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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.
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>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"

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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).
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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.

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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.

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This is true for finite sets. For infinite sets, the Sierpinski triangle is a counterexample.
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It’s that the line passes through exactly two points, which if you think about it is not exactly obvious.
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> 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.

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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.

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