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I am trying to learn axioms of geometry, and I can not seem to find any proof to the following theorem that doesn't use circular reasoning:

If π is a plane and l is a line on that plane, then all the points in πl can be divided into two sets S1 and S2 such that if two points A and B are members of the same set, then then, the line segment defined by the two points doesn't intersect l, while if A and B are members of different sets, then the line segment will intersect l.

Here's how far I've come:

We can define an equivalence relation ~:

$$A sim BLeftrightarrow overleftrightarrow{AB} cap l = emptyset$$

It's easily proven that such relation is an equivalence relation.

We know that there exists at least one point C on the plane π that is not on the line l, therefor ~ has at least one equivalence class. We can also show that there is more than one class by taking a point D on the line l and constructing the line CD, then according to 2nd axiom of order there exists a point E on the line CE such that D is between C and E, and clearly the line segment CE intersects the line l and E is not in the same class as C.

The problem begins when I try to prove that only two such sets exists. I've seen the following done:

Let's assume there are three classes. Then we can take points A,B and C such that they are not related. We have a contradiction due to Pasch's theorem because l intersects all three line segments AB, BC and AC.

This would be fine, but wherever I see the proof of Pasch's theorem it uses the fact that there are exactly two half planes.

Thanks in advance :)

axiomatic-geometry

share|cite|improve this question

asked Mar 10 at 23:09

BoxonixBoxonix

74

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here, Pasch is in fact an axiom.
  $endgroup$
  – Hagen von Eitzen
  Mar 10 at 23:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Pasch's axiom states that if A,B and C are non collinear points and line p which does not contain A, B nor C and intersects the line segment AB, then it must intersect at least one of of the other two (AC or BC), but what I need is Pasch's theorem which can apparently be derived from the other axioms of order and it states that p can intersect either AC or BC, but not both.
  $endgroup$
  – Boxonix
  Mar 10 at 23:25
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does the answer here help?
  $endgroup$
  – Hagen von Eitzen
  Mar 10 at 23:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it does, thank you.
  $endgroup$
  – Boxonix
  Mar 11 at 0:21
  

  
  
  
  

add a comment | 

0

1

$begingroup$

I am trying to learn axioms of geometry, and I can not seem to find any proof to the following theorem that doesn't use circular reasoning:

If π is a plane and l is a line on that plane, then all the points in πl can be divided into two sets S1 and S2 such that if two points A and B are members of the same set, then then, the line segment defined by the two points doesn't intersect l, while if A and B are members of different sets, then the line segment will intersect l.

Here's how far I've come:

We can define an equivalence relation ~:

$$A sim BLeftrightarrow overleftrightarrow{AB} cap l = emptyset$$

It's easily proven that such relation is an equivalence relation.

We know that there exists at least one point C on the plane π that is not on the line l, therefor ~ has at least one equivalence class. We can also show that there is more than one class by taking a point D on the line l and constructing the line CD, then according to 2nd axiom of order there exists a point E on the line CE such that D is between C and E, and clearly the line segment CE intersects the line l and E is not in the same class as C.

The problem begins when I try to prove that only two such sets exists. I've seen the following done:

Let's assume there are three classes. Then we can take points A,B and C such that they are not related. We have a contradiction due to Pasch's theorem because l intersects all three line segments AB, BC and AC.

This would be fine, but wherever I see the proof of Pasch's theorem it uses the fact that there are exactly two half planes.

Thanks in advance :)

axiomatic-geometry

share|cite|improve this question

asked Mar 10 at 23:09

BoxonixBoxonix

74

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here, Pasch is in fact an axiom.
  $endgroup$
  – Hagen von Eitzen
  Mar 10 at 23:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Pasch's axiom states that if A,B and C are non collinear points and line p which does not contain A, B nor C and intersects the line segment AB, then it must intersect at least one of of the other two (AC or BC), but what I need is Pasch's theorem which can apparently be derived from the other axioms of order and it states that p can intersect either AC or BC, but not both.
  $endgroup$
  – Boxonix
  Mar 10 at 23:25
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does the answer here help?
  $endgroup$
  – Hagen von Eitzen
  Mar 10 at 23:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it does, thank you.
  $endgroup$
  – Boxonix
  Mar 11 at 0:21
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to learn axioms of geometry, and I can not seem to find any proof to the following theorem that doesn't use circular reasoning:

If π is a plane and l is a line on that plane, then all the points in πl can be divided into two sets S1 and S2 such that if two points A and B are members of the same set, then then, the line segment defined by the two points doesn't intersect l, while if A and B are members of different sets, then the line segment will intersect l.

Here's how far I've come:

We can define an equivalence relation ~:

$$A sim BLeftrightarrow overleftrightarrow{AB} cap l = emptyset$$

It's easily proven that such relation is an equivalence relation.

We know that there exists at least one point C on the plane π that is not on the line l, therefor ~ has at least one equivalence class. We can also show that there is more than one class by taking a point D on the line l and constructing the line CD, then according to 2nd axiom of order there exists a point E on the line CE such that D is between C and E, and clearly the line segment CE intersects the line l and E is not in the same class as C.

The problem begins when I try to prove that only two such sets exists. I've seen the following done:

Let's assume there are three classes. Then we can take points A,B and C such that they are not related. We have a contradiction due to Pasch's theorem because l intersects all three line segments AB, BC and AC.

This would be fine, but wherever I see the proof of Pasch's theorem it uses the fact that there are exactly two half planes.

Thanks in advance :)

axiomatic-geometry

share|cite|improve this question

asked Mar 10 at 23:09

BoxonixBoxonix

74

$endgroup$

share|cite|improve this question

asked Mar 10 at 23:09

BoxonixBoxonix

74

share|cite|improve this question

asked Mar 10 at 23:09

BoxonixBoxonix

74

asked Mar 10 at 23:09

BoxonixBoxonix

74

asked Mar 10 at 23:09

BoxonixBoxonix

74