Mixing
Relation between the Axiom of Choice and a the existence of a hyperplane not containing a vector
Clash Royale CLAN TAG#URR8PPP
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In a lot of problems in linear one uses the existence, for each $E$ vectorial space over a field $k$ , and each $xin E$, of a Hyperplane $H$ such that $E=kcdot x oplus H$ (Let us denote $mathcalP$ this property). With Zorn's Lemma, the existence of a such $H$ is trivial. However, as this seems weaker than the existence of a basis of $E$, maybe $mathcalP$ does not imply the Axiom of choice. Thus my question is : is $mathcalP$ equivalent to the Axiom of Choice and if not, is there a weaker form of the Axiom of Choice equivalent to $mathcalP$ ?
axiom-of-choice vector-spaces
asked 4 hours ago
Swann
404
add a comment |Â
up vote
4
down vote
favorite
In a lot of problems in linear one uses the existence, for each $E$ vectorial space over a field $k$ , and each $xin E$, of a Hyperplane $H$ such that $E=kcdot x oplus H$ (Let us denote $mathcalP$ this property). With Zorn's Lemma, the existence of a such $H$ is trivial. However, as this seems weaker than the existence of a basis of $E$, maybe $mathcalP$ does not imply the Axiom of choice. Thus my question is : is $mathcalP$ equivalent to the Axiom of Choice and if not, is there a weaker form of the Axiom of Choice equivalent to $mathcalP$ ?
axiom-of-choice vector-spaces
asked 4 hours ago
Swann
404
2
Just for the record, choice is definitely necessary to show this result: taking $E$ to be $mathbb R$ as $mathbb Q$-vector space, $H$ would have to be a subset of $mathbb R$ isomorphic to $mathbb R/mathbb Q$, but it is consistent that this set can't be linearly ordered.
– Wojowu
4 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In a lot of problems in linear one uses the existence, for each $E$ vectorial space over a field $k$ , and each $xin E$, of a Hyperplane $H$ such that $E=kcdot x oplus H$ (Let us denote $mathcalP$ this property). With Zorn's Lemma, the existence of a such $H$ is trivial. However, as this seems weaker than the existence of a basis of $E$, maybe $mathcalP$ does not imply the Axiom of choice. Thus my question is : is $mathcalP$ equivalent to the Axiom of Choice and if not, is there a weaker form of the Axiom of Choice equivalent to $mathcalP$ ?
axiom-of-choice vector-spaces
asked 4 hours ago
Swann
404
In a lot of problems in linear one uses the existence, for each $E$ vectorial space over a field $k$ , and each $xin E$, of a Hyperplane $H$ such that $E=kcdot x oplus H$ (Let us denote $mathcalP$ this property). With Zorn's Lemma, the existence of a such $H$ is trivial. However, as this seems weaker than the existence of a basis of $E$, maybe $mathcalP$ does not imply the Axiom of choice. Thus my question is : is $mathcalP$ equivalent to the Axiom of Choice and if not, is there a weaker form of the Axiom of Choice equivalent to $mathcalP$ ?
axiom-of-choice vector-spaces
axiom-of-choice vector-spaces
asked 4 hours ago
Swann
404
asked 4 hours ago
Swann
404
asked 4 hours ago
Swann
404
asked 4 hours ago
Swann
404
asked 4 hours ago
Swann
404
404
2
Just for the record, choice is definitely necessary to show this result: taking $E$ to be $mathbb R$ as $mathbb Q$-vector space, $H$ would have to be a subset of $mathbb R$ isomorphic to $mathbb R/mathbb Q$, but it is consistent that this set can't be linearly ordered.
– Wojowu
4 hours ago
add a comment |Â
2
Just for the record, choice is definitely necessary to show this result: taking $E$ to be $mathbb R$ as $mathbb Q$-vector space, $H$ would have to be a subset of $mathbb R$ isomorphic to $mathbb R/mathbb Q$, but it is consistent that this set can't be linearly ordered.
– Wojowu
4 hours ago
2
2
Just for the record, choice is definitely necessary to show this result: taking $E$ to be $mathbb R$ as $mathbb Q$-vector space, $H$ would have to be a subset of $mathbb R$ isomorphic to $mathbb R/mathbb Q$, but it is consistent that this set can't be linearly ordered.
– Wojowu
4 hours ago
Just for the record, choice is definitely necessary to show this result: taking $E$ to be $mathbb R$ as $mathbb Q$-vector space, $H$ would have to be a subset of $mathbb R$ isomorphic to $mathbb R/mathbb Q$, but it is consistent that this set can't be linearly ordered.
