Verifying if the 3 incidence axioms are true in a given incidence plane candidate

[mathinker]

Hello,

I'm asked to consider the following candidate to incidence plane: the points are the ones of a radius 1 sphere, {(x, y, z) : x^2 + y^2 + z^2 = 1}, and the lines are the circumferences of radius 1 contained in the said sphere.

This is how I solved, I don't know if I'm on the right track.

Let A1, A2 and A3 be, respectively, the three incidence axioms: for each two (distinct) points there is one and only one line that passes through; each line contains, at least, two (distinct) points and there are, at least, three non-collinear points.

A2 and A3 follow directly from the euclidean plane properties. The circumference is the geometric place of the points at the distance, in the case, 1 from its centre; it has infinite points. A circumference contained in the spherical surface is uniquely defined by three non-collinear points.

A1 is not verified: the points correspond to all of the surface in union with the ones within the sphere.
By A1, for any two distinct points there's only one line passing through, which doesn't happen in the case where one of the points is within the sphere. Moreover, for any given pair of points in opposite sides on the line, there is another, perpendicular to the main, passing through.

Thank you very much for helping me!

 

[stapel]

I'm asked to consider the following candidate to incidence plane: the points are the ones of a radius 1 sphere, {(x, y, z) : x^2 + y^2 + z^2 = 1}, and the lines are the circumferences of radius 1 contained in the said sphere.

This is how I solved, I don't know if I'm on the right track.

Let A1, A2 and A3 be, respectively, the three incidence axioms: for each two (distinct) points there is one and only one line that passes through; each line contains, at least, two (distinct) points and there are, at least, three non-collinear points.

A2 and A3 follow directly from the euclidean plane properties.

But you're not in the plane; you're on a sphere.

The circumference is the geometric place of the points at the distance, in the case, 1 from its centre; it has infinite points.

I'm not sure what you mean by this...? If you are referring to the "lines" of this space, they are "great circles"; that is, they are the circles that are as big as possible on the sphere, like the equator is on Earth. But the lines are *all* of the great circles.

Any line segment has infinitely-many points.

A circumference contained in the spherical surface is uniquely defined by three non-collinear points.

On what basis do you make this claim? Also, aren't you needing to show that TWO non-collinear points determine a line?

A1 is not verified: the points correspond to all of the surface in union with the ones within the sphere.

Where are you getting that the interior points of the sphere are part of the space in question?

By A1, for any two distinct points there's only one line passing through, which doesn't happen in the case where one of the points is within the sphere. Moreover, for any given pair of points in opposite sides on the line, there is another, perpendicular to the main, passing through.

How have you determined that any random pair of points, one each on either side of a line, will be endpoints of a segment perpendicular to that line? (Hint: This statement is not true.)

 

[Dr.Peterson]

You may have already responded, but in case you haven't, I'll add a slightly different perspective.

the points are the ones of a radius 1 sphere, {(x, y, z) : x^2 + y^2 + z^2 = 1}, and the lines are the circumferences of radius 1 contained in the said sphere.

A few of your terms are a little off, so I suspect you are translating from another language, and I'll restate what you say using better terms. Here:

Consider a geometry whose "points" are all the points of a unit sphere (note: this means the surface, not the interior), and whose "lines" are all the great circles of the sphere (circles with radius 1 lying on the sphere).​

Let A1, A2 and A3 be, respectively, the three incidence axioms: for each two (distinct) points there is one and only one line that passes through; each line contains, at least, two (distinct) points and there are, at least, three non-collinear points.

Your axioms are:

• A1: for any two (distinct) points, there is one and only one line that passes through both.
• A2: each line contains, at least two (distinct) points.
• A3: there are at least, three non-collinear points (points that are not on the same line).

A2 and A3 follow directly from the euclidean plane properties. The circumference is the geometric place of the points at the distance, in the case, 1 from its centre; it has infinite points. A circumference contained in the spherical surface is uniquely defined by three non-collinear points.

You claim, first, (A2):

• The great circle is the geometric locus of the points at distance 1 from centre of the sphere; it has infinitely many points.

No, the locus of all points at distance 1 from the center is the entire sphere.
But it is true that any "line" (great circle), being a circle, contains at least two points, so this axiom is satisfied.

Then (A3):

• A great circle contained in the sphere is uniquely defined by three non-collinear points

This would need to be proved, if it were relevant! But A3 says merely that there are three non-collinear points, not something about what is true of them. So you need to prove that there are three points not on the same great circle. This is not hard, but you need to state what you are proving, and give a reason.

