Can a vector space over an infinite field be a finite union of proper subspaces?

Let $V$ be the union $\cup_{i=1}^n V_i$, where the $V_i$ are proper subspaces and the ground field $k$ is infinite. Pick a non-zero vector $x\in V_1$. Pick $y\in V-V_1$, and note that there are infinitely many vectors of the form $x+\alpha y$, with $\alpha\in k^\ast$. Now $x+\alpha y$ is never in $V_1$, and so there is some $V_j$, $j\neq 1$, with infinitely many of these vectors, so it contains $y$, and thus contains $x$. Since $x$ was arbitrary, we see $V_1$ is contained in $\cup_{i=2}^n V_i$; clearly this process can be repeated to find a contradiction.

Steve

You can prove by induction on n that:

An affine space over an infinite field $F$ is not the union of $n$ proper affine subspaces.

The inductive step goes like this: Pick one of the affine subspaces $V$. Pick an affine subspace of codimension one which contains it, $W$. Look at all the translates of $W$. Since $F$ is infinite, some translate $W'$ of $W$ is not on your list. Now restrict all other subspaces down to $W'$ and apply the inductive hypothesis.

This gives the tight bound that an $F$ affine space is not the union of $n$ proper subspaces if $|F|>n$. For vector spaces, one can get the tight bound $|F|\geq n$ by doing the first step and then applying the affine bound.

I recently completed a short expository note on this subject, Covering Numbers in Linear Algebra. See:

http://math.uga.edu/~pete/coveringnumbersv2.pdf