Outsidedness

My proof of the Constructibility Theorem in a previous post has a flaw. A base region of a simplex is the inside with one boundary reversed. The Fourth Constructibility Lemma is false as stated because the regions may be in the base regions of the constructed simplex. Thus I need a corollary of the Simplex Theorem proving a space with a boundary dividing all base regions of a simplex is not linear. The proof of the Simplex Corollary will require the Migration Theorem, and the Outside Neighbor Theorem. The Migration Theorem requires the Corner Theorem.

MIGRATION

Define a neighbor of region R wrt boundary b as a region separated by and only by b from R. Regions R₀ and R₁ are neighbors iff they are neighbors wrt some boundary. A region R and boundary b are attached iff R has a neighbor wrt b. The corners of a set of boundaries B are the regions attached to every boundary in B. Region R₁ is the opposite of region R₀ iff R₁ is separated from R₀ by and only by boundaries attached to R₀. Spaces S₀ⁿ and S₁ⁿ are migrations of each other iff they are identical except for mutually opposite regions R₀ and R₁, such that R₀ is nonempty in and only in S₀ⁿ, and R₁ is nonempty in and only in S₁ⁿ.

CORNER THEOREM

A space Sⁿ with m boundaries is linear iff for each set of l boundaries B from Sⁿ, the number of corners of B is 2^lf(n-l,m-l).

The sections of Sⁿ with B removed by each boundary in B produce a space corresponding to f(n-l,m-l) regions in Sⁿ with B removed. The boundaries of B divide each of those regions into 2^l regions. Thus the only if part is proved. For the if part, proceed by induction on number of boundaries. The correct numbers of corners are preserved when removing a boundary, and the section by the removed boundary has the correct numbers of corners, so the space without the boundary removed is linear by the Antisection Theorem. To complete the proof, note that the empty space has the correct numbers of corners and is linear.

MIGRATION THEOREM

If S₀ⁿ and S₁ⁿ are migrations of each other, then S₀ⁿ is linear iff S₁ⁿ is linear.

The corners of a set of boundaries B are the same in S₀ⁿ and S₁ⁿ except for R₀, R₁, and neighbor regions of R₀ and R₁. R₀ and R₁ replace each other, and their neighbors replace each other. Thus the theorem is proved by the Corner Theorem.

OUTSIDE

A region is inside iff it is a subregion of the inside of a simplex subspace. A region that is not inside is outside. Intuitively, think of outside regions as peripheral regions with infinite extent. An outside neighbor of a region is a neighbor that is outside.

OUTSIDE NEIGHBOR THEOREM

In a space of dimension n, the number of outside neighbors of any region is not more than n+1.

A simplex has only n+1 outside neighbors, and adding a boundary does not increase the number of outside neighbors of the subregions.

SIMPLEX COROLLARY

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

By definition of linear, I can consider the simplex subspace. Assume Sⁿ extended by b is linear. Consider the section of Sⁿ by b. If b does not divide the inside of Sⁿ, b divides the vertex regions of the migration of Sⁿ. This contradicts the Simplex Theorem, so b does divide the inside of Sⁿ. In the section of Sⁿ by b, because of the Outside Neighbor Theorem, one of the regions, say R opposite boundary b₀ from the inside, is inside in the section. If R is a simplex, its opposite corresponds to another one of the base regions in Sⁿ, and is not empty. This contradiction implies that R is not a simplex, and has more than n neighbors. Then the section by b₀ of Sⁿ with b₀ removed has base regions divided by b. This contradicts induction on dimension. To complete the proof, note that in a 1 dimensional space, no boundary divides the base regions of a simplex because no boundary divides more than 1 region.