Van der Waerden type theorem

Question

I am trying to understand the proof of that theorem. However, I am not able to comprehend how to use the existense of $\hat{w}(k;r-1)$ and part (iv) of Theorem 2.5 in order to complete the proof.

Answer 1

At this point, we can assume that none of the points in the set $$ \{d, 2d, 3d, \dots, \hat{w}(k;r-1) \cdot d\} $$ are red. The appeal to Theorem 2.5 is kind of bogus since the statement of that theorem doesn't even talk about $\hat{w}$, only about $w$, but the idea behind that is to appeal to the argument that coloring this set is equivalent to coloring $$ \{1, 2, 3, \dots, \hat{w}(k,r-1)\}. $$ and any such coloring contains a monochromatic $k$-term AP of the same color as its difference.

---

Just to be clear about this, let's make that formal. We have an $(r-1)$-coloring of the original set: a function $\chi : \{d, 2d, 3d, \dots, \hat{w}(k;r-1) \cdot d\} \to \{1,2,\dots,r-1\}$. Use it to define a coloring $\chi' : \{1,2,\dots, \hat{w}(k;r-1)\} \to \{1,2,\dots,r-1\}$ by setting $\chi'(x) = \chi(x\cdot d)$.

By definition of $\hat{w}$, any coloring of $\{1,2,\dots, \hat{w}(k;r-1)\}$ by $r-1$ colors contains a monochromatic $k$-term AP of the same color as its difference: there are numbers $x$ and $y$ such that the arithmetic progression $x, x+y, x+2y, \dots, x+(k-1)y$ and its difference $y$ are all the same color (with respect to $\chi'$).

By how we defined $\chi'$ earlier, this means that $xd, (x+y)d, (x+2y)d, \dots, (x+(k-1)y)d$, and $yd$ are all the same color with respect to $\chi$. But since $xd, (x+y)d, (x+2y)d, \dots, (x+(k-1)y)d$ is an arithmetic progression of length $k$ with common difference $yd$, that's exactly what we wanted to find, and the proof is done.