Suppose we are given a smooth function $f\colon \mathbb{R}^n \rightarrow \mathbb{R}$ and some number $c$. What can be said about the preimage $f^{-1}(c)$.


There's the theorem on regular preimages, asserting that if $\nabla f$ is nowhere vanishing on $M=f^{-1}(c),$
then $M$ is a smooth submanifold of dimension $n-1$. 


But now lets assume that one does not have a regular preimage. Can one say anything about the preimage here (except what would already follow from continuity), or is this a hopeless case?



Preimage of a smooth function
I do not mean to spoil the exercise, but note that if $\phi$ is any smooth function on a Banach space $E$, with bounded derivatives of any order (say $\|D^j\phi\|\_\infty < \infty$ for any $j\ge0$), then $f(x):=\sum\_{k=1}^\infty d\_k^k \phi\big(\frac{x-a\_k}{d\_k} \big)$ certainly defines a smooth function for any choice of a sequence of vectors $ a\_k\in E$, and of numbers $0 < d\_k \le 1/2$. To be able to make the zero-set of $f$ whatever closed set $C\subset E$ you like, just requires $E$ to be separable, and possess a smooth $\phi$ with bounded derivatives and bounded support (and non-zero, of course, hence wlog also non-negative).


