If a field $F$ is such that $\left|F\right|>n-1$ why is $V$ a vector space over $F$ not equal to the union of $n$ proper subspaces of $V$ If $U_1$, $U_2,\ldots,U_n$ are proper subspaces of a vector space $V$ over a field $F$, and $|F|\gt n-1$, why is $V$ not equal to the union of the subspaces $U_1$, $U_2,\ldots,U_n$?
 A: If $|F|=q<\infty$, and $V$ is $m$-dimensional ($m<\infty$), then any proper subspace $U_i$ has at most $q^{m-1}-1$ non-zero elements. So to cover the $q^m-1$ non-zero vectors of $V\,$, the given $n\le q$ subspaces are not going to be enough, because 
$$n(q^{m-1}-1)\le q(q^{m-1}-1)<q^m-1.$$ So we need at least $|F|+1>n$ subspaces to get the job done.
If $m=\infty$, then we can extend all the subspaces to have codimension one (i.e. $\dim_F(V/U_i)=1$ for all $i$). In that case the intersection $U$ of all the $U_i$:s has finite codimension, and we can study $V/U$ instead of $V$ reducing the probelm to the previous case.
If $|F|=\infty, m<\infty$? Well, then we need some reinterpretation. The following argument shows that we need an infinite number of subspaces to cover $V$, and an uncountable number of subspaces to cover $\mathbf{R}^m$. Again, assume that all the subspaces have codimension one (w.l.o.g.), and that $m\geq 2$ (also w.l.o.g.). Identify $V$ with $F^m$, and consider the set
$$
S=\{(1,t,t^2,\ldots,t^{m-1})\in V\mid t\in F\}.
$$
Any $U_i$ is now a hyperplane and consists of zeros $(x_1,x_2,\ldots,x_m)$ of a single non-trivial homogeneous linear equation
$$a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_m=0.$$
Therefore the number of elements of the intersection $S\cap U_i$ is equal to the number of solutions $t\in F$ of $ a_{i1}+a_{i2}t+\cdots+a_{im}t^{m-1}=0$ and is thus $<m$, because a non-zero polynomial of degree $<m$ has less than $m$ solutions in a field. This shows that if $F$ is infinite, we need an infinite number of subspaces to cover all of $S$. Also, if $F$ is uncountable, then we need an uncountable number of subspaces to cover $S$. Obviously it is necessary to cover all of $S$ in order to cover all of $V$.
A: Hint $ $ Let $\rm\,U = U_1\! \cup \,\cdots\,\cup U_n,\,$ wlog irredundant (i.e. no $\rm\,U_i\,$ lies in the union of the others). Choose $\rm\,v\not\in U_1,$ $\rm\, u\in U_1,\, u\not\in U_{i>1}.\,$ Put $\rm\, L = v + u\:\! F.\,$ Then $\rm\,|L\cap U_1| = 0,\,$ $\rm |L\cap U_{i\,>1}| \le 1.\,$ Therefore $\rm\,|L\cap U| \le n-1 < |F| = |L|,\,$ so the "generic" line $\rm\,L\,$ has a point not in $\rm U.\ $
Proof $\  $ First,  $ $ notice $\rm\ |L\cap U_1| = 0\ $ since $\rm\, u,v+cu \in U_1 \Rightarrow\, (v+cu)-cu\, =\, v \in U_1\,$ contra choice of $\rm\,v.\,$ Second, $\rm\,|L\cap U_{i\,>1}| \le 1\, $ since if $\rm\,v+cu,\, v+du\in U_i$ then so too is their difference $\rm\,(c-d)u.\,$ Thus $\rm\,c = d\ $ (else scaling by $\rm\,(c-d)^{-1}$ $\Rightarrow$ $\rm\,u\in U_{i\,>1}\,$ contra choice of $\rm\,u).\,$ Finally $\rm\,v+cu\, =\, v+du\,$ $\Rightarrow$ $\rm\,(c-d)\,u = 0\,$ $\Rightarrow$ $\rm\,c=d,\,$ so $\rm\,c\,\mapsto\, v+c\,u\ $ is $\,1$-to-$1,\,$ thus $\rm\,|F| = |L|.$
