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We know if $\Omega \subset \mathbb{R}^{n}$ is a bounded $C^k$ domain, then its boundary $\partial\Omega$ is a $C^k$ compact hypersurface of dimension $n-1$.

Is it true that every $m-$dimensional compact hypersurface is the boundary of a bounded domain $\Omega \subset \mathbb{R}^{m+1}$??

Presumably the $C^k$ smoothness of the hypersurface implies $C^k$ smoothness of the domain.

Definition of hypersurface: A set $\Gamma \subset \mathbb{R}^{n+1}$ is a $C^k$-hypersurface if for each $x \in \Gamma$, there is an open set $U \in \mathbb{R}^{n+1}$ containing $x$ and a $C^k$ function $\phi$ such that $$U \cap \Gamma = \{ y \in U \mid \phi(y) = 0\}$$ and $\nabla \phi(y) \neq 0$ for all $y \in U \cap \Gamma$.

Definition of $C^k$-domain: enter image description here

Abbre
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Here is a proof when the compact hypersurface is connected.

A compact hypersurface $H$ in $\mathbb R^n$ is orientable, hence if it is connected then $$H_{n-1}(H)\cong \mathbb Z$$ By Alexander duality, the zeroeth reduced cohomology group of the complement of $H$ in $S^n$ is free abelian of rank $1$, hence the hypersurface divides $S^n$ (considered as $\mathbb R^n$ with an additional point) into two components. $H$ is equal to the boundary of each of these components, which are both open sets. When we pass back to $\mathbb R^n$ via stereographic projection, one of these components will be bounded and the other unbounded. The hypersurface bounds the bounded component, so we are done.

Matt Samuel
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  • Any particular reason for the downvote? – Matt Samuel Jan 15 '15 at 21:23
  • I didn't downvote; but perhaps it was felt the smoothness issue wasn't (adequately) addressed? – Andrew D. Hwang Jan 15 '15 at 21:35
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    @user86418 It's pretty clear that a domain with a $C^k$ boundary is a $C^k$ domain from the definition. – Matt Samuel Jan 15 '15 at 21:53
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    Hey thanks for the answer. It is hard to understand due to unfamiliar content but I just wanted confirmation anyway. – Abbre Jan 15 '15 at 22:03
  • @Abbre you're welcome. You can come back and look at it after you've learned some more algebraic topology. I'm pretty sure there's no easy way to prove it without some amount of algebraic topology, it's a notoriously difficult problem that world famous mathematicians in the 19th century published incorrect proofs for for $n=2$. – Matt Samuel Jan 15 '15 at 22:11
  • @MattSamuel Yes I think you're right, there is no easy proof. – Abbre Jan 15 '15 at 22:20
  • @MattSamuel: It was rash (of me) to attempt to guess the mind of a down-voter. :) – Andrew D. Hwang Jan 15 '15 at 22:36
  • @user86418 Here is a more reasonable guess: I forgot to specify that for this proof to work the hypersurface has to be connected. – Matt Samuel Jan 15 '15 at 23:53
  • Every manifold wich is boundary of an open set is orientable, as i was reading here https://math.stackexchange.com/questions/377236/inducing-orientations-on-boundary-manifolds, so do you think should we also ask for the manifold to be oriented? – Alicia Basilio Oct 21 '17 at 18:38
  • @Alicia Sad thing is that this stuff is growing foggy in my memory. I was a year out of grad school when I wrote this and my focus was not topology. I understand it at least (it hasn't been that long!) but I wouldn't have written this now. You're free to edit my ancient answer. – Matt Samuel Oct 22 '17 at 02:50