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












3












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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.






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  • $begingroup$
    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
    3 hours ago










  • $begingroup$
    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♦
    52 mins ago
















3












$begingroup$


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.






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.







$endgroup$












  • $begingroup$
    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
    3 hours ago










  • $begingroup$
    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♦
    52 mins ago














3












3








3





$begingroup$


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.






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.






set-theory axiom-of-choice




set-theory axiom-of-choice






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edited 54 mins ago









Asaf Karagila♦

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  • $begingroup$
    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
    3 hours ago










  • $begingroup$
    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♦
    52 mins ago


















  • $begingroup$
    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
    3 hours ago










  • $begingroup$
    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♦
    52 mins ago
















$begingroup$
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
3 hours ago




$begingroup$
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
3 hours ago












$begingroup$
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♦
52 mins ago




$begingroup$
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♦
52 mins ago










2 Answers
2






active

oldest

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5












$begingroup$

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









$endgroup$





















    5












    $begingroup$

    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♦
      55 mins ago












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    2 Answers
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    $begingroup$

    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









    $endgroup$


















      5












      $begingroup$

      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









      $endgroup$
















        5












        5








        5





        $begingroup$

        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 2 hours ago









        Asaf Karagila♦Asaf Karagila

        308k33441775




        308k33441775























            5












            $begingroup$

            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♦
              55 mins ago
















            5












            $begingroup$

            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♦
              55 mins ago














            5












            5








            5





            $begingroup$

            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 2 hours ago









            Andrés E. CaicedoAndrés E. Caicedo

            66.1k8160252




            66.1k8160252








            • 1




              $begingroup$
              Good examples, albeit significantly more complicated! :-)
              $endgroup$
              – Asaf Karagila♦
              55 mins ago














            • 1




              $begingroup$
              Good examples, albeit significantly more complicated! :-)
              $endgroup$
              – Asaf Karagila♦
              55 mins ago








            1




            1




            $begingroup$
            Good examples, albeit significantly more complicated! :-)
            $endgroup$
            – Asaf Karagila♦
            55 mins ago




            $begingroup$
            Good examples, albeit significantly more complicated! :-)
            $endgroup$
            – Asaf Karagila♦
            55 mins ago










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