It is true that the axiom of choice is equivalent to the statement that every linear space has a Hamel basis. There are some linear spaces which definitely don't need axiom of choice to possess (rather canonical) basis: for example $c\_{00}$, the space of all sequences with compact supports. Is it possible to give an example of *one* linear space $V$ with the property: the existence of basis in $V$ implies the axiom of choice?



Linear space with (Hamel) basis and the axiom of choice
No. It is not possible.


Suppose that $V$ is a specified vector space, then it is consistent that the axiom of choice fails very far above $V$ in the hierarchy of sets (the von Neumann hierarchy). In particular it would mean that $V$ has a basis, but still the axiom of choice fails, as it fails very far above $V$.


Generally speaking, no particular set can witness the axiom of choice. It is possible, that one set can decide the axiom of choice (so it is possible that one vector space's basis decides the axiom of choice), but this requires additional assumptions. More specifically, this assumption was formulated by Andreas Blass as "Small Violations of Choice" (SVC). It means that there is a set which "more or less" decides a lot of the information regarding the axiom of choice in the universe. From this set we can create a vector space, that has a basis if and only if the original set can be well-ordered and that would imply the axiom of choice in full.


