to parse. (And it's a good exercise for understanding how classes 'actually work'. You mention Devlin's book here and if I recall right, he has a pretty good discussion on this.)
answered Jul 17 at 2:04
spaceisdarkgreen
27.6k21547
edited Jul 17 at 2:16
answered Jul 17 at 2:04
spaceisdarkgreen
27.6k21547
answered Jul 17 at 2:04
spaceisdarkgreen
27.6k21547
answered Jul 17 at 2:04
spaceisdarkgreen
27.6k21547
27.6k21547
Classes can be considered to be just abbreviations for formulas. For example when we write $xin$ On (the class of ordinals) it's an abbreviation for "$x$ is transitive and well-ordered by $in$."
– DanielWainfleet
Jul 17 at 3:07
@DanielWainfleet not sure if you're addressing me or OP with that, but I say as much (though perhaps not very clearly).
– spaceisdarkgreen
Jul 17 at 3:10
For the OP. I usually make that clear in a comment but this time I forgot to.
– DanielWainfleet
Jul 17 at 5:44
add a comment |Â
Classes can be considered to be just abbreviations for formulas. For example when we write $xin$ On (the class of ordinals) it's an abbreviation for "$x$ is transitive and well-ordered by $in$."
– DanielWainfleet
Jul 17 at 3:07
@DanielWainfleet not sure if you're addressing me or OP with that, but I say as much (though perhaps not very clearly).
– spaceisdarkgreen
Jul 17 at 3:10
For the OP. I usually make that clear in a comment but this time I forgot to.
– DanielWainfleet
Jul 17 at 5:44
Classes can be considered to be just abbreviations for formulas. For example when we write $xin$ On (the class of ordinals) it's an abbreviation for "$x$ is transitive and well-ordered by $in$."
– DanielWainfleet
Jul 17 at 3:07
Classes can be considered to be just abbreviations for formulas. For example when we write $xin$ On (the class of ordinals) it's an abbreviation for "$x$ is transitive and well-ordered by $in$."
– DanielWainfleet
Jul 17 at 3:07
@DanielWainfleet not sure if you're addressing me or OP with that, but I say as much (though perhaps not very clearly).
– spaceisdarkgreen
Jul 17 at 3:10
@DanielWainfleet not sure if you're addressing me or OP with that, but I say as much (though perhaps not very clearly).
– spaceisdarkgreen
Jul 17 at 3:10
For the OP. I usually make that clear in a comment but this time I forgot to.
– DanielWainfleet
Jul 17 at 5:44
For the OP. I usually make that clear in a comment but this time I forgot to.
– DanielWainfleet
Jul 17 at 5:44
add a comment |Â
up vote
0
down vote
($Ord=$ collection of all ordinals)
($Ord_lim=$ collection of all limit ordinals)
(Class function= same as function but does not require the domain to be a set)
Here is how I saw it for the first time(note, this is not a definition, but a theorem):
Transfinite recursion 1: Suppose that $X$ is some class(collection of sets). Let $D$ be the class of functions $f:alphato X$ where $alphain Ord$, and suppose that $G:Dto X$ is some function. Then there is a (class) function $F:Ordto X$ such that for any ordinal $alpha$ we have $F(alpha)=G(Frestrictionalpha)$.
And we use this to define the alephs(or more accurate: to define the aleph function, $aleph:Ordto Card, aleph_alpha=aleph(alpha)$).
But there is another version I saw that may be more along the lines what you searched:
Transfinite recursion 2: given $X$ a class, $G_1$ set and $G_2,3$ are (class) functions then there exists an unique (class) function $F:Ordto X$ such that:
1: $F(0)=G_1$
2: $forallalphain Ord(F(alpha+1)=G_2(F(alpha)))$
3: $forall betain Ord_lim(F(beta)=G_3(Frestriction beta))$
Using the second theorem it may be easier to understand as it is a lot more similar to classic recursion and to induction.
answered Jul 17 at 0:13
Holo
4,2512629
add a comment |Â
up vote
0
down vote
($Ord=$ collection of all ordinals)
($Ord_lim=$ collection of all limit ordinals)
(Class function= same as function but does not require the domain to be a set)
Here is how I saw it for the first time(note, this is not a definition, but a theorem):
Transfinite recursion 1: Suppose that $X$ is some class(collection of sets). Let $D$ be the class of functions $f:alphato X$ where $alphain Ord$, and suppose that $G:Dto X$ is some function. Then there is a (class) function $F:Ordto X$ such that for any ordinal $alpha$ we have $F(alpha)=G(Frestrictionalpha)$.
