How strong is the axiom of well-ordered choice?












9












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    ...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
    5 hours ago






  • 2




    $begingroup$
    That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
    $endgroup$
    – Mike Battaglia
    5 hours ago










  • $begingroup$
    Ahh, I missed that distinction. Thank you!
    $endgroup$
    – Steven Stadnicki
    5 hours ago










  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
    $endgroup$
    – Carl Mummert
    5 hours ago
















9












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    ...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
    5 hours ago






  • 2




    $begingroup$
    That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
    $endgroup$
    – Mike Battaglia
    5 hours ago










  • $begingroup$
    Ahh, I missed that distinction. Thank you!
    $endgroup$
    – Steven Stadnicki
    5 hours ago










  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
    $endgroup$
    – Carl Mummert
    5 hours ago














9












9








9


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?










share|cite|improve this question











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago







Mike Battaglia

















asked 6 hours ago









Mike BattagliaMike Battaglia

1,4321126




1,4321126












  • $begingroup$
    ...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
    5 hours ago






  • 2




    $begingroup$
    That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
    $endgroup$
    – Mike Battaglia
    5 hours ago










  • $begingroup$
    Ahh, I missed that distinction. Thank you!
    $endgroup$
    – Steven Stadnicki
    5 hours ago










  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
    $endgroup$
    – Carl Mummert
    5 hours ago


















  • $begingroup$
    ...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
    5 hours ago






  • 2




    $begingroup$
    That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
    $endgroup$
    – Mike Battaglia
    5 hours ago










  • $begingroup$
    Ahh, I missed that distinction. Thank you!
    $endgroup$
    – Steven Stadnicki
    5 hours ago










  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
    $endgroup$
    – Carl Mummert
    5 hours ago
















$begingroup$
...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
5 hours ago




$begingroup$
...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
5 hours ago




2




2




$begingroup$
That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
$endgroup$
– Mike Battaglia
5 hours ago




$begingroup$
That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
$endgroup$
– Mike Battaglia
5 hours ago












$begingroup$
Ahh, I missed that distinction. Thank you!
$endgroup$
– Steven Stadnicki
5 hours ago




$begingroup$
Ahh, I missed that distinction. Thank you!
$endgroup$
– Steven Stadnicki
5 hours ago












$begingroup$
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
5 hours ago




$begingroup$
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
5 hours ago












$begingroup$
Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
$endgroup$
– Carl Mummert
5 hours ago




$begingroup$
Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
$endgroup$
– Carl Mummert
5 hours ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

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









$endgroup$













  • $begingroup$
    Do you have a source for a proof of ${sf AC_{rm WO}}iff forall x(aleph(x)=aleph^*(x))$?
    $endgroup$
    – Holo
    3 hours ago












  • $begingroup$
    @Holo karagila.org/2014/on-the-partition-principle
    $endgroup$
    – Asaf Karagila
    3 hours ago













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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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









$endgroup$













  • $begingroup$
    Do you have a source for a proof of ${sf AC_{rm WO}}iff forall x(aleph(x)=aleph^*(x))$?
    $endgroup$
    – Holo
    3 hours ago












  • $begingroup$
    @Holo karagila.org/2014/on-the-partition-principle
    $endgroup$
    – Asaf Karagila
    3 hours ago


















6












$begingroup$

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









$endgroup$













  • $begingroup$
    Do you have a source for a proof of ${sf AC_{rm WO}}iff forall x(aleph(x)=aleph^*(x))$?
    $endgroup$
    – Holo
    3 hours ago












  • $begingroup$
    @Holo karagila.org/2014/on-the-partition-principle
    $endgroup$
    – Asaf Karagila
    3 hours ago
















6












6








6





$begingroup$

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









$endgroup$



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












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









Asaf KaragilaAsaf Karagila

306k33437768




306k33437768












  • $begingroup$
    Do you have a source for a proof of ${sf AC_{rm WO}}iff forall x(aleph(x)=aleph^*(x))$?
    $endgroup$
    – Holo
    3 hours ago












  • $begingroup$
    @Holo karagila.org/2014/on-the-partition-principle
    $endgroup$
    – Asaf Karagila
    3 hours ago




















  • $begingroup$
    Do you have a source for a proof of ${sf AC_{rm WO}}iff forall x(aleph(x)=aleph^*(x))$?
    $endgroup$
    – Holo
    3 hours ago












  • $begingroup$
    @Holo karagila.org/2014/on-the-partition-principle
    $endgroup$
    – Asaf Karagila
    3 hours ago


















$begingroup$
Do you have a source for a proof of ${sf AC_{rm WO}}iff forall x(aleph(x)=aleph^*(x))$?
$endgroup$
– Holo
3 hours ago






$begingroup$
Do you have a source for a proof of ${sf AC_{rm WO}}iff forall x(aleph(x)=aleph^*(x))$?
$endgroup$
– Holo
3 hours ago














$begingroup$
@Holo karagila.org/2014/on-the-partition-principle
$endgroup$
– Asaf Karagila
3 hours ago






$begingroup$
@Holo karagila.org/2014/on-the-partition-principle
$endgroup$
– Asaf Karagila
3 hours ago




















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