In their 1935 paper, A combinatorial problem in geometry, Erdos and Szekeres prove Ramsey's Theorem. One of the cases is:
If $i = 1$, the theorem holds for every $k$ and $l$. For if we select out of $m$ some determined elements (combinations of order $1$) as the class $\alpha$, so that every $k$-gon must contain at least one of the $\alpha$ elements, there are at most $(k-l)$ elements which do not belong to the class $\alpha$. Then there must be at least $(m-k+l)$ elements of $\alpha$. If $(m-k+1) \ge l$, then there must be an $l$-gon of the $\alpha$ elements and thus $m \le k+l-2$ which is evidently false for sufficiently great $m$.
How does the math work out in the last few lines? I have that if $(m-k+1)\ge l$, $m\ge k-l+1$. Can someone help me understand how they arrive at $m \le k+l-2$?
The idea is that if $m - k + 1 \ge l$, then there's an $l$-gon, contradicting our assumption.
So this cannot be, and we have $m-k+1 \le l-1$ instead. (The $-1$ appears because both sides are integers.) This can be rearranged to $m \le k+l-2$.