Let $A=\{x\ |\ x\in\mathbb Z\_{\ge 0},\ x\ $ with some conditions$\ \}$.  

Let $B=\mathbb Z\_{\ge 0}-A$.  

Define $\ 2A= \{a+b : a \in A,\ b \in A\}$.  

Define $\ 2B=\{a+b : a \in B,\ b \in B\}$.  

Then the set $\ \{n,\ n+1k ,\ n+2k, \ ...\}\ \subseteq\ 2A\ $or $\ 2B$ for some positive integers $n,k$?



A general question on nonnegative integer sequence
No. As suggested by S. Carnahan (with the exact numbers tweaked),
$$
A = \bigcup\_{k=0}^\infty [3^{2k},3^{2k+1}) \quad\text{and}\quad B = \Bbb N\setminus A = \bigcup\_{k=0}^\infty [3^{2k+1},3^{2k+2}).
$$
Then
$$
2A \subset \bigcup\_{k=0}^\infty [3^{2k},2\cdot3^{2k+1}) \quad\text{and}\quad 2B \subset \bigcup\_{k=0}^\infty [3^{2k+1},2\cdot3^{2k+2}),
$$
and hence both $2A$ and $2B$ have arbitrarily large gaps; this prohibits either set from containing an infinite arithmetic progression.


