Return to Revisions

No. There are countably many arithmetic progressions $\{n,n+k,\dots\}$; enumerate them in some order. Then recursively prescribe elements of $A$ and $B$: given the $j$th arithmetic progression, insist that two large integers (larger than anything in $A$ or $B$ already) whose sum lies in the AP must be in $A$, and two other large integers whose sum lies in the AP must be in $B$. The result will be disjoint infinite subsets $A_0$ and $B_0$; set $A=A_0$ and $B=\Bbb N\setminus A$, and neither $2A$ nor $2B$ will contain any full arithmetic progression.