Partition of positive reals with each part closed under addition without choice
$begingroup$
It is an easy exercise using transfinite recursion to prove the following (in ZFC):
There exists sets $S,T$ that partition $mathbb{R}_{>0}$ such that each of $S$ and $T$ is closed under addition.
It is equally easy, using a transcendence base of $mathbb{R}$ over $mathbb{Q}$, to prove the following generalization:
For any cardinal $k < #(mathbb{R})$, it is possible to partition $mathbb{R}_{>0}$ into $k$ parts each of which is closed under addition.
The problem, of course, is that the transfinite recursion depends on AC. My question is, can the first weak version be proven in ZF (no choice)? If so, I would suspect that ZF can prove:
For any finite $k$, it is possible to partition $mathbb{R}_{>0}$ into $k$ parts each of which is closed under addition.
I am also curious to know whether ZF can prove any of the following:
$mathbb{R}_{>0}$ can be partitioned into countably many parts each of which is closed under addition.
$mathbb{R}_{>0}$ can be partitioned into uncountably many parts each of which is closed under addition.
I am unable to think of any algebraic way to split the positive reals in the desired fashion. Clearly, one of them must be uncountable, but I do not even see an obvious uncountable subset that is closed under addition. I would guess that ZF cannot prove any of them, but maybe that is just because I cannot see how to do it.
set-theory real-numbers axiom-of-choice set-partition provability
asked Dec 21 '18 at 11:48
user21820user21820
40.1k544162
$endgroup$
|
show 4 more comments
$begingroup$
It is an easy exercise using transfinite recursion to prove the following (in ZFC):
There exists sets $S,T$ that partition $mathbb{R}_{>0}$ such that each of $S$ and $T$ is closed under addition.
It is equally easy, using a transcendence base of $mathbb{R}$ over $mathbb{Q}$, to prove the following generalization:
For any cardinal $k < #(mathbb{R})$, it is possible to partition $mathbb{R}_{>0}$ into $k$ parts each of which is closed under addition.
The problem, of course, is that the transfinite recursion depends on AC. My question is, can the first weak version be proven in ZF (no choice)? If so, I would suspect that ZF can prove:
For any finite $k$, it is possible to partition $mathbb{R}_{>0}$ into $k$ parts each of which is closed under addition.
I am also curious to know whether ZF can prove any of the following:
$mathbb{R}_{>0}$ can be partitioned into countably many parts each of which is closed under addition.
$mathbb{R}_{>0}$ can be partitioned into uncountably many parts each of which is closed under addition.
I am unable to think of any algebraic way to split the positive reals in the desired fashion. Clearly, one of them must be uncountable, but I do not even see an obvious uncountable subset that is closed under addition. I would guess that ZF cannot prove any of them, but maybe that is just because I cannot see how to do it.
set-theory real-numbers axiom-of-choice set-partition provability
asked Dec 21 '18 at 11:48
user21820user21820
40.1k544162
$endgroup$
1
$begingroup$
I have a feeling this was discussed here before.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:50
$begingroup$
I couldn't find it. Please link any that you find? =)
$endgroup$
– user21820
Dec 21 '18 at 11:51
$begingroup$
Why do you need a transcendence basis? Wouldn't you need a Hamel basis?
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:51
$begingroup$
Yes algebraic independence is unnecessary; I was just whacking with the usual hammers I wield.
$endgroup$
– user21820
Dec 21 '18 at 11:52
$begingroup$
I am guessing that math.stackexchange.com/a/360828/622 can be extended to countably many as well. Or in general to "less than the additivity of null/meager ideals".
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:55
|
show 4 more comments
$begingroup$
It is an easy exercise using transfinite recursion to prove the following (in ZFC):
There exists sets $S,T$ that partition $mathbb{R}_{>0}$ such that each of $S$ and $T$ is closed under addition.
It is equally easy, using a transcendence base of $mathbb{R}$ over $mathbb{Q}$, to prove the following generalization:
For any cardinal $k < #(mathbb{R})$, it is possible to partition $mathbb{R}_{>0}$ into $k$ parts each of which is closed under addition.
