# Triangulating open subsets of $\mathbb{R}^n$

This seems like something that is probably obvious to geometric topologists but I'm not so familiar with the theory of triangulation. Internet searches have not proved as fruitful as I had imagined. Open subsets of $$\mathbb{R}^n$$ are really nice" manifolds so surely the answer to the following question is yes, but I'd like to see a procedure for how to do it without appealing to much heavier machinery.

Q: Can all connected open subsets of $$\mathbb{R}^n$$ be triangulated? Can they be PL-triangulated?

I know that arbitrary 2-manifolds and 3-manifolds can be triangulated (implying a positive answer to my question in these dimensions) but in higher dimensions things get more subtle. I'm hoping to hear that I can PL-triangulate any connected open subset $$M$$ of $$\mathbb{R}^n$$, $$n\geq 2$$ so that as points in $$M$$ approach $$\partial M$$, the diameter of the simplices containing these points tends to $$0$$.

A reference, specific to this fundamental case would be ideal.

• A guy named Brian removed my answer. The guy must have problems. Anyhow, yes, every open set can be triangulated. You can always subdivide into almost disjoint dyadic cubes (there is a Wikipedia page on that). Then everything drops down to triangulate a cube. And this is easy. Best. – Kosh Feb 6 '20 at 21:44
• @kosh I understand what a dyadic cube is and it feels like it's in the right direction but the wiki page on dydaic cubes doesn't say much about triangulation. I've asked a very specific question and an answer should come with a specific construction. Sure, if you have a cover of $M$ by closed cubes (dyadic or not) with disjoint interiors and where each cube intersects at most finitely other cubes, then the conclusion is obvious. So how do you propose to build such a cover? – J.K.T. Feb 6 '20 at 22:26
• – Kosh Feb 6 '20 at 23:05
• What do you mean by "PL-triangulation"? This is a terminology I have not heard before, and when I search for it, I get "triangulation in Polish". But actually, what do you mean by "triangulation" itself. The method @kosh suggests provides a covering of the set by a countably infinite number of geometric hyper-triangles, but then you talk about triangulation on manifolds, where the edges are allowed to be curved hyper-surfaces. And if you can do that, then you can do open sets in $\Bbb R^n$ in a finite triangulation. – Paul Sinclair Feb 7 '20 at 2:31
• @PaulSinclair en.wikipedia.org/wiki/Triangulation_(topology) A triangulation is a choice of homeomorphism with a simplicial complex. It's PL if all links are spheres. This is standard terminology in topology. – J.K.T. Feb 7 '20 at 3:34

## 1 Answer

Yes, every open subset $$\Omega$$ of $$\mathbb{R}^n$$ can be triangulated. Indeed, it is always possible to subdivide $$\Omega$$ into closed cubes with disjoint interiors and where each cube intersects at most finitely other cubes. Then everything drops down to triangulate cubes.

On can take a look at the Wikipedia page on the Whitney covering lemma and to the references therein: