Let ${f_1,f_2,\ldots,f_{n-1}}$ be lineary independent set of vectors in $\mathbb{C}^n$. I finish to proof that there exists open sets $V_1,\ldots,V_{n-1}$ in $\mathbb C^n$ and $v_i\in V_i$ $i=1,2,\ldots,n-1$ such that $f_i\in V_i$ and the set ${v_1,\ldots,v_{n-1}}$ linearly independent using a determinant trick. Now, Let $\{p_1,\ldots,p_k\}$ be a set of points in $\mathbb{C}^n$.
I try to show that the set
$$A_i=\{(v_1,\ldots,v_{n-1}\in V_1\times V_2\times\cdots\times V_{n-1})\mid p_i\notin L(v_1,v_2,\ldots,v_{n-1})\}$$
Is open a dense. where $L(v_1,\ldots,v_{n-1})$ is the span of the vectors $v_1,\ldots,v_{n-1}$. I have only been able to prove that it is different from empty, using the fact that in the previous construction at least I can find a hyperplane that passes through the origin that does not contain the point, but I have not been able to see that it is open or dense.
I appreciate the collaboration that you can give me in this problem.
I assume that, for every $i$, $p_i\not= 0$.
We choose the open sets $(V_i)$ small enough for the following to be true:
for every $v_j\in V_j,i=j,\cdots,n-1$, the set $(v_j)_j$ is linearly independent. Let $W=\Pi_{j\leq n-1}V_j$.
Proposition. $\cap_{i\leq k}A_i$ is open dense in $W$.
Proof. It suffices to show that $\cup_{i\leq k}(W\setminus A_i)$ is Zariski closed and not $W$ or, for every $i\leq k$, $W\setminus A_i$ is Zariski closed and not $W$.
Note that $(v_j)_j\in W\setminus A_i$ iff $\det(p_i,v_1,\cdots,v_{n-1})=0$ (since the set $(v_j)_j$ is free) and, consequently, $W\setminus A_i$ is Zariski closed.
It remains to see that there is $(v_i)_i\in W$ s.t. $\det(p_i,v_1,\cdots,v_{n-1})\not= 0$; otherwise, the algebraic function $f((v_j))=\det(p_i,v_1,\cdots,v_{n-1})$ is locally $0$, then globally $0$ on $(\mathbb{C}^n)^{n-1}$. Of course, it's false; just complete $p_i$ in a basis of $\mathbb{C}^n$.