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Setting:

I consider a two times continuously differentiable function $H:\mathbb R^{2n}\to \mathbb R$ and a regular value $\beta\in\mathbb R$ such that $N=H^{-1}(\beta)$ is compact and connected.

Claim:

$\mathbb R^{2n}\setminus N$ has a bounded component $B$ and an unbounded component $A$.

My attempt:

It is well known, that $N$ is a $2n-1$ dimensional submanifold of $\mathbb R^{2n}$, i.e. using the implicit function theorem. And since $N$ is connected, we can see it as one piece.

But I'm curious why $\mathbb R^{2n}\setminus N$ has two components. Why can't $N$ be a manifold with boundary like $[0,1]\times\{0\}$ for the case $n=1$? Or some strange construction, since $N$ is just connected but not path connected?

We can also consider a compact connected $m-1$ dimensional submanifold of a $m$ dimensional space. Can we conclude that it has to separate the space in a bounded and an unbounded set? Since it is used, without any explanation or arguments, in a paper it seems to be obvious for the author, but not for me.

Mundron Schmidt
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    Manifold with boundary is not a manifold in a strict sense. The regular value theorem gives a submanifold (no boundary). Every compact connected codimension 1 submanifold in $E^k$ separates $E^k$ into exactly two components: One bounded and one unbounded. – Moishe Kohan Nov 03 '17 at 15:37
  • @Moishe Cohen I see that it can't be a manifold with boundary. But do you have a reference that a compact connected codimension 1 submanifold separate space in a bounded and an unbounded set? – Mundron Schmidt Nov 03 '17 at 16:27
  • https://math.stackexchange.com/questions/1712071/differentiable-version-of-the-jordan-brouwer-separation-theorem, https://math.stackexchange.com/questions/1485090/generalized-jordan-brower-separation-theorem?rq=1, https://math.stackexchange.com/questions/85168/higher-dimensional-jordan-brower-separation-theorem?rq=1 (and many more). – Moishe Kohan Nov 03 '17 at 16:32

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