Existence of complement of a subspace without Zorn's lemma [duplicate]
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This question already has an answer here:
Existence of vector space complement and axiom of choice
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Let $E$ be a $mathbb{K}$-vector space. I have seen that every subspace $F subset E$ has an (algebraic) complement $F'$ ($F+F'=E$ and $F cap F'={0}$).
One proof (using that every vector space admits a basis) is found here, and another using directly Zorn's lemma in here.
What I would like to know is if it is possible to prove that statement without using the axiom of choice.
One idea that came to me was this: in the finite case, the complement of a subspace $F subset E$ is isomorphic to the quotient space $E/F$. Is this true in the general case? If there were a way to embed $E/F$ in $E$ in such a way that it is the complement of $F$, we could prove the statement.
On the other hand, a way to prove that this is not possible to achieve could be to deduce that every vector space admits a basis (without using AC) from the fact that every subspace admits a complement. But I can't see how this could be done.
Thanks in advance for any help
linear-algebra axiom-of-choice
asked Jan 2 at 14:03
AcasAcas
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Jan 2 at 14:37
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:
Existence of vector space complement and axiom of choice
1 answer
Let $E$ be a $mathbb{K}$-vector space. I have seen that every subspace $F subset E$ has an (algebraic) complement $F'$ ($F+F'=E$ and $F cap F'={0}$).
One proof (using that every vector space admits a basis) is found here, and another using directly Zorn's lemma in here.
What I would like to know is if it is possible to prove that statement without using the axiom of choice.
One idea that came to me was this: in the finite case, the complement of a subspace $F subset E$ is isomorphic to the quotient space $E/F$. Is this true in the general case? If there were a way to embed $E/F$ in $E$ in such a way that it is the complement of $F$, we could prove the statement.
On the other hand, a way to prove that this is not possible to achieve could be to deduce that every vector space admits a basis (without using AC) from the fact that every subspace admits a complement. But I can't see how this could be done.
Thanks in advance for any help
linear-algebra axiom-of-choice
asked Jan 2 at 14:03
AcasAcas
6010
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marked as duplicate by Asaf Karagila♦ axiom-of-choice
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Jan 2 at 14:37
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.
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There are probably a lot of related questions. Try looking at the conjunction of the tags you've used. math.stackexchange.com/questions/tagged/…
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– Asaf Karagila♦
Jan 2 at 14:40
add a comment |
$begingroup$
This question already has an answer here:
Existence of vector space complement and axiom of choice
1 answer
Let $E$ be a $mathbb{K}$-vector space. I have seen that every subspace $F subset E$ has an (algebraic) complement $F'$ ($F+F'=E$ and $F cap F'={0}$).
One proof (using that every vector space admits a basis) is found here, and another using directly Zorn's lemma in here.
What I would like to know is if it is possible to prove that statement without using the axiom of choice.
One idea that came to me was this: in the finite case, the complement of a subspace $F subset E$ is isomorphic to the quotient space $E/F$. Is this true in the general case? If there were a way to embed $E/F$ in $E$ in such a way that it is the complement of $F$, we could prove the statement.
On the other hand, a way to prove that this is not possible to achieve could be to deduce that every vector space admits a basis (without using AC) from the fact that every subspace admits a complement. But I can't see how this could be done.
Thanks in advance for any help
linear-algebra axiom-of-choice
asked Jan 2 at 14:03
AcasAcas
6010
$endgroup$
This question already has an answer here:
Existence of vector space complement and axiom of choice
1 answer
Let $E$ be a $mathbb{K}$-vector space. I have seen that every subspace $F subset E$ has an (algebraic) complement $F'$ ($F+F'=E$ and $F cap F'={0}$).
One proof (using that every vector space admits a basis) is found here, and another using directly Zorn's lemma in here.
What I would like to know is if it is possible to prove that statement without using the axiom of choice.
One idea that came to me was this: in the finite case, the complement of a subspace $F subset E$ is isomorphic to the quotient space $E/F$. Is this true in the general case? If there were a way to embed $E/F$ in $E$ in such a way that it is the complement of $F$, we could prove the statement.
On the other hand, a way to prove that this is not possible to achieve could be to deduce that every vector space admits a basis (without using AC) from the fact that every subspace admits a complement. But I can't see how this could be done.
Thanks in advance for any help
This question already has an answer here:
Existence of vector space complement and axiom of choice
1 answer
linear-algebra axiom-of-choice
linear-algebra axiom-of-choice
asked Jan 2 at 14:03
AcasAcas
6010
asked Jan 2 at 14:03
AcasAcas
6010
asked Jan 2 at 14:03
AcasAcas
6010
asked Jan 2 at 14:03
AcasAcas
6010
asked Jan 2 at 14:03
AcasAcas
6010
6010
marked as duplicate by Asaf Karagila♦ axiom-of-choice
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Jan 2 at 14:37
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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Jan 2 at 14:37
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.
