Non-progressing sets

$S_1 = \{ n \in\mathbb{N} :$ if $\lfloor\log_{10} (1+\log_{10} n)\rfloor$ even $\}$

$S_2 = \mathbb{N} \setminus S_1$

Disclaimer: This is by no means a rigorous proof but just an intuition on which a proof can be built upon.

The first condition can be satisfied by having 'gaps' in both $S_1$ and $S_2$ that grow faster in size than linearly. An example of such a construction is having $S_1$ be the set of integers whose number of digits is odd and $S_2$ be the set of integers whose number of digits is even. The intuition is that for any arithmetic progression $a_n = a_0 + kn$ , where $k$ is the common difference, there would exist an $n'$ such that for all $n > n'$ , the gap is larger than $k$ and so one of $a_n$ is going to have to fall through the gap and belong to the other set.

That said, the 2nd condition can be seen similarly by noticing that for any geometric progression $g_n$ , the number of prime factors counted with multiplicity follows an arithmetic progression. A construction that disallows such would satisfy the 2nd condition.

Hence the construction below satisfies both conditions:

$S_1$ is the set of integers whose number of prime factors, counted with multiplicity, has an even number of digits. $S_2$ is the set of integers whose number of prime factors, counted with multiplicity, has an odd number of digits.

Not sure but $S_1$ can be assigned any random natural numbers that follow the given description, which is possible because I can design an example, and $S_2 = n$ $\epsilon$ $\N - S_1$