The following statement seems to be taken as given in papers I'm reading:



> 
> Let $\mathcal{M}^n$ be a compact, embedded hypersurface in $\mathbb{R}^{n+1}$. Then $\mathcal{M}$ is the boundary of some compact domain $\Omega \subset \mathbb{R}^
> {n+1}$.
> 
> 
> 


Is this an elementary result? I feel there must be some algebraic topology argument here. Any suggestions?



Compact Hypersurfaces Bounding Compact Domains
There are several ways to see this fact, which is a simple instance of [Alexander duality](https://secure.wikimedia.org/wikipedia/en/wiki/Alexander_duality).
Here is the simplest I know.


Let $H$ be a compact smooth hypersurface in $\mathbb{R}^n$, whithout boundary, and $x\in \mathbb{R}^n\setminus H$.
Then the radial projection $p\_x : H \to \mathbb{S}^{n-1}$, $y\mapsto (y-x)/\|y-x\|$ has a degree mod $2$, say $d\_x$, which may be defined as the number of elements mod $2$ of $p\_x^{-1}(u)$ for almost all $u\in\mathbb{S}^{n-1}$ (this is well-defined by transversality theory). Then the subset of $x\in \mathbb{R}^n$ such that $d\_x=1$ is your $\Omega$, a relatively compact open set with boundary $H$.
EDIT : here one must assume $n>1$, as in Mohan's answer, otherwise the relative compactness might not hold.


