An example showing that van der Waerden's theorem is not true for infinite arithmetic progressions

Solution 1:

Color the integers black or white according to whether they have an even or an odd number of digits.

Solution 2:

I will add an example which is taken from the book Beautiful Mathematics by Martin Erickson, page 72. And it is rather similar to some of the answer which have already been posted.

$$\begin{array}{cccccccccccccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\ \circ & \bullet & \bullet & \circ & \circ & \circ & \bullet & \bullet & \bullet & \bullet & \circ & \circ & \circ & \circ & \circ & \bullet & \bullet & \bullet & \bullet & \bullet \end{array} $$

We start by coloring $1$ by one color. Then we color two following numbers by the other one. Next segment consist of three numbers of the first color, etc.

In this way we get longer and longer segments of each color.

If $a+nd$ is any arithmetic progression, then it will contain some numbers from both colors. Indeed there are two consecutive segments of lengths $d$ and $d+1$. Should this arithmetic progression be monochromatic, it would have to "skip over" one of those segments. But this is not possible, since the difference is $d$ and the length of each of these two segments is at least $d$.