Nov. 6th, 2011 After several long months a post on MathOverflow pushed me to reconsider this math, and I have found a mistake. The claim was still true, as shown by Läuchli $\small[1]$, however despite trying to do my best to understand the argument for this specific claim, it eluded me for several days. I then proceeded to construct my own proof, this time errors free - or so I hope. While at it, I am revising the writing style.
Jul. 21st, 2012 While reviewing this proof again it was apparent that its most prominent use in generating such space over the field of two elements fails, as the third lemma implicitly assumed $x+x\neq x$. Now this has been corrected and the proof is truly complete.
$\newcommand{\sym}{\operatorname{sym}}
\newcommand{\fix}{\operatorname{fix}}
\newcommand{\span}{\operatorname{span}}
\newcommand{\im}{\operatorname{Im}}
\newcommand{\Id}{\operatorname{Id}}
$
I got it! The answer is that you can construct such vector space.
I will assume that you are familiar with ZFA and the construction of permutation models, references can be found in Jech's Set Theory $\small[2, \text{Ch}. 15]$ as well The Axiom of Choice $\small{[3]}$. Any questions are welcomed.
Some notations, if $x\in V$ which is assumed to be a model of ZFC+Atoms then:
- $\sym(x) =\{\pi\in\mathscr{G} \mid \pi x = x\}$, and
- $\fix(x) = \{\pi\in\mathscr{G} \mid \forall y\in x:\ \pi y = y\}$
Definition: Suppose $G$ is a group, $\mathcal{F}\subseteq\mathcal{P}(G)$ is a normal subgroups filter if:
- $G\in\mathcal{F}$;
- $H,K$ are subgroups of $G$ such that $H\subseteq K$, then $H\in\mathcal{F}$ implies $K\in\mathcal{F}$;
- $H,K$ are subgroups of $G$ such that $H,K\in\mathcal{F}$ then $H\cap K\in\mathcal{F}$;
- ${1}\notin\mathcal{F}$ (non-triviality);
- For every $H\in\mathcal{F}$ and $g\in G$ then $g^{-1}Hg\in\mathcal{F}$ (normality).
Now consider the normal subgroups-filter $\mathcal{F}$ to be generated by the subgroups $\fix(E)$ for $E\in I$, where $I$ is an ideal of sets of atoms (closed under finite unions, intersections and subsets).
Basics of permutation models:
A permutation model is a transitive subclass of the universe $V$ that for every ordinal $\alpha$, we have $x\in\mathfrak{U}\cap V_{\alpha+1}$ if and only if $x\subseteq\mathfrak{U}\cap V_\alpha$ and $\sym(x)\in\mathcal{F}$.
The latter property is known as being symmetric (with respect to $\mathcal{F}$) and $x$ being in the permutation model means that $x$ is hereditarily symmetric. (Of course at limit stages take limits, and start with the empty set)
If $\mathcal{F}$ was generated by some ideal of sets $I$, then if $x$ is symmetric with respect to $\mathcal{F}$ it means that for some $E\in I$ we have $\fix(E)\subseteq\sym(x)$. In this case we say that $E$ is a support of $x$.
Note that if $E$ is a support of $x$ and $E\subseteq E'$ then $E'$ is also a support of $x$, since $\fix(E')\subseteq\fix(E)$.
Lastly if $f$ is a function in $\mathfrak{U}$ and $\pi$ is a permutation in $G$ then $\pi(f(x)) = (\pi f)(\pi x)$.
Start with $V$ a model of ZFC+Atoms, assuming there are infinitely (countably should be enough) many atoms. $A$ is the set of atoms, endow it with operations that make it a vector space over a field $\mathbb{F}$ (If we only assume countably many atoms, we should assume the field is countable too. Since we are interested in $\mathbb F_2$ this assertion is not a big hassle). Now consider $\mathscr{G}$ the group of all linear automorphisms of $A$, each can be extended uniquely to an automorphism of $V$.
Now consider the normal subgroups-filter $\mathcal{F}$ to be generated by the subgroups $\fix(E)$ for $E\in I$, where $E$ a finite set of atoms. Note that since all the permutations are linear they extend unique to $\span(E)$. In the case where $\mathbb F$, our field, is finite then so is this span.
Let $\mathfrak{U}$ be the permutation model generated by $\mathscr{G}$ and $\mathcal{F}$.
