# Vector space bases without axiom of choice

I want to find an example of a vector space with no base if we assume that axiom of choice is incorrect. This question might be duplicate so please alert me. Thanks.

• I can't prove it, but I'd be surprised if one could establish a basis for the space of all (real- or complex-valued) sequences without the axiom of choice. – Daniel Fischer Aug 6 '13 at 22:06
• @DanielFischer DERP. – Pedro Tamaroff Aug 6 '13 at 22:09
• @PeterTamaroff Nope, COFFEE! – Daniel Fischer Aug 6 '13 at 22:09
• @DanielFischer Why no both? – Pedro Tamaroff Aug 6 '13 at 22:10
• Inspired by the comment of @ZhenLin below: If the axiom of choice is incorrect, the only thing we know is that there exists a setting where it fails. But we don't know which one, which probably makes it impossible to pin down a concrete vector space without a base. So probably, you should reformulate this part to something like "if we cannot invoke the axiom of choice". – azimut Aug 6 '13 at 22:24

The proof is not as constructive as one would expect.

I suggest that you take a look at Andreas Blass' paper in which he proved that the existence of bases implies the axiom of choice. From the proof it is easy to construct a counterexample.

The proof idea is to take a family of non-empty sets, and to show that there exists a choice function (not quite exactly, Blass goes through another equivalence first). Assuming the axiom of choice fails, there is a family of non-empty sets which doesn't have a choice function. From this you can construct a counterexample to the principle that Blass is using in his equivalence, and then you can easily construct the vector space which doesn't have a basis.

On the other hand, it is not very hard to construct specific models in which there are specific vector spaces without bases. $\ell_2$ doesn't have a basis in Solovay's model, and generally in models of $\sf ZF+DC+BP$ (where $\sf BP$ is the statement that every set of real numbers have the Baire property).

I think for any field $F$, the $F$-vector space $F^{\mathbb{N}}$ of all sequences over $F$ provides a suitable example. (Note that the unit vectors only generate the subspace consisting of all vectors with finitely many non-zero entries.)
• Just because we can't describe the basis doesn't mean it doesn't exist! For this particular example, if the axiom of choice holds up to a sufficiently large cardinal, then $F^{\mathbb{N}}$ would have a basis. But the axiom of choice could still fail higher up. – Zhen Lin Aug 6 '13 at 22:21