A set $S\subseteq\mathbb{A}^n$ is quasi-affine iff $S=Z\setminus V$ for closed $Z$ and $U$?

Question

I'm confused by a remark in note I'm reading. It essentially says,

Let $S\subseteq\mathbb{A}^n$ be a subset of affine $n$-space over an algebraically closed field. It's clear that $S$ is quasi-affine iff $S$ is a difference of closed sets, that is, $S=Z\setminus U$ for $U$ and $Z$ closed in $\mathbb{A}^n$.

Part of what troubles me is that $S$ is a subset, not necessarily an affine variety. Can anyone elucidate why quasi-affine sets are precisely differences of closed sets?

Answer 1

I guess this depends on your definition of a quasi-affine set. One definition is a quasi-affine set is an open subset of an affine variety. Since a variety $Z$ is a closed subset of $\mathbb{A}^n$, a quasi-affine set is an open subset of $Z$. With the subspace topology, this means a quasi-affine set is the intersection of $Z$ with an open set $Y$. Or in other words, a quasi-affine set has the form

$$Z\cap Y=Z\setminus Y^c$$

Setting $U=Y^c$ gives the result.