Models of set theory where not every set can be linearly ordered












5












$begingroup$


Can anybody point me towards a model of set theory where not every set can be linearly ordered, and a corresponding proof. I have seen it claimed that in Fraenkels second permutation model that there is a set that cannot be linearly ordered, but cannot find a proof.



Essentially, I am asking for a proof that without choice sometimes the linear ordering principle fails.






set-theory axiom-of-choice







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  • 1




    $begingroup$
    In the case of the Fraenkel model, would this just come down to saying that any linear ordering would have a finite support, and then we just consider a permutation of two atoms outside of said support?
    $endgroup$
    – LGar
    Apr 15 at 19:23






  • 1




    $begingroup$
    Yes, by the way, a direct argument in both the models of Fraenkel is that any linear order would have a finite support and we can find a permutation that moves some things in an incongruous way.
    $endgroup$
    – Asaf Karagila♦
    Apr 15 at 21:35






  • 1




    $begingroup$
    Possible duplicate of Proving "every set can be totally ordered" without using Axiom of Choice
    $endgroup$
    – YuiTo Cheng
    Apr 15 at 23:04










  • $begingroup$
    This question is not as far as I can tell a duplicate - that question is asking for a proof of the linear ordering principle without choice, while I was asking for a proof that the linear ordering principle can sometimes fail in the abscence of choice.
    $endgroup$
    – LGar
    Apr 16 at 0:21










  • $begingroup$
    What about Is every set linearly ordered in ZF
    $endgroup$
    – YuiTo Cheng
    Apr 16 at 0:57


















5












$begingroup$


Can anybody point me towards a model of set theory where not every set can be linearly ordered, and a corresponding proof. I have seen it claimed that in Fraenkels second permutation model that there is a set that cannot be linearly ordered, but cannot find a proof.



Essentially, I am asking for a proof that without choice sometimes the linear ordering principle fails.






set-theory axiom-of-choice







share|cite|improve this question









New contributor




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







$endgroup$








  • 1




    $begingroup$
    In the case of the Fraenkel model, would this just come down to saying that any linear ordering would have a finite support, and then we just consider a permutation of two atoms outside of said support?
    $endgroup$
    – LGar
    Apr 15 at 19:23






  • 1




    $begingroup$
    Yes, by the way, a direct argument in both the models of Fraenkel is that any linear order would have a finite support and we can find a permutation that moves some things in an incongruous way.
    $endgroup$
    – Asaf Karagila♦
    Apr 15 at 21:35






  • 1




    $begingroup$
    Possible duplicate of Proving "every set can be totally ordered" without using Axiom of Choice
    $endgroup$
    – YuiTo Cheng
    Apr 15 at 23:04










  • $begingroup$
    This question is not as far as I can tell a duplicate - that question is asking for a proof of the linear ordering principle without choice, while I was asking for a proof that the linear ordering principle can sometimes fail in the abscence of choice.
    $endgroup$
    – LGar
    Apr 16 at 0:21










  • $begingroup$
    What about Is every set linearly ordered in ZF
    $endgroup$
    – YuiTo Cheng
    Apr 16 at 0:57
















5












5








5


1



$begingroup$


Can anybody point me towards a model of set theory where not every set can be linearly ordered, and a corresponding proof. I have seen it claimed that in Fraenkels second permutation model that there is a set that cannot be linearly ordered, but cannot find a proof.



Essentially, I am asking for a proof that without choice sometimes the linear ordering principle fails.






set-theory axiom-of-choice







share|cite|improve this question









New contributor




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







$endgroup$




Can anybody point me towards a model of set theory where not every set can be linearly ordered, and a corresponding proof. I have seen it claimed that in Fraenkels second permutation model that there is a set that cannot be linearly ordered, but cannot find a proof.



Essentially, I am asking for a proof that without choice sometimes the linear ordering principle fails.






set-theory axiom-of-choice




set-theory axiom-of-choice






share|cite|improve this question









New contributor




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











share|cite|improve this question









New contributor




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









share|cite|improve this question




share|cite|improve this question








edited Apr 16 at 0:22







LGar













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









asked Apr 15 at 19:21









LGarLGar

506




506




New contributor




LGar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





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






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








  • 1




    $begingroup$
    In the case of the Fraenkel model, would this just come down to saying that any linear ordering would have a finite support, and then we just consider a permutation of two atoms outside of said support?
    $endgroup$
    – LGar
    Apr 15 at 19:23






