Main Question or Discussion Point
Is the existence of basis in all vector space equivalent to the axiom of choice?
What is P(C)? The power set of C? I don't understand. And isn't this theorem supposed to rely on Zorn's lemma?Hurkyl said:Every element of B is a linear combination of finitely many elements of C, and this defines a function f:B --> P(C)
Apparently, you need the ultrafilter principle, which follows from the AC, but is not equivalent to it. So I was right that the uniqueness of cardinality of basis doesn't follow from ZF, but wrong in assuming that it requires choice. See Schechter HAFHurkyl said:I'm pretty sure you don't need the axiom of choice to prove that if you have two bases, their cardinalities are equal.
the support of a function is the closure of the set where the function is nonzero. unless we're talking about topological vector spaces, I don't think we can apply that term.Hurkyl said:Is "support" the word I'm looking for? I.E. "f(b) is the support of b in the basis C." (or something like that?)