# 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. – Allen Hatcher May 3 '15 at 17:33

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.