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Finite dimensional subspaces of a linear space

I know that "every vector space has a basis" is equivalent to the "Axiom of Choice".

My question: Can I prove that $\mathbb{R}^k$ has a basis (where $k\in \mathbb{N}$) only with ZF? If so, how?

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marked as duplicate by Asaf Karagila, Andres Caicedo, BenjaLim, Nate Eldredge, J. M. Jul 15 '12 at 4:24

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The result you want to ask about is that a finitely-generated vector space (a quotient of $\mathbb{R}^k$ for some $k$) has a basis. The proof is very easy: pick a set of generators, and if they are not linearly independent discard one. Repeat. This requires no choice. –  Qiaochu Yuan Jul 14 '12 at 5:49
@QiaochuYuan: I think "no choice" is stretching it -- after all you're choosing a generator to discard. It's just that Finite Choice is a theorem of ZF. –  Henning Makholm Jul 14 '12 at 17:03
Okay, sure. When I say "no choice" I mean "no use of the axiom of choice." –  Qiaochu Yuan Jul 14 '12 at 17:05

1 Answer 1

up vote 6 down vote accepted

For $\mathbb R^k$ you can exhibit a basis, namely the vectors $(1,0,0,0,\ldots,0), (0,1,0,0,\ldots,0), (0,0,1,0,\ldots,0),\ldots ,(0,0,0,0,\ldots,1)$.

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You're right.. More precisely, i just proved it by induction thanks –  Katlus Jul 14 '12 at 4:46

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