Simplex

A linear surface is constructible through any n regions in Rⁿ. According to the proof in the last post, the only reason a linear surface is not constructible through n+1 regions is that the regions are vertex regions of a simplex. To me that reinforces the importance of simplexes. The proof I supplied for the "Simplex Theorem" in the first post was insufficiently precise. Here is a better proof, preceded by supportive definition, lemma, and theorem. Subspace and region were defined in the first post. Here I define subregion.

If Sⁿ is a subspace of Qⁿ, and R is a region of Sⁿ, then region P, in Qⁿ, is a subregion of R iff R(b)=P(b) for each boundary b in Sⁿ.

Intermediate Region Lemma

If Sⁿ is a subspace of linear space Qⁿ, R is a region in Sⁿ, b extends Qⁿ to a linear space, and b divides R, then b divides a subregion of R in Qⁿ.

Consider the section of Sⁿ by b, S^n-1. Let P be the region in S^n-1 isomorphic to R. By transitivity of isomorphism, the section of Qⁿ by b is isomorphic to an extension of S^n-1. Thus, the subregions of R are isomorphic to the subregions of P. Since at least R is a subregion of R, there is a subregion of P. Therefore, b divides a subregion of R.

Intermediate Region Theorem

If linear space Sⁿ extends to Qⁿ, region R in Sⁿ extends to regions {R₀, R₁, R₂, ...} in Qⁿ, R_a and R_b in {R₀, R₁, R₂, ...} are on opposite sides of both b₀ and b₁, Qⁿ extended by b₂ is linear, and b₂ divides R_a and R_b, then there exists R_c in {R₀, R₁, R₂, ...}, such that b₂ divides R_c, R_c is opposite one of b₀ or b₁ from R_a, and R_c is opposite the other of b₀ or b₁ from R_b.

Suppose b₀ divides R into R₁ and R₂, and b₁ divides R. By the lemma, b₁ divides R₁ or R₂. First suppose b₁ divides only one of R₁ or R₂. Suppose without loss of generality that b₁ divides R₁ into R₃ and R₄. Again without loss of generality, assume R_a is a subregion of R₂ and R_b is a subregion of R₄. Consider the section, S^n-1, of Sⁿ by b₂. Let P, P₂, P₄ in S^n-1 be isomorphic to R, R₂, R₄ in Sⁿ, respectively. Since P is divided by two boundaries, isomorphic to b₀ and b₁, it has at least three subregions, two of which are P₂ and P₄, and the third is isomorphic to R₃. Therefore, b₂ divides R₃. By the lemma, there is subregion R_c of R₃ divided by b₂, completing the proof for the case that b₁ divides only one of R₁ or R₂. Next assume b₁ divides R₁ into R₃ and R₄, and b₁ divides R₂ into R₅ and R₆. Assume without loss of generality R_a is a subregion of R₃ and R_b is a subregion of R₆. Consider the section, S^n-1, of Sⁿ by b₂. Let P, P₃, P₆ in S^n-1 be isomorphic to R, R₃, R₆ in Sⁿ, respectively. Since P is divided by two boundaries, isomorphic to b₀ and b₁, it has at least three subregions, two of which are P₃ and P₆, and the third is isomorphic to either R₄ or R₅. Therefore, b₂ divides either R₄ or R₅. Assume without loss of generality that b₂ divides R₄. By the lemma, there is subregion R_c of R₄ divided by b₂, completing the proof.

Simplex Theorem

If b divides the vertex regions of a simplex in linear space, Sⁿ, then Sⁿ extended by b is not linear.

If the vertexes of a simplex are divided by a boundary, b, then b divides the vertexes of the restriction of Sⁿ to the simplex. Thus, I need only prove the theorem for Sⁿ consisting of a simplex alone. Proceed by induction on the dimension of Sⁿ. If n=1, no boundary divides the vertex regions because no boundary divides more than one region, and a simplex has two vertexes. Make the induction assumption that if the dimension is less than n, then any extension dividing vertex regions is not linear. Let V₀, V₁, ..., V_n be vertex regions of Sⁿ opposite the empty region from boundaries b₀, b₁, ..., b_n, respectively. Suppose for proof by contradiction, that V₀, V₁, ..., V_n are divided by b, and the extension by b is linear. For each b_i in {b₁, b₂, ..., b_n}, let R_i be the region of {b₁, b₂, ..., b_n}-{b_i} containing V_i. V₀ is a subregion of R_i because none of {b₁, b₂, ..., b_n}-{b_i} separates V_i from V₀. Since b divides V_i and V₀, and V_i and V₀ are subregions of R_i, the "Intermediate Region Theorem" applies. Thus, b divides a region U_i in R_i between b_i and b₀. Since only b₀ and b_i divide R_i, and the region opposite b_i from V_i is empty, R_i has only three subregions, V_i, V₀, and U_i. Consider S^n-1, the section by b₀ of Sⁿ with b₀ removed. Since S^n-1 has n boundaries, it is a simplex. For each i in {1, 2, ..., n}, let W_i be the vertex region of S^n-1 opposite b_i from the empty region, P_i be the region containing U_i and V_i as subregions, U_ia and U_ib be the subregions of U_i separated by b, V_ia and V_ib be the subregions of V_i separated by b, P_ia be the region with U_ia and V_ia as subregions, and P_ib be the region with U_ib and V_ib as subregions. Since the region opposite b_i from P_i is V₀, and b₀ does not divide V₀, P_i is isomorphic to W_i. Since b₀ divides both P_ia and P_ib, b divides W_i. By the induction assumption, S^n-1 extended by b is not linear. On the other hand, Sⁿ extended by b is linear by assumption. Thus by the "Section Theorem", S^n-1 extended by b is linear. This contradiction completes the proof.