The problem, of course, is that the transfinite recursion depends on AC. My question is, can the first weak version be proven in ZF (no choice)? If so, I would suspect that ZF can prove:
For any finite $k$, it is possible to partition $mathbb{R}_{>0}$ into $k$ parts each of which is closed under addition.
I am also curious to know whether ZF can prove any of the following:
$mathbb{R}_{>0}$ can be partitioned into countably many parts each of which is closed under addition.
$mathbb{R}_{>0}$ can be partitioned into uncountably many parts each of which is closed under addition.
I am unable to think of any algebraic way to split the positive reals in the desired fashion. Clearly, one of them must be uncountable, but I do not even see an obvious uncountable subset that is closed under addition. I would guess that ZF cannot prove any of them, but maybe that is just because I cannot see how to do it.
set-theory real-numbers axiom-of-choice set-partition provability
asked Dec 21 '18 at 11:48
user21820user21820
40.1k544162
$endgroup$
It is an easy exercise using transfinite recursion to prove the following (in ZFC):
There exists sets $S,T$ that partition $mathbb{R}_{>0}$ such that each of $S$ and $T$ is closed under addition.
It is equally easy, using a transcendence base of $mathbb{R}$ over $mathbb{Q}$, to prove the following generalization:
For any cardinal $k < #(mathbb{R})$, it is possible to partition $mathbb{R}_{>0}$ into $k$ parts each of which is closed under addition.
The problem, of course, is that the transfinite recursion depends on AC. My question is, can the first weak version be proven in ZF (no choice)? If so, I would suspect that ZF can prove:
For any finite $k$, it is possible to partition $mathbb{R}_{>0}$ into $k$ parts each of which is closed under addition.
I am also curious to know whether ZF can prove any of the following:
$mathbb{R}_{>0}$ can be partitioned into countably many parts each of which is closed under addition.
$mathbb{R}_{>0}$ can be partitioned into uncountably many parts each of which is closed under addition.
I am unable to think of any algebraic way to split the positive reals in the desired fashion. Clearly, one of them must be uncountable, but I do not even see an obvious uncountable subset that is closed under addition. I would guess that ZF cannot prove any of them, but maybe that is just because I cannot see how to do it.
set-theory real-numbers axiom-of-choice set-partition provability
set-theory real-numbers axiom-of-choice set-partition provability
asked Dec 21 '18 at 11:48
user21820user21820
40.1k544162
asked Dec 21 '18 at 11:48
user21820user21820
40.1k544162
asked Dec 21 '18 at 11:48
user21820user21820
40.1k544162
asked Dec 21 '18 at 11:48
user21820user21820
40.1k544162
asked Dec 21 '18 at 11:48
user21820user21820
40.1k544162
40.1k544162
1
$begingroup$
I have a feeling this was discussed here before.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:50
$begingroup$
I couldn't find it. Please link any that you find? =)
$endgroup$
– user21820
Dec 21 '18 at 11:51
$begingroup$
Why do you need a transcendence basis? Wouldn't you need a Hamel basis?
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:51
$begingroup$
Yes algebraic independence is unnecessary; I was just whacking with the usual hammers I wield.
$endgroup$
– user21820
Dec 21 '18 at 11:52
$begingroup$
I am guessing that math.stackexchange.com/a/360828/622 can be extended to countably many as well. Or in general to "less than the additivity of null/meager ideals".
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:55
|
show 4 more comments
1
$begingroup$
I have a feeling this was discussed here before.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:50
$begingroup$
I couldn't find it. Please link any that you find? =)
$endgroup$
– user21820
Dec 21 '18 at 11:51
$begingroup$
Why do you need a transcendence basis? Wouldn't you need a Hamel basis?
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:51
$begingroup$
Yes algebraic independence is unnecessary; I was just whacking with the usual hammers I wield.
$endgroup$
– user21820
Dec 21 '18 at 11:52
$begingroup$
I am guessing that math.stackexchange.com/a/360828/622 can be extended to countably many as well. Or in general to "less than the additivity of null/meager ideals".
