is it consistent with AC that every set is measurable? [duplicate]
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This question already has an answer here:
Lebesgue-measurable sets requiring the Axiom of Choice to construct
1 answer
Does every set is measurable follow from AC, or from it's negation?
I think that by Vitali's construction from the AC follows that some set is not masurable. But here in the 1st comment they claim that it may be consistent with ZFC that every set is measurable. How is it?
set-theory axiom-of-choice measurable-functions
edited Dec 22 '18 at 22:13
Andrés E. Caicedo
65.6k8159250
asked Dec 22 '18 at 20:27
user122424user122424
1,1482716
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marked as duplicate by Asaf Karagila♦ axiom-of-choice
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Dec 22 '18 at 20:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Lebesgue-measurable sets requiring the Axiom of Choice to construct
1 answer
Does every set is measurable follow from AC, or from it's negation?
I think that by Vitali's construction from the AC follows that some set is not masurable. But here in the 1st comment they claim that it may be consistent with ZFC that every set is measurable. How is it?
set-theory axiom-of-choice measurable-functions
edited Dec 22 '18 at 22:13
Andrés E. Caicedo
65.6k8159250
asked Dec 22 '18 at 20:27
user122424user122424
1,1482716
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marked as duplicate by Asaf Karagila♦ axiom-of-choice
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Dec 22 '18 at 20:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Vitali constructs a not Lebesgue-measurable set with the axiom of choice.
$endgroup$
– Viktor Glombik
Dec 22 '18 at 20:32
3
$begingroup$
How could you first comment claim that it is consistent that every set is measurable with ZFC? It is exactly not what the comment says.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 20:37
add a comment |
$begingroup$
This question already has an answer here:
Lebesgue-measurable sets requiring the Axiom of Choice to construct
1 answer
Does every set is measurable follow from AC, or from it's negation?
I think that by Vitali's construction from the AC follows that some set is not masurable. But here in the 1st comment they claim that it may be consistent with ZFC that every set is measurable. How is it?
set-theory axiom-of-choice measurable-functions
edited Dec 22 '18 at 22:13
Andrés E. Caicedo
65.6k8159250
asked Dec 22 '18 at 20:27
user122424user122424
1,1482716
$endgroup$
This question already has an answer here:
Lebesgue-measurable sets requiring the Axiom of Choice to construct
1 answer
Does every set is measurable follow from AC, or from it's negation?
I think that by Vitali's construction from the AC follows that some set is not masurable. But here in the 1st comment they claim that it may be consistent with ZFC that every set is measurable. How is it?
This question already has an answer here:
Lebesgue-measurable sets requiring the Axiom of Choice to construct
1 answer
set-theory axiom-of-choice measurable-functions
set-theory axiom-of-choice measurable-functions
edited Dec 22 '18 at 22:13
Andrés E. Caicedo
65.6k8159250
asked Dec 22 '18 at 20:27
user122424user122424
1,1482716
edited Dec 22 '18 at 22:13
Andrés E. Caicedo
65.6k8159250
asked Dec 22 '18 at 20:27
user122424user122424
1,1482716
edited Dec 22 '18 at 22:13
Andrés E. Caicedo
65.6k8159250
edited Dec 22 '18 at 22:13
Andrés E. Caicedo
65.6k8159250
edited Dec 22 '18 at 22:13
Andrés E. Caicedo
65.6k8159250
65.6k8159250
asked Dec 22 '18 at 20:27
user122424user122424
1,1482716
asked Dec 22 '18 at 20:27
user122424user122424
1,1482716
asked Dec 22 '18 at 20:27
user122424user122424
1,1482716
1,1482716
marked as duplicate by Asaf Karagila♦ axiom-of-choice
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Dec 22 '18 at 20:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Asaf Karagila♦ axiom-of-choice
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Dec 22 '18 at 20:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Vitali constructs a not Lebesgue-measurable set with the axiom of choice.
$endgroup$
– Viktor Glombik
Dec 22 '18 at 20:32
3
$begingroup$
How could you first comment claim that it is consistent that every set is measurable with ZFC? It is exactly not what the comment says.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 20:37
add a comment |
1
$begingroup$
Vitali constructs a not Lebesgue-measurable set with the axiom of choice.
$endgroup$
– Viktor Glombik
Dec 22 '18 at 20:32
3
$begingroup$
How could you first comment claim that it is consistent that every set is measurable with ZFC? It is exactly not what the comment says.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 20:37
1
1
$begingroup$
Vitali constructs a not Lebesgue-measurable set with the axiom of choice.
$endgroup$
– Viktor Glombik
Dec 22 '18 at 20:32
$begingroup$
Vitali constructs a not Lebesgue-measurable set with the axiom of choice.
$endgroup$
– Viktor Glombik
Dec 22 '18 at 20:32
3
3
$begingroup$
How could you first comment claim that it is consistent that every set is measurable with ZFC? It is exactly not what the comment says.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 20:37
$begingroup$
How could you first comment claim that it is consistent that every set is measurable with ZFC? It is exactly not what the comment says.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 20:37
add a comment |
1 Answer
1
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oldest
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$begingroup$
It doesn't follow from AC or from its negation.
In fact it is refuted by AC, as you pointed out, e.g. the Vitali set.
It doesn't follow from the negation, because a section of $mathbb{R}tomathbb{R/Q}$ is enough to perform the Vitali construction and certainly AC doesn't follow from such a section existing.
