Is it possible to draw a finite or infinite set S of points in the plane, such that any line drawn in this plane neither intersects with exactly one point in S or an infinite number of points in S?
$\begingroup$
$\endgroup$
1
-
10$\begingroup$ S being the empty set satisfies this. $\endgroup$– hexominoJun 12, 2021 at 22:04
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
1
I believe the set of points (x,y) such that $x y = ±1$ satisfies the conditions.
The set consists of 4 segments of hyperbolas. Any straight line crosses at least 2 of these segments resulting in 2 to 4 intersections. Except for the lines x=0 or y=0 which cross none.
Here's a graph:
answered Jun 12, 2021 at 21:11
-
1
$\begingroup$
$\endgroup$
3
It is not possible with any finite or countably infinite set.
Take any point P in the set. For each other point Q in the set, draw a line between P and Q. Then take the angle between line PQ and the horizontal axis. This forms a set of angles A.
If S is countable, then the set of angles A must also be countable (since there is at most once angle per point in S). However, since angles are real numbers, and the real numbers are uncountable, by Cantor's diagonal argument we can construct some angle x not in our (countable) set A.
If we take the line going through the point P at the angle x, then this new line will not pass through any other point in the set; if it did, then its angle would have been in set A. Therefore, we have constructed a line that intersects exactly one point of the set S.
Since this is not allowed, and the only assumption we made about S was that it was countable, it must be the case that S is infinite and uncountable.
answered Jun 13, 2021 at 3:23
-
$\begingroup$ When you say "since there is at most once angle per point in S", don't you mean "at most one angle per pair of points in S"? And that's justified because the Cartesian product of a countable number of countable sets is countable. $\endgroup$ Feb 18 at 19:27
-
$\begingroup$ @LieutenantZipp No, we are only considering lines that pass through the fixed point P, so each point in S other than P defines a line. If S is finite then there are |S|-1 lines. However I think your variation also results in a working proof. $\endgroup$ Feb 18 at 19:42
-
$\begingroup$ Ah, gotcha. Apologies for the misreading. I thought you were taking all lines through all pairs of points, which now seems to me just an unnecessarily thorough variant of what you actually did. $\endgroup$ Feb 18 at 20:20
$\begingroup$
$\endgroup$
3
To rule out @hexomino's trivial solution (empty set) let us require that every straight line intersects S in fintitely many and at least 2 points.
Then one simple way to construct S is
using two shifted disjoint copies of the graph of the function y=x3, for example y±(x) = x3±1. Any straight line is either vertical and will intersect both graphs at the same x-offset or of the form yL(x) = ax+b. As each of yL(x)=y+(x) and yL(x)=y-(x) has between 1 and 3 solutions we get between 2 and 6 intersections.
answered Jun 13, 2021 at 4:10
-
$\begingroup$ I’m not sure how this rules out S as the empty set. The empty set is a set containing zero points. Any line drawn in the plane intersects with exactly zero points in S (not exactly one, and not infinite) since there are zero points in the set. Your answer constructs a distinctly non-empty set S, which doesn’t do anything to disprove that the empty set doesn’t work. $\endgroup$– El-GuestJun 13, 2021 at 6:25
-
3$\begingroup$ My argument is as follows: OP as it stands allows for the empty set as a valid solution. As this is probably not what OP intended I changed the question to explicitly rule out empty intersections. $\endgroup$ Jun 13, 2021 at 7:22
-
1$\begingroup$ Oh, you’ve changed the question, and provided an answer to the changed question. Fair enough. $\endgroup$– El-GuestJun 13, 2021 at 11:33