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:55
1
1
$begingroup$
I have a feeling this was discussed here before.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:50
$begingroup$
I have a feeling this was discussed here before.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:50
$begingroup$
I couldn't find it. Please link any that you find? =)
$endgroup$
– user21820
Dec 21 '18 at 11:51
$begingroup$
I couldn't find it. Please link any that you find? =)
$endgroup$
– user21820
Dec 21 '18 at 11:51
$begingroup$
Why do you need a transcendence basis? Wouldn't you need a Hamel basis?
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:51
$begingroup$
Why do you need a transcendence basis? Wouldn't you need a Hamel basis?
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:51
$begingroup$
Yes algebraic independence is unnecessary; I was just whacking with the usual hammers I wield.
$endgroup$
– user21820
Dec 21 '18 at 11:52
$begingroup$
Yes algebraic independence is unnecessary; I was just whacking with the usual hammers I wield.
$endgroup$
– user21820
Dec 21 '18 at 11:52
$begingroup$
I am guessing that math.stackexchange.com/a/360828/622 can be extended to countably many as well. Or in general to "less than the additivity of null/meager ideals".
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:55
$begingroup$
I am guessing that math.stackexchange.com/a/360828/622 can be extended to countably many as well. Or in general to "less than the additivity of null/meager ideals".
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 11:55
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $(A_imid i<alpha)$ is such partition. I claim that $alpha$ is at least the size of the additivity of the null ideal, or one of the $A_i$'s is not Lebesgue measurable.
To see that this is indeed the case, recall that $A+A={a+bmid a,bin A}$ contains an interval for any $A$ of positive measure. If each $A_i$ is null, and $alpha$ is less than the additivity of the null ideal, then $bigcup A_i$ is null, in which case we have a contradiction. At the same time, if one of the $A_i$'s got positive measure, then $A_i+A_i$ contains an interval, which is impossible.
The same can be said about the meager ideal and the Baire Property.
(The above is based on https://math.stackexchange.com/a/360828/622)
In particular, at least under $sf ZF+DC$, it is consistent that there is no finite or countable partition of this sort (e.g. if all sets are Lebesgue measurable or have the Baire Property).
answered Dec 21 '18 at 12:14
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
$begingroup$
To clarify, what assumptions do you need over ZF to prove that there is a model of ZF plus DC plus no countable partition of this sort? Is Con(ZF) enough, or do we need an inaccessible cardinal? Also, do you know the answer for whether ZF proves an uncountable partition of this sort exists?
$endgroup$
– user21820
Dec 21 '18 at 12:41
$begingroup$
Lebesgue = inaccessible; Baire = ZF.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 12:42
$begingroup$
Holo has given the obvious simple answer for the uncountable partition. I was blur. So I guess your answer completely addresses the non-trivial part of my question. Thanks! =)
$endgroup$
– user21820
Dec 21 '18 at 14:54
$begingroup$
Interestingly, $Bbb{R/Q}$ might have strictly larger cardinality in models of ZF+DC+BP/LM, which raises the interesting question about exactly $2^{aleph_0}$ sets, and requiring them all to have size continuum too.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 18:25
$begingroup$
Hmm interesting. ZFC can construct a partition of size $#(mathbb{R})$ each of whose parts has size $#(mathbb{R})$, by dovetailing. But can ZF+DC do it even if we drop the size requirement, and simply want a partition of size $#(mathbb{R})$?
$endgroup$
– user21820
Dec 22 '18 at 7:42
|
show 1 more comment
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$begingroup$
Suppose that $(A_imid i<alpha)$ is such partition. I claim that $alpha$ is at least the size of the additivity of the null ideal, or one of the $A_i$'s is not Lebesgue measurable.
To see that this is indeed the case, recall that $A+A={a+bmid a,bin A}$ contains an interval for any $A$ of positive measure. If each $A_i$ is null, and $alpha$ is less than the additivity of the null ideal, then $bigcup A_i$ is null, in which case we have a contradiction. At the same time, if one of the $A_i$'s got positive measure, then $A_i+A_i$ contains an interval, which is impossible.
