I am finally able to prove the theorems I left unproved in the first posting of this blog. I'll start with a new approach to the "Independent Boundary Theorem". Then I'll present a few lemmas for the "Constructibility Theorem", introduced as the "if" part of the "Linear Space Theorem" in the first posting. In the following, all regions in are bounded by linear surfaces, so they are closed and convex.

*R*

^{n}Independent Boundary Theorem

If ** S** extended by

^{n}

**is linear and**b

_{0}

**extended by**S

^{n}

**is linear, then there exists a linear superspace of**b

_{1}

**extended by**S

^{n}

**and**b

_{0}

**.**b

_{1}

If **n=1**, and ** b** and

_{0}

**divide different regions, then add points such that all regions identical, except for**b

_{1}

**and**b

_{0}

**, to the regions divided by**b

_{1}

**and**b

_{0}

**are nonempty. If**b

_{1}

**n=1**, and

**and**b

_{0}

**divide the same region, then add points such that 3 out of 4 of the regions identical, except for**b

_{1}

**and**b

_{0}

**, to the region divided by**b

_{1}

**and**b

_{0}

**are nonempty. If**b

_{1}

**n>1**, Proceed by induction on number of boundaries. If

**is empty, construct all 4 possible regions. Otherwise, choose**S

^{n}

**from**b

_{2}

**. Consider the section**S

^{n}

**of**S

^{n-1}

**by**S

^{n}-{b

_{2}}

**. By the "Section Theorem",**b

_{2}

**is linear. Apply the induction hypothesis to get linear**S

^{n-1}

**. Apply the induction hypothesis again to extend**S

^{n}-{b

_{2}}+{b

_{0},b

_{1}}

**by the sections of**S

^{n-1}

**and**b

_{0}

**by**b

_{1}

**. Choose sides of**b

_{2}

**for points in**b

_{2}

**such that**S

^{n}-{b

_{2}}+{b

_{0},b

_{1}}

**divides regions isomorphic to**b

_{2}

**extended by the sections of**S

^{n-1}

**and**b

_{0}

**by**b

_{1}

**. Use the "Antisection Theorem" to prove that**b

_{2}

**is linear.**S

^{n}+{b

_{0},b

_{1}}

First Constructibility Lemma

If linear surface ** s** goes through (tangent or dividing) all regions

**, and no linear surface divides every region from**

*{R*

_{1}, R_{2}, ...}**, then**

*{R*

_{1}, R_{2}, ...}**is tangent to at least**

*s***n+1**of

**.**

*{R*

_{1}, R_{2}, ...}If ** s** is tangent to fewer than

**n+1**of

**, then a linear surface dividing all of**

*{R*

_{1}, R_{2}, ...}**can be constructed by moving**

*{R*

_{1}, R_{2}, ...}**an arbitrarily small amount into the regions from**

*s***it is tangent to. This contradiction proves that**

*{R*

_{1}, R_{2}, ...}**is tangent to at least**

*s***n+1**of

**.**

*{R*

_{1}, R_{2}, ...}Second Constructibility Lemma

If ** b** is a boundary for all of the regions divided by

_{0}

**in a linear space**b

_{1}

**,**S

^{n}

**is isomorphic to linear surfaces in**S

^{n}

**,**

*R*

^{n}**is isomorphic to**b

_{0}

**in**

*s*

_{0}**,**

*R*

^{n}**is the section of**S

^{n-1}

**by**S

^{n}

**,**b

_{1}

**in**b

_{0}

**is isomorphic to linear subsurface**S

^{n-1}

**in**

*l*

_{0}**, then there exists linear surface**

*s*

_{0}**in**

*s*

_{1}**isomorphic to**

*R*

^{n}**.**b

_{1}

Choose ** R_{0}** in

**isomorphic to some**

*R*

^{n}**divided by**R

_{0}

**. Construct**b

_{1}

**arbitrarily close to**

*s*

_{1}**, through**

*s*

_{0}**, dividing**

*l*

_{0}**. I must show that the isomorphism between**

*R*_{0}**and**S

^{n}

**maps the regions divided by**

*R*

^{n}**to the regions divided by**b

_{1}

**. Suppose**

*s*

_{1}**divides**b

_{1}

**in**R

_{1}

**isomorphic to**S

^{n}

**in**

*R*_{1}**. By definition of section,**

*R*

^{n}**and**R

_{0}

**in**R

_{1}

**on the same side of**S

^{n}

**implies**b

_{0}

**and**R

_{0}

**in**R

_{1}

**are on the same side of**S

^{n-1}

**. Let**b

_{0}

**be**Q

^{n}

**with**S

^{n}

**removed. Since**b

_{0}

