TITLE: Does every smooth manifold of infinite topological type admit a complete Riemannian metric?
QUESTION [6 upvotes]: To elaborate a bit, I should say that the question of the existence of a complete metric is only of interest in the case of manifolds of infinite topological type; if a manifold is compact, any metric is complete, and if a noncompact manifold has finite topological type(ie is diffeomorphic to the interior of a compact manifold with boundary,) one can contruct a complete metric on the manifold with boundary via a partition of unity, and then divide by the square of a defining function to get an complete asymptotically metric on the interior.
I have absolutely no intuition for how "wild" these manifolds can be. The only examples I can think of are infinite connected sums and quotients of negatively curved symmetric spaces by sufficiently complicated groups, but I'd imagine that one can construct some pathological examples by limiting arguments.

REPLY [3 votes]: We can even prove more: every smooth manifold can be equipped with a Riemannian metric of bounded geometry (positive injectivity radius, bounded curvature tensor and all derivatives of the curvature tensor are also bounded).
This is Theorem 2' in "R. E. Greene, Complete metrics of bounded curvature on noncompact manifolds, Archiv der Mathematik 31 (1978), no. 1, 89-95".