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How strong is the axiom of well-ordered choice?

Why is the Axiom of Well-ordered Choice not strong enough to prove Zorn's Lemma?No uncountable ordinals without the axiom of choice?Axiom of choice , Hartogs ordinals, well-ordering principleWhich sets are well-orderable without Axiom of Choice?Well-orderings of $mathbb R$ without Choice“There is no well-ordered uncountable set of real numbers”Axiom of choice and the well ordering principleThe class of well-founded sets satisfies the axiom of foundation and the axiom of choiceAxiom of Choice is equivalent to Well-ordering Theorem: Hrbacek, Jech - “Introduction to Set Theory”the power set of every well-ordered set is well-ordered implies well orderingHow strong are weak choice principles?

11

1

$begingroup$

I sometimes see references to the "Axiom of Well-Ordered Choice," but I'm not sure how strong it is. It states that every well-ordered family of sets has a choice function.

By "well-ordered family," I don't mean that the sets within the family are well-ordered, but that the family must index all the sets within the family by some ordinal.

How strong is this axiom? Can it prove the Hahn-Banach theorem, the ultrafilter lemma, anything about measurable sets, etc? Does it have any implications about what sets can be well-ordered (the reals for instance), or perhaps prove anything about the Hartogs number of sets, etc?

Does anyone have a reference for this?

set-theory axiom-of-choice foundations well-orders

share|cite|improve this question

edited Mar 8 at 22:00

Mike Battaglia

asked Mar 8 at 20:22

Mike BattagliaMike Battaglia

1,5421128

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ...isn't this axiom a consequence of ZF? One can choose the (well-defined, since it's well-ordered) lexicographically least element of each set in the family...
  $endgroup$
  – Steven Stadnicki
  Mar 8 at 20:46
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
  $endgroup$
  – Mike Battaglia
  Mar 8 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ahh, I missed that distinction. Thank you!
  $endgroup$
  – Steven Stadnicki
  Mar 8 at 20:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've never seen this axiom before. Does the family itself need to be a set or can it be a proper class?
  $endgroup$
  – Robert Shore
  Mar 8 at 20:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
  $endgroup$
  – Carl Mummert
  Mar 8 at 20:56
  

  
  
  
  

 | 
show 2 more comments

11

1

$begingroup$

I sometimes see references to the "Axiom of Well-Ordered Choice," but I'm not sure how strong it is. It states that every well-ordered family of sets has a choice function.

By "well-ordered family," I don't mean that the sets within the family are well-ordered, but that the family must index all the sets within the family by some ordinal.

How strong is this axiom? Can it prove the Hahn-Banach theorem, the ultrafilter lemma, anything about measurable sets, etc? Does it have any implications about what sets can be well-ordered (the reals for instance), or perhaps prove anything about the Hartogs number of sets, etc?

Does anyone have a reference for this?

set-theory axiom-of-choice foundations well-orders

share|cite|improve this question

edited Mar 8 at 22:00

Mike Battaglia

asked Mar 8 at 20:22

Mike BattagliaMike Battaglia

1,5421128

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ...isn't this axiom a consequence of ZF? One can choose the (well-defined, since it's well-ordered) lexicographically least element of each set in the family...
  $endgroup$
  – Steven Stadnicki
  Mar 8 at 20:46
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
  $endgroup$
  – Mike Battaglia
  Mar 8 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ahh, I missed that distinction. Thank you!
  $endgroup$
  – Steven Stadnicki
  Mar 8 at 20:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've never seen this axiom before. Does the family itself need to be a set or can it be a proper class?
  $endgroup$
  – Robert Shore
  Mar 8 at 20:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
  $endgroup$
  – Carl Mummert
  Mar 8 at 20:56
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

I sometimes see references to the "Axiom of Well-Ordered Choice," but I'm not sure how strong it is. It states that every well-ordered family of sets has a choice function.

By "well-ordered family," I don't mean that the sets within the family are well-ordered, but that the family must index all the sets within the family by some ordinal.

How strong is this axiom? Can it prove the Hahn-Banach theorem, the ultrafilter lemma, anything about measurable sets, etc? Does it have any implications about what sets can be well-ordered (the reals for instance), or perhaps prove anything about the Hartogs number of sets, etc?

