Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a intuitive way to understand "the existence of a basis for every vector space is equivalent to the Axiom of Choice"?

share|improve this question
What is and what is not clear to you about this statement? –  Joshua Biderman Mar 11 '14 at 5:37
@Stella: Are you serious? –  Asaf Karagila Mar 11 '14 at 5:41
I just asked what their current level of understanding is so one would know where to start? So, yes? –  Joshua Biderman Mar 11 '14 at 5:44
@Stella, I understand what AC is, and of course what vector space and its basis is, but it's hard to see the said equivalence or even start to appreciate it after being told so. –  qazwsx Mar 11 '14 at 5:45
@Stella: The statement itself is very simple to understand. The intuition as for why the proof works is not. –  Asaf Karagila Mar 11 '14 at 5:48

1 Answer 1

up vote 6 down vote accepted

Caveat lector: There are plenty of easily understandable statements in mathematics, whose proofs are far from trivial or intuitive. This is not exactly the case here, but it is not the case that it's not the case here either. I will try to give a handwavy explanation as to why this works, but this is far from trivial.

Well, the proof itself goes through some other equivalents as well. So it might be slightly easier to understand the full circle:

  1. The axiom of choice.
  2. Zorn's lemma.
  3. Every vector space has a basis.
  4. The axiom of multiple choice (every family of non-empty sets of size at least $2$ has a function which chooses from each set a finite proper subset).

$(1)\implies(2)$ is a well-known argument of transfinite induction, and $(2)\implies(3)$ is the classical use of Zorn's lemma. $(3)\implies(4)$ can be intuitively comes from the fact that linear combinations are finite, so we can define a vector space and from its basis choose the finite subsets of the family. Finally, $(4)\implies(1)$ can be handwavingly described as reiterating the subset process until we are left with just singletons from which the choice is canonical.

The proofs themselves of course require a lot more details, and the above intuitive and handwaving explanation is far from sufficient to fully get them. But that's the main intuition, I think.

share|improve this answer
Don't you think Andreas Blass should be answering this? ;-) –  Arthur Fischer Mar 11 '14 at 8:10
Don't you think it's time for a Fischer-Fischer paper? ;-) –  Asaf Karagila Mar 11 '14 at 8:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.