# Existence of complement of a subspace without Zorn's lemma [duplicate]

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

## marked as duplicate by Asaf Karagila♦ axiom-of-choice StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 2 at 14:37

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.

• The additive projection might be zero. – red_trumpet Jan 2 at 14:36
• 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. – Asaf Karagila Jan 2 at 14:38
• @red_trumpet: note the word "onto" in there? – GEdgar Jan 2 at 14:42
• Right missed that. – red_trumpet Jan 2 at 14:42