Given a sequence $S$ with $21$ different numbers. It is known that there isn't any monotone subsequence in the length of $6$. Prove that there exists $2$ monotone subsequences, one decreasing and the other increasing, in the length of $5$.
Solution: $4\cdot5+1=21$
So according to Erdős-Szekeres we know that there exists an increasing monotone subsequence in the length of $5+1$ or decreasing subsequence in the length of $4+1$, or the other way around. And it's given there is no monotonic subsequence in the length of $6$. Therefore we know that if there isn't a increasing subsequence in the length of $6$ then there is a decreasing subsequence in the length of $5$ and, and if there isn't a decreasing subseqeuence in the length of $6$ there exists an increasing subsequence in the length of $5$. Therefore we have two subseqeuences , one decreasing one increasing in the length of $5$.
Can someone verify if this solution is correct. I am not sure if I understood the theorem correctly. Thank you in advance!
