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> Every finite set of points in the Euclidean plane that is not collinear has a line that passes through exactly two of the points. I can't make out the point

> Every finite set of points in the Euclidean plane that is not collinear has a line that passes through exactly two of the points. I can't make out the point here (no pun). Of course a line can pass through any two points. It could pass through three if those points were collinear but the statement says they're not. So what is the new fact?

chenb4425 1mo ago
It’s that the line passes through exactly two points, which if you think about it is not exactly obvious.
math_loser 1mo ago
> So what is the new fact? For all arbitrarily sized (but finite) sets of not collinear points, there's always a line that passes through exactly two points in the set.
tzs 1mo ago
It can help to think about theorems like this by restating them as a puzzle asking for a counterexample. Given N points, N > 2, can you arrange them in a Euclidean plane so that (1) they are not all on the same line, and (2) every line that goes through two of the points must also go through at least one more of the points? The theorem says that you cannot do this.
- squirrel 1mo ago
  Well put. A similar way to say it: ask the 9 year old to draw ALL of the possible lines connecting two points in the set — of course this is possible, just might take awhile. The theorem says that at least one of the lines drawn must hit ONLY the two points the 9 year old used when drawing that particular line, not any others.
  - squirrel 1mo ago
    I just thought of another way to restate. Suppose you give me any finite set of points, any finite set at all, and you also tell me that when the 9 year old draws any line through any two of them, she will always hit a third. Then I can immediately conclude all the points lie on a single line, that is, that any line the 9 year old draws will hit all the points.
glimshe 1mo ago
Try to come up with a set non-colinear points where NO line passes through two and ONLY TWO points and you'll see the value of the statement. You may think "I'm sure I can arrange these points in a way where EVERY line will cross three or more points" but you will fail if you try unless ALL points are colinear.
- scythe 1mo ago
  This is true for finite sets. For infinite sets, the Sierpinski triangle is a counterexample.
  - ky3 1mo ago
    > the Sierpinski triangle is a counterexample How so? It's bounded by the large initial triangle. The line containing any two of the vertices doesn't intersect any other point.
  - tzs 1mo ago
    That’s an uncountable set. If we want a counter example for uncountable sets a simpler example is a circular area. Anyone happen to know if it is true for countably infinite sets?
    - namibj 1mo ago
      Just a regular grid? That's countable you just go in a spiral.
- JetSetWilly 1mo ago
  You can always find a line that passes between two points. Why would you even try to find a line that doesn’t pass between two points? What is the difficult part here.
  - ky3 1mo ago
    Find one that bridges only two points, no more. Unless ofc you're given a configuration that's obviously impossible: all of them in a line.
Sniffnoy 1mo ago
I think what's going on here is that you've misunderstood the theorem's hypothesis. The hypothesis isn't that no three of the points are collinear; rather, it's the weaker statement that there isn't any one single line that all the points lie on. It's true that with your version of the hypothesis the theorem would be trivial; but with the actual hypothesis it is is nontrivial.
- stackghost 1mo ago
  >rather, it's the weaker statement that there isn't any one single line that all the points lie on ... of course there's no single line that all the points lie on. They've been defined to be non-collinear. Edit: can't reply because of HN's stupid rate-limit mechanism, but to this: >So the theorem proves that no matter which way you arrange any finite set of points, except for all on the same line, then you can always find a line with exactly two points. Of course you can. It's absolutely implied by the problem definition. My 9 year old could do this, given a ruler and a pencil, with 100% success rate. I absolutely do not believe this is a novel "theorem"
  - hyperhello 1mo ago
    His statement helped me. It's not that every three points are non-collinear, it's that any three points are non-collinear. A set of points all lying on a line is the only exception; you can have every point lying on a line except for one, or two, or whatever you want. In a square grid of sixteen points, there are lots of sets of four collinear points for example, but not all sixteen, and that's what counts. So the theorem proves that no matter which way you arrange any finite set of points, except for all on the same line, then you can always find a line with exactly two points.
  - chenb4425 1mo ago
    I think you are mistaking the fact that you can easily find an example satisfying the theorem’s statement with the proof that the statement is always true. Of course given any set of points that aren’t all on the same line, your nine your old could find a line passing through only two points. But could they explain to you why this is always possible, no matter the configuration of points? You can’t just say “I draw a line between two points and that’s it.” You must also explain why there isn’t a third point on the line, and why that line’s existence is guaranteed, which is not obvious (at least to me).
ky3 1mo ago
Maybe this'll help: Every bunch of points on a plane has a lonely bridge. A lonely bridge is a line connecting only two points, no more. Obvious caveat: The points can't all lie in the same line (the collinear condition). Not-so-obvious caveat: There can't be an infinite number of points.
fxwin 1mo ago
> It could pass through three if those points were collinear but the statement says they're not The theorem doesn't presume that no three points are collinear, it presumes that the set as a whole isn't collinear, which is a much weaker statement.

