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Let $M$ be a compact ($\mathcal{C}^2$) hypersurface of $\mathbb{R}^k$. Then $M$ is the global level set of a function $f$ having $0$ as a regular value (see this related question).

I think that $f$ can be constructed using the Jordan-Brouwer theorem and the tubular neighborhood theorem. (Define $f$ to be a signed distance to $M$, negative in the bounded connected component of $\mathbb{R}^k \backslash M$.)

But is there a more elementary proof of this?

Olivier
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  • Well, you can't get it in too elementary a fashion. Obviously, you can construct $f$ locally and use a partition of unity to glue the local functions together, but you need to know the (co)normal bundle is trivial to be guaranteed that the exterior derivative of the resulting function is everywhere nonzero. – Ted Shifrin Apr 20 '17 at 18:55
  • By "hypersurface" do you mean dimension $k-1$? If so, then I think you are right, the answer is yes. – Lee Mosher Apr 20 '17 at 20:06
  • Yes, the dimension of $M$ is $k-1$. – Olivier Apr 20 '17 at 21:10

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No, there isn't a more elementary proof. This is because this because this fact and the Jordan-Brouwer theorem can somewhat easily be obtained one from the other. Here is a sketch of the Jordan-Brouwer theorem obtained from $M = f^{-1}(0)$.

Let $M = f^{-1}(0)$ for some smooth function $f : \mathbb{R}^k \rightarrow \mathbb{R}$ having regular value $0$, with $M$ connected. Using the tubular neighborhood theorem, it can be shown that $\mathbb{R}^k$ has at most $2$ connected components. Since $f$ has regular value $0$, the disjoint open sets $B = \{x \,|\, f(x) < 0\}$ and $N = \{x \,|\, f(x) > 0\}$ are non-empty. Thus they are the two connected components of $\mathbb{R}^k \backslash M$. Since $M$ is compact, it can be verified that one of them is bounded.

Olivier
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