Geometric statement of Prime Avoidance?

The contrapositive has a nice geometric interpretation. The algebraic statement is:

If $I\nsubseteq \mathfrak{p}_{i}$ for any $i$, then $I\nsubseteq \cup_{i} \mathfrak{p}_{i}$, i.e. there is some $x\in I$ which is not in any of the $\mathfrak{p}_{i}$.

Now consider the affine scheme $X=\operatorname{Spec}R$. Then $I\nsubseteq \mathfrak{p}_{i}$ for any $i$ translates into $\mathfrak{p}_{i}\in X\setminus V(I)$ for all $i$. That is, the points $\mathfrak{p}_{i}$ are all contained in the open set $X\setminus V(I)$. What the prime avoidance tells us is that we can now find an element $x\in I$ which is not in any of the prime ideals $\mathfrak{p}_{i}$, which translates into saying that $$ \mathfrak{p}_{i} \in D(x)\subseteq X\setminus V(I)$$ So if we have finitely many points contained in an open set of an affine scheme, we can always find a smaller principal open set containing them.

Reference: I found this on the Stacks Project.