When can an open surface be embedded into a smooth compact surface? Why can't a double covering of a complex line $C$ with ramification order 2 at all integer points of $C$ be embedded into a smooth compact surface? I think we have infinitely many pairs of "intersections" above $C$, but what does this contradict? Some finite property of smooth compact surfaces?
 A: The property that is violated is that a compact surface has finite genus. By definition, the genus of a surface $S$ is the maximal cardinality of a set $\mathcal C$ of pairwise disjoint simple closed curves such that $S - \bigcup_{c \in \mathcal C} c$ is connected. For a compact, connected surface-with-boundary $S$ one can work out a formula for the genus $g$ expressed in terms of the Euler characteristic $\chi$, the number $b$ of components of $\partial F$, and the orientability of $S$; for instance if $S$ is orientable then its genus is $g = \frac{2-\chi-b}{2}$.
It follows that if $F$ is a subsurface of $S$ then the genus of $F$ is no larger than the genus of $S$, and so in particular if $S$ has finite genus then so does $F$. But the doubly ramified cover of $\mathbb C$ over the integer points has infinite genus.
A: A map of algebraic curves $X\to Y$ can only be ramified at finitely many points in characteristic zero, since the sheaf of relative differentials $\Omega_{X/Y}$ is zero at the generic point. If your double covering could be embedded in a compact Riemann surface, you would obtain a map of algebraic curves which is ramified at infinitely many points which cannot happen.