Lemma I: Suppose $E$ is a finite set, and $u,v$ are two vectors such that $v\notin\span(E\cup\{u\})$ and $u\notin\span(E\cup\{v\})$ (in which case we say that $u$ and $v$ are linearly independent over $E$), then there is a permutation which fixes $E$ and permutes $u$ with $v$.
Proof: Without loss of generality we can assume that $E$ is linearly independent, otherwise take a subset of $E$ which is. Since $E\cup\{u,v\}$ is linearly independent we can (in $V$) extend it to a base of $A$, and define a permutation of this base which fixes $E$, permutes $u$ and $v$. This extends uniquely to a linear permutation $\pi\in\fix(E)$ as needed. $\square$
Lemma II: In $\mathfrak{U}$, $A$ is a vector space over $\mathbb F$, and if $W\in\mathfrak{U}$ is a linear proper subspace then $W$ has a finite dimension.
Proof: Suppose $W$ is as above, let $E$ be a support of $W$. If $W\subseteq\span(E)$ then we are done. Otherwise take $u\notin W\cup \span(E)$ and $v\in W\setminus \span(E)$ and permute $u$ and $v$ while fixing $E$, denote the linear permutation with $\pi$. It is clear that $\pi\in\fix(E)$ but $\pi(W)\neq W$, in contradiction. $\square$
Lemma III: If $T\in\mathfrak{U}$ is a linear endomorphism of $A$, and $E$ is a support of $T$ then $x\in\span(E)\Leftrightarrow Tx\in\span(E)$, or $Tx=0$.
Proof: First for $x\in \span(E)$, if $Tx\notin\span(E)$ for some $Tx\neq u\notin\span(E)$ let $\pi$ be a linear automorphism of $A$ which fixes $E$ and $\pi(Tx)=u$. We have, if so:
$$u=\pi(Tx)=(\pi T)(\pi x) = Tx\neq u$$
On the other hand, if $x\notin\span(E)$ and $Tx\in\span(E)$ and if $Tx=Tu$ for some $x\neq u$ for $u\notin\span(E)$, in which case we have that $x+u\neq x$ set $\pi$ an automorphism which fixes $E$ and $\pi(x)=x+u$, now we have: $$Tx = \pi(Tx) = (\pi T)(\pi x) = T(x+u) = Tx+Tu$$ Therefore $Tx=0$.
Otherwise for all $u\neq x$ we have $Tu\neq Tx$. Let $\pi$ be an automorphism fixing $E$ such that $\pi(x)=u$ for some $u\notin\span(E)$, and we have: $$Tx=\pi(Tx)=(\pi T)(\pi x) = Tu$$ this is a contradiction, so this case is impossible. $\square$
Theorem: if $T\in\mathfrak{U}$ is an endomorphism of $A$ then for some $\lambda\in\mathbb F$ we have $Tx=\lambda x$ for all $x\in A$.
Proof:
Assume that $T\neq 0$, so it has a nontrivial image. Let $E$ be a support of $T$. If $\ker(T)$ is nontrivial then it is a proper subspace, thus for a finite set of atoms $B$ we have $\span(B)=\ker(T)$. Without loss of generality, $B\subseteq E$, otherwise $E\cup B$ is also a support of $T$.
For every $v\notin\span(E)$ we have $Tv\notin\span(E)$. However, $E_v = E\cup\{v\}$ is also a support of $T$. Therefore restricting $T$ to $E_v$ yields that $Tv=\lambda v$ for some $\lambda\in\mathbb F$.
Let $v,u\notin\span(E)$ linearly independent over $\span(E)$. We have that: $Tu=\alpha u, Tv=\mu v$, and $v+u\notin\span(E)$ so $T(v+u)=\lambda(v+u)$, for $\lambda\in\mathbb F$.
$$\begin{align}
0&=T(0) \\ &= T(u+v-u-v)\\
&=T(u+v)-Tu-Tv \\ &=\lambda(u+v)-\alpha u-\mu v=(\lambda-\alpha)u+(\lambda-\mu)v
\end{align}$$ Since $u,v$ are linearly independent we have $\alpha=\lambda=\mu$. Due to the fact that for every $u,v\notin\span(E)$ we can find $x$ which is linearly independent over $\span(E)$ both with $u$ and $v$ we can conclude that for $x\notin E$ we have $Tx=\lambda x$.
