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.