I was recently looking at the happy ending problem, and on the linked Wikipedia page, there is the following conjecture (the Erdos-Szekeres conjecture), which states the following:
The smallest number of points for which any general point arrangement contains a convex subset of $n$ points is $2^{n-2}+1$.
General point arrangement just means that there are no three collinear points.
It is known that at least $2^{n-2}+1$ points are needed, but I had the following question. If we assume the truth of this conjecture, can we show the following claim?
For $n\ge 4$, if a set of $2^{n-2}$ points in general point arrangement does not have a convex subset of $n$ points, then the convex hull is an $(n-1)$-gon.
Looking at the pictures on the Wikipedia page, this seems to be the case, but I wasn't able to prove it. My thoughts/what I've found are below.
It's clearly true for $n=4$, since if a set of $2^{4-2}=4$ points didn't have a convex subset of $4$ points, the convex hull would have to be a triangle.
With $n=5$, if we had a set of $8$ points, and the convex hull was a triangle, there would have to be $3$ points for the vertices of the triangle, and $5$ points in the inside. The $5$ points wouldn't make a convex pentagon, so they're either a triangle with two points inside or a (convex) quadrilateral with a point on the inside. So overall, it's either two points inside a triangle inside a triangle or one point inside a quadrilateral inside a triangle.
The first case is considered as "type (3,3,2)" on this page, and it's shown that there exists a convex pentagon. So this isn't possible. For the second case, I think some similar argument could apply.
But this approach doesn't seem very generalizable to higher $n$ and never even assumes the truth of the Erdos-Szekeres conjecture. If we assume the conjecture, I feel like there's some more slick argument that I'm just not seeing to show this is true.