I find it easier to adjust the proof than
to find all such cases.

acco

I sought this proof because it only uses the axioms of incidence which are the first step when one tries to build projective spaces from axioms. The incidence laws for a plane projective space are that any two different points lie on a just one line and that any two different lines meet in just one point. As far as I know, Desargue's theorem cannot be proved using those axioms only. However, the incidence axioms for 3-dimensional projective geometry are all that is required for Desargue's theorem even in the case when the triangles lie in a plane.

A 2-dimensional projective space with a pair of triangles that didn't satisfy Desargue's theorem couldn't be embedded as a plane in a higher dimensional projective space.

All the best,
Hubert

But if it is decided, that the version of Desargues theorem and the proof should be more general, then it would not be a problem,
if A1 may be equal to B1 or line A1B1 may be equal to A2B2.

Indeed this will require the page rewritten starting with

Desargues Theorem
Let A1B1C1 and A2B2C2 be two triangles. Consider two conditions:
1. Lines A1A2, B2B2, C1C2 through the corresponding vertices can be chosen concurrent.
2. Collinear points ab, bc, ca can be chosen on the (extended) sides A1B1 and A2B2, B1C1 and B2C2, C1A1 and C2A2, respectively.

If you and Alex are satisfied with the current restrictions
the suggestion is to write near the top of the web page:

It is assumed, that none of the corresponding vertices and sides are equal.

All the best to you and Alex,
acco

My feeling is that an appearance of two exclusive conditions needs to be explained. Without an example, the reader will be left wondering why the conditions have been introduced and where they have been used.

I would much prefer a statement proved of course from incidence axioms as formulated by Hubert: if the pair of C-vertices leads to coplanar triangles, then either the A- or B-pair will work.

Yes, I was rather careless with the formulation. As I implied in my first reply, my interest in the proof was to illustrate that it could be done and I seem to have left the special cases to the reader.

All the best,

Hubert

Therefore two suggestions
1. State as an assumption, that none of the corresponding vertices and sides are equal.
OR
2. You may use my note, formulating a proof in a symmetrical way, at
https://lanco.host22.com/
as you like.

All the best,
acco

The fact that lines, points or planes that are not uniquely defined usually means that any one satisfying the conditions will serve.

I suppose you know that each field F gives a projective space FP^n for each dimension n. They all satisfy Desargue's although it is rather meaningless for F = Z/2 since there are only 2 points on each line.

Thanks again for the comments,

Hubert

The proof I suggest is found in a version 2 at
https://lanco.host22.com/
and is also valid for the Fano plane, which may be mentioned, if adapted. As you will see, the extra work is not much.

Kind regards
acco

Yes indeed, there are three points on each line. Sorry.

Is there an example of a two dimensional space where the incidence axioms hold but Desargue's theorem is false?

Regards,

Hubert

Many regards
Allan Cortzen

Allan Cortzen

As I have mentioned before, many places (10 or more) in the proof else are misleading.

Alone from the first line of the proof:
1. "O be the common point of A1A2, B1B2, C1C2 " is without meaning is the triangles are identical.
2. Suppose O=A1=A2 and the triangle are in the same plane.
That "C1 is not in the plane OA1B1" is unclear, as many planes contains the line OA1B1, and we may or may not think of a plane containing C1.

It continues.
Let me just mention in the converse proof dually it may happen, that E=O'. Now from E the point O' is projected onto the original plane. This is meaningless.

Kind Regards
Allan Cortzen

>And nothing about exceptions,

Why? I think it is useful to point out that the proof fails in some cases. Especially because it never explicitly makes use of the "general position" assumption.

>as they would require a lot of
>work to handle.

That's is correct. And I have no intention to dig into that. But the lack of motivation to consider the exceptions is no reason not to mention their existence.

>As I have mentioned before, many places (10 or more) in the
>proof else are misleading.
>
>Alone from the first line of the proof:
>1. "O be the common point of A1A2, B1B2, C1C2 " is without
>meaning is the triangles are identical.
>2. Suppose O=A1=A2 and the triangle are in the same plane.
>That "C1 is not in the plane OA1B1" is unclear, as many
>planes contains the line OA1B1, and we may or may not think
>of a plane containing C1.
>
>It continues.

I agree with that. These are the examples where the proof fails. So what's wrong with alerting the reader up front to the fact that in some circumstances the proof fails?

>Let me just mention in the converse proof dually it may
>happen, that E=O'. Now from E the point O' is projected onto
>the original plane. This is meaningless.

A construct invoked by the proof is meaningless in some circumstances. This in particular means that in these circumstances the proof fails.

Actually, I am not sure I am getting your point.

If we assume the general position the proof goes through. I believe we both agree on that point. We diverge in regard to my mention of the existence of special cases. You appear to claim that under the general position assumption there are no special cases. I agree with that. The special cases violate that assumption and, in addition, make some steps in the proof meaningless. Why this can't be mentioned is beyond me.

The reasons are
1. The theorem is OK, only the proof is executed under limitations
2. In general position e.g. A2=B1 is allowed
3. In general position e.g. A1A2 and B1B2 should intersect in one point, so A1A2!=B1B2
4. The triangle concept has not yet been discussed, and has the whole time been three non-collinear points.

About the exceptions:
"These are simple enough to be treated individually and will not be considered here."
I agree, but it is not so simple to find these cases, because of the language in proof. (Remember your first response)
The general position is used again and again in the proof without explicitly mentioning it.
Two cases already in the first line has been pointed out. It is not only cases similar to the case where A1,B1,A2,B2 are collinear and C1=C2.

Greetings
Allan

Alex