The same can be said about the meager ideal and the Baire Property.
(The above is based on https://math.stackexchange.com/a/360828/622)
In particular, at least under $sf ZF+DC$, it is consistent that there is no finite or countable partition of this sort (e.g. if all sets are Lebesgue measurable or have the Baire Property).
answered Dec 21 '18 at 12:14
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
$begingroup$
To clarify, what assumptions do you need over ZF to prove that there is a model of ZF plus DC plus no countable partition of this sort? Is Con(ZF) enough, or do we need an inaccessible cardinal? Also, do you know the answer for whether ZF proves an uncountable partition of this sort exists?
$endgroup$
– user21820
Dec 21 '18 at 12:41
$begingroup$
Lebesgue = inaccessible; Baire = ZF.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 12:42
$begingroup$
Holo has given the obvious simple answer for the uncountable partition. I was blur. So I guess your answer completely addresses the non-trivial part of my question. Thanks! =)
$endgroup$
– user21820
Dec 21 '18 at 14:54
$begingroup$
Interestingly, $Bbb{R/Q}$ might have strictly larger cardinality in models of ZF+DC+BP/LM, which raises the interesting question about exactly $2^{aleph_0}$ sets, and requiring them all to have size continuum too.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 18:25
$begingroup$
Hmm interesting. ZFC can construct a partition of size $#(mathbb{R})$ each of whose parts has size $#(mathbb{R})$, by dovetailing. But can ZF+DC do it even if we drop the size requirement, and simply want a partition of size $#(mathbb{R})$?
$endgroup$
– user21820
Dec 22 '18 at 7:42
|
show 1 more comment
$begingroup$
Suppose that $(A_imid i<alpha)$ is such partition. I claim that $alpha$ is at least the size of the additivity of the null ideal, or one of the $A_i$'s is not Lebesgue measurable.
To see that this is indeed the case, recall that $A+A={a+bmid a,bin A}$ contains an interval for any $A$ of positive measure. If each $A_i$ is null, and $alpha$ is less than the additivity of the null ideal, then $bigcup A_i$ is null, in which case we have a contradiction. At the same time, if one of the $A_i$'s got positive measure, then $A_i+A_i$ contains an interval, which is impossible.
The same can be said about the meager ideal and the Baire Property.
(The above is based on https://math.stackexchange.com/a/360828/622)
In particular, at least under $sf ZF+DC$, it is consistent that there is no finite or countable partition of this sort (e.g. if all sets are Lebesgue measurable or have the Baire Property).
answered Dec 21 '18 at 12:14
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
$begingroup$
To clarify, what assumptions do you need over ZF to prove that there is a model of ZF plus DC plus no countable partition of this sort? Is Con(ZF) enough, or do we need an inaccessible cardinal? Also, do you know the answer for whether ZF proves an uncountable partition of this sort exists?
$endgroup$
– user21820
Dec 21 '18 at 12:41
$begingroup$
Lebesgue = inaccessible; Baire = ZF.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 12:42
$begingroup$
Holo has given the obvious simple answer for the uncountable partition. I was blur. So I guess your answer completely addresses the non-trivial part of my question. Thanks! =)
$endgroup$
– user21820
Dec 21 '18 at 14:54
$begingroup$
Interestingly, $Bbb{R/Q}$ might have strictly larger cardinality in models of ZF+DC+BP/LM, which raises the interesting question about exactly $2^{aleph_0}$ sets, and requiring them all to have size continuum too.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 18:25
$begingroup$
Hmm interesting. ZFC can construct a partition of size $#(mathbb{R})$ each of whose parts has size $#(mathbb{R})$, by dovetailing. But can ZF+DC do it even if we drop the size requirement, and simply want a partition of size $#(mathbb{R})$?
