But anyway, lets go back to our tet. problem. In deed there is a reg, tet. that can fill space by it?s self, it is the dysphenoid, and it is 1/4 of the oblate octahedron which also can fill space by it? self. This however is a unique case, we see that it is possible to fill space simply with a single kind of block, rather it is a tet. or a fraction of octahedron or a space filling octahedron on it?s own, the building blocks are one and the same unique to this case.
I come to realize, that if a tetrahedron any infinite number of tet. is divided in to 8 vol. identical tet. as it is also true in the UNIQUE case I just have described, there always going to be a inner octahedron equaling to 4 tet. so 8 total, and if I can have a infinite amount of these tet. we would fill space, and this space would be a infinite large tet. , since every tet. has a inner octahedron the large picture would have to be a infinitely large tetrahedron. The point is that I CAN FILL SPACE WITH ALL THE INFINITE NUMBER OF TETHEDRONS, this method, obeys the question. Please if anyone would comment or correct me if I?m wrong, but it seams to me There is actually no tetrahedron than can NOT fill space in this fashion. It seams harder to solve when we reverse the question and state ?Is there a tet. that can NOT fill space?? I think, NO THERE IS ACTUALLY NO TETRAHEDRON THAT CAN NOT FILL SPACE, because all tetrahedrons contain a inner octahedron thus containing ! a companion tetrahedron that allows space filling.