How strong is the axiom of well-ordered choice?
$begingroup$
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 5 hours ago
Mike BattagliaMike Battaglia
1,4271126
$endgroup$
|
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 5 hours ago
Mike BattagliaMike Battaglia
1,4271126
$endgroup$
$begingroup$
...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
4 hours ago
2
$begingroup$
That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
$endgroup$
– Mike Battaglia
4 hours ago
$begingroup$
Ahh, I missed that distinction. Thank you!
$endgroup$
– Steven Stadnicki
4 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
4 hours ago
$begingroup$
Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
$endgroup$
– Carl Mummert
4 hours ago
|
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 5 hours ago
Mike BattagliaMike Battaglia
1,4271126
$endgroup$
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 5 hours ago
Mike BattagliaMike Battaglia
1,4271126
asked 5 hours ago
Mike BattagliaMike Battaglia
1,4271126
edited 3 hours ago
Mike Battaglia
asked 5 hours ago
Mike BattagliaMike Battaglia
1,4271126
asked 5 hours ago
Mike BattagliaMike Battaglia
1,4271126
asked 5 hours ago
Mike BattagliaMike Battaglia
1,4271126
1,4271126
$begingroup$
...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
4 hours ago
2
$begingroup$
That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
$endgroup$
– Mike Battaglia
4 hours ago
$begingroup$
Ahh, I missed that distinction. Thank you!
$endgroup$
– Steven Stadnicki
4 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
4 hours ago
$begingroup$
Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
$endgroup$
– Carl Mummert
4 hours ago
|
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
4 hours ago
2
$begingroup$
That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
$endgroup$
– Mike Battaglia
4 hours ago
$begingroup$
Ahh, I missed that distinction. Thank you!
$endgroup$
– Steven Stadnicki
4 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
4 hours ago
$begingroup$
Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
$endgroup$
– Carl Mummert
4 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
4 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
4 hours ago
2
2
$begingroup$
That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
$endgroup$
– Mike Battaglia
4 hours ago
$begingroup$
That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
$endgroup$
– Mike Battaglia
4 hours ago
$begingroup$
Ahh, I missed that distinction. Thank you!
$endgroup$
– Steven Stadnicki
4 hours ago
$begingroup$
Ahh, I missed that distinction. Thank you!
$endgroup$
– Steven Stadnicki
4 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
4 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
4 hours ago
$begingroup$
Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
$endgroup$
– Carl Mummert
4 hours ago
$begingroup$
Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
$endgroup$
– Carl Mummert
4 hours ago
|
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 3 hours ago
Asaf Karagila♦Asaf Karagila
306k33437768
$endgroup$
$begingroup$
Do you have a source for a proof of ${sf AC_{rm WO}}iff forall x(aleph(x)=aleph^*(x))$?
$endgroup$
– Holo
2 hours ago
$begingroup$
@Holo karagila.org/2014/on-the-partition-principle
$endgroup$
– Asaf Karagila♦
2 hours ago
add a comment |
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1 Answer
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active
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 3 hours ago
Asaf Karagila♦Asaf Karagila
306k33437768
$endgroup$
$begingroup$
Do you have a source for a proof of ${sf AC_{rm WO}}iff forall x(aleph(x)=aleph^*(x))$?
$endgroup$
– Holo
2 hours ago
$begingroup$
@Holo karagila.org/2014/on-the-partition-principle
$endgroup$
– Asaf Karagila♦
2 hours ago
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 3 hours ago
Asaf Karagila♦Asaf Karagila
306k33437768
$endgroup$
$begingroup$
Do you have a source for a proof of ${sf AC_{rm WO}}iff forall x(aleph(x)=aleph^*(x))$?
$endgroup$
– Holo
2 hours ago
$begingroup$
@Holo karagila.org/2014/on-the-partition-principle
$endgroup$
– Asaf Karagila♦
2 hours ago
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 3 hours ago
Asaf Karagila♦Asaf Karagila
306k33437768
$endgroup$
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 3 hours ago
Asaf Karagila♦Asaf Karagila
306k33437768
answered 3 hours ago
Asaf Karagila♦Asaf Karagila
306k33437768
answered 3 hours ago
Asaf Karagila♦Asaf Karagila
306k33437768
answered 3 hours ago
Asaf Karagila♦Asaf 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
2 hours ago
$begingroup$
@Holo karagila.org/2014/on-the-partition-principle
$endgroup$
– Asaf Karagila♦
2 hours ago
add a comment |
$begingroup$
Do you have a source for a proof of ${sf AC_{rm WO}}iff forall x(aleph(x)=aleph^*(x))$?
$endgroup$
– Holo
2 hours ago
$begingroup$
@Holo karagila.org/2014/on-the-partition-principle
$endgroup$
– Asaf Karagila♦
2 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
2 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
2 hours ago
$begingroup$
@Holo karagila.org/2014/on-the-partition-principle
$endgroup$
– Asaf Karagila♦
2 hours ago
$begingroup$
@Holo karagila.org/2014/on-the-partition-principle
$endgroup$
– Asaf Karagila♦
2 hours ago
add a comment |
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$begingroup$
...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
4 hours ago
2
$begingroup$
That is a family of well-ordered sets, not a well-ordered family of (arbitrary sets).
$endgroup$
– Mike Battaglia
4 hours ago
$begingroup$
Ahh, I missed that distinction. Thank you!
$endgroup$
– Steven Stadnicki
4 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
4 hours ago
$begingroup$
Google suggests this: settheory.mathtalks.org/andreas-blass-well-ordered-choice
$endgroup$
– Carl Mummert
4 hours ago