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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    $\begingroup$ Covering Numbers in Linear Algebra by Pete L. Clark might help. $\endgroup$ – Pierre-Yves Gaillard Aug 30 '11 at 13:42
  • $\begingroup$ @Pierre-Yves: yes, the answer to the question is contained in my note. $\endgroup$ – Pete L. Clark Sep 4 '11 at 23:59
  • $\begingroup$ I referred to this question in another thread, and wanted to make the title clearer. Sorry about the bump. $\endgroup$ – Jyrki Lahtonen Nov 4 '18 at 5:39

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 $V$.


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 $\rm\,|F| = |L|\,.$


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