Remark 3 in this text claims that every polyhedron admits a triangulation, but this is not obvious to me and I was not able to find a proof that I could understand (only some complicated computer science algorithms).
Does someone know an easy proof?
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Have you found an answer since the time you asked this? If so, would you mind posting it? – Blazej May 08 '20 at 20:20
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@Blazej afraid not. I ended up assuming that $K$ was compact which was always the case for the application I needed it for. That way we can cover $K$ by a finite number of simplices and then use barycentric subdivision to turn that collection into a simplicial complex. – user388557 May 09 '20 at 09:11