Does anyone have a reference for this?

set-theory axiom-of-choice foundations well-orders

share|cite|improve this question

edited Mar 8 at 22:00

Mike Battaglia

asked Mar 8 at 20:22

Mike BattagliaMike Battaglia

1,5421128

$endgroup$

share|cite|improve this question

edited Mar 8 at 22:00

Mike Battaglia

asked Mar 8 at 20:22

Mike BattagliaMike Battaglia

1,5421128

share|cite|improve this question

edited Mar 8 at 22:00

Mike Battaglia

asked Mar 8 at 20:22

Mike BattagliaMike Battaglia

1,5421128

asked Mar 8 at 20:22

Mike BattagliaMike Battaglia

1,5421128

asked Mar 8 at 20:22

Mike BattagliaMike Battaglia

1,5421128

1 Answer
1

active

oldest

votes

9

$begingroup$

The axiom of well-ordered choice, or $sf AC_rm WO$, is strictly weaker than the axiom of choice itself. If we start with $L$ and add $omega_1$ Cohen reals, then go to $L(Bbb R)$, one can show that $sf AC_rm WO$ holds, while $Bbb R$ cannot be well-ordered there.

Pincus proved in the 1970s that this is equivalent to the following statement on Hartogs and Lindenbaum numbers:

$sf AC_rm WO$ is equivalent to the statement $forall x.aleph(x)=aleph^*(x)$.

Here, the Lindenbaum number, $aleph^*(x)$, is the least ordinal which $x$ cannot be mapped onto. One obvious fact is that $aleph(x)leqaleph^*(x)$.

In the late 1950s or early 1960s Jensen proved that this assumption also implies $sf DC$. This is also a very clever proof.

The conjunction of these two consequences gives us that $aleph_1leq 2^aleph_0$, as a result of a theorem of Shelah from the 1980s, this implies there is a non-measurable set of reals.

As far as Hahn–Banach, or other things of that sort, I do not believe that much is known on the topic. But to sum up, this axiom does not imply that the reals are well-ordered, but it does imply there is a non-measurable set of reals because there is a set of reals of size $aleph_1$ and $sf DC$ holds. Moreover, it is equivalent to saying that the Hartogs and Lindenbaum numbers are equal for all sets.

share|cite|improve this answer

answered Mar 8 at 22:15

Asaf Karagila♦Asaf Karagila

307k33441774

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have a source for a proof of $sf AC_rm WOiff forall x(aleph(x)=aleph^*(x))$?
  $endgroup$
  – Holo
  Mar 8 at 23:14
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Holo karagila.org/2014/on-the-partition-principle
  $endgroup$
  – Asaf Karagila♦
  Mar 8 at 23:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (The proof there assumes PP, but really you only use that for maps onto ordinals, which is the same as saying $aleph=aleph^*$.)
  $endgroup$
  – Asaf Karagila♦
  Mar 9 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you very much!
  $endgroup$
  – Holo
  Mar 9 at 16:43
  

  
  
  
  

add a comment | 

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9

$begingroup$

The axiom of well-ordered choice, or $sf AC_rm WO$, is strictly weaker than the axiom of choice itself. If we start with $L$ and add $omega_1$ Cohen reals, then go to $L(Bbb R)$, one can show that $sf AC_rm WO$ holds, while $Bbb R$ cannot be well-ordered there.

Pincus proved in the 1970s that this is equivalent to the following statement on Hartogs and Lindenbaum numbers:

$sf AC_rm WO$ is equivalent to the statement $forall x.aleph(x)=aleph^*(x)$.

Here, the Lindenbaum number, $aleph^*(x)$, is the least ordinal which $x$ cannot be mapped onto. One obvious fact is that $aleph(x)leqaleph^*(x)$.

In the late 1950s or early 1960s Jensen proved that this assumption also implies $sf DC$. This is also a very clever proof.

The conjunction of these two consequences gives us that $aleph_1leq 2^aleph_0$, as a result of a theorem of Shelah from the 1980s, this implies there is a non-measurable set of reals.

As far as Hahn–Banach, or other things of that sort, I do not believe that much is known on the topic. But to sum up, this axiom does not imply that the reals are well-ordered, but it does imply there is a non-measurable set of reals because there is a set of reals of size $aleph_1$ and $sf DC$ holds. Moreover, it is equivalent to saying that the Hartogs and Lindenbaum numbers are equal for all sets.

