Colouring of $\mathbb{N}$ that avoids all non-constant infinite arithmetic progressions? Can you color every positive integer either black or white such that there are no entirely white or entirely black non-constant infinite arithmetic progressions?



How about switching color every power of 2?

1,4,5,6,7,16,...,31,64,...

2,3,8,9,10,11,12,13,14,15,32,...

Any attempt at an AP would eventually get cut off.
Color the first integer white, the next two integers black, the next three integers white, etc. We will call these sections of successive integers with the same colors, "blocks". Thus we have successively blocks of length 1, 2, 3, 4, etc.

Now consider an arithmetic progression $a_n = a_0 + nd$ for some positive integers $a_0, d$ ($n = 0, 1, 2, \dots$). Let $k$ be a positive integer such that the block of length $kd$ comes after $a_0$. There is at least one element from the progression that lies in this block. Suppose that $a_n$ is the last element from the progression that lies in this block; then $a_{n+1}$ lies in the next block, of length $kd+1$, and therefore has the opposite color. Thus any arithmetic progression contains successive elements $a_n, a_{n+1}$ of opposite color, as we needed to prove.
If I Consider the irrational number $z=\sum_{i=1}^{\infty} 10^{-i!}=0,110001000...$ as a representation of natural numbers, where 1 means white and 0 means black. Let $\{a_n\}$ be an arithmetic progression, where $a_{n+1}-a_n=k$: if a similar series existed, it would mean that, in the decimal represntation of z there is the same digit repeating each k digits, but this is impossible because z is irrational. Therefore, this progression doesn't exist.