# Definition of a polyhedral region

I believe the following two conditions on a subset $S$ of $\mathbb{R}^3$ may be equivalent. I would like to know if they are equivalent, and where I can find either a counterexample or a proof of their equivalence.

(1) $S$ is a finite union of closed (solid) tetrahedra.

(2) $S$ is a bounded set equal to the closure of its interior, whose boundary consists of a finite union of triangles.

If possible, I would like to find a proof in a reference. If it turns out that (1) and (2) are not equivalent, but they can be corrected slightly so as to become equivalent, then please say so.

Edit. I'd like to thank the two people who have already answered my question, who clearly put a lot of thought into their answers. However, as I stated above, I would really like to find a proof (ideally a detailed one) in a published reference.

• In (1) I guess by a tetrahedron you mean a solid tetrahedron. If so, then they seem equivalent: Roughly speaking, the finiteness condition in (1) corresponds to the compactness in (2), and the condition that the interior is dense in (2) corresponds to the union in (1). Note that dense interior means that $S$ contains no triangle which is not part of the boundary of a tetrahedron. By the way, you might want to check out simplexes and simplicial complexes from, for instance, Armstrong's Basic Topology. – Alp Uzman Jan 4 '16 at 22:31
• @A.AlpUzman Thanks. I've edited the question to clarify that in (1) I have solid tetrahedra in mind. – David Jan 4 '16 at 22:40

One clean way to show (1) from (2) is to show that any set satisfying (2) is the finite union of shapes which are like cylinders where the end faces are not assumed parallel - but which is a simple polyhedron easily writable as a union of tetrahedra. In particular, take $T$ to be a finite set of (closed) triangles composing the boundary of $S$. For convenience, let us assume that the elements of $T$ has disjoint interiors since, if they do not, we can choose a different collection $T$ by dividing each triangle into several pieces, split along its intersection with other triangles' interiors.

Next, let us take some vector $v$ not parallel to any face and corresponding projection $\pi:\mathbb R^3\rightarrow\mathbb R^2$ with rank $2$ taking $v$ to $0$. This projection takes every triangle in $T$ to a non-degenerate triangle in the plane. Let us define, for convenience, the edges of our shape as $\partial T = \bigcup_{t\in T}\partial t$. Now, consider the projection $\pi(\partial T)$ and the set $K$ of bounded, connected components of $\mathbb R^2\setminus \pi(\partial T)$.

We do this because these components $k\in K$ have useful properties. Firstly, it is obvious that each $k$ is a polygon so $\pi^{-1}(k)$ is a polygonal prism (of infinite length). Secondly, for any distinct triangles $t,t'\in T$ we may define an order by $t>t'$ if every element of $t\cap \pi^{-1}(k)$ is writable as an element of $t'\cap\pi^{-1}(k)$ plus a positive multiple of $v$. This works since one may always write an element of $t\cap \pi^{-1}(k)$ as an element of $t'\cap\pi^{-1}(k)$ plus a multiple of $v$, and if the sign of this multiple were not constant over $k$, at some point it would be $0$ indicating an intersection of $t$ and $t'$ in their interiors, which we forbade.

Finally, take every element $k\in K$ and the sequence $t_1,t_2,\ldots,t_n\in T$ to enumerate the members of $T$ intersecting $\pi^{-1}(k)$. This is ordered such that $t_1>t_2>\ldots>t_n$. All of the following discussion will assume $\pi^{-1}(k)$ as the domain rather than $\mathbb R^3$. Firstly, one may quickly verify that the set $\pi^{-1}(k)\setminus \partial S$ then has $n+1$ components: the points above $t_1$ (i.e. writable as a member of $t_1$ plus a positive multiple of $v$), the points below $t_1$ but above $t_2$, the points below $t_2$ but above $t_3$, and so on until we get to the points below $t_n$. Moreover, if any part of such a component is within $S$, the whole component must be as they are connected sets disjoint from the boundary. Furthermore, the first and last sets clearly are not in $S$ being unbounded. However, if the set of points below $t_i$ but above $t_{i+1}$ was in $S$ then set of points below $t_{i+1}$ but above $t_{i+2}$ must not be in $S$ - and vice versa. This is necessary to ensure that the boundary truly is the boundary of the closure of the interior.

