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



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> 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?
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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?



Nash isometric embedding for noncompact manifolds
Nash's theorem states that any smooth embedding $f$ with Lipschitz constant less than 1 can be approximated by smooth isometric embedding (if the dimension of Euclidean space is sufficiently large).
So you only need to find a proper embedding $f$.


Take any embedding with Lipshitz constant $<\tfrac13$, and add one more coordinte $z$ that is a proper function with Lipshitz constant $<\tfrac13$.
You get a smooth proper embedding $f$ with Lipschitz constant at most $\tfrac23$.


To construct $z$, you may smooth a distance function $\tfrac14\cdot \mathrm{dist}\_p$; for sure you need completeness.


