Friday, August 29, 2014


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.


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.


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 2lf(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 2l 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.


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.


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.


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.


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.

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