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It's well known that smooth embedded submanifold in $\Bbb{R}^n$ locally a level set (and locally is a graph), as in the thread Is every embedded submanifold globally a level set? setting up.

What if we consider only topological submanifold in $\Bbb{R}^n$? For the "locally graph" statement I see rectangles is a counterexample for the corner point cannot be locally graph. But for the "locally level set" statement I find no obivious counterexample.

onRiv
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    Every closed subset is a level set of some continuous function. – Moishe Kohan Nov 14 '23 at 17:18
  • @MoisheKohan Thank you very much. I get it now. (The linked thread contains a comment mentioning the strong Uryshon lemma too. I didn’t notice that) – onRiv Nov 15 '23 at 05:44

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Recall that given a topological spare $X$, a subset in $X$ is said to be precompact in $X$ if its closure in $X$ is compact.

And Lemma 1.10 in [LeeSM]: Every topological manifold has a countable basis of preompact coordinate balls.

Now given a topological manifold $M \subseteq \mathbb{R}^N$. From the above lemma, we have (countable many) open sets $U \subseteq \mathbb{R}^N$ covering $M$ such that $U \cap M$ is heomorphic to $\mathbb{R}^m$ and $\overline{U \cap M}$ is compact in $M$.

Then $\overline{U \cap M}$ is also compact in $\mathbb{R}^N$.

Hence $\overline{U \cap M}$ is closed (and bounded) in $\mathbb{R}^N$. By the strong Urysohn lemma, there is continuous function $$ f: \mathbb{R}^N \rightarrow[0,1] $$ such that $f(M)=0$ and $f>0$ otherwise.

Hence $U \cap M$ is a level set of $f$ in $U$.

onRiv
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