Is there a surface in $\mathbb R^3$ which is a closed subset and whose curvature is negative and bounded away from zero?

And the small-print...

By surface I mean smooth surface without boundary, and by smooth I mean at least $C^2$. If one allows a boundary the question becomes silly, as a closed disc in a catenoid will do. The smoothness requirement is subtler, but we all know about the Nash-Kuiper theorem which gives, among many things, isometric embeddings of compact sufaces of negative curvature in $\mathbb R^3$ of class $C^1$.

I am looking for surfaces which are

*closed subsets*of $\mathbb R^3$. They will not be*closed surfaces*, though: pretty much every single textbook on the differential geometry of surfaces includes an exercise to the point that a closed surface in $\mathbb R^3$ has a point of positive curvature.Ideally, the surface is embedded. At least, though, it should be inmmersed, for otherwise one can easily find examples which are even of

*constant*negative curvature.Finally, the question is only interesting if the curvature is bounded

*away from zero*, for it is easy to produce examples of surfaces of negative curvature, like the catenoid.

Navigating between the Scylla and Charybdis of uninteresting cases is a pain :-)

closedin he sense that the surface is a closed subset of $\mathbb R^3$, not the usual one (compact and boundaryless) – Mariano Suárez-Alvarez♦ Nov 1 '12 at 3:58