  • 1




    $begingroup$
    Yes, by the way, a direct argument in both the models of Fraenkel is that any linear order would have a finite support and we can find a permutation that moves some things in an incongruous way.
    $endgroup$
    – Asaf Karagila♦
    Apr 15 at 21:35






  • 1




    $begingroup$
    Possible duplicate of Proving "every set can be totally ordered" without using Axiom of Choice
    $endgroup$
    – YuiTo Cheng
    Apr 15 at 23:04










  • $begingroup$
    This question is not as far as I can tell a duplicate - that question is asking for a proof of the linear ordering principle without choice, while I was asking for a proof that the linear ordering principle can sometimes fail in the abscence of choice.
    $endgroup$
    – LGar
    Apr 16 at 0:21










  • $begingroup$
    What about Is every set linearly ordered in ZF
    $endgroup$
    – YuiTo Cheng
    Apr 16 at 0:57
















  • 1




    $begingroup$
    In the case of the Fraenkel model, would this just come down to saying that any linear ordering would have a finite support, and then we just consider a permutation of two atoms outside of said support?
    $endgroup$
    – LGar
    Apr 15 at 19:23






  • 1




    $begingroup$
    Yes, by the way, a direct argument in both the models of Fraenkel is that any linear order would have a finite support and we can find a permutation that moves some things in an incongruous way.
    $endgroup$
    – Asaf Karagila♦
    Apr 15 at 21:35






  • 1




    $begingroup$
    Possible duplicate of Proving "every set can be totally ordered" without using Axiom of Choice
    $endgroup$
    – YuiTo Cheng
    Apr 15 at 23:04










  • $begingroup$
    This question is not as far as I can tell a duplicate - that question is asking for a proof of the linear ordering principle without choice, while I was asking for a proof that the linear ordering principle can sometimes fail in the abscence of choice.
    $endgroup$
    – LGar
    Apr 16 at 0:21










  • $begingroup$
    What about Is every set linearly ordered in ZF
    $endgroup$
    – YuiTo Cheng
    Apr 16 at 0:57










1




1




$begingroup$
In the case of the Fraenkel model, would this just come down to saying that any linear ordering would have a finite support, and then we just consider a permutation of two atoms outside of said support?
$endgroup$
– LGar
Apr 15 at 19:23




$begingroup$
In the case of the Fraenkel model, would this just come down to saying that any linear ordering would have a finite support, and then we just consider a permutation of two atoms outside of said support?
$endgroup$
– LGar
Apr 15 at 19:23




1




1




$begingroup$
Yes, by the way, a direct argument in both the models of Fraenkel is that any linear order would have a finite support and we can find a permutation that moves some things in an incongruous way.
$endgroup$
– Asaf Karagila♦
Apr 15 at 21:35




$begingroup$
Yes, by the way, a direct argument in both the models of Fraenkel is that any linear order would have a finite support and we can find a permutation that moves some things in an incongruous way.
$endgroup$
– Asaf Karagila♦
Apr 15 at 21:35




1




1




$begingroup$
Possible duplicate of Proving "every set can be totally ordered" without using Axiom of Choice
$endgroup$
– YuiTo Cheng
Apr 15 at 23:04




$begingroup$
Possible duplicate of Proving "every set can be totally ordered" without using Axiom of Choice
$endgroup$
– YuiTo Cheng
Apr 15 at 23:04












$begingroup$
This question is not as far as I can tell a duplicate - that question is asking for a proof of the linear ordering principle without choice, while I was asking for a proof that the linear ordering principle can sometimes fail in the abscence of choice.
$endgroup$
– LGar
Apr 16 at 0:21




$begingroup$
This question is not as far as I can tell a duplicate - that question is asking for a proof of the linear ordering principle without choice, while I was asking for a proof that the linear ordering principle can sometimes fail in the abscence of choice.
$endgroup$
– LGar
Apr 16 at 0:21












$begingroup$
What about Is every set linearly ordered in ZF
$endgroup$
– YuiTo Cheng
Apr 16 at 0:57






$begingroup$
What about Is every set linearly ordered in ZF
$endgroup$
– YuiTo Cheng
Apr 16 at 0:57












2 Answers
2






active

oldest

votes


















9












$begingroup$

Yes, both of Fraenkel's models are examples of such models. To see why note that:




  1. In the first model, the atoms are an amorphous set. Namely, there cannot be split into two infinite sets. An amorphous set cannot be linearly ordered. To see why, note that ${ain Amid atext{ defines a finite initial segment}}$ is either finite or co-finite. Assume it's co-finite, otherwise take the reverse order, then by removing finitely many elements we have a linear ordering where every proper initial segment is finite. This defines a bijection with $omega$, of course. So the set can be split into two infinite sets after all.


