# A Jordan Separation Theorem for Polyhedral Surfaces

Let me begin by defining what a polyhedral surface is.

A path-connected subset $P$ of $\mathbb{R}^{3}$ is called a polyhedral surface iff it is the union of a finite collection $\mathcal{C}$ of polygons (possibly non-convex) that satisfies the following three conditions:

• The intersection of any pair of distinct polygons in $\mathcal{C}$ is exactly one of three things: (i) a common edge, (ii) a common vertex or (iii) $\varnothing$. (This implies that the interiors of the polygons in $\mathcal{C}$ are disjoint.)
• Any edge of any polygon in $\mathcal{C}$ is an edge of exactly one other polygon in $\mathcal{C}$.
• If an edge of a polygon in $\mathcal{C}$ intersects an edge of another polygon in $\mathcal{C}$ in a common vertex, then the two edges are also edges of a third polygon in $\mathcal{C}$.

I believe this to be a ubiquitously understood precise statement of what a polyhedral surface should be.

Question. Let $P$ be a polyhedral surface. Is there an analogue of the Jordan Separation Theorem that states that $P$ is the boundary of two path-connected open subsets of $\mathbb{R}^{3}$, one of which is bounded and the other unbounded?

The definition given above allows for a polyhedral surface that is homeomorphic to $\mathbb{S}^{1} \times \mathbb{S}^{1}$, the boundary of a $2$-torus, which is why I am imposing only path-connectedness and not simple-connectedness also.

I suspect that the way to proceed is to first prove that a polyhedral surface is indeed a topological surface (i.e., a topological $2$-manifold) that is closed (i.e., compact and without boundary). Then one can apply the Classification Theorem for Surfaces to prove that it must be homeomorphic to either a $2$-sphere or a finite connected sum of $2$-tori. However, I feel that I may be missing something out.

Thank you very much for your help!

• The third condition in the definition of a polyhedral surface currently reads: "If an edge of a polygon in C intersects an edge of another polygon in C in a common vertex, then the two edges are also edges of a third polygon in C." This seems too restrictive since it excludes vertices of valence greater than three. Lee Mosher's answer gives the correct condition. May 3, 2015 at 17:33

## 2 Answers

The magic words are: Alexander Duality. This was initially proved in the simplicial category, just as you state it, so all is well. Since Alexander's original (1915) paper is six pages long, you should just read it.

What you are calling a "polyhedral surface" allows for objects that are not topological 2-manifolds, because you have not imposed any condition on the link of a vertex. For instance, your definition allows for $P$ to be a union of two topological 2-spheres touching at exactly one point. This $P$ is a counterexample to your question.

If you impose the additional condition that the link of a vertex is a circle, then $P$ is indeed a topological 2-manifold and the answer of @IgorRivin kicks in.

• Indeed, I missed that! Apr 19, 2015 at 15:39
• Hi Lee. I noticed that! I was going to edit my post. Apr 19, 2015 at 15:58