In Harris' algebraic geometry book, $p_{1},\ldots,p_{r}\in\mathbb{P}^{n}$ are said to be in general position if no $n+1$ or fewer of them are dependent.
I want to prove that, if $p_{1},\ldots,p_{r}\in\mathbb{P}^{n}$ are in general position, then they are generic points, i.e., there exists a Zariski open dense subset $\mathcal{U}\subseteq(\mathbb{P}^{n})^{r}$ such that $$ \mathcal{U}\subseteq\{(p_{1},\ldots,p_{r})\in(\mathbb{P}^{n})^{r}:p_{1},\ldots,p_{r} \text{ are in general position}\}. $$ I do not know if it is true, but I think so.
By the way, since $(\mathbb{P}^{n})^{r}$ is irreducible, every open subset of $(\mathbb{P}^{n})^{r}$ is dense, so, if we find an open subset $\mathcal{U}$ satisfying that property, we won't have to prove that it is dense.
Sure this is true. In fact, you can choose $\mathcal U$ to be the set of $r$-tuples of points in general position itself.
Let $p_i=[x_i]$ with $x_i\in\mathbb C^{n+1}$ and let $x\in\mathbb C^{(n+1)\times r}$ be the matrix whose columns are the $x_i$. Then, the points $p_i$ are in general position if and only if the maximal minors of this matrix do not vanish. These maximal minors are homogeneous polynomials in the entries of the $x_i$, so their zero sets define closed subsets of $(\mathbb P^n)^r$. Their union is also closed, and their complement is $\mathcal U$.