I am writing a short (30-50 pages) report on AC for an exam. I really would like to include the proof that "Every vector space has a basis $\implies$ AC". Actually, every proof I could find proves that AMC (the Axiom of Multiple Choices) holds, not AC. Since ZF $\vdash$ "AC $\iff$ AMC", that would be formally fine, but then I would like to include the proof of "AMC $\implies$ AC". But every proof I could find proves it by passing through "$\cal P(\alpha)$ is well-orderable for every ordinal $\alpha \implies$ AC".

The proof of the latter seems to use von Neumann hierarchy. The professor I will present my report to is not a logician (he's an Algebra professor), and I'm pretty sure that he wouldn't like to read too much set-theory stuff.

Therefore my question is: do you know any way to prove "Every vector space has a basis $\implies$ AC" without mentioning von Neumann hierarchy?


  • $\begingroup$ I think this is still open, but I'm not sure. $\endgroup$ – Asaf Karagila Jul 30 '14 at 13:48
  • $\begingroup$ What do you mean precisely? I mean, in ZF every implication above is a (known) equivalence, therefore from a mathematical p.o.w. I can't see any open problem...or do you mean that it's still unknown whether there is any "shorter" proof? $\endgroup$ – aerdna91 Jul 30 '14 at 13:54
  • $\begingroup$ You are asking "Is it true in $\sf ZFA$ that "Every vector space has a basis implies the axiom of choice". Are you not? This is a problem, and I think it is open. What else would you mean "without the von Neumann hierarchy?" Besides that, I think that you should tell your professor (in a polite way, of course) that in order to talk about linear transformations you should know some linear algebra, and in order to talk about set theoretic axioms you should know some set theory. There's ultimately no escape from that. $\endgroup$ – Asaf Karagila Jul 30 '14 at 13:57
  • $\begingroup$ Probably I wasn't clear enough, sorry! I just want a proof that works in ZF but doesn't explicitly mention the von Neumann hierarchy. $\endgroup$ – aerdna91 Jul 30 '14 at 14:00
  • $\begingroup$ I see. Okay then. $\endgroup$ – Asaf Karagila Jul 30 '14 at 14:01

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