Irreducible algebraic curve contain another one? Let $C$ be an an irreducible algebraic curve, then can there be another algebraic curve $C' \subset C$ such that $C \backslash C'$ is infinite?
 A: Elaborating a bit further on Qiaochu Yuan's comment: Any curve $C'$ contained in an irreducible algebraic curve $C$ is obtained by removing a finite number of points. Any such $C'$ is an open, dense subvariety of $C$. If you were to remove more, i.e. require that $C'$ is a proper closed subset of $C$, then it would force $C'$ to have lower dimension than $C$, making it, well, not a curve.
Proof. Let $f: C'\hookrightarrow C$ be an immersion of curves (over an algebraically closed field $\Bbbk$) with $C$ irreducible. Since $\dim(C')=1=\dim(C)$, the image of $f$ is dense (its closure is a proper closed subset, by reasons of dimension it must be equal to $C$). This means that $f$ is dominant, and since $f$ is also an immersion, it must be a birational equivalence. Hence (e.g. Corollary I.4.5 in Hartshorne), there are open subsets $U'\subseteq C'$ and $U\subseteq C$ such that $f$ yields an isomorphism $U'\cong U$. Now both $U$ and $U'$ are the complement of finitely many points. Hence, the image of $f$ must be the complement of finitely many points, too.
