A simple application of the Hales-Jewett theorem give the following result: if $r$, $k$ are natural numbers, then there is a finite set $S$ of natural numbers which does not contain a $(k+1)$-AP, yet every coloring with $r$ colors gives a monochromatic $k$-AP.
Let $N$ be sufficiently large. The HJ theorem says that if we color all sequences of length $N$, consisting of numbers between 0 and $k-1$, with $r$ colors, there is a monocolored combinatorial line, that is, a configuration of the form $f_0,\dots,f_{k-1}$, such that for a nonempty subset $T\subseteq \{0,\dots,N-1\}$ $f_i$ is $i$ in $T$, and the same outside it.
Now the set $S$ above consists of the sums $x_0+x_1M+x_2M^2+\cdots+x_{N-1}M^{N-1}$,
($0\leq x_j\leq k-1$), where $M$ is very large. That is, we map the HJ-hypercube via the mapping $(x_0,x_1,\dots,x_{N-1})\mapsto x_0+x_1M+x_2M^2+\cdots+x_{N-1}M^{N-1}$. If $M$ is large enough, then no linear combination with small ($\lt k$) coefficients that gives 0, unless the same linear combination gives 0 in the left hand side, i.e., in the hypercube, in which there is no $(k+1)$-AP.