  2. In the second model, the atoms can be written as a countable union of pairs which do not have a choice function. If the atoms were linearly orderable in that model, then we could have defined a choice function from the pairs: take the smallest one.



For models of $sf ZF$ one can imitate Fraenkel's construction using sets-of-sets-of Cohen reals as your atoms. This can be found in Jech's "Axiom of Choice" book in Chapter 5, as Cohen's second model.






share|cite|improve this answer









$endgroup$





















    7












    $begingroup$

    An interesting example of a different kind is any model where all sets of reals have the Baire property. In any such set the quotient of $mathbb R$ by the Vitali equivalence relation is not linearly orderable. See here for a sketch.



    Examples of such models are Solovay's model where all sets of reals are Lebesgue measurable, or natural models of the axiom of determinacy, or Shelah's model from section 7 of




    MR0768264 (86g:03082a). Shelah, Saharon. Can you take Solovay's inaccessible away? Israel J. Math. 48 (1984), no. 1, 1–47.







    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Good examples, albeit significantly more complicated! :-)
      $endgroup$
      – Asaf Karagila♦
      Apr 15 at 21:33












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    2 Answers
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    2 Answers
    2






    active

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    active

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    active

    oldest

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    9












    $begingroup$

    Yes, both of Fraenkel's models are examples of such models. To see why note that:




    1. In the first model, the atoms are an amorphous set. Namely, there cannot be split into two infinite sets. An amorphous set cannot be linearly ordered. To see why, note that ${ain Amid atext{ defines a finite initial segment}}$ is either finite or co-finite. Assume it's co-finite, otherwise take the reverse order, then by removing finitely many elements we have a linear ordering where every proper initial segment is finite. This defines a bijection with $omega$, of course. So the set can be split into two infinite sets after all.


    2. In the second model, the atoms can be written as a countable union of pairs which do not have a choice function. If the atoms were linearly orderable in that model, then we could have defined a choice function from the pairs: take the smallest one.



    For models of $sf ZF$ one can imitate Fraenkel's construction using sets-of-sets-of Cohen reals as your atoms. This can be found in Jech's "Axiom of Choice" book in Chapter 5, as Cohen's second model.






    share|cite|improve this answer









    $endgroup$


















      9












      $begingroup$

      Yes, both of Fraenkel's models are examples of such models. To see why note that:




      1. In the first model, the atoms are an amorphous set. Namely, there cannot be split into two infinite sets. An amorphous set cannot be linearly ordered. To see why, note that ${ain Amid atext{ defines a finite initial segment}}$ is either finite or co-finite. Assume it's co-finite, otherwise take the reverse order, then by removing finitely many elements we have a linear ordering where every proper initial segment is finite. This defines a bijection with $omega$, of course. So the set can be split into two infinite sets after all.


      2. In the second model, the atoms can be written as a countable union of pairs which do not have a choice function. If the atoms were linearly orderable in that model, then we could have defined a choice function from the pairs: take the smallest one.



      For models of $sf ZF$ one can imitate Fraenkel's construction using sets-of-sets-of Cohen reals as your atoms. This can be found in Jech's "Axiom of Choice" book in Chapter 5, as Cohen's second model.






      share|cite|improve this answer









      $endgroup$
















        9












        9








        9





        $begingroup$

        Yes, both of Fraenkel's models are examples of such models. To see why note that:




        1. In the first model, the atoms are an amorphous set. Namely, there cannot be split into two infinite sets. An amorphous set cannot be linearly ordered. To see why, note that ${ain Amid atext{ defines a finite initial segment}}$ is either finite or co-finite. Assume it's co-finite, otherwise take the reverse order, then by removing finitely many elements we have a linear ordering where every proper initial segment is finite. This defines a bijection with $omega$, of course. So the set can be split into two infinite sets after all.


        2. In the second model, the atoms can be written as a countable union of pairs which do not have a choice function. If the atoms were linearly orderable in that model, then we could have defined a choice function from the pairs: take the smallest one.



