Complement of compact subspace of surface Let $X$ be a smooth 2-manifold, $K$ be a compact subset of $X$, such that only one component of $X\backslash K$ does not have compact closure, call this component $U$ (there may be other components). Must $X\backslash U$ be compact?
The reason I am asking this is that Donaldson's 'Riemann surfaces' seem to rely extensively on this face in chapter 10 without any sort of explanation. This is obvious for $X=\mathbb{R}^2$, but I am having difficulty in proving or providing a counterexample for the general case.
 A: Yes, $X \setminus U$ must be compact.
To prove this, first let's enlarge the set $K$ somewhat: we may include $K$ into a subset $K \subset Y$ such that $Y$ is a smooth, compact, surface-with-boundary. This is not too hard to see: take a collection of smooth balls whose interiors cover $K$; by compactness finitely many of these balls suffices; we may perturb the balls so that their boundaries intersect in general position; the union of these balls is a compact surface-with-boundary having corners on the boundary; we can then smooth the corners.
The surface $Y$ has finitely many boundary components and hence $X \setminus Y$ has only finitely many components. One of these components, denoted $V$, contains $U$. Every component of $X \setminus Y$ except $V$ is contained in a component of $X \setminus K$ having compact closure. Thus each components of $X \setminus Y$ except $V$ has compact closure. It follows that $L = X \setminus (Y \cup V)$ is a union of finitely many sets with compact closure and therefore has compact closure. Since $X \setminus U \subset X \setminus V = Y \cup L$, it follows that $X \setminus U$ has compact closure.
