Sidedness Geometry

Linear Space

Take points and boundaries as undefined. Define a sidedness relation S:B×P→{0,1}, where B is a finite set of boundaries and P is a finite set of points. Two points p₀, p₁ are said to be on the same side of a boundary b iff S(b,p₀) = S(b,p₁), and they are said to be opposite b iff S(b,p₀) ≠ S(b,p₁). Define a space as Sⁿ = (B,P,S,n), where S is a sidedness relation on B×P, and integer n>0 is called the dimension of the space. Define a region as R:B→{0,1}. Note that there are 2^m regions in a space with m boundaries. A point p∈P is said to be in region R iff S(b,p) = R(b) for each b∈B. A region is said to be empty iff no points are in it. Define a subspace of Sⁿ = (B,P,S,n) to be S₀ⁿ = (B₀,P,S₀,n) such that B₀⊂B, and S₀(b,p) = S(b,p) for each b∈B₀, and p∈P. A space Sⁿ is said to be linear iff in each subspace the number of nonempty regions is equal to f(n,m), where m is the number of boundaries in the subspace, f(1,m) = m+1, f(n,0) = 1, and f(n+1,m+1) = f(n+1,m)+f(n,m). A sidedness relation S:B×P→{0,1}, where B is a finite set of linear surfaces in Euclidean space Rⁿ and P⊂Rⁿ is a finite set of points not on the surfaces in B, is given by n_b∙(p-p_b)>0, where n_b is a normal to b∈B, p_b is a point on b, and p∈P. A set of n linear surfaces in Rⁿ determines an intersection point. A set of linear surfaces in Rⁿ is associated with a set of intersection points determined by all subsets of n linear surfaces. A finite set of linear surfaces in Rⁿ is noncoincidental iff no two surfaces in it are parallel, no n+1 surfaces in it share a point, and no linear surface goes through more than n intersection points. A space Sⁿ is isomorphic with a finite set of noncoincidental linear surfaces in Rⁿ iff there exist one to one mappings g:B→B, h:P→P, and s_b:{0,1}→{0,1}, such that S(b,p) = s_b(S(g(b),h(p))) for each b∈B and p∈P.

Linear Space Theorem

A finite set of noncoincidental linear surfaces in Rⁿ exists if and only if isomorphic linear space Sⁿ exists.
First I will prove that a finite set of linear surfaces in Rⁿ exists only if isomorphic linear space Sⁿ exists. Each point in R divides one of a finite set of regions spanning R, so the number of regions in R is f(1,m) = m+1, where m is the number of points dividing R. Thus, the "only if" part of the theorem is true for n = 1. There exist any number of points in Rⁿ on either side of a linear surface, so f(n,1) = 2, for all n. Thus, the "only if" part of the theorem is true for m = 1. Suppose the "only if" part of the theorem is true for n < i. Also suppose that if n = i, then the "only if" part of the theorem is true for m < j. I must show that the "only if" part of the theorem is true for n = i and m = j. Let L be a set of j-1 noncoincidental linear surfaces in Rⁱ, and let l be a linear surface in Rⁱ, noncoincidental with L. By supposition, L divides Rⁱ into f(i,j-1) regions. The intersections of l with the surfaces in L are isomorphic with linear surfaces in R^i-1. By supposition, l divides f(i-1,j-1) of the regions formed by L in Rⁱ. Therefore, there are f(i,j-1)+f(i-1,j-1) = f(i,j) regions formed by L∪{l} in Rⁱ, and the "only if" part of the theorem is proved. The "if" part I will prove below after proving some supportive theorems.

Section

A boundary b is said to extend S₀ⁿ = (B₀,P₀,S₀,n) to S₁ⁿ = (B₁,P₁,S₁,n) iff S₀ⁿ is a subspace of S₁ⁿ, and B₁ = B₀∪{b}. If b extends S₀ⁿ = (B₀,P₀,S₀,n) to S₁ⁿ, then b divides region R₀ in S₀ⁿ into regions R_1a and R_1b in S₁ⁿ iff R_1a and R_1b are not empty, R_1a ≠ R_1b, and R_1a(b₀) = R_1b(b₀) = R₀(b₀) for each b₀∈B₀. If b extends S₀ⁿ = (B₀,P₀,S₀,n), then S₁^n-1 is the section of S₀ⁿ by b iff S₁^n-1 is isomorphic with S₁ⁿ = (B₀,P⊂P₀,S₀,n), where p∈P iff p is in a region divided by b.

Section Theorem

If S^n-1 is a section of S₀ⁿ by b, b extends S₀ⁿ to S₁ⁿ, S₀ⁿ is linear, and S₁ⁿ is linear, then S^n-1 is linear.
The number of regions in S^n-1 is the number of regions divided by b. The number of regions divided by b is the number of regions in S₀ⁿ subtracted from the number of regions in S₁ⁿ. Since S₀ⁿ and S₁ⁿ are linear, the number of regions in S^n-1 is f(n,m+1)-f(n,m) = (f(n,m)+f(n-1,m))-f(n,m) = f(n-1,m). The same reasoning applies to each subspace of S^n-1, because each subspace of S^n-1 is the section of a subspace of S₀ⁿ by b. Therefore, S^n-1 is linear.

Antisection Theorem

If S^n-1 is a section of S₀ⁿ by b, b extends S₀ⁿ to S₁ⁿ, S^n-1 is linear, and S₀ⁿ is linear, then S₁ⁿ is linear.
The number of regions in S^n-1 is the number of regions divided by b. The number of regions in S₁ⁿ is the number of regions divided by b added to the number of regions in S₀ⁿ. Since S^n-1 and S₀ⁿ are linear, the number of regions in S₁ⁿ is f(n-1,m)+f(n,m) = f(n,m+1). The same reasoning applies to each subspace of S₁ⁿ, because each subspace of S₁ⁿ is the extension of a subspace of S₀ⁿ by b. Therefore, S₁ⁿ is linear.

