Jordan–Brouwer separation theorem

Remarks

The proof uses Alexander duality in $S^{n+1} = \R^{n+1} \cup \set{\infty}$. It gives $\tilde{H}_0(S^{n+1} \setminus X; \Z) \cong \tilde{H}^n(X; \Z) \cong \Z$, with Čech cohomology on the right. So $S^{n+1} \setminus X$ has two path components, which are open, hence also its connected components. Removing the point $\infty$ from the component that contains it does not disconnect it, since $n + 1 \ge 2$. For $n = 1$ the theorem is the Jordan curve theorem in the plane.

Example: the unit sphere $S^n$ itself separates $\R^{n+1}$ into the open unit ball and the set $\set{x : \norm{x} > 1}$.

For $n \ge 2$ the theorem says nothing about the shape of the components. For the Alexander horned sphere in $\R^3$, the unbounded component is not simply connected, unlike the outside of a round sphere. The Schoenflies theorem, which says that the bounded component is an open disc, holds for $n = 1$ but needs extra hypotheses, such as local flatness, for $n \ge 2$. The case $n = 0$ is excluded: two points on the line $\R$ leave three components.

Going deeper

For a student

The theorem is one of the first applications of homology to geometry. It is used to prove invariance of domain, and through it the topological invariance of dimension. The contrast with the Schoenflies problem shows that homology can detect the number of components but not their homeomorphism type.