So Im working through my notes which prove Van der Waerden's Theorem for the case $m=3$. The method my lecturer has chosen is to first prove the Lemma below. The Lemma is proved by induction but I can't understand the base case.
The definition of a $focus$ is:
Can someone explain how this base case ($k=1$) works. I understand why we get a MAP of length 2 but my lecturer seems to chose $n$ as $r+1$ to begin with but then ends up concluding that $n$ is at most $2r+1$
The formulation "$n(r,3,1)$ exists and is at most $2r+1$" is misleading, as it talks about $n(r,3,1)$ as if this denoted a minimal number, whereas $n=n(r,m,k)$ was introduced as any number with the given property, not necessarily minimal. It should say "we can choose $n(r,3,1)=2r+1$" instead.
The lecturer didn't choose $n$ as $r+1$. Without fixing $n$, we can prove that any $r$-colouring of $\{1,2,\ldots,r+1\}$ contains a monochromatic arithmetic progression of length $2$ and thus any $r$-colouring of $\{1,2,\ldots,2r+1\}$ contains a focussed monochromatic arithmetic progression. Thus we can choose $n(r,3,1)=2r+1$.