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:
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.