$endgroup$
– user21820
Dec 22 '18 at 7:42
|
show 1 more comment
$begingroup$
Suppose that $(A_imid i<alpha)$ is such partition. I claim that $alpha$ is at least the size of the additivity of the null ideal, or one of the $A_i$'s is not Lebesgue measurable.
To see that this is indeed the case, recall that $A+A={a+bmid a,bin A}$ contains an interval for any $A$ of positive measure. If each $A_i$ is null, and $alpha$ is less than the additivity of the null ideal, then $bigcup A_i$ is null, in which case we have a contradiction. At the same time, if one of the $A_i$'s got positive measure, then $A_i+A_i$ contains an interval, which is impossible.
The same can be said about the meager ideal and the Baire Property.
(The above is based on https://math.stackexchange.com/a/360828/622)
In particular, at least under $sf ZF+DC$, it is consistent that there is no finite or countable partition of this sort (e.g. if all sets are Lebesgue measurable or have the Baire Property).
answered Dec 21 '18 at 12:14
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
Suppose that $(A_imid i<alpha)$ is such partition. I claim that $alpha$ is at least the size of the additivity of the null ideal, or one of the $A_i$'s is not Lebesgue measurable.
To see that this is indeed the case, recall that $A+A={a+bmid a,bin A}$ contains an interval for any $A$ of positive measure. If each $A_i$ is null, and $alpha$ is less than the additivity of the null ideal, then $bigcup A_i$ is null, in which case we have a contradiction. At the same time, if one of the $A_i$'s got positive measure, then $A_i+A_i$ contains an interval, which is impossible.
The same can be said about the meager ideal and the Baire Property.
(The above is based on https://math.stackexchange.com/a/360828/622)
In particular, at least under $sf ZF+DC$, it is consistent that there is no finite or countable partition of this sort (e.g. if all sets are Lebesgue measurable or have the Baire Property).
answered Dec 21 '18 at 12:14
Asaf Karagila♦Asaf Karagila
308k33441775
answered Dec 21 '18 at 12:14
Asaf Karagila♦Asaf Karagila
308k33441775
answered Dec 21 '18 at 12:14
Asaf Karagila♦Asaf Karagila
308k33441775
answered Dec 21 '18 at 12:14
Asaf Karagila♦Asaf Karagila
308k33441775
308k33441775
$begingroup$
To clarify, what assumptions do you need over ZF to prove that there is a model of ZF plus DC plus no countable partition of this sort? Is Con(ZF) enough, or do we need an inaccessible cardinal? Also, do you know the answer for whether ZF proves an uncountable partition of this sort exists?
$endgroup$
– user21820
Dec 21 '18 at 12:41
$begingroup$
Lebesgue = inaccessible; Baire = ZF.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 12:42
$begingroup$
Holo has given the obvious simple answer for the uncountable partition. I was blur. So I guess your answer completely addresses the non-trivial part of my question. Thanks! =)
$endgroup$
– user21820
Dec 21 '18 at 14:54
$begingroup$
Interestingly, $Bbb{R/Q}$ might have strictly larger cardinality in models of ZF+DC+BP/LM, which raises the interesting question about exactly $2^{aleph_0}$ sets, and requiring them all to have size continuum too.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 18:25
$begingroup$
Hmm interesting. ZFC can construct a partition of size $#(mathbb{R})$ each of whose parts has size $#(mathbb{R})$, by dovetailing. But can ZF+DC do it even if we drop the size requirement, and simply want a partition of size $#(mathbb{R})$?
$endgroup$
– user21820
Dec 22 '18 at 7:42
|
show 1 more comment
$begingroup$
To clarify, what assumptions do you need over ZF to prove that there is a model of ZF plus DC plus no countable partition of this sort? Is Con(ZF) enough, or do we need an inaccessible cardinal? Also, do you know the answer for whether ZF proves an uncountable partition of this sort exists?
