Suppose 55 students registered in a course line up arbitrarily. Prove that I can go through the line-up and identify 8 students (not necessarily consecutive) who are either in increasing order of height or decreasing order of height. Assume that no two students are exactly the same height.
I've tried using Mirsky's theorem to prove find the max chain but I'm stuck on finding the minimum number of possible antichains.
This is a direct application of the Erdős–Szekeres theorem.
Proof: Give each student $i$ in line a pair of values $(x_i,y_i)$ where $x_i$ represents the length of the longest increasing chain of students that ends with student $i$ and $y_i$ represents the length of the longest decreasing chain of students that ends with student $i$. Observe that no two students have the same pair of values: given any $j>k$ if student $j$ is taller than student $k$ then $x_j>x_k$ and otherwise $y_j>y_k$. There are only 49 pairs of numbers between $(1,1)$ and $(7,7)$, so among 55 students there must be one with one value or the other over 7. This represents the existence of a monotone chain that is 8 or more students as desired.