Models of set theory where not every set can be linearly ordered
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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.
Essentially, I am asking for a proof that without choice sometimes the linear ordering principle fails.
set-theory axiom-of-choice
asked Apr 15 at 19:21
LGarLGar
506
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add a comment |
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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.
Essentially, I am asking for a proof that without choice sometimes the linear ordering principle fails.
set-theory axiom-of-choice
asked Apr 15 at 19:21
LGarLGar
506
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
add a comment |
$begingroup$
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
asked Apr 15 at 19:21
LGarLGar
506
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
asked Apr 15 at 19:21
LGarLGar
506
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.
asked Apr 15 at 19:21
LGarLGar
506
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.
edited Apr 16 at 0:22
LGar
asked Apr 15 at 19:21
LGarLGar
506
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.
asked Apr 15 at 19:21
LGarLGar
506
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.
Check out our Code of Conduct.
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
add a comment |
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
add a comment |
2 Answers
2
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oldest
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$begingroup$
Yes, both of Fraenkel's models are examples of such models. To see why note that:
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.
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.
answered Apr 15 at 19:30
Asaf Karagila♦Asaf Karagila
309k33441775
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add a comment |
$begingroup$
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.
answered Apr 15 at 19:39
Andrés E. CaicedoAndrés E. Caicedo
66.1k8160252
$endgroup$
1
$begingroup$
Good examples, albeit significantly more complicated! :-)
$endgroup$
– Asaf Karagila♦
Apr 15 at 21:33
add a comment |
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2 Answers
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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:
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.
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.
answered Apr 15 at 19:30
Asaf Karagila♦Asaf Karagila
309k33441775
$endgroup$
add a comment |
$begingroup$
Yes, both of Fraenkel's models are examples of such models. To see why note that:
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.
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.
answered Apr 15 at 19:30
Asaf Karagila♦Asaf Karagila
309k33441775
$endgroup$
add a comment |
$begingroup$
Yes, both of Fraenkel's models are examples of such models. To see why note that:
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.
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.
answered Apr 15 at 19:30
Asaf Karagila♦Asaf Karagila
309k33441775
$endgroup$
Yes, both of Fraenkel's models are examples of such models. To see why note that:
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.
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.
answered Apr 15 at 19:30
Asaf Karagila♦Asaf Karagila
309k33441775
answered Apr 15 at 19:30
Asaf Karagila♦Asaf Karagila
309k33441775
answered Apr 15 at 19:30
Asaf Karagila♦Asaf Karagila
309k33441775
answered Apr 15 at 19:30
Asaf Karagila♦Asaf Karagila
309k33441775
309k33441775
add a comment |
add a comment |
$begingroup$
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.
answered Apr 15 at 19:39
Andrés E. CaicedoAndrés E. Caicedo
66.1k8160252
$endgroup$
1
$begingroup$
Good examples, albeit significantly more complicated! :-)
$endgroup$
– Asaf Karagila♦
Apr 15 at 21:33
add a comment |
$begingroup$
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.
answered Apr 15 at 19:39
Andrés E. CaicedoAndrés E. Caicedo
66.1k8160252
$endgroup$
1
$begingroup$
Good examples, albeit significantly more complicated! :-)
$endgroup$
– Asaf Karagila♦
Apr 15 at 21:33
add a comment |
$begingroup$
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.
answered Apr 15 at 19:39
Andrés E. CaicedoAndrés E. Caicedo
66.1k8160252
$endgroup$
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.
answered Apr 15 at 19:39
Andrés E. CaicedoAndrés E. Caicedo
66.1k8160252
answered Apr 15 at 19:39
Andrés E. CaicedoAndrés E. Caicedo
66.1k8160252
answered Apr 15 at 19:39
Andrés E. CaicedoAndrés E. Caicedo
66.1k8160252
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
add a comment |
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
add a comment |
LGar is a new contributor. Be nice, and check out our Code of Conduct.
LGar is a new contributor. Be nice, and check out our Code of Conduct.
LGar is a new contributor. Be nice, and check out our Code of Conduct.
LGar is a new contributor. Be nice, and check out our Code of Conduct.
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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