– Wojowu
4 hours ago
add a comment |Â
1 Answer
1
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up vote
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It is not hard to see that this statement is equivalent to "In every vector space, for every vector $v$ there is a functional $f$ such that $f(v)=1$".
If $cal P$ holds, then the projection onto $kcdot x$ is a linear functional which is nontrivial; if there is a nontrivial functional then choose $x$ which is mapped to $1_k$, and consider $H$ as the kernel of the functional. Some $cal P$ is equivalent to "If $V$ is nontrivial, then $V^*$ is nontrivial".
As of October 2018, it is still unknown whether or not this is equivalent to the axiom of choice in full. But there are some intermediate results, for example it is consistent that there is a vector space over a field $k$ whose dual is trivial, for any fixed $k$. Moreover, this is consistent with $sf DC_kappa$ for any prescribed $kappa$.
One should remark that restricting to Banach spaces and continuous functionals, the nontriviality of the [topological] dual is equivalent to the Hahn–Banach theorem.
Let me also add that Marianne Morillon showed that in $sf ZFA$, $cal P(Bbb Q)$, namely restricting our attention to vector spaces over $Bbb Q$, is not enough to deduce that every vector space over $Bbb Q$ admits a basis, and that a weaker form of choice called "Axiom of Multiple Choice" implies $cal P$ for fields of characteristics $0$.
Morillon, Marianne, Linear forms and axioms of choice., Commentat. Math. Univ. Carol. 50, No. 3, 421-431 (2009). ZBL1212.03034.
These two facts are not telling us a whole lot about what happens in $sf ZF$, since in $sf ZF$ the axiom of multiple choice implies full choice and existence of bases over $Bbb Q$ implies full choice.
answered 3 hours ago
Asaf Karagila
20.4k673171
Would not the dual of a line bundle without nonzero sections provide an example of a nonzero (in fact, 1-dimensional) vector space without nonzero functionals on it? More precisely, the space of such functionals would be 1-dimensional itself but would fail to have any (global) elements.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
3 hours ago
I have literally no idea what any of that means.
– Asaf Karagila
3 hours ago
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე 1-dimensional vector space always has nonzero functionals - if $v$ is nonzero vector, then every vector is of the form $alpha v$ and $alpha vmapstoalpha$ is a nonzero functional. I think your confusion might arise from misunderstanding what "nonzero" here is supposed to mean - it just means any functional which is not zero everywhere. This is different from vector bundle terminology, where "nonzero" means that it's everywhere nonzero.
– Wojowu
3 hours ago
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე: Again, I am not sure what you mean. But feel free to send me an email with details. My address can be found through my profile.
– Asaf Karagila
2 hours ago
1
The statement in the question was "for each $x$ ... there is $H$ ...." So I think the equivalent statement should not be just that there is a non-zero linear functional but that, for each vector $xneq 0$, there is a linear functional $f$ with $f(x)neq0$.
– Andreas Blass
1 hour ago
 |Â
show 12 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
It is not hard to see that this statement is equivalent to "In every vector space, for every vector $v$ there is a functional $f$ such that $f(v)=1$".
If $cal P$ holds, then the projection onto $kcdot x$ is a linear functional which is nontrivial; if there is a nontrivial functional then choose $x$ which is mapped to $1_k$, and consider $H$ as the kernel of the functional. Some $cal P$ is equivalent to "If $V$ is nontrivial, then $V^*$ is nontrivial".
As of October 2018, it is still unknown whether or not this is equivalent to the axiom of choice in full. But there are some intermediate results, for example it is consistent that there is a vector space over a field $k$ whose dual is trivial, for any fixed $k$. Moreover, this is consistent with $sf DC_kappa$ for any prescribed $kappa$.
One should remark that restricting to Banach spaces and continuous functionals, the nontriviality of the [topological] dual is equivalent to the Hahn–Banach theorem.
Let me also add that Marianne Morillon showed that in $sf ZFA$, $cal P(Bbb Q)$, namely restricting our attention to vector spaces over $Bbb Q$, is not enough to deduce that every vector space over $Bbb Q$ admits a basis, and that a weaker form of choice called "Axiom of Multiple Choice" implies $cal P$ for fields of characteristics $0$.
Morillon, Marianne, Linear forms and axioms of choice., Commentat. Math. Univ. Carol. 50, No. 3, 421-431 (2009). ZBL1212.03034.