share|cite|improve this answer

answered Mar 8 at 22:15

Asaf Karagila♦Asaf Karagila

307k33441774

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have a source for a proof of $sf AC_rm WOiff forall x(aleph(x)=aleph^*(x))$?
  $endgroup$
  – Holo
  Mar 8 at 23:14
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Holo karagila.org/2014/on-the-partition-principle
  $endgroup$
  – Asaf Karagila♦
  Mar 8 at 23:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (The proof there assumes PP, but really you only use that for maps onto ordinals, which is the same as saying $aleph=aleph^*$.)
  $endgroup$
  – Asaf Karagila♦
  Mar 9 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you very much!
  $endgroup$
  – Holo
  Mar 9 at 16:43
  

  
  
  
  

add a comment | 

9

$begingroup$

The axiom of well-ordered choice, or $sf AC_rm WO$, is strictly weaker than the axiom of choice itself. If we start with $L$ and add $omega_1$ Cohen reals, then go to $L(Bbb R)$, one can show that $sf AC_rm WO$ holds, while $Bbb R$ cannot be well-ordered there.

Pincus proved in the 1970s that this is equivalent to the following statement on Hartogs and Lindenbaum numbers:

$sf AC_rm WO$ is equivalent to the statement $forall x.aleph(x)=aleph^*(x)$.

Here, the Lindenbaum number, $aleph^*(x)$, is the least ordinal which $x$ cannot be mapped onto. One obvious fact is that $aleph(x)leqaleph^*(x)$.

In the late 1950s or early 1960s Jensen proved that this assumption also implies $sf DC$. This is also a very clever proof.

The conjunction of these two consequences gives us that $aleph_1leq 2^aleph_0$, as a result of a theorem of Shelah from the 1980s, this implies there is a non-measurable set of reals.

As far as Hahn–Banach, or other things of that sort, I do not believe that much is known on the topic. But to sum up, this axiom does not imply that the reals are well-ordered, but it does imply there is a non-measurable set of reals because there is a set of reals of size $aleph_1$ and $sf DC$ holds. Moreover, it is equivalent to saying that the Hartogs and Lindenbaum numbers are equal for all sets.

share|cite|improve this answer

answered Mar 8 at 22:15

Asaf Karagila♦Asaf Karagila

307k33441774

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have a source for a proof of $sf AC_rm WOiff forall x(aleph(x)=aleph^*(x))$?
  $endgroup$
  – Holo
  Mar 8 at 23:14
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Holo karagila.org/2014/on-the-partition-principle
  $endgroup$
  – Asaf Karagila♦
  Mar 8 at 23:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (The proof there assumes PP, but really you only use that for maps onto ordinals, which is the same as saying $aleph=aleph^*$.)
  $endgroup$
  – Asaf Karagila♦
  Mar 9 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you very much!
  $endgroup$
  – Holo
  Mar 9 at 16:43
  

  
  
  
  

add a comment | 

$begingroup$

The axiom of well-ordered choice, or $sf AC_rm WO$, is strictly weaker than the axiom of choice itself. If we start with $L$ and add $omega_1$ Cohen reals, then go to $L(Bbb R)$, one can show that $sf AC_rm WO$ holds, while $Bbb R$ cannot be well-ordered there.

Pincus proved in the 1970s that this is equivalent to the following statement on Hartogs and Lindenbaum numbers:

$sf AC_rm WO$ is equivalent to the statement $forall x.aleph(x)=aleph^*(x)$.

Here, the Lindenbaum number, $aleph^*(x)$, is the least ordinal which $x$ cannot be mapped onto. One obvious fact is that $aleph(x)leqaleph^*(x)$.

In the late 1950s or early 1960s Jensen proved that this assumption also implies $sf DC$. This is also a very clever proof.

The conjunction of these two consequences gives us that $aleph_1leq 2^aleph_0$, as a result of a theorem of Shelah from the 1980s, this implies there is a non-measurable set of reals.

As far as Hahn–Banach, or other things of that sort, I do not believe that much is known on the topic. But to sum up, this axiom does not imply that the reals are well-ordered, but it does imply there is a non-measurable set of reals because there is a set of reals of size $aleph_1$ and $sf DC$ holds. Moreover, it is equivalent to saying that the Hartogs and Lindenbaum numbers are equal for all sets.

share|cite|improve this answer

answered Mar 8 at 22:15

Asaf Karagila♦Asaf Karagila

307k33441774

$endgroup$

share|cite|improve this answer

answered Mar 8 at 22:15

Asaf Karagila♦Asaf Karagila

307k33441774

answered Mar 8 at 22:15

Asaf Karagila♦Asaf Karagila

307k33441774

answered Mar 8 at 22:15

Asaf Karagila♦Asaf Karagila

307k33441774