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This question already has an answer here:

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

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marked as duplicate by Asaf Karagila axiom-of-choice 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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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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