It seems that the smooth isometric embedding theorem by Nash is true also for noncompact manifolds.

Is it true that any (complete, connected) Riemannian manifold $(M^n,g)$ admits a proper smooth isometric embedding $\iota:M^n\hookrightarrow\mathbb R^N$ into some Euclidean space?

This is equivalent to ask that $\iota(M)$ is a closed subset of $\mathbb R^N$.

Remark. It is pretty trivial to modify the proof of Whitney's embedding theorem to obtain that any manifold admits a proper embedding into some Euclidean space (if we don't care about obtaining the best dimension for which this is possible...). But of course completeness is a necessary condition to find a proper isometric embedding$\iota$. Is completeness also sufficient?

Are weaker results known? E.g. is it true if $\operatorname{inj}(M)>0$ and/or the sectional curvature $\operatorname{sec}(M)$ is bounded?

It seems that the smooth isometric embedding theorem by Nash is true also for noncompact manifolds.

Is it true that any (complete, connected) Riemannian manifold $(M^n,g)$ admits a proper smooth isometric embedding $\iota:M^n\hookrightarrow\mathbb R^N$ into some Euclidean space?

This is equivalent to ask that $\iota(M)$ is a closed subset of $\mathbb R^N$.

Remark. It is pretty trivial to modify the proof of Whitney's embedding theorem to obtain that any manifold admits a proper embedding into some Euclidean space (if we don't care about obtaining the best dimension for which this is possible...). But of course completeness is a necessary condition to find a proper isometric embedding  $\iota$. Is completeness also sufficient?

Are weaker results known? E.g. is it true if $\operatorname{inj}(M)>0$ and/or the sectional curvature $\operatorname{sec}(M)$ is bounded?