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.
ClosedsurfacesSurfaces in $\mathbb R^3$ with negative curvature bounded away from zero
Is there a surface in $\mathbb R^3$ which is a closed (butnotcompact) subset and whose curvature is negative and bounded away from zero?
BysurfaceImeansmoothsurfacewithoutboundary, andbysmoothImeanatleast$C^2$.Ifoneallowsaboundarythequestionbecomessilly, as Willremindedme a littlewhileagocloseddiscinacatenoidwilldo.Thesmoothnessrequirementissubtler, butweallknowabout the catenoidNash-Kuipertheoremwhichgives,amongmanythings,isometricembeddingsofcompactsufacesofnegativecurvaturein$\mathbbR^3$ofclass$C^1$.
Iamlookingforsurfaceswhichareclosedsubsetsof$\mathbbR^3$.Theywillnotbeclosedsurfaces,though:prettymucheverysingletextbookonthedifferentialgeometryofsurfacesincludesanexercisetothepointthataclosedsurfacein$\mathbbR^3$ has apointofpositivecurvature.