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Can someone please give me the proof that if you don't assume the axiom of choice, you can prove that there exists a vector space without a basis. I have the book, The Axiom of Choice by T.J.Jech, but don't really understand his proof. So any help in understanding it will be great,

Thanks.

Pedro
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B. Drennan
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    It's obviously not enough to not assume the Axiom of Choice: that won't let you prove anything that you can't prove with the Axiom of Choice. Instead, you have to assume the negation of the Axiom of Choice. – TonyK Jan 06 '17 at 15:17
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    Related. –  Jan 06 '17 at 15:18
  • The book you mention predates this result by more than a decade. So it is unlikely to appear there. What you can find there is an example of a vector space without a basis. And no, this construction cannot be simplified by much. You need to know about forcing, or ZFA and permutation models, or something equivalent, in order to understand it. – Asaf Karagila Jan 06 '17 at 15:33

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