A finite-dimensional vector space cannot be covered by finitely many proper subspaces? Let $V$ be a finite-dimensional vector space, $V_i$ is a proper subspace of $V$ for every $1\leq i\leq m$ for some integer $m$. In my linear algebra text, I've seen a result that $V$ can never be covered by $\{V_i\}$, but I don't know how to prove it correctly. I've written down my false proof below:
First we may prove the result when $V_i$ is a codimension-1 subspace. Since $codim(V_i)=1$, we can pick a vector $e_i\in V$ s.t. $V_i\oplus\mathcal{L}(e_i)=V$, where $\mathcal{L}(v)$ is the linear subspace span by $v$. Then we choose $e=e_1+\cdots+e_m$, I want to show that none of $V_i$ contains $e$ but I failed.
Could you tell me a simple and corrected proof to this result? Ideas of proof are also welcome~
Remark: As @Jim Conant mentioned that this is possible for finite field, I assume the base field of $V$ to be a number field.
 A: This question was asked on MathOverflow several years ago and received many answers: please see here.
One of these answers was mine.  I referred to this expository note, which has since appeared in the January 2012 issue of the American Mathematical Monthly.
A: This is a special case of a fact that an affine space over an infinite field is irreducible.
The proof can be found in most books on elementary algebraic geometry(see for example Fulton's algebraic curves).
A: Having been thinking about functional analysis for the past week, Baire's category theorem came to mind, but unfortunately this assumes the field is $\mathbb{R}$ or $\mathbb{C}$:
A finite dimensional linear subspace is closed, and a proper linear subspace has empty interior. So by Baire, a countable union of  finite dimensional linear subspaces again has empty interior; in particular it is not the whole space.
(I believe the first sentence is still true in the generality of $V$ being a topological vector space. However to apply Baire to $V$ we need it to be locally compact Hausdorff (i.e. finite dimensional, by Riesz) or completely metrizable (e.g. a Frechet or even F- space).
A: [Edit] This answer is contained in another answer of mine. Sorry about that. Switching to CW[/Edit]
Pick a basis and a system of coordinates $x_1,\ldots,x_n$ for $V$. WLOG assume that $n \geq 2$. As you observed,
without loss of generality we can assume that the subspaces are all of codimension one, i.e. spaces of solutions of a single homogeneous equation 
$$
a_1x_1+a_2x_2+\cdots +a_nx_n=0
$$
in the coordinates $x_i,i=1,\ldots,n$. Therefore a single subspace will intersect the infinite set
$$
S=\{(1,t,t^2,\ldots,t^{n-1})\mid t\in k\}
$$
at finitely many points, because the polynomial $a_1+a_2t+a_3t^2+\cdots+a_nt^{n-1}$ has at most $n-1$ zeros.
Therefore it is impossible to cover all of $S$, hence all of $V$, with finitely many subspaces.
Note that if $k$ is uncountable, then this argument shows that we need uncountably many subspaces. 
A: Do you know the proof that the union of two subspaces of a vector space is a subspace if and only if one of the two subspaces is contained in the other? If the field is infinte one can come up with a similar proof for your statement.
Assume $V$ is covered by finitely many $V_i$, and assume that the cover is minimal. Then there is wlog a $v\in V_1$ which is not in any other $V_i$ and there is also wlog a $w\in V_2-V_1$. Then the vectors $av+w$ for $0\neq a\in k$ (where $k$ is the base field) are in pairwise different spaces $V_i$. Indeed if $av+w$ and $bv+w$ both are in $V_i$, then so is $(a-b)v$ which is a contradiction. Since $k$ is infinite this proves your statement.
A: Let me prove that that if $k$ an infinite field, a finite number of hyperplanes $H_1,\ldots, H_r$ can't cover the vector space $k^n$ .   
If we had $k^n=\bigcup_{i=1}^{r} H_i$ where $H_i$ is the kernel of the non-zero linear form $l^{(i)}(x_1,\ldots,x_n)=\sum_{j=1}^{n}a_j^{(i)}x_j\in (k^n)^\ast$ , the degree $r$ polynomial  $P(x_1,\ldots,x_n)=\prod_{i=1}^{r} l^{(i)}(x_1,\ldots,x_n)\in k[x_1,\ldots,x_n]$ would vanish at all points of  $k^n$ without being the zero polynomial .
This is well known to be impossible if $k$ is infinite: Jacobson Theorem 2.19, page 136.
