I am reading Görtz and Wedhorn's "Algebraic Geometry I: Schemes with Examples and Exercises". In Proposition 3.23, it is proven that there is a bijection between points of the scheme and closed irreducible subsets. In one moment in the proof, it goes:
"Let $Z$ be an irreducible closed subset. Let $U$ be an affine open subset such that $Z\cap U \neq \emptyset$. Then the closure of $Z \cap U$ is Z, because Z is irreducible."
I tried proving that last line, and I got to a strange result. I thought $Z = \overline{Z \cap U} \cup (Z\cap U^c)$. But this implies that $Z \cap U^c = \emptyset$, which looks strange to me.