03 Jun 08
Originally posted by David113My first instinct is to say no. I think I can prove that no continuous curve works, and clearly no bounded set works. But there could be some bizarre collection of discrete points that does the job - I can't see how to disprove that quite yet.
I don't have a solution...
Is there a set of points S in the plane, such that every straight line in the plane contains exactly 2 points of S?
03 Jun 08
Originally posted by JirakonA circumference is the immediate intuitive answer, but I fail to see what sense it makes to consider these intersections at infinity.
What about the points that define a circle with a radius of infinity?
That's the first thing that came to mind. It would have to be infinite, since if it were a finite number of points, one could easily draw a horizontal line far above the last point. I'm pretty sure that's the solution.
Give me a functional form taking the limit of r to infinity and I'll give you an epsilon that proves that there is at least a line that doesn't intersect on two points.
03 Jun 08
Originally posted by JirakonCare to mention one of the points in that set then?
What about the points that define a circle with a radius of infinity?
That's the first thing that came to mind. It would have to be infinite, since if it were a finite number of points, one could easily draw a horizontal line far above the last point. I'm pretty sure that's the solution.
03 Jun 08
Originally posted by FabianFnasA point is an ordered pair of two REAL numbers. No infinity.
What about the points that define a circle with a radius of infinity?
Then every line should intersect this circle in exactly two points, at the polar coordinates r1=r2=inf, and phi1=f and phi2=f+pi.
Is this the answer?
03 Jun 08
3 edits
Originally posted by mtthwI don't think it can be discrete points. Imagine if you have a set of discrete points and pick a line that intersects only two points. If you pick one of those points as center and keep rotating that line infinitesimally, then I don't see how it's possible to cover all possibilities with any mesh of discrete points.
My first instinct is to say no. I think I can prove that no continuous curve works, and clearly no bounded set works. But there could be some bizarre collection of discrete points that does the job - I can't see how to disprove that quite yet.
Edit - So I'd say no, there is no such set S.
Edit 2 - Which probably means I'm wrong and missing an ingenious answer.
03 Jun 08
Originally posted by Palynka
A circumference is the immediate intuitive answer, but I fail to see what sense it makes to consider these intersections at infinity.
.
if you do include any trans finite numbers in the plane; then ithink that you need to include all trans finite numvbers.
another quwstion is: can you define a plane, and lines on that plane, so that set s exists.
03 Jun 08
A point is an ordered pair of two REAL numbers. No infinity.
If that's the case, then there is no solution:
Suppose such a set exists in which all points have finite coordinates. Take the point(s) with the highest y-value (y_max). Now imagine the line y = y_max + 1. This line does not intersect any of the points. Therefore no such set exists.
03 Jun 08
Originally posted by JirakonWell, it was a wild guess, the closest I can get to an answer.
What about the points that define a circle with a radius of infinity?
That's the first thing that came to mind. It would have to be infinite, since if it were a finite number of points, one could easily draw a horizontal line far above the last point. I'm pretty sure that's the solution.
Is it possible to construct a plane where inf really is included? As you can with R, i.e. R* ?