$W_1,...,W_n$ be any $1$-dimensional subspaces , is there a $2$-dimensional subspace $W$ with $W \cap (\cup_{i=1}^n W_i)=\{0\}$?

Question

Let $V$ be a vector space of dimension $\ge 3$ over an infinite field $k$. Let $W_1,...,W_n$ be any $1$-dimensional subspaces of $V$. Then must there exist a two dimensional subspace $W$ of $V$ such that $W \cap (\cup_{i=1}^n W_i)=\{0\}$ ?

Answer 1

Yes.

Start with any two-dimansional subspace $U$. We can pick a basis $(v_1,v_2)$ of $U$ such that $v_1$ is in none of the $W_i$: We have infinitely many choices to replace $v_1$ with $v_1+\alpha v_2$, $\alpha\in k$, and each $W_i$ prohibits at most one value of $\alpha$.

Now pick $v_3\notin U$. and consider the vector spaces $U_\beta=\langle v_1,v_2+\beta v_3\rangle$, $\beta\in k$. As $U_\beta\cap U_{\beta'}=\langle v_1\rangle$, each $W_i$ is contained in at most one $U_\beta$. Hence by a suitable choice of $\beta$, we find that $$ U_\beta\cap \bigcup_{i=1}^nW_i=\{0\}.$$

Remark: Apparently the claim also hold for finite fields $k$, as long as $n<|k|$. And vice versa, it also works for infinite $n$, as long as $k$ is "more infinite" (e.g., for $k=\Bbb R$, we can have countably many $W_i$).

Answer 2

A nonconstructive solution:

Convention, in the following argument, $W$ cross $W_i$ at some point means the two subspaces have a common nonzero vector.

Suppose for the sake of contradiction that, all 2-dim subspace crosses at least a point on one $W_i$.

Since there is finitely many $W_i$, there must be one $W_k$ that crosses infinitely many 2-dim subspaces, among which there are infinitely many that do not contains any $W_i$ (otherwise there are only finitely many 2-dim subspaces whose orthogonal complement is not contains any $W_i$, this can't be true since finitely many 1-dim subspaces can't cover a 2-dim space, nor to say a 3-d space).

But this can't be true that (the statement before the previous long brackets: there are infinitely many subspaces that do not contains any $W_i$(among which is $W_k$) but crosses a certain $W_k$, in fact, there are none).

A contradiction.

This proof is quite strange, I admit.