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There are probably a lot of related questions. Try looking at the conjunction of the tags you've used. math.stackexchange.com/questions/tagged/…
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– Asaf Karagila♦
Jan 2 at 14:40
add a comment |
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There are probably a lot of related questions. Try looking at the conjunction of the tags you've used. math.stackexchange.com/questions/tagged/…
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– Asaf Karagila♦
Jan 2 at 14:40
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There are probably a lot of related questions. Try looking at the conjunction of the tags you've used. math.stackexchange.com/questions/tagged/…
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– Asaf Karagila♦
Jan 2 at 14:40
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There are probably a lot of related questions. Try looking at the conjunction of the tags you've used. math.stackexchange.com/questions/tagged/…
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Jan 2 at 14:40
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1 Answer
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No, it is not a theorem of ZF (provided ZF is consistent). For information of this kind you may consult Rubin & Rubin, Equivalents of the Axiom of Choice, I and II.
One case with no complement constructible in ZF is this one: $mathbb Q subset mathbb R$, vector spaces over $mathbb Q$. There is a model for ZF due to Solovay (and Shelah?) where every subset of $mathbb R$ has the property of Baire. But a group homomorphism $f : mathbb R to mathbb R$ such that $f^{-1}(U)$ has the property of Baire for every open set $U$ must be continuous. But any additive projection of $mathbb R$ onto the subspace $mathbb Q$ is discontinuous.
answered Jan 2 at 14:26
GEdgarGEdgar
62.2k267168
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The additive projection might be zero.
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– red_trumpet
Jan 2 at 14:36
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Solovay was the first to prove the consistency of the statement, Shelah showed that the use of an inaccessible cardinal was unnecessary if one is willing to have a much more complicated construction. At least for the case of the Baire property. If one wants Lebesgue measurability (which also gives the wanted result), then one needs the inaccessible cardinal.
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– Asaf Karagila♦
Jan 2 at 14:38
2
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@red_trumpet: note the word "onto" in there?
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– GEdgar
Jan 2 at 14:42
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Right missed that.
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– red_trumpet
Jan 2 at 14:42
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, it is not a theorem of ZF (provided ZF is consistent). For information of this kind you may consult Rubin & Rubin, Equivalents of the Axiom of Choice, I and II.
One case with no complement constructible in ZF is this one: $mathbb Q subset mathbb R$, vector spaces over $mathbb Q$. There is a model for ZF due to Solovay (and Shelah?) where every subset of $mathbb R$ has the property of Baire. But a group homomorphism $f : mathbb R to mathbb R$ such that $f^{-1}(U)$ has the property of Baire for every open set $U$ must be continuous. But any additive projection of $mathbb R$ onto the subspace $mathbb Q$ is discontinuous.
answered Jan 2 at 14:26
GEdgarGEdgar
62.2k267168
$endgroup$
$begingroup$
The additive projection might be zero.
$endgroup$
– red_trumpet
Jan 2 at 14:36
$begingroup$
Solovay was the first to prove the consistency of the statement, Shelah showed that the use of an inaccessible cardinal was unnecessary if one is willing to have a much more complicated construction. At least for the case of the Baire property. If one wants Lebesgue measurability (which also gives the wanted result), then one needs the inaccessible cardinal.
$endgroup$
– Asaf Karagila♦
Jan 2 at 14:38
2
$begingroup$
@red_trumpet: note the word "onto" in there?
$endgroup$
– GEdgar
Jan 2 at 14:42
$begingroup$
Right missed that.
$endgroup$
– red_trumpet
Jan 2 at 14:42
add a comment |
$begingroup$
No, it is not a theorem of ZF (provided ZF is consistent). For information of this kind you may consult Rubin & Rubin, Equivalents of the Axiom of Choice, I and II.
One case with no complement constructible in ZF is this one: $mathbb Q subset mathbb R$, vector spaces over $mathbb Q$. There is a model for ZF due to Solovay (and Shelah?) where every subset of $mathbb R$ has the property of Baire. But a group homomorphism $f : mathbb R to mathbb R$ such that $f^{-1}(U)$ has the property of Baire for every open set $U$ must be continuous. But any additive projection of $mathbb R$ onto the subspace $mathbb Q$ is discontinuous.
answered Jan 2 at 14:26
GEdgarGEdgar
62.2k267168
$endgroup$
$begingroup$
The additive projection might be zero.
$endgroup$
– red_trumpet
Jan 2 at 14:36
$begingroup$
Solovay was the first to prove the consistency of the statement, Shelah showed that the use of an inaccessible cardinal was unnecessary if one is willing to have a much more complicated construction. At least for the case of the Baire property. If one wants Lebesgue measurability (which also gives the wanted result), then one needs the inaccessible cardinal.
$endgroup$
– Asaf Karagila♦
Jan 2 at 14:38
2
$begingroup$
@red_trumpet: note the word "onto" in there?
$endgroup$
– GEdgar
Jan 2 at 14:42
$begingroup$
Right missed that.