Simplex

A simplex in linear space Sⁿ is any n+1 boundaries in Sⁿ. Note that because n linear surfaces are linearly independent in Rⁿ, and the "only if" part of the Linear Space Theorem is true, f(n,n) = 2ⁿ. By definition of linear space, f(1,2) = 2²-1, and if f(n-1,n) = 2ⁿ-1, then f(n,n+1) = f(n,n)+f(n-1,n) = 2ⁿ+2ⁿ-1 = 2ⁿ⁺¹-1, so the number of nonempty regions in a simplex is 2ⁿ⁺¹-1. Consider the empty region of a simplex in space Sⁿ. Reverse any one sidedness of the empty region, and call the resultant region a vertex region of the simplex. Since there are n+1 boundaries in a simplex, there are n+1 vertex regions in it. Reverse all of the sidednesses of the empty region and call the resultant region the inside of the simplex. Since there is only one empty region in a simplex, there is only one inside in it.

Simplex Theorem

There exists no boundary dividing all vertex regions of a simplex.
Suppose there does exist boundary b dividing all n+1 vertex regions of simplex b₁, b₂, ... b_n+1. Consider the section of b₁, b₂, ... b_n by b. That section is a simplex with vertices sectioned from the vertices opposite b₁, b₂, ... b_n. Since the vertex region opposite b_n+1 is opposite one of b₁, b₂, ... b_n from each of the sectioned vertex regions, it is not sectioned to any one region of the sectioned simplex. This contradicts the supposition of b, so no such b exists, and the theorem is proved.

Independent Boundary Theorem

If Sⁿ extended by b₀ is linear, and Sⁿ extended by b₁ is linear, then there exists a linear superspace of Sⁿ extended by b₀ and Sⁿ extended by b₁.
(This proof attempt is wrong. I am working on a correction.) For S¹ it is true because b₀ and b₁ divide different regions, so points on one or the other side of b₀ in the region it divides will all be on one side of b₁. Suppose it is true for S^n-1. I must prove it for Sⁿ. Suppose Sⁿ is not linear if it is extended by both b₀ and b₁ together, but Sⁿ is linear if it is extended by b₀ or b₁ individually. Discard boundaries from Sⁿ until the reduced Sⁿ is linear when extended by both b₀ and b₁. This is possible because b₀ and b₁ alone in Sⁿ is a linear space. Let b₂ be the last boundary discarded from Sⁿ. Consider the section by b₂, S^n-1. By the Section Theorem, S^n-1 is linear because Sⁿ is linear and Sⁿ extended by b₂ is linear. Similarly, S^n-1 extended by the section of b₀, or by the section of b₁, is linear. Thus, by supposition, S^n-1 extended by the sections of both b₀ and b₁ is linear. By the Antisection Theorem, since Sⁿ extended by b₀ and b₁ is linear, and S^n-1 extended by the sections of b₀ and b₁ is linear, Sⁿ extended by b₀, b₁, and b₂ is linear. This contradicts the supposition, so the theorem is proved.

"If" Part of Linear Space Theorem

If linear space Sⁿ exists, then a finite set of linear surfaces in Rⁿ isomorphic with Sⁿ exist.
(This proof attempt, in particular the struckout text, is wrong. I am working on a correction.) Suppose no R₀ⁿ exits with boundaries and sidednesses isomorphic to some linear space S₀ⁿ. Let S₁ⁿ be boundaries of S₀ⁿ which are isomorphic with the boundaries and sidednesses of some R₁ⁿ. Until impossible, extend S₁ⁿ by boundaries from S₀ⁿ such that S₁ⁿ is still isomorphic with an extension of R₁ⁿ. By supposition, there exists boundary b₀ in S₀ⁿ such that b₀ is not in S₁ⁿ. Let R₁, R₂, ... R_n+1 be the n+1 regions of S₁ⁿ divided by b₀ such that no b₀ exists which divides those regions in R₁ⁿ. Let p₁ and q₁ be in R₁, opposite b₀, and similarly for (p₂,q₂), ... (p_n+1,q_n+1). Construct b₁, b₂, ... b_n+1 such that p₁ and q₁ are in one vertex region of the simplex formed by b₁, b₂, ... b_n+1, and (p₂,q₂), ... (p_n+1,q_n+1) are in the other vertex regions. The prior construction is possible because, given n+1 points in Rⁿ a simplex can be constructed with those points as vertices; a point can be chosen in each of R₁, R₂, ... R_n+1; the vertex region of each point at least partially overlaps the region the point is in; and without loss of generality, we may assume p_i and q_i are in the portion of R_i which overlaps the vertex region. Extend R₁ⁿ by b₁, b₂, ... b_n+1 to R₂ⁿ. Since R₂ⁿ obeys the "only if" part of the Linear Space Theorem, S₁ⁿ can be similarly extended to S₂ⁿ such that S₂ⁿ is isomorphic to R₂ⁿ. Since S₁ⁿ can be extended by b₀, and b₀ divides different regions from b₁, b₂, ... b_n+1, S₂ⁿ can be extended by b₀ to S₃ⁿ because of the Independent Boundary Theorem. Now, b₀ goes through each of the vertex regions of b₁, b₂, ... b_n+1 in S₃ⁿ. This violates the Simplex Theorem. Therefore the supposition was false, and there does exist R₀ⁿ with boundaries and sidednesses isomorphic with S₀ⁿ.