Proving a monotonic subsequence exists CONTEXT
I have been studying about finding the limit of a sequence lately, and it becomes apparent that one way for a sequence to have limit is that it is monotonic and has a lower (or upper) bound. I, then, recognized that a sequence can also be convergent at many points.
And this problem comes to my mind. It is not any part of the convergent issue but rather a combinatoric one.
I had answered this, but I want to see if any of you do anything different.
Any solution is appreciated.
PROBLEM
Let there be a numerical sequence consists of $(m\times n+1)$ numbers $a_1, a_2,a_3,...,a_{mn+1}$, with $i \in \mathbb{N}$, and $a_i \in \mathbb{R}$.
Prove that:$\exists$ an increasing subsequence of $m+1$ numbers or a decreasing subsequence of $n+1$ numbers
 A: @Gerry Myerson This is my way for the problem. I don't know if it has anything to do with Erdos-Szekeres, but still check this out.
Solutions
Each number $x$ in the sequence belongs to some sub-sequences. Note that the sub-sequence have $\ge 1$ numbers.
With each number $x$, denote $P(x)$ as the longest ascending sub-sequence that includes $x$, and the number of numbers in $P(x)$ is k.
Thus, we have $\exists x_1 < x_2 < x_3 < ... < x_k = x$
Assume that for all x, we have $\vert P(x) \vert \le m \tag{1}$
Consider all the following sets: $A_i = \left\{x \vert P(x)=i \right\}; i=1,2,...,m$
Because we have $mn+1$ numbers, so according to the Pigeonhole Principle, there exists one set $A_k$ such that $\vert A_k \vert \ge n+1$
But with each number $a_i$ and $a_j$ belong to the same $A_k$, if, $i<j$ and $a_i<a_j$, then by definition, since $a_i$ and $a_j$ are already the largest numbers in their sub-sequences, $a_i=a_j$, which is a clear contradiction. Thus, the numbers in $A_k$ forms a descending sequence of $n+1$ numbers. $\tag{2}$
(1) and (2) $\Rightarrow$Q.E.D
Post script...
While demanding that there is another approach to this problem, I think this is the most elementary and basic solution that any beginner at combinatorics can think of.
Still, I appreciate your ways of thinking and introduction of new theories that I haven't known.
