Models of set theory where not every set can be linearly ordered Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How can there be genuine models of set theory?Some questions concerning set-theoretic models of first-order theoriesExample of a model in set theory where the axiom of extensionality does not hold?Zorn's lemma and maximal linearly ordered subsetsCan, or should, we ignore uncountable models of set theory?Models in set theory and continuum hypothesisCan every non-empty set satisfying the axioms of $sfZF$ be totally ordered?Every transitive $in$-linearly ordered set is $in$-well ordered without axiom of foundationConfusion about countable models of ZF set theory.Every countable linearly ordered set is similar to one of its subsets
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Models of set theory where not every set can be linearly ordered
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How can there be genuine models of set theory?Some questions concerning set-theoretic models of first-order theoriesExample of a model in set theory where the axiom of extensionality does not hold?Zorn's lemma and maximal linearly ordered subsetsCan, or should, we ignore uncountable models of set theory?Models in set theory and continuum hypothesisCan every non-empty set satisfying the axioms of $sfZF$ be totally ordered?Every transitive $in$-linearly ordered set is $in$-well ordered without axiom of foundationConfusion about countable models of ZF set theory.Every countable linearly ordered set is similar to one of its subsets
$begingroup$
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
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LGar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.
set-theory axiom-of-choice
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
2 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♦
10 mins ago
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.
set-theory axiom-of-choice
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
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
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Check out our Code of Conduct.
edited 11 mins ago
Asaf Karagila♦
308k33441775
308k33441775
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asked 2 hours ago
LGarLGar
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385
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LGar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
2 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♦
10 mins ago
add a comment |
$begingroup$
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
2 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♦
10 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
2 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
2 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♦
10 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♦
10 mins ago
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.
$endgroup$
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.
$endgroup$
1
$begingroup$
Good examples, albeit significantly more complicated! :-)
$endgroup$
– Asaf Karagila♦
12 mins ago
add a comment |
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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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 2 hours ago
Asaf Karagila♦Asaf Karagila
308k33441775
308k33441775
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.
$endgroup$
1
$begingroup$
Good examples, albeit significantly more complicated! :-)
$endgroup$
– Asaf Karagila♦
12 mins ago
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.
$endgroup$
1
$begingroup$
Good examples, albeit significantly more complicated! :-)
$endgroup$
– Asaf Karagila♦
12 mins ago
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.
$endgroup$
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 2 hours ago
Andrés E. CaicedoAndrés E. Caicedo
66.1k8160252
66.1k8160252
1
$begingroup$
Good examples, albeit significantly more complicated! :-)
$endgroup$
– Asaf Karagila♦
12 mins ago
add a comment |
1
$begingroup$
Good examples, albeit significantly more complicated! :-)
$endgroup$
– Asaf Karagila♦
12 mins ago
1
1
$begingroup$
Good examples, albeit significantly more complicated! :-)
$endgroup$
– Asaf Karagila♦
12 mins ago
$begingroup$
Good examples, albeit significantly more complicated! :-)
$endgroup$
– Asaf Karagila♦
12 mins ago
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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$begingroup$
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
2 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♦
10 mins ago