Is axiom of choice required for there to be an infinite linearly independent set in a (non-finite-dimensional) vector space?

In discussing this answer, I noted that while the statement:

Any vector space has a basis

is equivalent to the axiom of choice, I wondered if the statement that:

Any vector space either has a finite basis or an infinite set of linear independent vectors

was weaker than the axiom of choice. It feels weaker – it feels like you can inductively define a countably infinite set of linearly independent vectors without full choice, but I’m not sure if that is possible with ZF alone, or requires all of the Axiom of Choice, or is weaker but requires some form of choice.

Solutions Collecting From Web of "Is axiom of choice required for there to be an infinite linearly independent set in a (non-finite-dimensional) vector space?"

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.