Injection into a proper class and choice without regularity
$begingroup$
In $sf ZF$, we have that the axiom of choice is equivalent to:
For all sets $X$, and for all proper classes $Y$, $X$ inject into $Y$
and
For all sets $X$, and for all proper classes $Y$, $Y$ surject onto $X$
To see that those are indeed equivalent to choice we have for one direction to inject a set $X$ into $Ord$ and this will give well ordering for $X$(and because $Ord$ well ordered, we can easily construct an injective from $X$ to $Ord$ using a surjective from $Ord$ to $X$)
To see that the other direction is true, take a set $α$ and a class $Y$, because we are assuming $sf AC$ we may assume WLOG that $α∈Ord$. Then we may use induction to create a sequence $(x_β)$ of ordinals such that for $β<γ$ we have $Y∩V_{x_β}subsetneq Y∩V_{x_γ}$, then we look at $V_{x_α}$, and by well ordering it find an injective $α→Y$(and surjective $Y→α$).
In the proof use relied heavily on the axiom of foundation, so we can ask are those 3 equivalent in $sf ZF^-$?
When talking with @Wojowu he told me that his intuition told him that $sf AC$ is not equivalent to the other 2, saying that he thinks that there is a model of $sf ZFC^-+mbox{a proper class of atoms}+mbox{only finite sets of atoms}$, in which case no infinite set inject into the class of atoms, but after searching I couldn't find any reference to such model. My questions:
If such model exists, can someone direct me to a reference, or explain it's construction? If not, how those 2 behave in $sf ZF^-$?
What about the other 2? Does the surjective version implies the injective version in $sf ZF^-$?
reference-request set-theory lo.logic axiom-of-choice
New contributor
$endgroup$
add a comment |
$begingroup$
In $sf ZF$, we have that the axiom of choice is equivalent to:
For all sets $X$, and for all proper classes $Y$, $X$ inject into $Y$
and
For all sets $X$, and for all proper classes $Y$, $Y$ surject onto $X$
To see that those are indeed equivalent to choice we have for one direction to inject a set $X$ into $Ord$ and this will give well ordering for $X$(and because $Ord$ well ordered, we can easily construct an injective from $X$ to $Ord$ using a surjective from $Ord$ to $X$)
To see that the other direction is true, take a set $α$ and a class $Y$, because we are assuming $sf AC$ we may assume WLOG that $α∈Ord$. Then we may use induction to create a sequence $(x_β)$ of ordinals such that for $β<γ$ we have $Y∩V_{x_β}subsetneq Y∩V_{x_γ}$, then we look at $V_{x_α}$, and by well ordering it find an injective $α→Y$(and surjective $Y→α$).
In the proof use relied heavily on the axiom of foundation, so we can ask are those 3 equivalent in $sf ZF^-$?
When talking with @Wojowu he told me that his intuition told him that $sf AC$ is not equivalent to the other 2, saying that he thinks that there is a model of $sf ZFC^-+mbox{a proper class of atoms}+mbox{only finite sets of atoms}$, in which case no infinite set inject into the class of atoms, but after searching I couldn't find any reference to such model. My questions:
If such model exists, can someone direct me to a reference, or explain it's construction? If not, how those 2 behave in $sf ZF^-$?
What about the other 2? Does the surjective version implies the injective version in $sf ZF^-$?
reference-request set-theory lo.logic axiom-of-choice
New contributor
$endgroup$
1
$begingroup$
math.stackexchange.com/questions/1337583/… might be helpful?
