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 two 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} $.

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!

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!