$endgroup$
– red_trumpet
Jan 2 at 14:42
add a comment |
$begingroup$
No, it is not a theorem of ZF (provided ZF is consistent). For information of this kind you may consult Rubin & Rubin, Equivalents of the Axiom of Choice, I and II.
One case with no complement constructible in ZF is this one: $mathbb Q subset mathbb R$, vector spaces over $mathbb Q$. There is a model for ZF due to Solovay (and Shelah?) where every subset of $mathbb R$ has the property of Baire. But a group homomorphism $f : mathbb R to mathbb R$ such that $f^{-1}(U)$ has the property of Baire for every open set $U$ must be continuous. But any additive projection of $mathbb R$ onto the subspace $mathbb Q$ is discontinuous.
answered Jan 2 at 14:26
GEdgarGEdgar
62.2k267168
$endgroup$
No, it is not a theorem of ZF (provided ZF is consistent). For information of this kind you may consult Rubin & Rubin, Equivalents of the Axiom of Choice, I and II.
One case with no complement constructible in ZF is this one: $mathbb Q subset mathbb R$, vector spaces over $mathbb Q$. There is a model for ZF due to Solovay (and Shelah?) where every subset of $mathbb R$ has the property of Baire. But a group homomorphism $f : mathbb R to mathbb R$ such that $f^{-1}(U)$ has the property of Baire for every open set $U$ must be continuous. But any additive projection of $mathbb R$ onto the subspace $mathbb Q$ is discontinuous.
answered Jan 2 at 14:26
GEdgarGEdgar
62.2k267168
answered Jan 2 at 14:26
GEdgarGEdgar
62.2k267168
answered Jan 2 at 14:26
GEdgarGEdgar
62.2k267168
answered Jan 2 at 14:26
GEdgarGEdgar
62.2k267168
62.2k267168
$begingroup$
The additive projection might be zero.
$endgroup$
– red_trumpet
Jan 2 at 14:36
$begingroup$
Solovay was the first to prove the consistency of the statement, Shelah showed that the use of an inaccessible cardinal was unnecessary if one is willing to have a much more complicated construction. At least for the case of the Baire property. If one wants Lebesgue measurability (which also gives the wanted result), then one needs the inaccessible cardinal.
$endgroup$
– Asaf Karagila♦
Jan 2 at 14:38
2
$begingroup$
@red_trumpet: note the word "onto" in there?
$endgroup$
– GEdgar
Jan 2 at 14:42
$begingroup$
Right missed that.
$endgroup$
– red_trumpet
Jan 2 at 14:42
add a comment |
$begingroup$
The additive projection might be zero.
$endgroup$
– red_trumpet
Jan 2 at 14:36
$begingroup$
Solovay was the first to prove the consistency of the statement, Shelah showed that the use of an inaccessible cardinal was unnecessary if one is willing to have a much more complicated construction. At least for the case of the Baire property. If one wants Lebesgue measurability (which also gives the wanted result), then one needs the inaccessible cardinal.
$endgroup$
– Asaf Karagila♦
Jan 2 at 14:38
2
$begingroup$
@red_trumpet: note the word "onto" in there?
$endgroup$
– GEdgar
Jan 2 at 14:42
$begingroup$
Right missed that.
$endgroup$
– red_trumpet
Jan 2 at 14:42
$begingroup$
The additive projection might be zero.
$endgroup$
– red_trumpet
Jan 2 at 14:36
$begingroup$
The additive projection might be zero.
$endgroup$
– red_trumpet
Jan 2 at 14:36
$begingroup$
Solovay was the first to prove the consistency of the statement, Shelah showed that the use of an inaccessible cardinal was unnecessary if one is willing to have a much more complicated construction. At least for the case of the Baire property. If one wants Lebesgue measurability (which also gives the wanted result), then one needs the inaccessible cardinal.
$endgroup$
– Asaf Karagila♦
Jan 2 at 14:38
$begingroup$
Solovay was the first to prove the consistency of the statement, Shelah showed that the use of an inaccessible cardinal was unnecessary if one is willing to have a much more complicated construction. At least for the case of the Baire property. If one wants Lebesgue measurability (which also gives the wanted result), then one needs the inaccessible cardinal.
$endgroup$
– Asaf Karagila♦
Jan 2 at 14:38
2
2
$begingroup$
@red_trumpet: note the word "onto" in there?
$endgroup$
– GEdgar
Jan 2 at 14:42
$begingroup$
@red_trumpet: note the word "onto" in there?
$endgroup$
– GEdgar
Jan 2 at 14:42
$begingroup$
Right missed that.
$endgroup$
– red_trumpet
Jan 2 at 14:42
$begingroup$
Right missed that.
$endgroup$
– red_trumpet
Jan 2 at 14:42
add a comment |
$begingroup$
There are probably a lot of related questions. Try looking at the conjunction of the tags you've used. math.stackexchange.com/questions/tagged/…
$endgroup$
– Asaf Karagila♦
Jan 2 at 14:40