A linear surface is constructible through any **n** regions in *R ^{n}*

**n+1**

If **S ^{n}**

**Q**

^{n}**R**is a region of

**S**

^{n}**P**, in

**Q**

^{n}**R**iff

**R(b)=P(b)**

**b**in

**S**

^{n}## Intermediate Region Lemma

If **S ^{n}**

**Q**

^{n}**R**is a region in

**S**

^{n}**b**extends

**Q**

^{n}**b**divides

**R**, then

**b**divides a subregion of

**R**in

**Q**

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

**b**,

**S**

^{n-1}**P**be the region in

**S**

^{n-1}**R**. By transitivity of isomorphism, the section of

**Q**

^{n}**b**is isomorphic to an extension of

**S**

^{n-1}**R**are isomorphic to the subregions of

**P**. Since at least

**R**is a subregion of

**R**, there is a subregion of

**P**. Therefore,

**b**divides a subregion of

**R**.

## Intermediate Region Theorem

If linear space **S ^{n}**

**Q**

^{n}**R**in

**S**

^{n}**{R**

_{0}, R_{1}, R_{2}, ...}**Q**

^{n}**R**

_{a}**R**

_{b}**{R**

_{0}, R_{1}, R_{2}, ...}**b**

_{0}**b**

_{1}**Q**

^{n}**b**

_{2}**b**

_{2}**R**

_{a}**R**

_{b}**R**

_{c}**{R**

_{0}, R_{1}, R_{2}, ...}**b**

_{2}**R**

_{c}**R**

_{c}**b**

_{0}**b**

_{1}**R**

_{a}**R**

_{c}**b**

_{0}**b**

_{1}**R**

_{b}Suppose **b _{0}**

**R**into

**R**

_{1}**R**

_{2}**b**

_{1}**R**. By the lemma,

**b**

_{1}**R**

_{1}**R**

_{2}**b**

_{1}**R**

_{1}**R**

_{2}**b**

_{1}**R**

_{1}**R**

_{3}**R**

_{4}**R**

_{a}**R**

_{2}**R**

_{b}**R**

_{4}**S**

^{n-1}**S**

^{n}**b**

_{2}**P**,

**P**

_{2}**P**

_{4}**S**

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

**R**

_{2}**R**

_{4}**S**

^{n}**P**is divided by two boundaries, isomorphic to

**b**

_{0}**b**

_{1}**P**

_{2}**P**

_{4}**R**

_{3}**b**

_{2}**R**

_{3}**R**

_{c}**R**

_{3}**b**

_{2}**b**

_{1}**R**

_{1}**R**

_{2}**b**

_{1}**R**

_{1}**R**

_{3}**R**

_{4}**b**

_{1}**R**

_{2}**R**

_{5}**R**

_{6}**R**

_{a}**R**

_{3}**R**

_{b}**R**

_{6}**S**

^{n-1}**S**

^{n}**b**

_{2}**P**,

**P**

_{3}**P**

_{6}**S**

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

**R**

_{3}**R**

_{6}**S**

^{n}**P**is divided by two boundaries, isomorphic to

**b**

_{0}**b**

_{1}**P**

_{3}**P**

_{6}**R**

_{4}**R**

_{5}**b**

_{2}**R**

_{4}**R**

_{5}**b**

_{2}**R**

_{4}**R**

_{c}**R**

_{4}**b**

_{2}## Simplex Theorem

If **b** divides the vertex regions of a simplex in linear space, **S ^{n}**

**S**

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

If the vertexes of a simplex are divided by a boundary, **b**, then **b** divides the vertexes of the restriction of **S ^{n}**

**S**

^{n}**S**

^{n}**n=1**

**n**, then any extension dividing vertex regions is not linear. Let

**V**

_{0}**V**

_{1}**V**

_{n}**S**

^{n}**b**

_{0}**b**

_{1}**b**

_{n}**V**

_{0}**V**

_{1}**V**

_{n}**b**, and the extension by

**b**is linear. For each

**b**

_{i}**{b**

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

**b**

_{n}}**R**

_{i}**{b**

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

**b**

_{n}**}-**

**{**

**b**

_{i}**}**

**V**

_{i}**V**

_{0}**R**

_{i}**{b**

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

**b**

_{n}**}-**

**{**

**b**

_{i}**}**

**V**

_{i}**V**

_{0}**b**divides

**V**

_{i}**V**

_{0}**V**

_{i}**V**

_{0}**R**

_{i}**b**divides a region

**U**

_{i}**R**

_{i}**b**

_{i}**b**

_{0}**b**

_{0}**b**

_{i}**R**

_{i}**b**

_{i}**V**

_{i}**R**

_{i}**V**

_{i}**V**

_{0}**U**

_{i}**S**

^{n-1}**b**

_{0}**S**

^{n}**b**

_{0}**S**

^{n-1}**n**boundaries, it is a simplex. For each

**i**in

**{1, 2**

**, ...,**

**n}**

**W**

_{i}**S**

^{n-1}**b**

_{i}**P**

_{i}**U**

_{i}**V**

_{i}**U**

_{ia}**U**

_{ib}**U**

_{i}**b**,

**V**

_{ia}**V**

_{ib}**V**

_{i}**b**,

**P**

_{ia}**U**

_{ia}**V**

_{ia}**P**

_{ib}**U**

_{ib}**V**

_{ib}**b**

_{i}**P**

_{i}**V**, and

_{0}**b**does not divide

_{0}**V**,

_{0}**P**

_{i}**W**

_{i}**b**

_{0}**P**

_{ia}**P**

_{ib}**b**divides

**W**

_{i}**S**

^{n-1}**b**is not linear. On the other hand,

**S**

^{n}**b**is linear by assumption. Thus by the "Section Theorem",

**S**

^{n-1}**b**is linear. This contradiction completes the proof.

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