Must a proper curve minus a point be affine? Let $C$ be a proper smooth geometrically connected curve over a field $K$, and let $P\in C(K)$ be a point. Must $C - P$ be affine?
EDIT: By Riemann-Roch, you can definitely find functions $f_1,\ldots,f_r : C-P\longrightarrow\mathbb{A}^n_K$, but how do you guarantee that for some $n$, you can find enough such $f_i$'s such that this gives you an embedding?
EDIT: Is the same true with $C$ not smooth?
 A: Theorem
If a nonzero finite number of points $p_1,\dots, p_r $ are deleted from $C$ the resulting curve will be affine.    
Indeed consider the divisor $D=p_1+\dots+ p_r $ on $C$.
Since it has positive degree some positive multiple $nD$ of it will be very ample.
Thus we get an embedding of $j:C\to \mathbb P^N$ (for some huge $N$) and a hyperplane section divisor $\Delta =H\cap j(C)$ on $j(C)$ such that $j^*\Delta=nD$.
But then $C\setminus \{p_1,\dots, p_r\}$ is isomorphic to $j(C)\cap (\mathbb P^N\setminus H)\cong j(C)\cap \mathbb A^N$ (the complement of a hyperplane in projective space is affine space) and since this last variety $j(C)\cap \mathbb A^N$  is clearly affine, so is  $C\setminus \{p_1,\dots, p_r\}$.
Edit
The theorem is valid even if $C$ is singular.
To see  that, consider the finite normalization morphism $n:\tilde C\to C$ and delete the inverse image of $\{p_1,\dots, p_r\}$, obtaining the smooth curve $C'=\tilde C\setminus n^{-1}(\{p_1,\dots, p_r\})$ which is affine by the result already proved for smooth curves.
Now consider the restricted finite  morphism $n':C'\to C\setminus \{p_1,\dots, p_r\}$.
Since $C'$ is affine and the finite morphism $n'$ is surjective the curve $C\setminus \{p_1,\dots, p_r\}$ will also be affine by Chevalley's Theorem (EGA $_{II}$, Théorème (6.7.1), page 136),  and we are done.
