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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 $U_1$, $U_2,\ldots,U_n$?

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Covering Numbers in Linear Algebra by Pete L. Clark might help. –  Pierre-Yves Gaillard Aug 30 '11 at 13:42
    
@Pierre-Yves: yes, the answer to the question is contained in my note. –  Pete L. Clark Sep 4 '11 at 23:59

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up vote 7 down vote accepted

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.). 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 $V$.

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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 $\rm\: u,\:v+c\:u \in U_1\: \Rightarrow\ (v+c\:u)-c\:u\: =\: v \in U_1\:$ contra choice of $\rm\:v\:.\:$ Second $\rm\:|L\cap U_{i\:>1}| \le 1\: $ since if $\rm\:v+c\:u,\ v+d\:u\in U_i\:$ 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 $\rm\:v+c\:u\: =\: v+d\:u\:$ $\Rightarrow$ $\rm\:(c-d)\:u = 0\:$ $\Rightarrow$ $\rm\:c=d\:,\:$ so $\rm\:c\:\mapsto\: v+c\:u\ $ is $\:1$-to-$1\:,\: $ hence $\rm\:|F| = |L|\:.$

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