Logical strength of a statement about vector spaces

Question

[Apologies if this is a really trivial question, I know virtually nothing about set theory, and the following came up while preparing undergraduate linear algebra lectures.]

I'm asking about the following statement concerning vector spaces:

If $V$ is a vector space having no finite spanning set, then $V$ contains an infinite linearly-independent set.

This is easy to prove using the axiom of dependent choice: formally, if $S$ is the set of finite subsets of $V$, and $\prec$ is the relation "$a \prec b$ if $a \subset b$ and $|b \setminus a| = 1$", then $R$ is a total relation, so DC says that we can find an infinite sequence $(a_n)_{n \in \mathbb{N}}$ with $a_n \prec a_{n+1} \forall n$, and then $\bigcup_{n \in \mathbb{N}} a_n$ is an infinite LI set.

Is this statement equivalent to DC? Or can it be proved in ZF, or maybe ZF + countable (non-dependent) choice?

Answer 1

This can be proved with just countable choice, so it is not equivalent to DC, since countable choice is known to be weaker than DC.

For each $n$, let $B_n$ consist of all $n$-tuples of independent vectors. By countable choice, we can choose an element of each $B_n$. But now we have an $\omega$-sequence of vectors. It may not be linearly independent, but we can thin it out to a linearly independent set by systematically removing vectors in the span of the previous vectors on the list. We will be left with infinitely many, since the span of the whole collection cannot be finite dimensional.

This argument is similar to how one proves from $\text{AC}_\omega$ that there is no infinite Dedekind-finite set. The natural argument would use DC, since one wants to pick more and more elements from what's left, but one can get away with just countable choice by picking from the $n$-tuples of distinct elements, and then putting these together (and removing duplicates) to make an $\omega$-sequence.