How strong is the axiom of well-ordered choice? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)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?
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How strong is the axiom of well-ordered choice?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)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?
$begingroup$
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
asked Mar 8 at 20:22
Mike BattagliaMike Battaglia
1,6491230
$endgroup$
|
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
asked Mar 8 at 20:22
Mike BattagliaMike Battaglia
1,6491230
$endgroup$
$begingroup$
...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
asked Mar 8 at 20:22
Mike BattagliaMike Battaglia
1,6491230
$endgroup$
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
set-theory axiom-of-choice foundations well-orders
asked Mar 8 at 20:22
Mike BattagliaMike Battaglia
1,6491230
asked Mar 8 at 20:22
Mike BattagliaMike Battaglia
1,6491230
edited Mar 8 at 22:00
Mike Battaglia
asked Mar 8 at 20:22
Mike BattagliaMike Battaglia
1,6491230
asked Mar 8 at 20:22
Mike BattagliaMike Battaglia
1,6491230
asked Mar 8 at 20:22
Mike BattagliaMike Battaglia
1,6491230
1,6491230
$begingroup$
...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
$begingroup$
...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
$begingroup$
...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
$begingroup$
...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
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
$endgroup$
– Carl Mummert
Mar 8 at 20:56
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Mar 8 at 22:15
Asaf Karagila♦Asaf Karagila
309k33441775
$endgroup$
$begingroup$
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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1 Answer
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1 Answer
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oldest
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$begingroup$
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.
answered Mar 8 at 22:15
Asaf Karagila♦Asaf Karagila
309k33441775
$endgroup$
$begingroup$
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.
answered Mar 8 at 22:15
Asaf Karagila♦Asaf Karagila
309k33441775
$endgroup$
$begingroup$
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.
answered Mar 8 at 22:15
Asaf Karagila♦Asaf Karagila
309k33441775
$endgroup$
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.
answered Mar 8 at 22:15
Asaf Karagila♦Asaf Karagila
309k33441775
answered Mar 8 at 22:15
Asaf Karagila♦Asaf Karagila
309k33441775
answered Mar 8 at 22:15
Asaf Karagila♦Asaf Karagila
309k33441775
answered Mar 8 at 22:15
Asaf Karagila♦Asaf Karagila
309k33441775
309k33441775
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
2
$begingroup$
@Holo karagila.org/2014/on-the-partition-principle
$endgroup$
– Asaf Karagila♦
Mar 8 at 23:26
$begingroup$
@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
$begingroup$
(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
$begingroup$
thank you very much!
$endgroup$
– Holo
Mar 9 at 16:43
add a comment |
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Required, but never shown
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...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...
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– Steven Stadnicki
Mar 8 at 20:46
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That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
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– Mike Battaglia
Mar 8 at 20:47
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Ahh, I missed that distinction. Thank you!
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– Steven Stadnicki
Mar 8 at 20:50
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I've never seen this axiom before. Does the family itself need to be a set or can it be a proper class?
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– Robert Shore
Mar 8 at 20:55
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Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
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– Carl Mummert
Mar 8 at 20:56