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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?

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    $\begingroup$ The preimage $f^{-1}(c)$ is of course closed but can be rather wild. For instance, for $n=1$, an arbitrary closed subset in $\mathbb R$ has such a form. $\endgroup$
    – Sasha Anan'in
    Oct 11, 2013 at 21:13
  • $\begingroup$ If you mean in the case of $n=1$, let $M=f^{-1}(0)$, where $f:{\mathbb R}\to{\mathbb R}$ is $C^\infty$.Then the complement of $M$ is given by $f^2(x)>0$ $\dots$ $\endgroup$
    – Sasha Anan'in
    Oct 11, 2013 at 21:29
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    $\begingroup$ Well, the OP is changing his/her question more quickly than I am typing. I give up. $\endgroup$
    – Sasha Anan'in
    Oct 11, 2013 at 21:38
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    $\begingroup$ Anyway, the case is hopeless. The zero set of a $C^\infty$ function $f:\mathbb R^n\to\mathbb R$ can be any closed set, as Sasha said already (there is nothing special about dimension $1$ here). Should we close? $\endgroup$
    – fedja
    Oct 11, 2013 at 21:41
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    $\begingroup$ @fedja I guess, not many professionals know your answer. In this sense, the question is helpful. $\endgroup$
    – Sasha Anan'in
    Oct 11, 2013 at 21:52

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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).

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    $\begingroup$ Nice! (For $n=1$, the idea was similar, but your exposition is great and does not look as spoiling.) $\endgroup$
    – Sasha Anan'in
    Oct 12, 2013 at 15:33

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