About Whitney Theorem Note that $M$ of dimension $n$ can be imbedded differentiably as a closed submanifold of ${\bf R}^{N=2n+1}$. 
Here Let $f$ be an imbedding. $f$ is one-to-one immersion, that is, rank $n$, which is homeomorphic. 
Question : Here I have a question : If we allowed $N$ to be large, then $f : M\rightarrow f(M)$ can be isometric ? 
Thank you in advance. 
 A: First, to clear the terminological confusion: There is a theory of metric spaces and a theory of Riemannian manifolds (Riemannian geometry). Both have their own notion of a metric and an isometry and these two notions are quite different. See the discussion here about which distance functions are defined via Riemannian metrics. It is clear, however, from the question asked that it is about Riemannian metrics and Riemannian isometric embeddings. The answer is given by 
Nash embedding theorem.  If $g$ is a Riemannian metric of class $C^k$ on 
an $m$-dimensional $C^k$-smooth manifold, then there exists a $C^k$-smooth  isometric embedding $f: (M,g)\to R^N$, $N = m(m+1)(3m+11)/2$. Here $R^N$ is equipped with the standard flat Riemannian metric. 
The most subtle (and amazing) part of this theorem is that $f$ is as smooth as $g$ is. 
See here for the detailed discussion, including improvements and generalizations of this theorem. While the result itself, I believe, is not particularly useful  in Riemannian geometry, Nash's proof of his theorem was of tremendous importance in geometry and analysis (see here).  