$endgroup$
– Asaf Karagila
4 hours ago
add a comment |
$begingroup$
In $sf ZF$, we have that the axiom of choice is equivalent to:
For all sets $X$, and for all proper classes $Y$, $X$ inject into $Y$
and
For all sets $X$, and for all proper classes $Y$, $Y$ surject onto $X$
To see that those are indeed equivalent to choice we have for one direction to inject a set $X$ into $Ord$ and this will give well ordering for $X$(and because $Ord$ well ordered, we can easily construct an injective from $X$ to $Ord$ using a surjective from $Ord$ to $X$)
To see that the other direction is true, take a set $α$ and a class $Y$, because we are assuming $sf AC$ we may assume WLOG that $α∈Ord$. Then we may use induction to create a sequence $(x_β)$ of ordinals such that for $β<γ$ we have $Y∩V_{x_β}subsetneq Y∩V_{x_γ}$, then we look at $V_{x_α}$, and by well ordering it find an injective $α→Y$(and surjective $Y→α$).
In the proof use relied heavily on the axiom of foundation, so we can ask are those 3 equivalent in $sf ZF^-$?
When talking with @Wojowu he told me that his intuition told him that $sf AC$ is not equivalent to the other 2, saying that he thinks that there is a model of $sf ZFC^-+mbox{a proper class of atoms}+mbox{only finite sets of atoms}$, in which case no infinite set inject into the class of atoms, but after searching I couldn't find any reference to such model. My questions:
If such model exists, can someone direct me to a reference, or explain it's construction? If not, how those 2 behave in $sf ZF^-$?
What about the other 2? Does the surjective version implies the injective version in $sf ZF^-$?
reference-request set-theory lo.logic axiom-of-choice
New contributor
$endgroup$
In $sf ZF$, we have that the axiom of choice is equivalent to:
For all sets $X$, and for all proper classes $Y$, $X$ inject into $Y$
and
For all sets $X$, and for all proper classes $Y$, $Y$ surject onto $X$
To see that those are indeed equivalent to choice we have for one direction to inject a set $X$ into $Ord$ and this will give well ordering for $X$(and because $Ord$ well ordered, we can easily construct an injective from $X$ to $Ord$ using a surjective from $Ord$ to $X$)
To see that the other direction is true, take a set $α$ and a class $Y$, because we are assuming $sf AC$ we may assume WLOG that $α∈Ord$. Then we may use induction to create a sequence $(x_β)$ of ordinals such that for $β<γ$ we have $Y∩V_{x_β}subsetneq Y∩V_{x_γ}$, then we look at $V_{x_α}$, and by well ordering it find an injective $α→Y$(and surjective $Y→α$).
In the proof use relied heavily on the axiom of foundation, so we can ask are those 3 equivalent in $sf ZF^-$?
When talking with @Wojowu he told me that his intuition told him that $sf AC$ is not equivalent to the other 2, saying that he thinks that there is a model of $sf ZFC^-+mbox{a proper class of atoms}+mbox{only finite sets of atoms}$, in which case no infinite set inject into the class of atoms, but after searching I couldn't find any reference to such model. My questions:
If such model exists, can someone direct me to a reference, or explain it's construction? If not, how those 2 behave in $sf ZF^-$?
What about the other 2? Does the surjective version implies the injective version in $sf ZF^-$?
reference-request set-theory lo.logic axiom-of-choice
reference-request set-theory lo.logic axiom-of-choice
New contributor
New contributor
edited 4 hours ago
András Bátkai
3,85142342
3,85142342
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asked 5 hours ago
HoloHolo
1263
1263
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math.stackexchange.com/questions/1337583/… might be helpful?
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– Asaf Karagila
4 hours ago
add a comment |
1
$begingroup$
math.stackexchange.com/questions/1337583/… might be helpful?
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– Asaf Karagila
4 hours ago
1
1
$begingroup$
math.stackexchange.com/questions/1337583/… might be helpful?
$endgroup$
– Asaf Karagila
4 hours ago
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math.stackexchange.com/questions/1337583/… might be helpful?
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– Asaf Karagila
4 hours ago
add a comment |
1 Answer
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$begingroup$
The results appear in Jech's "The Axiom of Choice" in the problem section of Chapter 9 (Problems 2,3, and 4).
Indeed, it is easy to see that the injections into classes imply the surjections from classes which imply choice. Exactly by means of the class of ordinals. So the point is to separate the others.
