# Redunduncy of Pasch's Axiom of Hilbert's Foundations of Geometry

I am baffled by Hilbert's Foundation of Geometry's Axioms because the relations are not defined. They are only described, not defined. For example, the notions of 'betweeness' or 'congruency' are any relations whatsoever, provided only that they satisfy certain axioms with respect to lines and points.

I was closely examining those axioms that define the notions, waiting until I somehow persuade myself that there is only one possible interpretation as to the definitions of the relations of 'betweeness', 'congruency', 'containing', etc, but I ended up discovering that, for 'betweeness' in particular, there is another interpretation that is possible.

What I mean is, according to me, if aliens came down from the sky to examine Hilbert's Foundation of Geometry, they would not understand what exactly is meant by 'betweeness', because there is more than one interpretation that satisfies the axioms of order.

For example, let me interpret 'between' informally: that for any three points on the same line, the point which is farthest from the other two points is the one we'll call 'between' the other two, and only that point. If the points happen to be the same distance apart, the point in the middle will be the one 'between', as in our original understanding, and only that point. Let's assume this is what 'between' means to aliens.

Let 'between' be the new interpretation, and 'betweeness' be the original interpretation.

There is nothing wrong with our notion of 'between', in fact the original notion, of 'betweeness', is informal: that point which is bound by the other two, for example. The axioms are made to explicitly state what the properties of betweeness are, so that we aren't working with informal relations in geometry. The goal here is to check whether the axioms of betweeness are true for the alien interpretation or not. Here are the axioms...

II,1. If $$A, B, C$$ are points of a straight line and $$B$$ lies between $$A$$ and $$C$$, then $$B$$ lies also between $$C$$ and $$A$$.

II,2.If $$A$$ and $$C$$ are two points of a straight line, then there exists at least one point $$B$$ lying between $$A$$ and $$C$$ and at least one point $$D$$ so situated that $$C$$ lies between $$A$$ and $$D$$.

II,3.Of any three points situated on a straight line, there is always one and only one which lies between the other two.

I am aware these are not all the axioms, but the 4th is called Hilbert's Discarded Axiom, apparantely because it was proved to be redundant(source: Wikipedia). Also there is a 5th axiom, also known as Pasch Axiom, but I will derive it from the previous axioms down in this question, to show that it is also redundant. I am not sure why even though it is derivable from the previous axioms it is not discarded yet.

Returning to our 'between' idea, we can see that it satisfies the axioms. Axiom II,1 directly follows from our idea of 'between' because our idea does not depend on the order. That is if $$B$$ is 'between' $$A$$ and $$C$$, it is also 'between' $$C$$ and $$A$$, because $$B$$ is simply the farthest from the other two.

Axiom II,2 is trivially true because we know we can always add a point $$B$$ far enough from $$A$$ and $$C$$ , so that $$B$$ is 'between' $$A$$ and $$C$$, and we can add a point $$D$$ close enough to $$A$$, such that $$C$$ is 'between' $$A$$ and $$D$$.

Axiom III,3 is true because we defined our 'between' as always exactly one point at a time: that which is farthest.

Now you can see that these are not formal proofs that the axioms follow from the idea of 'between', but it should be clear why. The goal is to show that those axioms that describe 'betweeness' can also be interpreted to describe 'between', so we can't decide what betweeness is.

To clarify, we can for example, define '+' as being any operation whatsoever, as long it's commutative, associative, and having a certain distributive property with respect to '$$\cdot$$'. That doesn't automatically define '+', unless we can show that the usual arithmetical addition is the only possible interpretation.

Theorem 5 is not derived from Pasch's Axiom, as stated in Hilbert's Foundation of Geometry.

Theorem 5. Every straight line $$a$$, which lies in a plane $$α$$ , divides the remaining points of this plane into two regions having the following properties: Every point $$A$$ of the one region determines with each point $$B$$ of the other region a segment $$AB$$ containing a point of the straight line $$a$$. On the other hand, any two points $$A,A′$$ of the same region determine a segment $$AA′$$ containing no point of $$a$$

Pasch's Axiom. Let $$A, B, C$$ be three points not lying in the same straight line and let $$a$$ be a straight line lying in the plane $$ABC$$ and not passing through any of the points $$A, B, C$$. Then, if the straight line $$a$$ passes through a point of the segment $$AB$$, it will also pass through either a point of the segment $$BC$$ or a point of the segment $$AC$$.

Here's the proof for Pasch's Axiom:

From the hypothesis we have that $$a$$ passes through a point of the segment $$AB$$. Therefore, the segment $$AB$$ contains a point from $$a$$.

From the hypothesis we know that $$a$$ does not pass through $$A$$ ,$$B$$, or $$C$$, therefore the points don't lie in $$a$$. We also have from the hypothesis that $$a$$ is a line lying in the plane $$ABC$$, and therefore by theorem 5, it divides the remaining points into two regions.

the segment $$AB$$ contains a point from $$a$$. therefore the points $$A$$ and $$B$$ are on different regions, because if they were on the same region, then from theorem 5 we would have that segment $$AB$$ does not contain a point from $$a$$, but this results in a contradiction. So $$A$$ and $$B$$ are on different regions.

