Question about the proof of Ramsey's Theorem

Question

I was reading up on a proof of Ramsey's Theorem and I can't understand this part of the proof:

Pick a vertex $v$ from the graph, and partition the remaining vertices into two sets $M$ and $N$, such that for every vertex $w$, $w$ is in $M$ if $(v, w)$ is blue, and $w$ is in $N$ if $(v, w)$ is red. Because the graph has $R(r - 1, s) + R(r, s - 1) = |M| + |N| + 1$ vertices, it follows that either $|M| \geq R(r − 1, s)$ or $|N| \geq R(r, s − 1)$.

Why is that $|M| \geq R(r − 1, s)$ or $|N| \geq R(r, s − 1)$? I know from the proof of $R(3,3) = 6$ that given a vertex, at least $3$ are guaranteed to be either red or blue, so what is the pigeonhole argument for the general case?

Answer 1

Because if it wasn't true that either $|M|\ge R(r-1,s)$ or $|N|\ge R(r,s-1)$, then we would have both $|M|\le R(r-1,s)-1$ and $|N|\le R(r,s-1)-1$, which would imply $$|M|+|N|\le R(r-1,s)+R(r,s-1)-2.$$ But this contradicts the fact that $|M|+|N|+1 = R(r-1,s)+R(r,s-1)$.

Answer 2

If $\left\vert{M}\right\vert \ngeq R(r-1,s)$ and $\left\vert{N}\right\vert \ngeq R(r,s-1)$, then $\left\vert{M}\right\vert \leq R(r-1,s)-1$ and $\left\vert{N}\right\vert \leq R(r,s-1)-1$. Therefore, number of vertices $=\left\vert{M}\right\vert + \left\vert{N}\right\vert +1 \leq R(r-1,s)-1 +R(r,s-1)-1 +1$. That is,$$\left\vert{M}\right\vert + \left\vert{N}\right\vert +1 \leq R(r-1,s) +R(r,s-1)-1$$ which is a contradiction to $$\left\vert{M}\right\vert + \left\vert{N}\right\vert +1 = R(r-1,s) +R(r,s-1)$$