TITLE: Does a Riemannian manifold with bounded geometry admit an isometric proper embedding into Euclidean space with uniformly thick tubular neighborhood
QUESTION [18 upvotes]: Suppose $(M,g)$ is an open Riemannian manifold with bounded geometry, i.e., the injectivity radius is $\ge \epsilon>0$ and each iterated covariant derivative of curvature is bounded with respect to $g$.
Question: Does there exist an embedding into some high dimensional $\mathbb R^N$ with the following properties:

$\bullet$ isometric

$\bullet$ proper (i.e., compact sets have compact inverse images)

$\bullet$ The normal tubular neighborhood contains a uniformly thick disk in each fiber.
The Gauss equations give a relationship between the (normal bundle valued) second fundamental form (or shape operator); but by embedding arc-length parameterized curves one can show that this is not enough to force focal points uniformly away. 
There is a related question 
(here), but it does not answer this question.
By taking a product with another Euclidean space one can then have embedding where the normal bundle is trivial. Then one can use such embedding to carry over results valid on $\mathbb R^n$ to Riemannian manifolds with bounded geometry. 
Edit: Many thanks for the comments and Anton for a very succinct answer.
I end with a wild guess:
Maybe, one can characterize the Riemannian manifolds of bounded geometry admitting such embeddings as those whose ends are asymptotically flat times something compact (by which I mean an bundle)?

REPLY [23 votes]: It does not hold for hyperbolic plane.
It follows since the volume growth of the hyperbolic plane is exponential, while volume growth of $\mathbb{R}^N$ is polynomial.