chenb4425 1mo ago

It’s that the line passes through exactly two points, which if you think about it is not exactly obvious.

math_loser 1mo ago

> So what is the new fact? For all arbitrarily sized (but finite) sets of not collinear points, there's always a line that passes through exactly two points in the set.

tzs 1mo ago

It can help to think about theorems like this by restating them as a puzzle asking for a counterexample. Given N points, N > 2, can you arrange them in a Euclidean plane so that (1) they are not all on the same line, and (2) every line that goes through two of the points must also go through at least one more of the points? The theorem says that you cannot do this.

squirrel 1mo ago
Well put. A similar way to say it: ask the 9 year old to draw ALL of the possible lines connecting two points in the set — of course this is possible, just might take awhile. The theorem says that at least one of the lines drawn must hit ONLY the two points the 9 year old used when drawing that particular line, not any others.
- squirrel 1mo ago
  I just thought of another way to restate. Suppose you give me any finite set of points, any finite set at all, and you also tell me that when the 9 year old draws any line through any two of them, she will always hit a third. Then I can immediately conclude all the points lie on a single line, that is, that any line the 9 year old draws will hit all the points.

glimshe 1mo ago

Try to come up with a set non-colinear points where NO line passes through two and ONLY TWO points and you'll see the value of the statement. You may think "I'm sure I can arrange these points in a way where EVERY line will cross three or more points" but you will fail if you try unless ALL points are colinear.

scythe 1mo ago
This is true for finite sets. For infinite sets, the Sierpinski triangle is a counterexample.
- ky3 1mo ago
  > the Sierpinski triangle is a counterexample How so? It's bounded by the large initial triangle. The line containing any two of the vertices doesn't intersect any other point.
- tzs 1mo ago
  That’s an uncountable set. If we want a counter example for uncountable sets a simpler example is a circular area. Anyone happen to know if it is true for countably infinite sets?
  - namibj 1mo ago
    Just a regular grid? That's countable you just go in a spiral.
JetSetWilly 1mo ago
You can always find a line that passes between two points. Why would you even try to find a line that doesn’t pass between two points? What is the difficult part here.
- ky3 1mo ago
  Find one that bridges only two points, no more. Unless ofc you're given a configuration that's obviously impossible: all of them in a line.

Sniffnoy 1mo ago

I think what's going on here is that you've misunderstood the theorem's hypothesis. The hypothesis isn't that no three of the points are collinear; rather, it's the weaker statement that there isn't any one single line that all the points lie on. It's true that with your version of the hypothesis the theorem would be trivial; but with the actual hypothesis it is is nontrivial.

stackghost 1mo ago
>rather, it's the weaker statement that there isn't any one single line that all the points lie on ... of course there's no single line that all the points lie on. They've been defined to be non-collinear. Edit: can't reply because of HN's stupid rate-limit mechanism, but to this: >So the theorem proves that no matter which way you arrange any finite set of points, except for all on the same line, then you can always find a line with exactly two points. Of course you can. It's absolutely implied by the problem definition. My 9 year old could do this, given a ruler and a pencil, with 100% success rate. I absolutely do not believe this is a novel "theorem"
- hyperhello 1mo ago
  His statement helped me. It's not that every three points are non-collinear, it's that any three points are non-collinear. A set of points all lying on a line is the only exception; you can have every point lying on a line except for one, or two, or whatever you want. In a square grid of sixteen points, there are lots of sets of four collinear points for example, but not all sixteen, and that's what counts. So the theorem proves that no matter which way you arrange any finite set of points, except for all on the same line, then you can always find a line with exactly two points.
- chenb4425 1mo ago
  I think you are mistaking the fact that you can easily find an example satisfying the theorem’s statement with the proof that the statement is always true. Of course given any set of points that aren’t all on the same line, your nine your old could find a line passing through only two points. But could they explain to you why this is always possible, no matter the configuration of points? You can’t just say “I draw a line between two points and that’s it.” You must also explain why there isn’t a third point on the line, and why that line’s existence is guaranteed, which is not obvious (at least to me).

ky3 1mo ago

Maybe this'll help: Every bunch of points on a plane has a lonely bridge. A lonely bridge is a line connecting only two points, no more. Obvious caveat: The points can't all lie in the same line (the collinear condition). Not-so-obvious caveat: There can't be an infinite number of points.

fxwin 1mo ago

> It could pass through three if those points were collinear but the statement says they're not The theorem doesn't presume that no three points are collinear, it presumes that the set as a whole isn't collinear, which is a much weaker statement.