Relationship between AC, WO and Zorns Lemma in ZF-PowersetAxiom of Choice and Order TypesAxiom of Choice in a weaker systemIs choice needed to establish the existence of idempotent ultrafilters?Does ZFC prove the universe is linearly orderable?Axiom of Choice and Number TheoryRelationship between fragments of the axiom of choice and the dependent choice principlesAbout the hypothesis of Zorn's lemmaZorn's lemma via Zermelo theoremRelation between the Axiom of Choice and a the existence of a hyperplane not containing a vectorHow is this fixed point theorem related to the axiom of choice?
Relationship between AC, WO and Zorns Lemma in ZF-Powerset
Axiom of Choice and Order TypesAxiom of Choice in a weaker systemIs choice needed to establish the existence of idempotent ultrafilters?Does ZFC prove the universe is linearly orderable?Axiom of Choice and Number TheoryRelationship between fragments of the axiom of choice and the dependent choice principlesAbout the hypothesis of Zorn's lemmaZorn's lemma via Zermelo theoremRelation between the Axiom of Choice and a the existence of a hyperplane not containing a vectorHow is this fixed point theorem related to the axiom of choice?
$begingroup$
In regular ZF, AC, WO and Zorn's Lemma are equivalent, but every proof I know (of the implication AC->WO and AC-> Zorn) uses the axiom of choice on the powerset of X (where X is the Set which is to be well-ordered). My question is, whether or not there is a proof of this equivalence that doesn't use the axiom of choice or whether there is a Model of ZF-Powerset in which AC holds but the well-ordering principle (or Zorns Lemma) fails.
set-theory axiom-of-choice independence-results
edited 13 hours ago
Martin Sleziak
3,13032231
asked 13 hours ago
Hannes JakobHannes Jakob
284
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Hannes Jakob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
In regular ZF, AC, WO and Zorn's Lemma are equivalent, but every proof I know (of the implication AC->WO and AC-> Zorn) uses the axiom of choice on the powerset of X (where X is the Set which is to be well-ordered). My question is, whether or not there is a proof of this equivalence that doesn't use the axiom of choice or whether there is a Model of ZF-Powerset in which AC holds but the well-ordering principle (or Zorns Lemma) fails.
set-theory axiom-of-choice independence-results
edited 13 hours ago
Martin Sleziak
3,13032231
asked 13 hours ago
Hannes JakobHannes Jakob
284
New contributor
Hannes Jakob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In regular ZF, AC, WO and Zorn's Lemma are equivalent, but every proof I know (of the implication AC->WO and AC-> Zorn) uses the axiom of choice on the powerset of X (where X is the Set which is to be well-ordered). My question is, whether or not there is a proof of this equivalence that doesn't use the axiom of choice or whether there is a Model of ZF-Powerset in which AC holds but the well-ordering principle (or Zorns Lemma) fails.
set-theory axiom-of-choice independence-results
edited 13 hours ago
Martin Sleziak
3,13032231
asked 13 hours ago
Hannes JakobHannes Jakob
284
New contributor
Hannes Jakob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In regular ZF, AC, WO and Zorn's Lemma are equivalent, but every proof I know (of the implication AC->WO and AC-> Zorn) uses the axiom of choice on the powerset of X (where X is the Set which is to be well-ordered). My question is, whether or not there is a proof of this equivalence that doesn't use the axiom of choice or whether there is a Model of ZF-Powerset in which AC holds but the well-ordering principle (or Zorns Lemma) fails.
set-theory axiom-of-choice independence-results
set-theory axiom-of-choice independence-results
edited 13 hours ago
Martin Sleziak
3,13032231
asked 13 hours ago
Hannes JakobHannes Jakob
284
New contributor
Hannes Jakob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
Martin Sleziak
3,13032231
asked 13 hours ago
Hannes JakobHannes Jakob
284
New contributor
Hannes Jakob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
Martin Sleziak
3,13032231
edited 13 hours ago
Martin Sleziak
3,13032231
edited 13 hours ago
Martin Sleziak
3,13032231
3,13032231
asked 13 hours ago
Hannes JakobHannes Jakob
284
New contributor
Hannes Jakob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 13 hours ago
Hannes JakobHannes Jakob
284
asked 13 hours ago
Hannes JakobHannes Jakob
284
284
New contributor
Hannes Jakob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Hannes Jakob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hannes Jakob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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$begingroup$
This is a classic theorem of Zarach, that it is consistent that $sf ZF^-$ holds with the Axiom of Choice, but not every set can be well-ordered.