Note, again, the sphere is a surface.

A1 is not verified: the points correspond to all of the surface in union with the ones within the sphere.
By A1, for any two distinct points there's only one line passing through, which doesn't happen in the case where one of the points is within the sphere. Moreover, for any given pair of points in opposite sides on the line, there is another, perpendicular to the main, passing through.

This is why I emphasized that the sphere is only the surface, and does not include its interior; so the "points" in your geometry are all on the surface. Here is your claim for A1, restated:

• for any two distinct points, there [is supposed to be] only one "line" [great circle] passing through both; but this doesn't happen in the case where one of the points is within the sphere.
• Moreover, for any given pair of points in opposite sides on the line, there is another, perpendicular to the main, passing through.

I don't know what you mean by that last sentence, but it seems irrelevant.

You need to consider any two points on the surface, and show that there is a great circle passing through both of them, and that there can't be more than one. Here's a hint: To find that circle, think about the center of the sphere, in addition to the two chosen points. What exactly is a great circle?

 

[mathinker]

Hello,

Thank you both Professors very much for helping me with the exercise and proper language use. I'm sorry for misspelling, I wrote the most similar sounding words.

I noticed I forgot to write it was an euclidean sphere, not just sphere. I'm not sure if that changes the main problem, if it implies the euclidean angle and distance definitions..

My attempt:

The lines of the geometry in study are concentric with a constant radius.

If we consider a point A on the sphere's surface, there are infinite-many great circles passing through it. Adding a second point, B, in the condition its distance to A is inferior to 2, it determines a single line l.

That because if two lines intersect, both will share in their sets two points of intersection, one each on either side of the sphere. These points are at the distance equal to the unit sphere's diameter, 2. There are infinitely many lines passing through any two points at this distance.

(?) A1 is satisfied except for the pairs of points in the condition above.

As for A3, let there be three non-collinear points A, B and C, such that there are the lines AB, BC and AC.

Each pair of lines cannot have more than one common point, or they'll represent the same set (by A1). In fact, since the three points are non-collinear, they cannot belong to a same line. So at least one of these is exterior to a great circle. The axiom is satisfied.


I hope I got a little closer to the right answer.
Thank you very much for helping me understand.

 

[Dr.Peterson]

You said the instructions were to "verify if the axioms are true"; it isn't clear how complete a proof is needed, but you are largely correct.

If we consider a point A on the sphere's surface, there are infinite-many great circles passing through it. Adding a second point, B, in the condition its distance to A is inferior to 2, it determines a single line l.

That because if two lines intersect, both will share in their sets two points of intersection, one each on either side of the sphere. These points are at the distance equal to the unit sphere's diameter, 2. There are infinitely many lines passing through any two points at this distance.

(?) A1 is satisfied except for the pairs of points in the condition above.

A1 is false because it is not true in the case that the two points are antipodal (like north and south poles); for any other pair of points, there is only one "line".

But the distance between antipodal points is 2 only in the Euclidean sense! If you were on the sphere, you would say their distance is pi. I found myself think as a resident of the globe, so I initially thought your statement was wrong.

The important thing is that one counterexample makes the axiom false; you didn't go that far in your answer.

As for A3, let there be three non-collinear points A, B and C, such that there are the lines AB, BC and AC.

Each pair of lines cannot have more than one common point, or they'll represent the same set (by A1). In fact, since the three points are non-collinear, they cannot belong to a same line. So at least one of these is exterior to a great circle. The axiom is satisfied.

A3 says that a set of three non-collinear points exists; so you can't start your explanation by saying "let there be three such points".

Again, since A1 is not true, you can't properly say "by A1". If you meant "by the reasoning I used for A1", you may be thinking correctly; but you need to say a lot more than you have. How can you construct a set of three non-collinear points, or at least show that they exist?

I think you're assuming that A2 has been fully dealt with; but it might be good to state clearly how you show that it is true.

 

[mathinker]

Hello,
Thank you very much Professor.

Continuing as oriented:

The lines of the geometry in study are concentric with a constant radius.

The spherical space has an infinity of points in its set. Being lines subsets of this space, they also have infinite-many elements. Axiom 2 is true.

If we consider a point A on the sphere's surface, there are infinitely many great circles passing through it. Adding a second point, B, as long as it is not antipodal, it determines, at first, a single line l.

When two lines intersect, it occurs at antipodal points, with the great circles sharing in their sets said points. There are infinitely many other circles passing through any pair of antipodal points of a given line.