        For models of $sf ZF$ one can imitate Fraenkel's construction using sets-of-sets-of Cohen reals as your atoms. This can be found in Jech's "Axiom of Choice" book in Chapter 5, as Cohen's second model.






        share|cite|improve this answer









        $endgroup$



        Yes, both of Fraenkel's models are examples of such models. To see why note that:




        1. In the first model, the atoms are an amorphous set. Namely, there cannot be split into two infinite sets. An amorphous set cannot be linearly ordered. To see why, note that ${ain Amid atext{ defines a finite initial segment}}$ is either finite or co-finite. Assume it's co-finite, otherwise take the reverse order, then by removing finitely many elements we have a linear ordering where every proper initial segment is finite. This defines a bijection with $omega$, of course. So the set can be split into two infinite sets after all.


        2. In the second model, the atoms can be written as a countable union of pairs which do not have a choice function. If the atoms were linearly orderable in that model, then we could have defined a choice function from the pairs: take the smallest one.



        For models of $sf ZF$ one can imitate Fraenkel's construction using sets-of-sets-of Cohen reals as your atoms. This can be found in Jech's "Axiom of Choice" book in Chapter 5, as Cohen's second model.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 15 at 19:30









        Asaf Karagila♦Asaf Karagila

        309k33441775




        309k33441775























            7












            $begingroup$

            An interesting example of a different kind is any model where all sets of reals have the Baire property. In any such set the quotient of $mathbb R$ by the Vitali equivalence relation is not linearly orderable. See here for a sketch.



            Examples of such models are Solovay's model where all sets of reals are Lebesgue measurable, or natural models of the axiom of determinacy, or Shelah's model from section 7 of




            MR0768264 (86g:03082a). Shelah, Saharon. Can you take Solovay's inaccessible away? Israel J. Math. 48 (1984), no. 1, 1–47.







            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Good examples, albeit significantly more complicated! :-)
              $endgroup$
              – Asaf Karagila♦
              Apr 15 at 21:33
















            7












            $begingroup$

            An interesting example of a different kind is any model where all sets of reals have the Baire property. In any such set the quotient of $mathbb R$ by the Vitali equivalence relation is not linearly orderable. See here for a sketch.



            Examples of such models are Solovay's model where all sets of reals are Lebesgue measurable, or natural models of the axiom of determinacy, or Shelah's model from section 7 of




            MR0768264 (86g:03082a). Shelah, Saharon. Can you take Solovay's inaccessible away? Israel J. Math. 48 (1984), no. 1, 1–47.







            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Good examples, albeit significantly more complicated! :-)
              $endgroup$
              – Asaf Karagila♦
              Apr 15 at 21:33














            7












            7








            7





            $begingroup$

            An interesting example of a different kind is any model where all sets of reals have the Baire property. In any such set the quotient of $mathbb R$ by the Vitali equivalence relation is not linearly orderable. See here for a sketch.



            Examples of such models are Solovay's model where all sets of reals are Lebesgue measurable, or natural models of the axiom of determinacy, or Shelah's model from section 7 of




            MR0768264 (86g:03082a). Shelah, Saharon. Can you take Solovay's inaccessible away? Israel J. Math. 48 (1984), no. 1, 1–47.







            share|cite|improve this answer









            $endgroup$



            An interesting example of a different kind is any model where all sets of reals have the Baire property. In any such set the quotient of $mathbb R$ by the Vitali equivalence relation is not linearly orderable. See here for a sketch.



            Examples of such models are Solovay's model where all sets of reals are Lebesgue measurable, or natural models of the axiom of determinacy, or Shelah's model from section 7 of




            MR0768264 (86g:03082a). Shelah, Saharon. Can you take Solovay's inaccessible away? Israel J. Math. 48 (1984), no. 1, 1–47.








            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 15 at 19:39









            Andrés E. CaicedoAndrés E. Caicedo

            66.1k8160252




            66.1k8160252








            • 1




              $begingroup$
              Good examples, albeit significantly more complicated! :-)
              $endgroup$
              – Asaf Karagila♦
              Apr 15 at 21:33














            • 1




              $begingroup$
              Good examples, albeit significantly more complicated! :-)
              $endgroup$
              – Asaf Karagila♦
              Apr 15 at 21:33








            1




            1




            $begingroup$
            Good examples, albeit significantly more complicated! :-)
            $endgroup$
            – Asaf Karagila♦
            Apr 15 at 21:33




            $begingroup$
            Good examples, albeit significantly more complicated! :-)
            $endgroup$
            – Asaf Karagila♦
            Apr 15 at 21:33










            LGar is a new contributor. Be nice, and check out our Code of Conduct.










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