I admit I normally assume AC, since it is equivalent to saying that any cartesian product of (> 0) nonempty sets is nonempty, which I can't not believe. So I do assume in this problem that every vector space has a basis (which not only implies, but is equivalent to, AC). * * * Beyond AC, however, no further assumption (like CH) is necessary for solving this problem. (Ordinary cardinal arithmetic suffices.) --Dan On 2012-08-26, at 4:54 PM, Michael Kleber wrote: << Well, you first need to know that the vector space has a basis in the first place. I mean, it does, if you assume the Axiom of Choice . . .. . . . . . . If you're willing to assume the Continuum Hypothesis, then we're done: there are only c many points, some subset is a basis, and no countable subset works. So c it is. But without CH, I don't know what else to say.