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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stackghost 28 minutes ago | parent [-]

>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 23 minutes ago \| parent [-]
		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.