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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...
Earlier quoted context omitted.
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.
>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…
I might be too stupid to understand why this is interesting and useful. If it helps I am a working physicist, and a lot of pure math is lost on me. I think I followed this, but I don't know why one would care or this would be interesting.
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.
> According to a strengthening of the theorem, every finite point set (not all on one line) has at least a linear number of ordinary lines. An algorithm can find an ordinary line in a set of n points in time O(n log n). [1] 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/Sy…
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"The set is not collinear" here means "there is no straight line passing through all the points simultaneously", not "there is no straight line passing through some three points".
... yes, I understand. There's nothing novel here. I feel like I'm taking fucking crazy pills.
EITHER all points are on the same line
OR 2 of the points are on the line only for them
Earlier quoted context omitted.
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.
>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…
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.
Earlier quoted context omitted.
That is not what was meant. Here is a better rephrasing: 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.
>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.
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.
> 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?
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.
> 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?
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.
Earlier quoted context omitted.
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.
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.