Show that every compact hypersurface in $\mathbb{R}^n$ is orientable. HINT: Jordan-Brouwer Separation Theorem.
This is an exercise from Guillemin and Pollack. So hypersurface means smooth hypersurface. Jordan's Theorem says that if $S$ is some hypersurface in $\mathbb{R}^n$, then $\mathbb{R}^n\setminus S$ has two connected components (and their common boundary is $S$).
But I don't really see how this implies that $S$ is orientable. Would anyone give me some help?
Thank you.
Let $x\in S$, and $u$ a vector of $T_xR^n$ orthogonal to $T_xS$, you can suppose that $\|u\|=1$ ( for the Euclidean norm) and $u$ points towards the component $U_1$ of $R^n-S$. This defines uniquely $u$ and you obtain a non trivial vector field $X(x)=u$. $i_X\omega$ restricted to $S$ is a volume form, where $\omega$ is the canonical volume form of $R^n$.