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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.

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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. – 2017-01-06
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    [Related](http://math.stackexchange.com/questions/207990/vector-spaces-and-ac?rq=1). – 2017-01-06
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    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. – 2017-01-06

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