These two facts are not telling us a whole lot about what happens in $sf ZF$, since in $sf ZF$ the axiom of multiple choice implies full choice and existence of bases over $Bbb Q$ implies full choice.
answered 3 hours ago
Asaf Karagila
20.4k673171
Would not the dual of a line bundle without nonzero sections provide an example of a nonzero (in fact, 1-dimensional) vector space without nonzero functionals on it? More precisely, the space of such functionals would be 1-dimensional itself but would fail to have any (global) elements.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
3 hours ago
I have literally no idea what any of that means.
– Asaf Karagila
3 hours ago
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე 1-dimensional vector space always has nonzero functionals - if $v$ is nonzero vector, then every vector is of the form $alpha v$ and $alpha vmapstoalpha$ is a nonzero functional. I think your confusion might arise from misunderstanding what "nonzero" here is supposed to mean - it just means any functional which is not zero everywhere. This is different from vector bundle terminology, where "nonzero" means that it's everywhere nonzero.
– Wojowu
3 hours ago
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე: Again, I am not sure what you mean. But feel free to send me an email with details. My address can be found through my profile.
– Asaf Karagila
2 hours ago
1
The statement in the question was "for each $x$ ... there is $H$ ...." So I think the equivalent statement should not be just that there is a non-zero linear functional but that, for each vector $xneq 0$, there is a linear functional $f$ with $f(x)neq0$.
– Andreas Blass
1 hour ago
 |Â
show 12 more comments
up vote
4
down vote
It is not hard to see that this statement is equivalent to "In every vector space, for every vector $v$ there is a functional $f$ such that $f(v)=1$".
If $cal P$ holds, then the projection onto $kcdot x$ is a linear functional which is nontrivial; if there is a nontrivial functional then choose $x$ which is mapped to $1_k$, and consider $H$ as the kernel of the functional. Some $cal P$ is equivalent to "If $V$ is nontrivial, then $V^*$ is nontrivial".
As of October 2018, it is still unknown whether or not this is equivalent to the axiom of choice in full. But there are some intermediate results, for example it is consistent that there is a vector space over a field $k$ whose dual is trivial, for any fixed $k$. Moreover, this is consistent with $sf DC_kappa$ for any prescribed $kappa$.
One should remark that restricting to Banach spaces and continuous functionals, the nontriviality of the [topological] dual is equivalent to the Hahn–Banach theorem.
Let me also add that Marianne Morillon showed that in $sf ZFA$, $cal P(Bbb Q)$, namely restricting our attention to vector spaces over $Bbb Q$, is not enough to deduce that every vector space over $Bbb Q$ admits a basis, and that a weaker form of choice called "Axiom of Multiple Choice" implies $cal P$ for fields of characteristics $0$.
Morillon, Marianne, Linear forms and axioms of choice., Commentat. Math. Univ. Carol. 50, No. 3, 421-431 (2009). ZBL1212.03034.
These two facts are not telling us a whole lot about what happens in $sf ZF$, since in $sf ZF$ the axiom of multiple choice implies full choice and existence of bases over $Bbb Q$ implies full choice.
answered 3 hours ago
Asaf Karagila
20.4k673171
Would not the dual of a line bundle without nonzero sections provide an example of a nonzero (in fact, 1-dimensional) vector space without nonzero functionals on it? More precisely, the space of such functionals would be 1-dimensional itself but would fail to have any (global) elements.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
3 hours ago
I have literally no idea what any of that means.
– Asaf Karagila
3 hours ago
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე 1-dimensional vector space always has nonzero functionals - if $v$ is nonzero vector, then every vector is of the form $alpha v$ and $alpha vmapstoalpha$ is a nonzero functional. I think your confusion might arise from misunderstanding what "nonzero" here is supposed to mean - it just means any functional which is not zero everywhere. This is different from vector bundle terminology, where "nonzero" means that it's everywhere nonzero.
– Wojowu
3 hours ago
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე: Again, I am not sure what you mean. But feel free to send me an email with details. My address can be found through my profile.
– Asaf Karagila
2 hours ago
1
The statement in the question was "for each $x$ ... there is $H$ ...." So I think the equivalent statement should not be just that there is a non-zero linear functional but that, for each vector $xneq 0$, there is a linear functional $f$ with $f(x)neq0$.