For $v\in\span(E)$ let $x\notin\span(E)$, we have that $v+x\notin\span(E)$ and therefore:
$$\begin{align}
Tx &= T(x+u - u)\\
&=T(x+u)-T(u)\\
&=\lambda(x+u)-\lambda u = \lambda x
\end{align}$$
We have concluded, if so that $T=\lambda x$ for some $\lambda\in\mathbb F$. $\square$
Set $\mathbb F=\mathbb F_2$ the field with two elements and we have created ourselves a vector space without any nontrivial automorphisms. However, one last problem remains. This construction was carried out in ZF+Atoms, while we want to have it without atoms. For this simply use the Jech-Sochor embedding theorem $\small[3, \text{Th}. 6.1, \text p. 85]$, and by setting $\alpha>4$ it should be that any endomorphism is transferred to the model of ZF created by this theorem.
(Many thanks to t.b. which helped me translating parts of the original paper of Läuchli.
Additional thanks to Uri Abraham for noting that an operator need not be injective in order to be surjective, resulting a shorter proof.)
Bibliography
Läuchli, H. Auswahlaxiom in der Algebra. Commentarii Mathematici Helvetici, vol 37, pp. 1-19.
Jech, T. Set Theory, 3rd millennium ed., Springer (2003).
Jech, T. The Axiom of Choice. North-Holland (1973).
Yes. The axiom of choice is needed in order to show that every vector space which is not finitely generated contains an infinite linearly independent subset.
The original consistency proof due to Lauchli (1962) was to construct a model in which there is a vector space that is not spanned by any finite set, but every proper subspace is finite.
Namely every collection of linearly independent vectors is finite.
If you have the axiom of dependent choice then you can actually perform the induction which you suggest and have a countably infinite set of independent vectors. But this is quite far from the full axiom of choice.
If one tries real hard, one can get away with just countable choice (which is strictly weaker than dependent choice). The argument is as follows:
Let $\cal A_n$ be the collection of all sets of linearly independent of size $n$, since the space is not finitely generated $\cal A_n$ is non-empty for all $n\neq 0$. Let $A_n\in\cal A_n$ be some chosen set. Again using countable choice let $A$ be the union of the $A_n$'s, and $A$ is countable so we can write it as $\{a_n\mid n\in\omega\}$.
Pick $v_0=a_0$, and proceed by induction to define $v_{n+1}$ to be $a_k$ whose index is the least $k$ such that $a_k$ not in the span of $\{v_0,\ldots, v_n\}$. This $a_k$ exists because $A_{n+1}$ spans a vector space of dimension $n+1$ so it cannot be a subset of $\operatorname{span}\{v_0,\ldots,v_n\}$.
The set $\{v_n\mid n\in\omega\}$ is linearly independent, which follows from the choice of the $v_n$'s.
Interestingly, Lauchli's example was of a space that every endomorphism is a scalar multiplication and as the above indicates this construction also contradicted $\mathsf{DC}$. In my masters thesis I showed that you can have $\mathsf{DC}_\kappa$ and still have a vector space which has no endomorphisms except scalar multiplication - even if it has relatively large subspaces.
Best Answer
If you take a look at the proof that the existence of a complement implies the axiom of choice then by a careful proof analysis one can construct a counterexample that is actually in a space which has a basis.
The proof of the algebraic principle goes through proving that the axiom of multiple choice holds, rather than the axiom of choice. The two are equivalent over $\sf ZF$, so it's fine. This version is as follows:
The proof of the direct complement begins by taking $X$ as above, and defining the following vector space: $$V=\bigoplus_{x\in X}F^{(x)}$$
Where $F^{(x)}$ is just the space of functions with finitely many non-zero values from $x$ into $F$. We also define $F^{(x)}_0$ to be the space $f\in F^{(x)}$ such that $\sum_{u\in x}f(u)=0$.
I will leave it to you to find a basis for this space (hint, $F^{(\bigcup X)}$). Now if we can find a complement to the subspace $W=\bigoplus_{x\in X}F^{(x)}_0$ then we can construct a function $f$ as above.
Begin now with a model where the axiom of choice fails, then the axiom of multiple choice fails, and we can find such $X$ where the vector space $V$ has a basis but $W$ doesn't have a direct complement.