Are the quasi-affine subsets of $\mathbb{A}^1_F$ necessarily open or closed?

Question

For what follows, my definition of a quasi-affine subset of $\mathbb{A}^n_F$ is one which can be written as $Z_1\setminus Z_2$, where $Z_1$ and $Z_2$ are closed subsets of $\mathbb{A}^n_F$. (I think this is also referred to being locally closed in some places?)

A few weeks ago I asked a question whether quasi-affine subsets of $\mathbb{A}^n_F$ ($F$ algebraically closed) are necessarily open or closed. This doesn't seem to hold for $n\geq 2$, so just as a follow up,

If $V=Z_1\setminus Z_2$ is a quasi-affine subset of $\mathbb{A}^1_F$, is $V$ necessarily open or closed in $\mathbb{A}^1_F$?

Answer 1

Hint: you can determine all closed subsets of $\mathbb{A}^1$ fairly easily.

How?

Use the fact that the coordinate ring is a PID.

Full solution:

$V$ must either be open or closed. Since $F[x]$ is a PID, all ideals are of the form $(p(x))$ for some polynomial $p$. Since $F$ is algebraically closed, if $p$ is nonconstant, $p$ factors as $p(x)=(x-c_1)^{n_1}(x-c_2)^{n_2}\cdots (x-c_i)^{n_i}$ where all $n_j>0$ and $c_j\in F$. But then $V((p(x))$ is just the collection of points $\{c_1,c_2,\cdots,c_i\}$. So every nonempty closed set in $\mathbb{A}^1$ is a finite union of points, or the entire space.

Conclusion (can't make the spoiler text work without this linebreak):

Omitting the trivial cases where $V$ is empty, we have two cases: the first is $Z_1=\mathbb{A}^1$ and $Z_2$ is some finite subset. Then $V$ is the complement of a finite subset and is open. The second is where $Z_1$ is a finite subset and $Z_2$ is a finite subset. Then $V$ is also a finite subset and is closed.