– Andreas Blass
1 hour ago
 |Â
show 12 more comments
up vote
4
down vote
up vote
4
down vote
It is not hard to see that this statement is equivalent to "In every vector space, for every vector $v$ there is a functional $f$ such that $f(v)=1$".
If $cal P$ holds, then the projection onto $kcdot x$ is a linear functional which is nontrivial; if there is a nontrivial functional then choose $x$ which is mapped to $1_k$, and consider $H$ as the kernel of the functional. Some $cal P$ is equivalent to "If $V$ is nontrivial, then $V^*$ is nontrivial".
As of October 2018, it is still unknown whether or not this is equivalent to the axiom of choice in full. But there are some intermediate results, for example it is consistent that there is a vector space over a field $k$ whose dual is trivial, for any fixed $k$. Moreover, this is consistent with $sf DC_kappa$ for any prescribed $kappa$.
One should remark that restricting to Banach spaces and continuous functionals, the nontriviality of the [topological] dual is equivalent to the Hahn–Banach theorem.
Let me also add that Marianne Morillon showed that in $sf ZFA$, $cal P(Bbb Q)$, namely restricting our attention to vector spaces over $Bbb Q$, is not enough to deduce that every vector space over $Bbb Q$ admits a basis, and that a weaker form of choice called "Axiom of Multiple Choice" implies $cal P$ for fields of characteristics $0$.
Morillon, Marianne, Linear forms and axioms of choice., Commentat. Math. Univ. Carol. 50, No. 3, 421-431 (2009). ZBL1212.03034.
These two facts are not telling us a whole lot about what happens in $sf ZF$, since in $sf ZF$ the axiom of multiple choice implies full choice and existence of bases over $Bbb Q$ implies full choice.
answered 3 hours ago
Asaf Karagila
20.4k673171
It is not hard to see that this statement is equivalent to "In every vector space, for every vector $v$ there is a functional $f$ such that $f(v)=1$".
If $cal P$ holds, then the projection onto $kcdot x$ is a linear functional which is nontrivial; if there is a nontrivial functional then choose $x$ which is mapped to $1_k$, and consider $H$ as the kernel of the functional. Some $cal P$ is equivalent to "If $V$ is nontrivial, then $V^*$ is nontrivial".
As of October 2018, it is still unknown whether or not this is equivalent to the axiom of choice in full. But there are some intermediate results, for example it is consistent that there is a vector space over a field $k$ whose dual is trivial, for any fixed $k$. Moreover, this is consistent with $sf DC_kappa$ for any prescribed $kappa$.
One should remark that restricting to Banach spaces and continuous functionals, the nontriviality of the [topological] dual is equivalent to the Hahn–Banach theorem.
Let me also add that Marianne Morillon showed that in $sf ZFA$, $cal P(Bbb Q)$, namely restricting our attention to vector spaces over $Bbb Q$, is not enough to deduce that every vector space over $Bbb Q$ admits a basis, and that a weaker form of choice called "Axiom of Multiple Choice" implies $cal P$ for fields of characteristics $0$.
Morillon, Marianne, Linear forms and axioms of choice., Commentat. Math. Univ. Carol. 50, No. 3, 421-431 (2009). ZBL1212.03034.
These two facts are not telling us a whole lot about what happens in $sf ZF$, since in $sf ZF$ the axiom of multiple choice implies full choice and existence of bases over $Bbb Q$ implies full choice.
answered 3 hours ago
Asaf Karagila
20.4k673171
edited 22 mins ago
answered 3 hours ago
Asaf Karagila
20.4k673171
answered 3 hours ago
Asaf Karagila
20.4k673171
answered 3 hours ago
Asaf Karagila
20.4k673171
20.4k673171
Would not the dual of a line bundle without nonzero sections provide an example of a nonzero (in fact, 1-dimensional) vector space without nonzero functionals on it? More precisely, the space of such functionals would be 1-dimensional itself but would fail to have any (global) elements.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
3 hours ago
I have literally no idea what any of that means.
– Asaf Karagila
3 hours ago
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე 1-dimensional vector space always has nonzero functionals - if $v$ is nonzero vector, then every vector is of the form $alpha v$ and $alpha vmapstoalpha$ is a nonzero functional. I think your confusion might arise from misunderstanding what "nonzero" here is supposed to mean - it just means any functional which is not zero everywhere. This is different from vector bundle terminology, where "nonzero" means that it's everywhere nonzero.
– Wojowu
3 hours ago
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე: Again, I am not sure what you mean. But feel free to send me an email with details. My address can be found through my profile.
