# Proof : Given a finite set of equidimensional proper subspaces of a vector space $V$, $\exists$ $x$ in $V$ that belongs to none of them

I am stuck at this statement in a book of linear algebra. Even though the author casually mentions it, I am having a hard time coming up with a proof, why it must be true. Can you guys help? I can think of a few cases where it is true. Like given a finite no of lines through centre in $$\mathbb{R}^2$$, we can always find a point, which doesn't lie on any of them and so on. But how to generalize it?

@paul garrett gave a counterexample for finite fields. For infinite fields, let $$V_1, ..., V_k$$ be proper subspaces of a vector space $$V$$, and let $$U=\bigcup_{i=1}^kV_k\subset V$$.
1. If $$U$$ is not a vector space, then it must be a proper subset of $$V$$, and thus $$V\setminus U$$ is nonempty.
2. If $$U$$ is a vector space, then, by this property (extended to a finite union of subspaces) there must be an $$i\in\left\{1,...,k\right\}$$ such that $$V_i$$ contains $$V_1, ..., V_k$$. But then $$U=V_i$$ so $$V\setminus U$$ is nonempty.
• This does use the fact/assumption that the field of scalars is infinite (like $\mathbb R$, $\mathbb Q$, or $\mathbb C$), as opposed to finite fields $\mathbb F_p\approx \mathbb Z/p$, for example. Jul 2, 2020 at 18:28
• Proceed by induction. It is clear that the property is holds for $k=2$. Now, assuming that it is true for $k\geq 2$, we will show that it is also true for $k+1$. Let $V_1, ..., V_{k+1}$ be subspaces of a vector space $V$ and assume that $U=\bigcup_{i=1}^{k+1}V_i$ is a vector space. Remark that $U=V_{k+1} \cup\bigcup_{i=1}^k V_i$ so either $\bigcup_{i=1}^k V_i$ is a subset of $V_{k+1}$, either $V_{k+1}$ is a subset of $\bigcup_{i=1}^k V_i$. If we are in the first case, we are finished. If we are in the second case, then $U=\bigcup_{i=1}^k V_i$ is a vector space and we can apply the property. Jul 2, 2020 at 18:47
• Over a finite field such as $\mathbb F_p$, a two-dimensional vector space $V$ is (in a silly but obvious way) the finite union of lines $\mathbb F_p\cdot v$, as $v$ runs over (the finite set) $V$. So the claim does not hold in this case... but for uninteresting reasons. Jul 2, 2020 at 18:48