I'm interested in proving that:
Given $ r $ points in the projective space, that is, $ p_1, ..., p_r \in \mathbb{P}_K^n $, show that there is an affine open $ U \subset \mathbb{P}_K^n $ such that $ p_1, ..., p_r \in U $.
My attempt:
Let us consider $ U_1, ..., U_r $ affine opens such that $ x_i \in U_i $, those that we know exist, in fact and the coordinates of the projective space are $ (x_0: ...: x_n) $ y $ p_i \in (x_j \neq 0) $ we can consider $ U_j = (x_j \neq 0) $, then my attempt is to consider $ U = \cup_{i = 1}^r U_r $.
The problem that I have with my argument is that if $ r = n + 1 $ and $ p_{i + 1} = x_i $ then the affine open are $ U_i = (x_i \neq 0) $ and $ U = \cup_{ i = 0}^n (x_i \neq 0) = \mathbb{P}_K^n $ is not affine open.
Any suggestions or other solutions to the problem?
Assuming $K=\bar{K}$ so we deal with single point rather than Galois orbits or suchlikes.
Turning this question the other way: given $r$ points $P_1,\dots,P_r\in\mathbb{P}^n_K$, we want to find a hyperplane $H=(\sum\lambda_i x_i=0)$ which does not pass through any of the $P_i$. (Then we have $\mathbb{P}^n_K-H$ is an affine open $\sum\lambda_ix_i\neq 0$ which contains all $P_1,\dots,P_r$.)
Or equivalently, by taking duals, we want to show that the union of $r$ hyperplanes is not the whole of $\mathbb{P}^n_K$.
But the last statement follows immediately from: A vector space over $K$ is not a finite union of proper linear subspaces when $K$ is infinite (why?).