– Asaf Karagila
2 hours ago
1
The statement in the question was "for each $x$ ... there is $H$ ...." So I think the equivalent statement should not be just that there is a non-zero linear functional but that, for each vector $xneq 0$, there is a linear functional $f$ with $f(x)neq0$.
– Andreas Blass
1 hour ago
 |Â
show 12 more comments
Would not the dual of a line bundle without nonzero sections provide an example of a nonzero (in fact, 1-dimensional) vector space without nonzero functionals on it? More precisely, the space of such functionals would be 1-dimensional itself but would fail to have any (global) elements.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
3 hours ago
I have literally no idea what any of that means.
– Asaf Karagila
3 hours ago
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე 1-dimensional vector space always has nonzero functionals - if $v$ is nonzero vector, then every vector is of the form $alpha v$ and $alpha vmapstoalpha$ is a nonzero functional. I think your confusion might arise from misunderstanding what "nonzero" here is supposed to mean - it just means any functional which is not zero everywhere. This is different from vector bundle terminology, where "nonzero" means that it's everywhere nonzero.
– Wojowu
3 hours ago
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე: Again, I am not sure what you mean. But feel free to send me an email with details. My address can be found through my profile.
– Asaf Karagila
2 hours ago
1
The statement in the question was "for each $x$ ... there is $H$ ...." So I think the equivalent statement should not be just that there is a non-zero linear functional but that, for each vector $xneq 0$, there is a linear functional $f$ with $f(x)neq0$.
– Andreas Blass
1 hour ago
Would not the dual of a line bundle without nonzero sections provide an example of a nonzero (in fact, 1-dimensional) vector space without nonzero functionals on it? More precisely, the space of such functionals would be 1-dimensional itself but would fail to have any (global) elements.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
3 hours ago
Would not the dual of a line bundle without nonzero sections provide an example of a nonzero (in fact, 1-dimensional) vector space without nonzero functionals on it? More precisely, the space of such functionals would be 1-dimensional itself but would fail to have any (global) elements.
– áƒ›áƒÂმუკრჯიბლáƒÂძე
3 hours ago
I have literally no idea what any of that means.
– Asaf Karagila
3 hours ago
I have literally no idea what any of that means.
– Asaf Karagila
3 hours ago
1
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე 1-dimensional vector space always has nonzero functionals - if $v$ is nonzero vector, then every vector is of the form $alpha v$ and $alpha vmapstoalpha$ is a nonzero functional. I think your confusion might arise from misunderstanding what "nonzero" here is supposed to mean - it just means any functional which is not zero everywhere. This is different from vector bundle terminology, where "nonzero" means that it's everywhere nonzero.
– Wojowu
3 hours ago
@მáƒÂმუკáƒÂჯიბლáƒÂძე 1-dimensional vector space always has nonzero functionals - if $v$ is nonzero vector, then every vector is of the form $alpha v$ and $alpha vmapstoalpha$ is a nonzero functional. I think your confusion might arise from misunderstanding what "nonzero" here is supposed to mean - it just means any functional which is not zero everywhere. This is different from vector bundle terminology, where "nonzero" means that it's everywhere nonzero.
– Wojowu
3 hours ago
1
1
@მáƒÂმუკáƒÂჯიბლáƒÂძე: Again, I am not sure what you mean. But feel free to send me an email with details. My address can be found through my profile.
– Asaf Karagila
2 hours ago
@მáƒÂმუკáƒÂჯიბლáƒÂძე: Again, I am not sure what you mean. But feel free to send me an email with details. My address can be found through my profile.
– Asaf Karagila
2 hours ago
1
1
The statement in the question was "for each $x$ ... there is $H$ ...." So I think the equivalent statement should not be just that there is a non-zero linear functional but that, for each vector $xneq 0$, there is a linear functional $f$ with $f(x)neq0$.
– Andreas Blass
1 hour ago
The statement in the question was "for each $x$ ... there is $H$ ...." So I think the equivalent statement should not be just that there is a non-zero linear functional but that, for each vector $xneq 0$, there is a linear functional $f$ with $f(x)neq0$.
– Andreas Blass
1 hour ago
 |Â
show 12 more comments
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2
Just for the record, choice is definitely necessary to show this result: taking $E$ to be $mathbb R$ as $mathbb Q$-vector space, $H$ would have to be a subset of $mathbb R$ isomorphic to $mathbb R/mathbb Q$, but it is consistent that this set can't be linearly ordered.
– Wojowu
4 hours ago