# A general question on nonnegative integer sequence [closed]

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 $n,k$?

• No: $A=\{ 2n: n\ge 0 \}$ – Christian Remling Jun 23 '14 at 6:01
• @ChristianRemling If $\ A=\{2n:n≥0\}\$,then $B=\{1,3,5,...\},2B = \mathbb Z_{\ge 2}$ for n=2? – Mike Jun 23 '14 at 6:08
• Certainly not, as the sum of two odd numbers is even. – Christian Remling Jun 23 '14 at 6:12
• Define $A$ to contain the interval $[2^{2^n},2^{2^{n+1}}]$ if and only if $n$ is odd and positive. We have recently suggested that you stop asking basic questions here, but you've just asked another one. – S. Carnahan Jun 23 '14 at 7:44
• What is a "consecutive set"? – S. Carnahan Jun 23 '14 at 8:17

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.
• Thanks,How about $A=\{x\ |\ x\in\mathbb Z_{\ge 0},\ x=\$some forms$\ \}$? – Mike Jun 23 '14 at 8:54