What is true is that under certain large cardinal hypotheses, it is consistent with ZF + Dependent choice that every set is measurable. And under no hypotheses at all it is consistent with ZF that every set is Borel (so measurable): indeed it is consistent with ZF that $mathbb{R}$ is a countable union of countable sets, and if this happens then so is every subset of $mathbb{R}$, and so every set is Borel as countable sets are Borel.
Finally, that's not what the linked 1st comment claims at all. It actually gives another proof that AC refutes said proposition : if every set were measurable, then the Banach Tarski paradox would actually be a paradox : it would be contradictory because the unit ball doesn't have measure $0$ or $infty$ and isometries preserve measure
answered Dec 22 '18 at 20:36
MaxMax
15k11143
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
It doesn't follow from AC or from its negation.
In fact it is refuted by AC, as you pointed out, e.g. the Vitali set.
It doesn't follow from the negation, because a section of $mathbb{R}tomathbb{R/Q}$ is enough to perform the Vitali construction and certainly AC doesn't follow from such a section existing.
What is true is that under certain large cardinal hypotheses, it is consistent with ZF + Dependent choice that every set is measurable. And under no hypotheses at all it is consistent with ZF that every set is Borel (so measurable): indeed it is consistent with ZF that $mathbb{R}$ is a countable union of countable sets, and if this happens then so is every subset of $mathbb{R}$, and so every set is Borel as countable sets are Borel.
Finally, that's not what the linked 1st comment claims at all. It actually gives another proof that AC refutes said proposition : if every set were measurable, then the Banach Tarski paradox would actually be a paradox : it would be contradictory because the unit ball doesn't have measure $0$ or $infty$ and isometries preserve measure
answered Dec 22 '18 at 20:36
MaxMax
15k11143
$endgroup$
add a comment |
$begingroup$
It doesn't follow from AC or from its negation.
In fact it is refuted by AC, as you pointed out, e.g. the Vitali set.
It doesn't follow from the negation, because a section of $mathbb{R}tomathbb{R/Q}$ is enough to perform the Vitali construction and certainly AC doesn't follow from such a section existing.
What is true is that under certain large cardinal hypotheses, it is consistent with ZF + Dependent choice that every set is measurable. And under no hypotheses at all it is consistent with ZF that every set is Borel (so measurable): indeed it is consistent with ZF that $mathbb{R}$ is a countable union of countable sets, and if this happens then so is every subset of $mathbb{R}$, and so every set is Borel as countable sets are Borel.
Finally, that's not what the linked 1st comment claims at all. It actually gives another proof that AC refutes said proposition : if every set were measurable, then the Banach Tarski paradox would actually be a paradox : it would be contradictory because the unit ball doesn't have measure $0$ or $infty$ and isometries preserve measure
answered Dec 22 '18 at 20:36
MaxMax
15k11143
$endgroup$
add a comment |
$begingroup$
It doesn't follow from AC or from its negation.
In fact it is refuted by AC, as you pointed out, e.g. the Vitali set.
It doesn't follow from the negation, because a section of $mathbb{R}tomathbb{R/Q}$ is enough to perform the Vitali construction and certainly AC doesn't follow from such a section existing.
What is true is that under certain large cardinal hypotheses, it is consistent with ZF + Dependent choice that every set is measurable. And under no hypotheses at all it is consistent with ZF that every set is Borel (so measurable): indeed it is consistent with ZF that $mathbb{R}$ is a countable union of countable sets, and if this happens then so is every subset of $mathbb{R}$, and so every set is Borel as countable sets are Borel.
Finally, that's not what the linked 1st comment claims at all. It actually gives another proof that AC refutes said proposition : if every set were measurable, then the Banach Tarski paradox would actually be a paradox : it would be contradictory because the unit ball doesn't have measure $0$ or $infty$ and isometries preserve measure
answered Dec 22 '18 at 20:36
MaxMax
15k11143
$endgroup$
It doesn't follow from AC or from its negation.
In fact it is refuted by AC, as you pointed out, e.g. the Vitali set.
It doesn't follow from the negation, because a section of $mathbb{R}tomathbb{R/Q}$ is enough to perform the Vitali construction and certainly AC doesn't follow from such a section existing.
What is true is that under certain large cardinal hypotheses, it is consistent with ZF + Dependent choice that every set is measurable. And under no hypotheses at all it is consistent with ZF that every set is Borel (so measurable): indeed it is consistent with ZF that $mathbb{R}$ is a countable union of countable sets, and if this happens then so is every subset of $mathbb{R}$, and so every set is Borel as countable sets are Borel.
Finally, that's not what the linked 1st comment claims at all. It actually gives another proof that AC refutes said proposition : if every set were measurable, then the Banach Tarski paradox would actually be a paradox : it would be contradictory because the unit ball doesn't have measure $0$ or $infty$ and isometries preserve measure
answered Dec 22 '18 at 20:36
MaxMax
15k11143
answered Dec 22 '18 at 20:36
MaxMax
15k11143
answered Dec 22 '18 at 20:36
MaxMax
15k11143
answered Dec 22 '18 at 20:36
MaxMax
15k11143
15k11143
add a comment |
add a comment |
1
$begingroup$
Vitali constructs a not Lebesgue-measurable set with the axiom of choice.
$endgroup$
– Viktor Glombik
Dec 22 '18 at 20:32
3
$begingroup$
How could you first comment claim that it is consistent that every set is measurable with ZFC? It is exactly not what the comment says.
$endgroup$
– Asaf Karagila♦
Dec 22 '18 at 20:37