This brings us to important fact: $S\cap \pi^{-1}(k)$ is the union of the set of points below $t_1$ but above $t_2$ with those below $t_3$ but above $t_4$ and so on (note that $n$ will necessarily be even). Each of these sets is merely the intersection of an infinite prism with two half planes - and, moreover, the closure of this set is within $S$ and is easily writable as a union of tetrahedra.

Then, we only need to assure ourselves that the union of these "prisms" is $S$. We already know that this is true, except possibly on $\pi^{-1}(\pi(\partial T))$ - but this is of no consequence since both the sets in question are the closure of their interior and $\pi^{-1}(\pi(\partial T))$ is a finite union of sections of planes, which will vanish if we take the interior of the complement of its union with or subtraction from any other set.

The proof to show (2) from (1) is more mundane so I won't go into as much detail - more or less, a reasonable strategy to take is to notice that the boundary of a union is a subset of the union of the boundaries, so the proof is basically to take every triangle on the boundary of a tetrahedron in the union and write out the constraints on which parts of the triangle will remain in the boundary after the union. The point then is that, each time we add a tetrahedron to the union, it may remove a region defined by finitely many half planes from any triangle on the boundary, but doing this to any triangle still leaves a finite union of triangles remaining on the boundary.

A possible approach for the direction (2) => (1): The idea is to formalise what you would geometrically do: draw connecting lines between boundary points until all of $S$ is covered with tetrahedrons (try it in two dimensions for example!)

Step one: Prove that the set of boundary triangles can be choosen to be a triangulation of the boundary, i.e. any two triangles either are equal, intersect in a common edge, a common vertex or not at all. I think this can be done by an induction over the original number of triangles + some fiddling with the triangles. The idea that I have in mind is that any other intersection that can occur gives rise to a possibility to divide one of the triangles into a finite number of smaller triangles that do satisfy our requirement.

Step two: Take all vertices of boundary triangles. This is a finite set $V=V_0$ of points. Take all connecting lines between any two of those points and add all new intersections to our set of points to obtain a new (finite) set $V_1$. Repeat until no new points were added. This is again a finite set and it gives us a triangulation of $conv(V_n)$.

Step three: $S\subseteq conv(V_n)=conv(V_0)=conv(\partial S)$. Now we have a triangulation of $conv(\partial S)$. Let $T$ be one of those tetrahedrons and consider $T\cap \partial S$. I claim that this is a (possibly empty) union of faces of $T$. Since $T$ is convex, if $T\cap \partial S$ contains any points at all, it contains the convex hull of those points too. It is now easy to see if $T\cap \partial S$ is not a union of faces of $T$, then the construction in step 2 would not have ended with $V_n$.

Step four: Now take all tetrahedrons that intersect the interior of $S$. I claim that their union is exactly $S$.

A fact almost identical to (2) -> (1) (except for some connectedness conditions, irrelevant here) is proved in outline by A.D. Aleksandrov in the expository (and polemical) article “Čto takoe mnogogrannik?” (What is a polyhedron?), appearing in Matematika v škole, 1981, no. 2. Here is the relevant part of the proof, on page 24.

If the polyhedron is convex, then it can be divided into convex pyramids by drawing, from any interior point, segments to all points of each face. Partitioning the faces into triangles, we in turn divide the pyramids into tetrahedra.

Now let a non-convex polyhedron be given. Draw planes containing all of its faces; they partition the polyhedron into parts each of which is a convex polyhedron, since each part is the intersection of finitely many half-spaces bounded by the planes drawn.

By dividing the faces of these polyhedra into triangles and applying the previous construction to each one, we obtain a partition of the entire original polyhedron into tetrahedra with the required properties.

Incidentally, the "required properties" are as follows: "Any two tetrahedra either have no points in common or have only one common vertex, one common edge, or one common face"; and a connectedness property not applicable under the hypotheses of the OP (namely, that one can pass from one tetrahedron to any other through a sequence of tetrahedra in which any two consecutive ones have a common face).