Zarach, Andrzej, Unions of $sf ZF^-$models which are themselves $sf ZF^-$ models, Logic colloquium ’80, Eur. Summer Meet., Prague 1980, Stud. Logic Found. Math. 108, 315-342 (1982). ZBL0524.03039.
answered 13 hours ago
Asaf KaragilaAsaf Karagila
22k681187
$endgroup$
1
$begingroup$
In the quoted paper (p.338), Zarach credits Zbigniew Szczepaniak with first having demonstrated in 1979 that in the ZF^- context, the axiom of choice does not imply that every set can be well-ordered; and he acknowledges how Szcepaniak's work relates to his.
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– Ali Enayat
4 hours ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is a classic theorem of Zarach, that it is consistent that $sf ZF^-$ holds with the Axiom of Choice, but not every set can be well-ordered.
Zarach, Andrzej, Unions of $sf ZF^-$models which are themselves $sf ZF^-$ models, Logic colloquium ’80, Eur. Summer Meet., Prague 1980, Stud. Logic Found. Math. 108, 315-342 (1982). ZBL0524.03039.
answered 13 hours ago
Asaf KaragilaAsaf Karagila
22k681187
$endgroup$
1
$begingroup$
In the quoted paper (p.338), Zarach credits Zbigniew Szczepaniak with first having demonstrated in 1979 that in the ZF^- context, the axiom of choice does not imply that every set can be well-ordered; and he acknowledges how Szcepaniak's work relates to his.
$endgroup$
– Ali Enayat
4 hours ago
add a comment |
$begingroup$
This is a classic theorem of Zarach, that it is consistent that $sf ZF^-$ holds with the Axiom of Choice, but not every set can be well-ordered.
Zarach, Andrzej, Unions of $sf ZF^-$models which are themselves $sf ZF^-$ models, Logic colloquium ’80, Eur. Summer Meet., Prague 1980, Stud. Logic Found. Math. 108, 315-342 (1982). ZBL0524.03039.
answered 13 hours ago
Asaf KaragilaAsaf Karagila
22k681187
$endgroup$
1
$begingroup$
In the quoted paper (p.338), Zarach credits Zbigniew Szczepaniak with first having demonstrated in 1979 that in the ZF^- context, the axiom of choice does not imply that every set can be well-ordered; and he acknowledges how Szcepaniak's work relates to his.
$endgroup$
– Ali Enayat
4 hours ago
add a comment |
$begingroup$
This is a classic theorem of Zarach, that it is consistent that $sf ZF^-$ holds with the Axiom of Choice, but not every set can be well-ordered.
Zarach, Andrzej, Unions of $sf ZF^-$models which are themselves $sf ZF^-$ models, Logic colloquium ’80, Eur. Summer Meet., Prague 1980, Stud. Logic Found. Math. 108, 315-342 (1982). ZBL0524.03039.
answered 13 hours ago
Asaf KaragilaAsaf Karagila
22k681187
$endgroup$
This is a classic theorem of Zarach, that it is consistent that $sf ZF^-$ holds with the Axiom of Choice, but not every set can be well-ordered.
Zarach, Andrzej, Unions of $sf ZF^-$models which are themselves $sf ZF^-$ models, Logic colloquium ’80, Eur. Summer Meet., Prague 1980, Stud. Logic Found. Math. 108, 315-342 (1982). ZBL0524.03039.
answered 13 hours ago
Asaf KaragilaAsaf Karagila
22k681187
answered 13 hours ago
Asaf KaragilaAsaf Karagila
22k681187
answered 13 hours ago
Asaf KaragilaAsaf Karagila
22k681187
answered 13 hours ago
Asaf KaragilaAsaf Karagila
22k681187
22k681187
1
$begingroup$
In the quoted paper (p.338), Zarach credits Zbigniew Szczepaniak with first having demonstrated in 1979 that in the ZF^- context, the axiom of choice does not imply that every set can be well-ordered; and he acknowledges how Szcepaniak's work relates to his.
$endgroup$
– Ali Enayat
4 hours ago
add a comment |
1
$begingroup$
In the quoted paper (p.338), Zarach credits Zbigniew Szczepaniak with first having demonstrated in 1979 that in the ZF^- context, the axiom of choice does not imply that every set can be well-ordered; and he acknowledges how Szcepaniak's work relates to his.
$endgroup$
– Ali Enayat
4 hours ago
1
1
$begingroup$
In the quoted paper (p.338), Zarach credits Zbigniew Szczepaniak with first having demonstrated in 1979 that in the ZF^- context, the axiom of choice does not imply that every set can be well-ordered; and he acknowledges how Szcepaniak's work relates to his.
$endgroup$
– Ali Enayat
4 hours ago
$begingroup$
In the quoted paper (p.338), Zarach credits Zbigniew Szczepaniak with first having demonstrated in 1979 that in the ZF^- context, the axiom of choice does not imply that every set can be well-ordered; and he acknowledges how Szcepaniak's work relates to his.
$endgroup$
– Ali Enayat
4 hours ago
add a comment |
Hannes Jakob is a new contributor. Be nice, and check out our Code of Conduct.
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