An example showing that van der Waerden's theorem is not true for infinite arithmetic progressions One of the possible formulations of Van der Waerden's theorem is the following:

If $\mathbb N=A_1\cup \dots\cup A_k$ is a partition of the set $\mathbb N$, then one of the sets $A_1,\dots,A_k$ contains finite arithmetic progressions of arbitrary length.

In the other words, if we color the set of all positive integers by finitely many colors, there must be a monochromatic set containing arbitrarily long finite arithmetic progressions.
I assume that the same result is not true if we require that one of the sets contain an infinite arithmetic progression.  (Otherwise this result would be well-known.)

What is an example showing that this stronger version of the above theorem is not true? (I.e., an example of  a coloring of $\mathbb N$ by finitely many colors with no monochromatic infinite arithmetic progression.)


Let's build an example with just two sets in the partition, $A$ and $B$.
Enumerate the set $\mathbb{N}\times\mathbb{N}^{>0}$ as $\{(n_k,d_k)\mid k\in \mathbb{N}\}$ (think of $n$ as the starting point of an arithmetic progression and $d$ as the constant difference between the terms). We start with $A = B = \emptyset$ and ensure that at every stage of the construction (indexed by $k\in \mathbb{N}$), $A$ and $B$ are finite subsets of $\mathbb{N}$.
At stage $k$ of the construction, look at the pair $(n_k,d_k)$. 
Case $1$: $n_k\in A$. Let $m$ be the least natural number such that $n_k + md_k\notin A$, and add $n_k+md_k$ to $B$.
Case $2$: $n_k\in B$. Repeat Case $1$, but with the roles of $A$ and $B$ reversed.
Case $3$: $n_k$ is not in $A$ or $B$. Add $n_k$ to $A$ and repeat Case $1$.
To complete stage $k$ of the construction, if the natural number $k$ is not in $A$ or $B$, add it to $A$.
At the end of the infinite construction, we have a partition of all of $\mathbb{N}$ into the two sets $A$ and $B$ (since each $k$ was either in $A$ or in $B$ at the end of stage $k$). Suppose there is an infinite arithmetic progression contained in $A$. It begins with some $n$ and the terms have constant difference $d$. Now $(n,d) = (n_k,d_k)$ for some $k\in \mathbb{N}$. But at the end of stage $k$, we ensured that some element $n+md$ of the arithemetic progression was in $B$, contradiction. 
The symmetric argument shows that there is no infinite arithmetic progression contained in $B$.
