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MIGRATION

Define a neighbor of region **R**wrt boundary

**b**as a region separated by and only by

**b**from

**R**. Regions

**R**and

_{0}**R**are neighbors iff they are neighbors wrt some boundary. A region

_{1}**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

**R**is the opposite of region

_{1}**R**iff

_{0}**R**is separated from

_{1}**R**by and only by boundaries attached to

_{0}**R**. Spaces

_{0}**S**and

_{0}^{n}**S**are migrations of each other iff they are identical except for mutually opposite regions

_{1}^{n}**R**and

_{0}**R**, such that

_{1}**R**is nonempty in and only in

_{0}**S**, and

_{0}^{n}**R**is nonempty in and only in

_{1}**S**.

_{1}^{n}##
CORNER THEOREM

A space **S**with

^{n}**m**boundaries is linear iff for each set of

**l**boundaries

**B**from

**S**, the number of corners of

^{n}**B**is

**2**.

^{l}f(n-l,m-l)The sections of

**S**with

^{n}**B**removed by each boundary in

**B**produce a space corresponding to

**f(n-l,m-l)**regions in

**S**with

^{n}**B**removed. The boundaries of

**B**divide each of those regions into

**2**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.

^{l}##
MIGRATION THEOREM

If **S**and

_{0}^{n}**S**are migrations of each other, then

_{1}^{n}**S**is linear iff

_{0}^{n}**S**is linear.

_{1}^{n}The corners of a set of boundaries

**B**are the same in

**S**and

_{0}^{n}**S**except for

_{1}^{n}**R**,

_{0}**R**, and neighbor regions of

_{1}**R**and

_{0}**R**.

_{1}**R**and

_{0}**R**replace each other, and their neighbors replace each other. Thus the theorem is proved by the Corner Theorem.

_{1}##
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 **, the number of outside neighbors of any region is not more than**n

**.**n+1

A simplex has only

**outside neighbors, and adding a boundary does not increase the number of outside neighbors of the subregions.**n+1

##
SIMPLEX COROLLARY

If **b**divides the base regions of a simplex in linear space

**S**

^{n}**S**

^{n}**b**is not linear.

By definition of linear, I can consider the simplex subspace. Assume

**S**

^{n}**b**is linear. Consider the section of

**S**

^{n}**b**. If

**b**does not divide the inside of

**S**

^{n}**b**divides the vertex regions of the migration of

**S**

^{n}**b**does divide the inside of

**S**

^{n}**S**

^{n}**b**, because of the Outside Neighbor Theorem, one of the regions, say

**opposite boundary**R

**from the inside, is inside in the section. If**b

_{0}

**is a simplex, its opposite corresponds to another one of the base regions in**R

**S**, and is not empty. This contradiction implies that

^{n}**is not a simplex, and has more than**R

**neighbors. Then the section by**n

**of**b

_{0}

**S**with

^{n}**removed has base regions divided by**b

_{0}

**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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