$endgroup$
– user21820
Dec 21 '18 at 12:41
$begingroup$
Lebesgue = inaccessible; Baire = ZF.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 12:42
$begingroup$
Holo has given the obvious simple answer for the uncountable partition. I was blur. So I guess your answer completely addresses the non-trivial part of my question. Thanks! =)
$endgroup$
– user21820
Dec 21 '18 at 14:54
$begingroup$
Interestingly, $Bbb{R/Q}$ might have strictly larger cardinality in models of ZF+DC+BP/LM, which raises the interesting question about exactly $2^{aleph_0}$ sets, and requiring them all to have size continuum too.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 18:25
$begingroup$
Hmm interesting. ZFC can construct a partition of size $#(mathbb{R})$ each of whose parts has size $#(mathbb{R})$, by dovetailing. But can ZF+DC do it even if we drop the size requirement, and simply want a partition of size $#(mathbb{R})$?
$endgroup$
– user21820
Dec 22 '18 at 7:42
$begingroup$
To clarify, what assumptions do you need over ZF to prove that there is a model of ZF plus DC plus no countable partition of this sort? Is Con(ZF) enough, or do we need an inaccessible cardinal? Also, do you know the answer for whether ZF proves an uncountable partition of this sort exists?
$endgroup$
– user21820
Dec 21 '18 at 12:41
$begingroup$
To clarify, what assumptions do you need over ZF to prove that there is a model of ZF plus DC plus no countable partition of this sort? Is Con(ZF) enough, or do we need an inaccessible cardinal? Also, do you know the answer for whether ZF proves an uncountable partition of this sort exists?
$endgroup$
– user21820
Dec 21 '18 at 12:41
$begingroup$
Lebesgue = inaccessible; Baire = ZF.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 12:42
$begingroup$
Lebesgue = inaccessible; Baire = ZF.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 12:42
$begingroup$
Holo has given the obvious simple answer for the uncountable partition. I was blur. So I guess your answer completely addresses the non-trivial part of my question. Thanks! =)
$endgroup$
– user21820
Dec 21 '18 at 14:54
$begingroup$
Holo has given the obvious simple answer for the uncountable partition. I was blur. So I guess your answer completely addresses the non-trivial part of my question. Thanks! =)
$endgroup$
– user21820
Dec 21 '18 at 14:54
$begingroup$
Interestingly, $Bbb{R/Q}$ might have strictly larger cardinality in models of ZF+DC+BP/LM, which raises the interesting question about exactly $2^{aleph_0}$ sets, and requiring them all to have size continuum too.
$endgroup$
– Asaf Karagila♦
Dec 21 '18 at 18:25
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Interestingly, $Bbb{R/Q}$ might have strictly larger cardinality in models of ZF+DC+BP/LM, which raises the interesting question about exactly $2^{aleph_0}$ sets, and requiring them all to have size continuum too.
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– Asaf Karagila♦
Dec 21 '18 at 18:25
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Hmm interesting. ZFC can construct a partition of size $#(mathbb{R})$ each of whose parts has size $#(mathbb{R})$, by dovetailing. But can ZF+DC do it even if we drop the size requirement, and simply want a partition of size $#(mathbb{R})$?
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– user21820
Dec 22 '18 at 7:42
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Hmm interesting. ZFC can construct a partition of size $#(mathbb{R})$ each of whose parts has size $#(mathbb{R})$, by dovetailing. But can ZF+DC do it even if we drop the size requirement, and simply want a partition of size $#(mathbb{R})$?
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– user21820
Dec 22 '18 at 7:42
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I have a feeling this was discussed here before.
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– Asaf Karagila♦
Dec 21 '18 at 11:50
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I couldn't find it. Please link any that you find? =)
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– user21820
Dec 21 '18 at 11:51
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Why do you need a transcendence basis? Wouldn't you need a Hamel basis?
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– Asaf Karagila♦
Dec 21 '18 at 11:51
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Yes algebraic independence is unnecessary; I was just whacking with the usual hammers I wield.
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– user21820
Dec 21 '18 at 11:52
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I am guessing that math.stackexchange.com/a/360828/622 can be extended to countably many as well. Or in general to "less than the additivity of null/meager ideals".
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– Asaf Karagila♦
Dec 21 '18 at 11:55