Let $Y$ be a quasi-affine variety and $Z_1 \subset Z_2$ be closed irreducible subsets of $Y$. I am trying to check that the inclusion ${\overline{Z_1}} \subset {\overline{Z_2}}$ is proper.
Here $Y \subseteq Z({\mathfrak{p}})$ is an open subset in ${\mathbb{A}}^n(k)$ (with ${\mathfrak{p}}$ prime). Now $Z_1 = C_1 \cap Y, Z_2 = C_2 \cap Y$ with $C_1, C_2$ closed in $Z({\mathfrak{p}})$ (and hence must be in ${\mathbb{A}}^n(k)$ as well). We can also choose these so that $C_1 \subset C_2$. However, in general, the closure doesn't respect intersection of sets (for instance take open intervals $(0,1)$ and $(1,2)$ in ${\mathbb{R}}$).
Does irreducibility plays a role here?
Thanks Kreiser: $Z_1 \subsetneq Z_2$ implies the choice $C_1 \subsetneq C_2$ is possible. Then $(Y \setminus C_1) \cap C_2 \neq \emptyset$. For any $x \in (Y \setminus C_1) \cap C_2$ we have $Y \setminus C_1$ is an open neighbourhood of $x$ and $x \in C_2 \cap Y \subseteq {\overline{C_2 \cap Y}}$. In case ${\overline{C_2 \cap Y}} = {\overline{C_1 \cap Y}}$, we must have $(C_1 \cap Y) \cap (Y \setminus C_1) \neq \emptyset$, which is absurd.