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Possible Duplicate:
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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    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.2012-07-14
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    @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.2012-07-14
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    Okay, sure. When I say "no choice" I mean "no use of the axiom of choice."2012-07-14

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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 thanks2012-07-14