Open Affine subsets of $\mathbb{A}^n_k$ are principal

Question

I have heard that affine open subsets of $\mathbb{A}^n_k$ are all principal (i.e of the form $D_f$). I know this is not true if one removes the assumption the open subset is affine. I have tried to prove this fact using quasi-compactedness of $\mathbb{A}_n^k$, but this is getting nowhere and I don't see how the assumption that the open is affine will be used with this method.

Answer 1

Lemma: Let $X$ be a normal Noetherian connected separated scheme, and $U\subset X$ a proper nonempty affine open subset. Then $X\setminus U$ has pure codimension one.

Proof: This is essentially algebraic Hartog's - if any part of $X\setminus U$ had codimension two or more, then regular functions would extend across it and thus these points would be inside $U$. For a full proof, see EGA IV part 4, corollary 21.12.7 or StacksProject 0BCQ.

Now to answer your question, let $U\subset \Bbb A^n$ be an affine open subset. If $U=\Bbb A^n$, then $U=D_1$. If $U$ is a proper affine open subset, then we may apply the lemma to get that $V:=\Bbb A^n\setminus U$ is a pure codimension one subvariety, or a divisor. Since the class group of $\Bbb A^n$ is zero, every divisor is principal, and so $V$ is the divisor associated to some function $f$. We may then take $U=D_f$.