Is the following true?


Let $\Sigma$ be a compact orientable hypersurface without boundary in $R^n$. Then $R^n\setminus\Sigma$ has at least two connected components. 



Alexander duality theorem
Yes: whenever $U \subset M$ is an open subspace with complement $Z$, then there is a long exact sequence
$$ \ldots\to H^\bullet\_c(U) \to H^\bullet\_c(M) \to H^\bullet\_c(Z) \to H^{\bullet+1}\_c(U)\to \ldots $$
which in this case gives 
$$ H^{n-1}\_c(\mathbf R^n) = 0 \to H^{n-1}\_c(\Sigma) \to H^n\_c(\mathbf R^n \setminus \Sigma) \to H^n\_c(\mathbf R^n) \to 0.$$
By Poincaré duality we have $H^n\_c(\mathbf R^n \setminus \Sigma) \cong H^0(\mathbf R^n \setminus \Sigma)^\vee$ and $H^{n-1}\_c(\Sigma) = H^0(\Sigma)^\vee$, so the ranks of these cohomology groups are just the numbers of connected components of $\mathbf R^n \setminus \Sigma$ resp. $\Sigma$. Since $H^n\_c(\mathbf R^n)$ is one-dimensional this shows that $\mathbf R^n \setminus \Sigma$ has exactly one more connected component than $\Sigma$. Compactness of $\Sigma$ was an unnecessary hypothesis.


