Let $K\subset \mathbb{R}^n$ be a closed set, then is there existing a smooth function $f\in C^{\infty}(\mathbb{R}^n,\mathbb{R})$, such that $$ (1)\quad f\ge 0, $$ $$ (2) \quad f^{-1}(0)=K. $$
Existence of a smooth function with a given kernel
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real-analysis
differential-topology
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0Yes: first deal with the case where $K$ is the complement of an open ball, then jump to the general case writing $K$ as a countable union of such sets. – 2012-10-12
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0@DavideGiraudo: Did you mean countable *intersection*? – 2012-10-12
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0@HaraldHanche-Olsen Yes (now I can't edit the commnet anymore). – 2012-10-12