Posted by surprisetalk 7 days ago
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?
I think you meant premise?
... 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"
There isn't a third point on the line you found because the problem stipulates that the set of points is not collinear.
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If you're given the above set of points, it's obviously not collinear due to the point at the top, but if you draw a line through any of the bottom two points, it will hit a third.
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.
In any finite set of points, either there is a line hitting all points, or there is a line hitting exactly 2 points.
It's nontrivial to prove.
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.
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.
How so? It's bounded by the large initial triangle. The line containing any two of the vertices doesn't intersect any other point.
Anyone happen to know if it is true for countably infinite sets?
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.
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.
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.
The base case is n=2. The line joining them passes through exactly two points because that is all you have.
Now we attempt the induction step. We have n+1 points. Leave one, p, out. We know that the theorem applies to the n points by the induction hypothesis. So we have points q and r that have a line going through them. And the point of the theorem is that the line goes through only q and r, exactly two points of the n. All we have to do is add in p, not on that line, and we are done.
But we are also stuck. Point p is not one of the n points participating in the induction hypothesis. Nothing tells us that p is not on the line joining q and r.
So how do we prove it? It is a good, intriguing puzzle, but in proof theory, not geometry.
The Kochen-Specker theorem is interesting (if not useful). The proofs have a similar flavour to Sylvester. One tries to make a set of projections 'compatible' and it turns out to be impossible. Maybe there's a deeper connection.
There are many such lines (think convex hull) and they are easy to find. This makes it hard to appreciate the theorem. You keep thinking oh whats the big deal.
[1] https://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theor...
However, this is not true as it is possible that P'=B. However it seems the proof still goes through (at least as depicted in the image, haven't thought hard about the general case).
Its mostly a question of whether you count the line defining a half-plane as belonging to the half plane or not, and clearly they do here
Isn't this a tautology?
The problem definition states that the set of points is in Euclidean space, which from Euclid's Axioms means we can draw a line between any two points. The set of points is defined to be not collinear, thus we cannot draw a line passing through more than two of them. This is just simple logic.
This is what is stated here: there always exists at least a pair where this does not happen.
This is wrong and you are misunderstanding what collinearity means. You could have a set of points where all but one are on the same line, and the set of a whole will be not collinear, while we can obviously draw a line that passes through more than two of them.
A different way of stating the theorem is that any finite set of points has either a line passing through all points (i.e. the set is collinear) or there exists a line that passes through exactly two points. This dichotomy (why two and not three? Why can't we construct a set where any line passes through at least three points?) is not immediately obvious.
Let X be a set of points not all of which are collinear. Then, there are two points a, b in X such that the line l passing through X only passes through a and b.
I don't see how this rephrasing changes anything. Of course there are two points a and b because again, the definition of the problem leads naturally, obviously, and definitionally to this result.
Consider a 3x3 grid. It satisfies this argument.
Can you have a set of points Y on a plane such that Y satisfies the following? Given any line passing through 2 points on Y, there is also a third point in Y that it passes through.
There's nothing novel here. I feel like I'm taking fucking crazy pills.
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. . .As soon as the set of points are defined to be non-collinear in Euclidean space, this property must be true, purely from the definition of the problem. To suggest otherwise would be to violate either the problem definition or the axioms of Euclidean geometry.
EITHER all points are on the same line
OR 2 of the points are on the line only for them