And if we have a proper class of atoms whose subsets are all finite, then it is a class which does not map onto $omega$, but every set has only finitely many in its transitive closure, so it can be well-ordered.
The last model is described well in Jech, this is Problem 4 in the aforementioned reference, and the key point is that the atoms are indexed by countable sequences of ordinals, so that there are always surjections onto every set, but there is no $omega$ sequence of atoms, which form a proper class, so there is no injection from any infinite set into the class of atoms. (And indeed, that implies all sets of atoms are finite.)
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$begingroup$
The results appear in Jech's "The Axiom of Choice" in the problem section of Chapter 9 (Problems 2,3, and 4).
Indeed, it is easy to see that the injections into classes imply the surjections from classes which imply choice. Exactly by means of the class of ordinals. So the point is to separate the others.
And if we have a proper class of atoms whose subsets are all finite, then it is a class which does not map onto $omega$, but every set has only finitely many in its transitive closure, so it can be well-ordered.
The last model is described well in Jech, this is Problem 4 in the aforementioned reference, and the key point is that the atoms are indexed by countable sequences of ordinals, so that there are always surjections onto every set, but there is no $omega$ sequence of atoms, which form a proper class, so there is no injection from any infinite set into the class of atoms. (And indeed, that implies all sets of atoms are finite.)
$endgroup$
add a comment |
$begingroup$
The results appear in Jech's "The Axiom of Choice" in the problem section of Chapter 9 (Problems 2,3, and 4).
Indeed, it is easy to see that the injections into classes imply the surjections from classes which imply choice. Exactly by means of the class of ordinals. So the point is to separate the others.
And if we have a proper class of atoms whose subsets are all finite, then it is a class which does not map onto $omega$, but every set has only finitely many in its transitive closure, so it can be well-ordered.
The last model is described well in Jech, this is Problem 4 in the aforementioned reference, and the key point is that the atoms are indexed by countable sequences of ordinals, so that there are always surjections onto every set, but there is no $omega$ sequence of atoms, which form a proper class, so there is no injection from any infinite set into the class of atoms. (And indeed, that implies all sets of atoms are finite.)
$endgroup$
add a comment |
$begingroup$
The results appear in Jech's "The Axiom of Choice" in the problem section of Chapter 9 (Problems 2,3, and 4).
Indeed, it is easy to see that the injections into classes imply the surjections from classes which imply choice. Exactly by means of the class of ordinals. So the point is to separate the others.
And if we have a proper class of atoms whose subsets are all finite, then it is a class which does not map onto $omega$, but every set has only finitely many in its transitive closure, so it can be well-ordered.
The last model is described well in Jech, this is Problem 4 in the aforementioned reference, and the key point is that the atoms are indexed by countable sequences of ordinals, so that there are always surjections onto every set, but there is no $omega$ sequence of atoms, which form a proper class, so there is no injection from any infinite set into the class of atoms. (And indeed, that implies all sets of atoms are finite.)
$endgroup$
The results appear in Jech's "The Axiom of Choice" in the problem section of Chapter 9 (Problems 2,3, and 4).
Indeed, it is easy to see that the injections into classes imply the surjections from classes which imply choice. Exactly by means of the class of ordinals. So the point is to separate the others.
And if we have a proper class of atoms whose subsets are all finite, then it is a class which does not map onto $omega$, but every set has only finitely many in its transitive closure, so it can be well-ordered.
The last model is described well in Jech, this is Problem 4 in the aforementioned reference, and the key point is that the atoms are indexed by countable sequences of ordinals, so that there are always surjections onto every set, but there is no $omega$ sequence of atoms, which form a proper class, so there is no injection from any infinite set into the class of atoms. (And indeed, that implies all sets of atoms are finite.)
answered 3 hours ago
Asaf KaragilaAsaf Karagila
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$begingroup$
math.stackexchange.com/questions/1337583/… might be helpful?
$endgroup$
– Asaf Karagila
4 hours ago