Injection into a proper class and choice without regularityAxiom of Choice and Order TypesProper class forcing vs forcing with a set of conditions bigger than one's modelHow big is the proper class of all sets?What is the order type of $L$ with Godel's well ordering?Minimal Generalized Continuum Hypothesis & Axiom of ChoiceOn the Axiom of Choice for Conglomerates and SkeletonsProper classes subnumerous to $V$ in a model of a Morse-Kelley related theoryAre classes still “larger” than sets without the axiom of choice?For which theories does ZFC without global choice prove the existence of a proper class monster model?“Surjective cardinals” - using surjections rather than injections to define isomorphism classes of sets

Injection into a proper class and choice without regularity


Axiom of Choice and Order TypesProper class forcing vs forcing with a set of conditions bigger than one's modelHow big is the proper class of all sets?What is the order type of $L$ with Godel's well ordering?Minimal Generalized Continuum Hypothesis & Axiom of ChoiceOn the Axiom of Choice for Conglomerates and SkeletonsProper classes subnumerous to $V$ in a model of a Morse-Kelley related theoryAre classes still “larger” than sets without the axiom of choice?For which theories does ZFC without global choice prove the existence of a proper class monster model?“Surjective cardinals” - using surjections rather than injections to define isomorphism classes of sets













5












$begingroup$


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 $Y∩V_x_βsubsetneq Y∩V_x_γ$, 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 $sf ZFC^-+mboxa proper class of atoms+mboxonly finite sets of atoms$, 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 $sf ZF^-$?










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  • 1




    $begingroup$
    math.stackexchange.com/questions/1337583/… might be helpful?
    $endgroup$
    – Asaf Karagila
    3 hours ago















5












$begingroup$


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 $Y∩V_x_βsubsetneq Y∩V_x_γ$, 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 $sf ZFC^-+mboxa proper class of atoms+mboxonly finite sets of atoms$, 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 $sf ZF^-$?










share|cite|improve this question









New contributor




Holo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 1




    $begingroup$
    math.stackexchange.com/questions/1337583/… might be helpful?
    $endgroup$
    – Asaf Karagila
    3 hours ago













5












5








5





$begingroup$


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 $Y∩V_x_βsubsetneq Y∩V_x_γ$, 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 $sf ZFC^-+mboxa proper class of atoms+mboxonly finite sets of atoms$, 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 $sf ZF^-$?










share|cite|improve this question









New contributor




Holo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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 $Y∩V_x_βsubsetneq Y∩V_x_γ$, 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 $sf ZFC^-+mboxa proper class of atoms+mboxonly finite sets of atoms$, 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






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edited 4 hours ago









András Bátkai

3,85142342




3,85142342






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asked 4 hours ago









HoloHolo

1263




1263




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New contributor





Holo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.







  • 1




    $begingroup$
    math.stackexchange.com/questions/1337583/… might be helpful?
    $endgroup$
    – Asaf Karagila
    3 hours ago












  • 1




    $begingroup$
    math.stackexchange.com/questions/1337583/… might be helpful?
    $endgroup$
    – Asaf Karagila
    3 hours ago







1




1




$begingroup$
math.stackexchange.com/questions/1337583/… might be helpful?
$endgroup$
– Asaf Karagila
3 hours ago




$begingroup$
math.stackexchange.com/questions/1337583/… might be helpful?
$endgroup$
– Asaf Karagila
3 hours ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

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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    3












    $begingroup$

    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.)






    share|cite|improve this answer









    $endgroup$

















      3












      $begingroup$

      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.)






      share|cite|improve this answer









      $endgroup$















        3












        3








        3





        $begingroup$

        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.)






        share|cite|improve this answer









        $endgroup$



        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.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        Asaf KaragilaAsaf Karagila

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