Proving that a vector space $\mathbb{R}^k, k\in \mathbb N$ has a basis with ZF (and no Axiom of Choice)

Question

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?

Answer 1

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