Every metric space has a completion, and a Riemannian manifold has a structure of metric space.
Is it possible to extend a Riemannian manifold an that this extension keeps being a Riemannian manifold?
I think that the answer is no, but is there any easy counterexample? At first, I thought about the cone without vertex as submanifold of $\mathbb{R}^3$ which is not complete and if you complete it you lose the smooth structure, but possibly this is not a valid one, because I thought about the cone (with vertex) keeps being a manifold regarded as an abstract one.