Using measure theory, one can see immediately that $\mathbb{R}^{n}$ is not the union of only finitely many (proper) affine subsets. Can we prove this using only algebraic methods? I know one way to do this using Vandermonde matrix and analyzing for each (proper) affine subset, points of the form $x_{0}+(1,r,r^{2},\ldots,r^{n-1})$ where $r$ varies in $\mathbb{R}$. Is there any other proof (hopefully) avoiding matrices and determinants? Moreover, what can we say when the ground field is finite?
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Do you know of affine varieties or algebraic geometry more generally? – Sep 01 '16 at 12:09
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Sadly, I know almost nothing about algebraic geometry. But any answer or comment is welcome. – Dilemian Sep 01 '16 at 12:10
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See this mathoverflow post for a very elementary proof of the following purely algebraic fact:
Theorem. If $F$ is an infinite field and $V$ is a vector space over $F$ then $V$ is not a union of finitely many proper affine subspaces of $V$.
Note that you do not even need to assume that $V$ is finite dimensional. In the case of finite $F$, the estimate is that the least number of proper affine subspaces needed to cover $V$ is $|F|$.
Moishe Kohan
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@user26857: I suspected this to be the case but did not see these proofs at MSE. Now, I see at least one: http://math.stackexchange.com/questions/145869/a-finite-dimensional-vector-space-cannot-be-covered-by-finitely-many-proper-subs/145876#145876. I will vote to close the question as a duplicate. – Moishe Kohan Sep 01 '16 at 14:31