# Finding a Vector not in a Collection of Proper Subspaces

Let's take $K$ to be an infinite field and $V$ a vector space over it. Allowing $U_1,...U_l$ to be $l$ proper subspaces of $V$ (none the same), and further assuming that for all $i$, $U_i$ is not in the union of the $U_j$ for $i\neq j$, I want to show that there exists a $v\in V$ not in any of the $U_i$.

Thus far, my course of action has been induction on $l$. The base case is trivial, and then for the induction step I let $v_1\in V$ be such that it's not in $U_i$ for $2\leq i\leq l+1$ and $v_{l+1}\in V$ such that it's not in $U_i$ for $1\leq i \leq l$. I want now that $v=v_1+av_{l+1}$ for $a\in K$ is the vector not in any of the $U_i$, but I'm stuck trying to show this. I'm assuming the hypothesis that $K$ is infinite comes into play but I can't quite see how.

Additionally, what more precise statement can I make in the case that $K$ is finite? Surely if $|K|\gt l$ then I can be more precise about the $a$ above?

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planetmath.org/encyclopedia/… – user26857 Oct 20 '12 at 17:20
Discussed in detail at mathoverflow.net/questions/26/… --- see Pete Clark's link to a paper of his for quantitative results in the finite case. – Gerry Myerson Oct 20 '12 at 23:37
@GerryMyerson Ah, both very helpful! I'm still a bit confused though. If I wanted to show that for $|K|\gt k$ there's such a $v$, how do I do that? I guess the paper you linked to shows me that it takes a union bigger than $k$ proper subspaces to cover $V$, but I'm still not sure how to find such a $v$. – AsinglePANCAKE Oct 22 '12 at 7:18
You seem to switch, several times, between $k$ and $l$. If they're meant to be the same thing, maybe you could pick one, and stick to it. I'm not familiar enough with the literature to know if there's a good construction for $v$, rather than just an existence proof. Of course if $V$ is finite-dimensional over a finite field then there is the uninspiring but effective systematic search through all the elements of $V$. – Gerry Myerson Oct 22 '12 at 11:32
@navigetor23 Thanks! But I'm a bit confused by the statement "one of the $V_i$ must contain infinitely many elements of $S$." How do we know this? – AsinglePANCAKE Oct 23 '12 at 2:04