Trying to prove dual of there are at least tree points on every line

[A_B]

trying to prove dual of "there are at least tree points on every line"

Hi,

Assuming the propositions of incidence:
(1) on any two distinct points is at least one line.
(2) on any two distinct points is at most one line.
(3) on any two distinct lines is at least one point.

and the extensions
(4) not all points are on the same line.
(5) There are at least three points on every line.

I am asked to prove the dual of (5), that is, "there are at least 3 lines on every point."
(The duals of all other statements above have already been proven.)

Perhaps there is something I'm missing (and that's the reason for posting) but it seems to me that if only one point exists (call it P) and no lines then (1), (2), (3), (4) and (5) are all vacuously true. So this would present a counterexample to the sought-after theorem.


thanks!
A_B

 

[mathwonk]

i agree with you.

 

[Hurkyl]

Can you word (4) more precisely?

 

[A_B]

Hi, thanks for the replies

(4) could be alternatively worded as

(4) There is no line such that all points lie on that line


And since in the case where there is only one point there is no line, this is vacuously true.

I've been looking at places other than the book I'm working through, and the condition

(A) there exist four points, no three of which are on the same line

is often added in a similar set of axioms. See for example http://www.math.cornell.edu/~web4520/CG3-0.pdf

Now it is obvious that (4) follow from (A), and according to the text linked to above, (5) will also follow (exercise 6 (a)).


I ran into this problem when beginning to work through "An Introduction to Projective Geometry" by O'Hara and Ward : http://archive.org/details/introductiontopr033583mbp.

Is anyone familiar with this book?
Would anyone suggest a better book?


Thanks,
A_B

 

[A_B]

rejoice!

If we take (1), (2) and (3) be be true and we consider the 'rival' extensions:

I. : (4) , (5) AND (6) THERE EXISTS AT LEAST ONE LINE

II. : (A)

Then I and II are equivalent.

I => II

since there is a line (a) (6), there are 3 points (A, B, C) on that line (5). But no line contains all points (4), so there must exist a fourth point (D) not on the line. But then there exists a line (b) on which lie the points A and D (1). The line (b) can not lie on either (B) or (C), since if it were, it would be equal to line (a) (2). So only points (A) and (D) lie on (b) so there must exist a point (E) distinct from all other points so far mentioned that is on (b). The points (B), (C), (D) and (E) satisfy II..



II => I

II => (4)
follows immediately

II => (5)
an excercise for the reader, hehe...
It is an excercise (6 a) here: http://www.math.cornell.edu/~web4520/CG3-0.pdf.

II => (6)
also follows immediately.


Clearly though, (6) is "yet another addition" to the extension, only there to exclude simple cases. The alternative II. seems preferable.


Still interested in better book suggestion though.

A_B

 

[Hurkyl]

Hi, thanks for the replies

(4) could be alternatively worded as

(4) There is no line such that all points lie on that line

That's one way, but what was intended by the textbook? Another way it could be worded is

(4) There exists a line and a point not on that line​

 

[A_B]

I have copied (4) "not all points are on the same line" directly from the book, I don't think that "There exists a line and a point not on that line" is equivalent to (4) as stated, thoughit does indeed solve the problem and it may indeed have been the meaning intended by the author.

Anyways, the book says "For the present they will suffice, but later more definite propositions will be substituted" about these extensions, so I suppose I shouldn't be too troubled by this problem.

A_B

 

[phinds]

I assume that

(3) on any two distinct lines is at least one point.

Is not saying that there is a point that exists on BOTH lines, since that would be incorrect for parallel lines (or any lines in 3D that do not intersect)

 

[A_B]

I assume that

(3) on any two distinct lines is at least one point.

Is not saying that there is a point that exists on BOTH lines, since that would be incorrect for parallel lines (or any lines in 3D that do not intersect)

That's exactly what it's saying. The notion of parallel lines is a euclidean (or affine) one, in projective geometry, all lines have a point in common. Think of a drawing of a railway track, while the rails never meet (Euclidean), the lines that represent them in the drawing do meet on the horizon. The drawing is a projection.

 

[HallsofIvy]

That's one way, but what was intended by the textbook? Another way it could be worded is

(4) There exists a line and a point not on that line​

That's not exactly the same thing since the original statement does NOT imply that a line exists. Perhaps "if L is a line then there exit point P not on line L".

 

[mathwonk]

i recommend you get a better book, hartshorne, hilbert...

 

[A_B]

Do you mean this book by Hartshorne: https://www.amazon.com/Foundations-Projective-Geometry-Robin-Hartshorne/dp/0805337571?

I can't find any book by Hilbert on projective geometry, do you mean "Foundations of Geometry"?

A_B