Equivalency of Pasch's Postulate and the plane separation postulate

by: Robert Hansen, proud member of the math squad.

 keyword: tutorial, geometry 

Introduction

Consider the following two statements:

1) Given a triangle ABC and line l which passes through no vertex of that triangle, if l passes through AB it must pass through AC or BC.

2) Given a line l and a plane P containing it, one can find two distinct sets $ H_1 $ and $ H_2 $ such that the union of $ H_1 $, $ H_2 $, and l is P; $ H_1 $, $ H_2 $ and l are disjoint; $ H_1 $ and $ H_2 $ are convex and given a point in $ H_1 $ and a point in $ H_2 $, the segment connecting them must intersect l.

These two statements are Pasch's Postulate and The Plane Separation Postulate, respectively. They are equivalent, and one can construct an if and only if relationship between them. Furthermore, they are both unprovable from more basic axioms, and are rather famous for showing the incompleteness of Euclid's axioms.

Background

Let's talk about some axioms and definitions we'll need to do this proof:

Axiom I1: Through any two different points, there exists exactly one line.

Axiom I2: Through any three non-collinear points there exists exactly one plane.

Axiom I3: Every plane contains at least three points non-collinear points

The next axioms talk about between-ness, another concept assumed in Euclid's geometry but never proved. To make our lives easier, instead of saying "B is between A and C", we simplify to "A-B-C".

Axiom B1: If A-B-C, then C-B-A.

Axiom B2: If A and C are two points, then there exists B and such that A-B-C and A-C-D.

Axiom B3: Given A, B, C, three collinear points, A-B-C, B-A-C or A-C-B.

Definition: A line segment AB is all points which lie between A and B.

Definition: A set S is convex if, for all points A, B in S, the line segment between them is also contained in S.

Definition: A triangle ABC are the three line segments AB, AC and BC and the points A, B and C.

And finally, a small theorem:

Theorem: Given two distinct lines l and m, $ l \cap m $ is at most one point.

Using these and only these, we can prove that Pasch's Postulate and the Plane Separation Postulate are equivalent.

However, to begin, let's turn Pasch's Postulate into something useful:

Pasch's Postulate

The astute reader may notice that Pasch's postulate does not preclude a line from intersecting both BC and AC. If such a thing were true, then everything we know and understand falls apart, so let us first prove the following:

Given a triangle ABC and a line l which does not pass through any vertex of the triangle, l cannot pass through AB, AC and BC.

To prove, suppose not. Let these intersections be D, E and F respectively. Of these, exactly one is between the other two.

Assume without loss of generality that that point is E.

Consider the triangle ADF. The line BC intersects DF, and therefore must pass through AD or DF.

But we know already that A-D-B and A-F-C, and the line containing AD and AF must intersect the line BC at at most one point each, which are B and C respectively. Therefore, A-B-D and A-C-F are impossible.

Which gives us a contradiction by Pasch's postulate.

Q.E.D.

This is what I'll call the strong version of Pasch's Postulate.

$ 1 \Rightarrow 2 $

I'll now show that Pasch's postulate implies the plane separation postulate.

This proof simply involves drawing lots of triangles.

Suppose Pasch's Postulate holds.

Let us consider a plane P and a line l contained within that plane. We know that the plane contains at least 3 non-collinear points, and the line two distinct points.

Therefore, we are guaranteed at least one point not on the line.

Let us call it C, and define the two planes described in the plane seperation postulate as follows: if the line between the point in question and A intersects l, then it belongs to set $ H_1 $, and otherwise it belongs to $ H_2 $. These sets are clearly disjoint, as the line segment can either intersect l or not intersect l.

Convexity:

Take B and C in $ H_1 $. Construct the triangle ABC. As l does not intersect AB or AC by construction, l cannot BC by Pasch's postulate.

Therefore, $ H_1 $ is convex.

Take B and C in $ H_2 $. Construct the triangle ABC. As l does intersect AB and AC, by the strong version of pasch's postulate, it does not intersect BC.

Therefore, $ H_2 $ is convex.

And finally, for the last property, consider B in $ H_1 $ and C in $ H_2 $. Construct the triangle ABC.

By construction, we know that l intersect AC and does not intersect AB. Therefore, by Pasch's postulate, it must intersect BC.

Q.E.D.

== $ 2 \Rightarrow 1 $

This one's easy.

Assume the plane separation postulate.

Given a triangle ABC and a line l intersecting AB, show that l must intersect BC or AC.

The line l divides the plane into two disjoint, convex sets $ H_1 $ and $ H_2 $. Assume A lies in $ H_1 $. Therefore, B lies in $ H_2 $.

C must lie in either $ H_1 $ or $ H_2 $. If it lies in $ H_1 $, then BC intersects l, and if it lies in $ H_2 $, then AC must intersect l.

Q.E.D.

Conclusion

This is just an introduction to the more rigorous version of Euclidean Geometry