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> Is there a surface in $\mathbb R^3$ which is a closed subset and whose curvature is negative and bounded away from zero?
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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 surfaces 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 immersed, 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 :-)



Surfaces in $\mathbb R^3$ with negative curvature bounded away from zero
Efimov proved that there are no $C^2$ isometric immersions of complete surfaces with negative Guassian curvature bounded away from zero.


N.V. Efimov, "Imposibility of a complete regular surface in euclidean 3-space whose Gaussian curvature has a negative upper bound" Soviet Math. Dokl. , 4 : 3 (1963) pp. 843–846 Dokl. Akad. Nauk SSSR , 150 : 6 (1963) pp. 1206–1209


One reference I know for this is chapter 10 of the book of Han and Hong, "Isometric Embedding of Riemannian Manifolds in Euclidean Spaces."


Edit: Tilla Klotz Milnor's [paper](http://www.ams.org/mathscinet-getitem?mr=301679) "Efimov's theorem about complete immersed surfaces of negative curvature" is an exposition of this theorem in English. I've only looked at the introduction so far but it looks rather thorough.


