Boundary triangulation induces triangulation In $R^n$ (the real space) we have an open connected set $D$, such that $\partial D$ is triangulable. Can we prove the closure $\bar{D}$ is triangulable or any counterexample?
Furthermore, the $\partial D$ are piecewise algebraic in the question I am considering, I do not know whether this would be helpful for the above statement.
Thanks for any help.
 A: In many categories, the answer is known to be yes, see
Emil Saucan, MR 2184196 Note on a theorem of Munkres, Mediterr. J. Math. 2 (2005), no. 2, 215--229.
A: Although you do not seem to require that the triangulation of the closure to
be compatible with the triangulation of the boundary, it is
true in $\mathbb{R}^3$ that a triangulated polyhedron $P$ has a compatible
interior tetrahedralization. Bern proved that, if $P$ has $n$ vertices,
such a tetrahedralization by $O(n^2)$ tetrahedra exists (and can be found quickly):

Bern, Marshall. "Compatible tetrahedralizations." Fundamenta Informaticae 22.4 (1995): 371-384. (ACM link.)

In fact, he proved all of $\mathbb{R}^3$ can be tetrahedralized compatible with
$P$'s surface triangulation (with some tetrahedra having a vertex at $\infty$).
It is interesting that if you change "triangulable" to "hexahedral-able,"
and ask if the surface mesh can be extended compatibly to an interior mesh, 
the answer is unknown:

"No algorithm is known to construct
  hexahedral meshes compatible with an arbitrary given quadrilateral
  mesh, or even to determine when a compatible hex mesh
  exists, even for the simple examples shown in Figure 1"


         



Erickson, Jeff. "Efficiently hex-meshing things with topology." Discrete & Computational Geometry 52.3 (2014): 427-449. (PDF download.)