And we use this to define the alephs(or more accurate: to define the aleph function, $aleph:Ordto Card, aleph_alpha=aleph(alpha)$).
But there is another version I saw that may be more along the lines what you searched:
Transfinite recursion 2: given $X$ a class, $G_1$ set and $G_2,3$ are (class) functions then there exists an unique (class) function $F:Ordto X$ such that:
1: $F(0)=G_1$
2: $forallalphain Ord(F(alpha+1)=G_2(F(alpha)))$
3: $forall betain Ord_lim(F(beta)=G_3(Frestriction beta))$
Using the second theorem it may be easier to understand as it is a lot more similar to classic recursion and to induction.
answered Jul 17 at 0:13
Holo
4,2512629
add a comment |Â
up vote
0
down vote
up vote
0
down vote
($Ord=$ collection of all ordinals)
($Ord_lim=$ collection of all limit ordinals)
(Class function= same as function but does not require the domain to be a set)
Here is how I saw it for the first time(note, this is not a definition, but a theorem):
Transfinite recursion 1: Suppose that $X$ is some class(collection of sets). Let $D$ be the class of functions $f:alphato X$ where $alphain Ord$, and suppose that $G:Dto X$ is some function. Then there is a (class) function $F:Ordto X$ such that for any ordinal $alpha$ we have $F(alpha)=G(Frestrictionalpha)$.
And we use this to define the alephs(or more accurate: to define the aleph function, $aleph:Ordto Card, aleph_alpha=aleph(alpha)$).
But there is another version I saw that may be more along the lines what you searched:
Transfinite recursion 2: given $X$ a class, $G_1$ set and $G_2,3$ are (class) functions then there exists an unique (class) function $F:Ordto X$ such that:
1: $F(0)=G_1$
2: $forallalphain Ord(F(alpha+1)=G_2(F(alpha)))$
3: $forall betain Ord_lim(F(beta)=G_3(Frestriction beta))$
Using the second theorem it may be easier to understand as it is a lot more similar to classic recursion and to induction.
answered Jul 17 at 0:13
Holo
4,2512629
($Ord=$ collection of all ordinals)
($Ord_lim=$ collection of all limit ordinals)
(Class function= same as function but does not require the domain to be a set)
Here is how I saw it for the first time(note, this is not a definition, but a theorem):
Transfinite recursion 1: Suppose that $X$ is some class(collection of sets). Let $D$ be the class of functions $f:alphato X$ where $alphain Ord$, and suppose that $G:Dto X$ is some function. Then there is a (class) function $F:Ordto X$ such that for any ordinal $alpha$ we have $F(alpha)=G(Frestrictionalpha)$.
And we use this to define the alephs(or more accurate: to define the aleph function, $aleph:Ordto Card, aleph_alpha=aleph(alpha)$).
But there is another version I saw that may be more along the lines what you searched:
Transfinite recursion 2: given $X$ a class, $G_1$ set and $G_2,3$ are (class) functions then there exists an unique (class) function $F:Ordto X$ such that:
1: $F(0)=G_1$
2: $forallalphain Ord(F(alpha+1)=G_2(F(alpha)))$
3: $forall betain Ord_lim(F(beta)=G_3(Frestriction beta))$
Using the second theorem it may be easier to understand as it is a lot more similar to classic recursion and to induction.
answered Jul 17 at 0:13
Holo
4,2512629
answered Jul 17 at 0:13
Holo
4,2512629
answered Jul 17 at 0:13
Holo
4,2512629
answered Jul 17 at 0:13
Holo
4,2512629
4,2512629
add a comment |Â
add a comment |Â
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