# Is the complement of an open subset a manifold with boundary?

Let $$U\subset M$$ be a proper open subset of a smooth manifold without boundary. Is then the complement $$M\setminus U$$ a smooth manifold with boundary?
A related question would be if the topological boundary $$\partial(M\setminus U)$$ is a smoth hypersurface.

• The complement of an open set is a closed set, so you're asking if every closed subset of a manifold is a manifold. The answer is no. For instance, remove each of the 4 open quadrants from the plane and you get a cross-shape which is not a manifold. Commented Apr 28, 2022 at 14:41
• Very well. How about the case where $U$ is connected, and its complement contains an open set? Commented Apr 28, 2022 at 15:22
• You're still way too general with this thinking. Being a manifold is a strong condition! As an example, take the complement of the disjoint union of a Cantor set and a closed disc in the plane. That's connected and contains an open subset in its complement. Or what about just an example where the complement is two manifolds of differing dimensions, the disjoint union of which is not itself a manifold. Commented Apr 28, 2022 at 15:34