**bounds the regions divided by**b

_{0}

**,**b

_{1}

**and**b

_{0}

**divide the same regions in**b

_{1}

**. Thus,**Q

^{n}

**in the section by**b

_{0}

**is isomorphic to**b

_{1}

**in the section by**b

_{1}

**. Therefore,**b

_{0}

**and**R

_{0}

**on the same side of**R

_{1}

**in**b

_{0}

**implies**b

_{1}

**and**R

_{0}

**are on the same side of**R

_{1}

**in**b

_{1}

**. Let**b

_{0}

**be**

*P*

^{n}**with**

*R*

^{n}**removed, then**

*s*

_{0}**is isomorphic to**

*P*

^{n}**. Thus,**Q

^{n}

**and**R

_{0}

**on the same side of**R

_{1}

**in**b

_{1}

**implies**b

_{0}

**and**

*R*_{0}**are on the same side of**

*R*_{1}**in**

*l*

_{0}**. By similarity of triangles,**

*s*

_{0}**and**

*R*_{0}**on the same side of**

*R*_{1}**in**

*l*

_{0}**,**

*s*

_{0}**and**

*R*_{0}**on the same side of**

*R*_{1}**in**

*s*

_{0}**,**

*R*

^{n}**arbitrarily close to**

*s*

_{1}**, and**

*s*

_{0}**dividing**

*s*

_{1}**implies**

*R*_{0}**divides**

*s*

_{1}**. A similar argument applies if**

*R*_{1}**and**R

_{0}

**in**R

_{1}

**are opposite**S

^{n}

**. Therefore,**b

_{0}

**dividing**b

_{1}

**implies**R

_{1}

**divides**

*s*

_{1}**isomorphic to**

*R*_{1}**, so**R

_{1}

**is isomorphic to**

*s*

_{1}**.**b

_{1}

Third Constructibility Lemma

If there is no linear surface through regions in

*{R*

_{1}, R_{2}, ...}**, and there is no linear surface tangent to more than**

*R*

^{n}**n**regions in

**, then there is a subset of**

*{R*

_{1}, R_{2}, ...}**n+1**regions

**from**

*{R*

_{3}, R_{4}, ...}**such that there is no linear surface through**

*{R*

_{1}, R_{2}, ...}**.**

*{R*

_{3}, R_{4}, ...}Any linear surface goes through an empty set of regions, so must be nonempty. Choose

*{R*

_{1}, R_{2}, ...}**and**

*{R*

_{5}, R_{6}, ...}**from**

*R***such that there is a linear surface through**

*{R*

_{1}, R_{2}, ...}**, but no linear surface through**

*{R*

_{5}, R_{6}, ...}**. Choose linear surface**

*{R, R*

_{5}, R_{6}, ...}**, closest to**

*s*_{0}**that goes through**

*R***. Because**

*{R*

_{5}, R_{6}, ...}**cannot be moved closer to**

*s*_{0}**without changing whether it goes through**

*R***, and**

*{R*

_{5}, R_{6}, ...}**is tangent to no more than**

*s*_{0}**n**regions from

**,**

*{R*

_{5}, R_{6}, ...}**is tangent to exactly**

*s*_{0}**n**regions from

**. Let**

*{R*

_{5}, R_{6}, ...}**be the**

*{R*

_{3}, R_{4}, ...}**n**regions from

**that**

*{R*

_{5}, R_{6}, ...}**is tangent to, together with**

*s*_{0}**. Suppose there is linear surface**

*R***that goes through all of**

*s*_{1}**. Move**

*{R*

_{3}, R_{4}, ...}**towards**

*s*_{0}**such that their intersection remains fixed. Since**

*s*_{1}**is closest to**

*s*_{0}**through**

*R***, it must leave one of**

*{R*

_{5}, R_{6}, ...}**before it gets closer to**

*{R*

_{5}, R_{6}, ...}**, and it must get closer to**

*R***to get closer to**

*R***. The only regions**

*s*_{1}**can leave immediately are the ones it is tangent to. Thus,**

*s*_{0}**must leave one of**

*s*_{0}**. Since**

*{R*

_{3}, R_{4}, ...}**only moves towards**

*s*_{0}**and the regions are convex,**

*s*_{1}**never returns to that region it leaves. Therefore**

*s*_{0}**does not go through all of**

*s*_{1}**, and there is no linear surface through all of**

*{R*

_{3}, R_{4}, ...}**.**

*{R*

_{3}, R_{4}, ...}Fourth Constructibility Lemma

If there is no linear surface through **n+1** regions in

*{R*

_{1}, R_{2}, ...}**, then there is a simplex with**

*R*

^{n}**in the vertex regions.**

*{R*

_{1}, R_{2}, ...}Fore each ** R** in

**choose**

*{R*

_{1}, R_{2}, ...}**through all of**

*s***except**

*{R*

_{1}, R_{2}, ...}**, closest to**

*R***. This is possible because it is always possible to construct a linear surface through**

