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Suppose that $f : \mathbb{R} \rightarrow \mathbb{R}$ is such that $f$ is continuously differentiable and non-constant. Is it true that for any $c \in \mathbb{R}$, the level set $\{f=c\}$ contains only isolated points? Is at most countable? I feel like the implicit function theorem has something to say about this, but I can't seem to connect the dots. Thanks!

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    $\begingroup$ It may be large too. Say $f(x) = x^2$ for positive $x$ and $0$ for negatives. $\endgroup$ – Orest Bucicovschi Oct 13 '17 at 17:07
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    $\begingroup$ By a theorem of Whitney, any closed subset of $\Bbb R$ is the zero set of some infinitely differentiable function. $\endgroup$ – Lord Shark the Unknown Oct 13 '17 at 17:11
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    $\begingroup$ Thanks a lot for the comments! $\endgroup$ – Mark Oct 13 '17 at 17:17
  • $\begingroup$ It can contain isolated points or be countably or uncountably infinite. See the last question of dpmms.cam.ac.uk/study/IA/AnalysisI/2008-2009/… $\endgroup$ – Teddy38 Oct 13 '17 at 17:26
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For isolated points and countably infinite ones I think you can find examples no problem. For the uncountably infinite one, try

$$f(x)=\exp(-1/x^2)\text{ if }x\geq 0\text{ and }f(x)=0\text{ if }x<0\ .$$

It shouldn't be too difficult to prove that the function is infinitely differentiable at $x=0$.

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