TITLE: 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 $V$
QUESTION [24 upvotes]: 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 $U_1$, $U_2,\ldots,U_n$?

REPLY [8 votes]: 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,  $ $ note $\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 $\rm\,|F| = |L|\,.$