*R***n**regions in

**. If**

*R*

^{n}**were not tangent to**

*s***n**regions from

**, there would be a linear surface through the same regions as**

*{R*

_{1}, R_{2}, ...}**, but closer to**

*s***. Thus,**

*R***is tangent to**

*s***n**regions from

**. For each**

*{R*

_{1}, R_{2}, ...}**, choose**

*s***arbitrarily close to**

*t***, not through the regions**

*s***is tangent to, and still not through**

*s***. Thus,**

*R***separates**

*t***from the other regions in**

*R***. In this way construct simplex**

*{R*

_{1}, R_{2}, ...}**with**

*{t*

_{1}, t_{2}, ...}**in the vertex regions.**

*{R*

_{1}, R_{2}, ...}Constructibility Theorem

If ** S** is a linear sidedness space, then there exist non-coincidental linear surfaces in

^{n}

**isomorphic to**

*R*

^{n }**.**S

^{n}

If **n=1**, then construct by choosing boundaries dividing the same regions in as divided in

*R*

^{1}**. This is possible because extensions to**S

^{1}

**divide nonempty regions. If**S

^{1}

**n>1**, proceed by induction on the number of boundaries in

**. If there are no boundaries in**S

^{n}

**, then the empty set of linear surfaces in**S

^{n}

**is isomorphic. Assume there is a set of non-coincidental linear surfaces isomorphic to every sidedness space with one fewer boundary than**

*R*

^{n}**. I must prove there is a set of non-coincidental linear surfaces isomorphic to**S

^{n}

**. Assume the contrary. Remove one boundary,**S

^{n}

**, from**b

**. There is a set,**S

^{n}

**, of non-coincidental linear surfaces isomorphic to**

*{s*

_{1}, s_{2}, ...}**with**S

^{n}

**removed. By assumption there is no linear surface dividing the regions,**b

**, isomorphic to the regions divided by**

*{R*

_{1}, R_{2}, ...}**. If there exists a linear surface,**b

**, through (tangent or dividing) all of**

*s***, then by the "First Constructibility Lemma",**

*{R*

_{1}, R_{2}, ...}**is tangent to at least**

*s***n+1**of

**. By the induction assumption, the boundaries of the regions are non-coincidental, so**

*{R*

_{1}, R_{2}, ...}**is tangent to no more than**

*s***n**regions unless it is a boundary for all of the regions it goes through. Thus,

**is a boundary for all of**

*s***, or there is**

*{R*

_{1}, R_{2}, ...}**in**

*R***that**

*{R*

_{1}, R_{2}, ...}**does not go through (neither tangent nor dividing). If**

*s***is a boundary, then since**

*s***n>1**, the boundary isomorphic to

**in the section of**

*s***by**S

^{n}

**is constructible by the induction assumption. By the "Second Constructibility Lemma", if**b

**is a boundary for the regions, then**

*s***is constructible. This contradiction shows that for each linear surface**b

**there is some**

*s***in**

*R***that**

*{R*

_{1}, R_{2}, ...}**does not go through. Since the boundaries of**

*s***are non-coincidental, no linear surface is tangent to more than**

*{R*

_{1}, R_{2}, ...}**n**of

**. By the "Third Constructibility Lemma", there is a subset of**

*{R*

_{1}, R_{2}, ...}**n+1**of

**, say**

*{R*

_{1}, R_{2}, ...}**, that no linear surface goes through. By the "Fourth Constructibility Lemma", there is a simplex, say**

*{R*

_{3}, R_{4}, ...}**of**

*{s*

_{3}, s_{4}, ...}**n+1**linear surfaces with

**in the vertex regions. By the "only if" part of the "Linearity Theorem", and the "Independent Boundary Theorem",**

*{R*

_{3}, R_{4}, ...}**with**S

^{n}

**removed can be extended by boundaries isomorphic to**b

**, to spaces extensible by**

*{s*

_{3}, s_{4}, ...}**. Thus,**b

**is extensible by a simplex with**S

^{n}

**dividing the vertex regions. This violates the "Simplex Theorem", so there is a linear surface,**b

**, that together with**

*s*_{0}**, are isomorphic to**

*{s*

_{1}, s_{2}, ...}**. If**S

^{n}

**are coincidental, then adjust**

*{s*

_{0}, s_{1}, s_{2}, ...}**by an arbitrarily small amount to find non-coincidental linear surfaces isomorphic to**

*s*_{0}**.**S

^{n}

Note that the proof extends an arbitrary set of linear surfaces with a new surface. Thus, every isomorphic set of linear surfaces can be extended to every superspace. In other words, to obtain a set of linear surfaces isomorphic to a given abstract linear space, linear surfaces can be added to the set in any order, and any linear surface that divides the correct regions suffices. To me this means linear spaces are very easy to construct. On the other hand, I could claim that linear spaces are very rigid. For example, no change in the position of a linear surface sufficient to close a line of sight is possible without changing which equivalence class the linear space is in.