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Let $S$ be a simplex in $\mathbf R^n$ and let $\{S_i\}$ be a triangulation of $S$.

  1. The boundary of $S$ is defined as the union of the boundary faces of $S$. Is this union equal to the topological boundary of $S$ (with respect to the Euclidean norm topology)? The answer seems to be an obvious yes but I'm having trouble proving it. Any tips or references?

  2. It seems to be a well established fact, although I have never seen it proved anywhere, that each boundary face of a subsimplex $S_i$ of the triangulation is either contained in a boundary face of $S$ or is shared by exactly one other subsimplex $S_j$. I have been trying to prove this for a while now and it seems highly nontrivial. The proof I am constructing relies on the answer to 1 being yes. Any tips or references are welcome.

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For 2, does it help to remember that in a triangulated space (or simplicial complex) every point is in the interior of exactly one simplex? – wckronholm May 26 '11 at 23:35
This is just my ignorance showing, but what is a triangulation of a simplex? For example, of a tetrahedron in $\mathbb{R}^3$? My confusion is that a simplex is already triangulated, under one interpretation of "triangulation." – Joseph O'Rourke May 27 '11 at 0:20
@Joseph A triangulation of an $n$-dimensional simplex $S$ is a finite collection of $n$-simplices, called the subsimplices of the triangulation, whose union equals S and such that the intersection of any two of them is a common face. – echoone May 27 '11 at 0:46
@wckronholm That cannot be true since two subsimplices may share a face. – echoone May 27 '11 at 0:47
@echoone Consider a single triangle (i.e. a 2-simplex). This space has one 2-simplex, three 1-simplices, and three 0-simplices. Then a point is either in the interior of the triangle (whence the interior of the 2-simplex), on one of the edges (whence the interior of one of the 1-simplices), or a vertex (which by convention is the interior of itself). Do you see how this extends to larger complexes? – wckronholm May 27 '11 at 1:06

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