## 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 2m 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 PRn is a finite set of points not on the surfaces in B, is given by nb∙(p-pb)>0, where nb is a normal to bB, pb is a point on b, and pP. 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 sb:{0,1}→{0,1}, such that S(b,p) = sb(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 Ri-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 R1a and R1b in S1n iff R1a and R1b are not empty, R1aR1b, and R1a(b0) = R1b(b0) = R0(b0) for each b0∈B0. If b extends S0n = (B0,P0,S0,n), then S1n-1 is the section of S0n by b iff S1n-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 Sn-1 is a section of S0n by b, b extends S0n to S1n, S0n is linear, and S1n is linear, then Sn-1 is linear.
The number of regions in Sn-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 Sn-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 Sn-1, because each subspace of Sn-1 is the section of a subspace of S0n by b. Therefore, Sn-1 is linear.

## Antisection Theorem

If Sn-1 is a section of S0n by b, b extends S0n to S1n, Sn-1 is linear, and S0n is linear, then S1n is linear.
The number of regions in Sn-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 Sn-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, ... bn+1. Consider the section of b1, b2, ... bn by b. That section is a simplex with vertices sectioned from the vertices opposite b1, b2, ... bn. Since the vertex region opposite bn+1 is opposite one of b1, b2, ... bn 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 Sn-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, Sn-1. By the Section Theorem, Sn-1 is linear because Sn is linear and Sn extended by b2 is linear. Similarly, Sn-1 extended by the section of b0, or by the section of b1, is linear. Thus, by supposition, Sn-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 Sn-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, ... Rn+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), ... (pn+1,qn+1). Construct b1, b2, ... bn+1 such that p1 and q1 are in one vertex region of the simplex formed by b1, b2, ... bn+1, and (p2,q2), ... (pn+1,qn+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, ... Rn+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 pi and qi are in the portion of Ri which overlaps the vertex region. Extend R1n by b1, b2, ... bn+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, ... bn+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, ... bn+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.