I have problems with the construction of $2^{n-2}$ points that contain no n-gon, particulary, the proof of the book "Open Problems in Mathematics".
The proof sais that:
For $i = 0, ..., n-2$ let $T_i$ be a set of ${{n-2}\choose i}$ points in general position in the plane, containing no $i+2$-cap and no $n-i$-cup and having the property that no two points in the set are connected by a line having slope of absolute value greater than 1. (The existence of $T_i$ follows from a previous theorem, and the condition on slopes can be satisfied by “compressing” $T_i$ vertically.) Place a small copy of $T_i$ in a neighborhood of the point on the unit circle making an angle of $\frac{\pi}{4} - \frac{i\pi}{(n-2)2}$ with the positive x-axis. Let $X$ be the union of $T_1, T_2, ..., T_{n-2}$. Clearly, $|X| = 2^{n-2}$. Suppose that $Y$ is a convexly independent subset of $X$. Let $k$ and $l$ be the smallest and the largest values of $i$ so that $Y\cap T_i\neq \emptyset$. If $k=l$, then $Y$ contains no $k+2$-cap and no $n-k$-cup. The construction guarantees that:
- $Y \cap T_k$ is a cap of at most $k + 1$ points,
- $Y \cap T_l$ is a cup of at most $n + l- 1$ points,
- $|Y \cap T_i|\leq 1$ for all $i = k + 1, ..., l-1$
Summing up, $|Y| = (k + 1) + (l + k - 1) + (n -l - 1) = n-1$.
I understood all the reasoning except the part that say:
$|Y\cap T_i|\leq 1$ for all $k+1\leq i\leq l-1$.
Why in the middle sets the intersection with $Y$ have at least one element?