In fact, considering the previous points A and B, these have, respectively, the antipodal points A' and B'. So for a same line l, it shares the 4 points with at least two other lines, one passing through {A,A'} and another through {B,B'}. So a line is not uniquely defined by a pair of distinct points; A1 is not satisfied.

Verification of A3:

Returning to the justification for A1. In this space, a line is necessarily constructed with two points, but that is not sufficient for its uniqueness.

If we consider a line l and all its pairs of antipodal points, it is true that each pair is also part of another set. For example, a line m, defined by two points such that l and m are different great circles.

Since these are different, there is at least one point in each which is non-collinear with the other, or else both would represent the same line.

Or rather, given the two sharing points, there has to be at least two other non-collinear points so to guarantee the existence of each of the intersecting lines in the example: one not collinear with l (defining m), and another not collinear with m (defining l), and therefore, three non-collinear points. The axiom is satisfied.

 

[Dr.Peterson]

Continuing as oriented:

I think you mean, "as directed". The words are similar in meaning, but are not synonyms!

As I said, I'm not at all sure what sort of explanations are expected; what you are saying doesn't really work as proofs, and is not quite clear even as informal justification.

In general, I would explain each axiom entirely in terms of Euclidean geometry and its theorems, and avoid using the terms of the incidence geometry.

The lines of the geometry in study are concentric with a constant radius.

The spherical space has an infinity of points in its set. Being lines subsets of this space, they also have infinite-many elements. Axiom 2 is true.

Technically, you are claiming that any subset of an infinite space is infinite; that is not true.

I would just say that the "lines" in this geometry are circles in Euclidean space, and any circle contains more than one point, so axiom A2 is true.

If we consider a point A on the sphere's surface, there are infinitely many great circles passing through it. Adding a second point, B, as long as it is not antipodal, it determines, at first, a single line l.

When two lines intersect, it occurs at antipodal points, with the great circles sharing in their sets said points. There are infinitely many other circles passing through any pair of antipodal points of a given line.

In fact, considering the previous points A and B, these have, respectively, the antipodal points A' and B'. So for a same line l, it shares the 4 points with at least two other lines, one passing through {A,A'} and another through {B,B'}. So a line is not uniquely defined by a pair of distinct points; A1 is not satisfied.

Everything you say here is true, but the last sentence does not follow. Saying more than you need to say will not convince a teacher that you understand the concepts!

I would just say that since there are more than one great circle ("line") through a pair of antipodal points A and A', A1 does not hold for all pairs. Any pair of points determine at least one "line", but it is not always unique.

Verification of A3:

Returning to the justification for A1. In this space, a line is necessarily constructed with two points, but that is not sufficient for its uniqueness.

If we consider a line l and all its pairs of antipodal points, it is true that each pair is also part of another set. For example, a line m, defined by two points such that l and m are different great circles.

Since these are different, there is at least one point in each which is non-collinear with the other, or else both would represent the same line.

Or rather, given the two sharing points, there has to be at least two other non-collinear points so to guarantee the existence of each of the intersecting lines in the example: one not collinear with l (defining m), and another not collinear with m (defining l), and therefore, three non-collinear points. The axiom is satisfied.

This is where I think it is important not to use the terms of the incidence geometry, but to express them in terms of the Euclidean geometry used to define it. This makes what you say clearer, and avoids the temptation to assume things that have not been proved. I've been trying to consistently put quotation marks around terms from the incidence geometry, like "lines", to distinguish them from actual Euclidean lines.

This axiom is far easier than you are making it. the truth is in there, but well-hidden by unnecessary ideas.

Axiom A3 requires merely showing that there are at least three points that are not on the same great circle. So I would just suppose we have chosen two non-antipodal points, and consider the unique great circle containing them. Then consider any point on the sphere that is not on that great circle. We know that a sphere is not just a single circle, so there must be such another point. Therefore, there are three "non-collinear" points.

Or, you could just point out that (0,0,1), (0,1,0), and (1,0,0) are three points on the unit sphere that are not on the same great circle. When a claim is about existence, the easiest way to prove it is to say, "Here is an example!" That's what I was hoping you would do, when I said,

How can you construct a set of three non-collinear points, or at least show that they exist?

I will say that discussion of incidence geometries is often a student's first exposure to these concepts of proof, so it is common not to see how simple a problem is.

 

[mathinker]

Hello,

Thank you very much Professor.

I've practising with similar exercises, trying to improve the answers with what I learnt.

I struggle with knowing what type of examples I should give, if it is enough to give something specific such as the points in this exercise, or a more generic example, so that any case can apply; or even just a sketch.

Thank you very much for helping me.