By the law of excluded middle, either $$C$$ is in a different region from $$A$$ or $$C$$ is not in a different region from $$A$$. if it is the former, then segment $$AC$$ contains a point from $$a$$, by theorem 5, and therefore, $$a$$ passes through a point in segment $$AC$$ and we're done. If it is the latter, then $$C$$ and $$A$$ are on the same region. But' $$C$$ and $$B$$ can't also be on the same region, because if that was the case, $$A$$ and $$B$$ would also be on the same region, and this results in a contradiction. Therefore $$C$$ and $$B$$ are on different regions, so by theorem 5, the segment $$BC$$ contains a point from $$a$$, and therefore $$a$$ passes through a point in the segment $$BC$$. QED.

Now, my interpretation 'between' is also true for Pasch's Axiom, because I showed that the axiom is redundant. I have to admit that my interpretation does not satisfy the 4th Axiom, but the 4th axiom is redundant, so I am left even more confused. That is, if the interpretation satisfies the first 3 axioms, which satisfies Pach's Axiom, and the 4th axiom can be proven by those axioms, why doesn't my interpretation satisfy it?

## So What?

You don't need to answer all of the questions, but...

Why is there another interpretation of 'betweeness' that satisfies Group II Order Axioms, and if there is only one interpretation, why doesn't the book explain how that is the case? What is the proof of the redundancy of the 4th Axiom, and simultaneously, how can the 4th Axiom be redundant when it is necessary for a single interpretation of betweeness? Why has this elementary proof of Pasch's Axiom been overlooked... No, in fact, wasn't Pasch's Axiom proven to be independent? I am surrounded by dark business and I need Sherlock Holmes to solve the problem.

• Hilbert's book is not easy reading, and he certainly did not tie up loose ends. I can recommend Hartshorne, Geometry: Euclid and Beyond. I had a bit of input to the fourth edition of M. J. Greenberg Euclidean and Non-Euclidean Geometry. You can download a related article by Greenberg from this page: maa.org/programs/maa-awards/writing-awards/… – Will Jagy Oct 18 '18 at 20:11
• I'll look into that. – Comparative Oct 18 '18 at 20:12
• The main point in Hilbert's point of view is exactly that objects and relations are not explicitly defined, but just described by the axioms. You can use whatever model you prefer, so long as it satisfies the axioms. Hilbert himself spoke about “beer mugs and benches” as a (not really) possible models for his axioms. – egreg Oct 18 '18 at 20:12
• Theorem 5 is equivalent to Pasch's axiom; I've seen books in which it is taken as the standard plane separation axiom in place of Pasch's axiom. Note that the first four axioms say nothing about how the betweenness relations on different lines interact, so there's no way you could get something like Theorem 5 (or Pasch) out of them. – Micah Oct 18 '18 at 20:23
• I think the Project Gutenberg edition of Hilbert is a little corrupted (possibly it mixed up two printings, one of which included the discarded axiom and one of which didn't). It gives five linear axioms, and then says "by aid of the four linear axioms II, 1-4", but really you need all five. – Micah Oct 18 '18 at 20:23

## 1 Answer

Contrary to what you say, Theorem 5 cannot be deduced without using Pasch's axiom. You cite Hilbert's Foundation of Geometry, and I'm guessing you are looking at this English translation. Unfortunately, that translation has several errors, and in particular the assertion on page 5 that the theorems rely on only axioms II,1-4 is totally wrong. No such assertion is found in the original German and so I suspect this error (like the error discussed in A confusion about the second connection axiom of Euclidean Geometry) was introduced by the translator.

In particular, your "alien" definition of betweenness does not satisfy Theorem 5. For instance, let $$a$$ be any line, and let $$A$$ and $$B$$ be two points that are very close and on the same side of $$a$$ (by the usual definition). Then by the alien definition, $$A$$ and $$B$$ must be on opposite sides of $$a$$, since there is a point between $$A$$ and $$B$$ on $$a$$ (since if line $$AB$$ intersects $$a$$ at $$C$$, then $$C$$ will be the farthest away of $$A, B,$$ and $$C$$). But now if you take a third point $$C$$ which is also very near $$A$$ and $$B$$, then $$C$$ must similarly be on the opposite side of $$a$$ from both $$A$$ and $$B$$. This is impossible since there are only two sides of $$a$$.

As a side comment, it's a feature, not a bug, that Hilbert's axioms describe the primitive notions rather than define them. You can think of the axioms as a set of rules that allow you to determine when a structure deserves the name "Euclidean geometry". You must provide the definition of your structure, and then the axioms tell you whether your definition really describes a model of Euclidean geometry or not. This approach is pervasive in modern mathematics: for instance, the axioms of a group don't tell you what the group elements or operation are; they just give conditions under which a set of elements equipped with an operation is called a "group".

Now, in the case of Hilbert's axioms, one can prove that up to isomorphism there is a unique structure which satisfies the axioms. So, up to isomorphism, there really is only way to define a structure which follows the axioms. But the "up to isomorphism" is crucial: the entire point of the axioms is that there are many different ways we could define what a "point" or "line" is or what "betweenness" means, and that geometry doesn't actually care exactly what definition we use. Geometry only cares about the structural relationships between all these primitive notions, which will be preserved by an isomorphism of models of the axioms.