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.
Best Answer
[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.