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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 $n,k$?

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  • $\begingroup$ No: $A=\{ 2n: n\ge 0 \}$ $\endgroup$ – Christian Remling Jun 23 '14 at 6:01
  • $\begingroup$ @ChristianRemling If $\ A=\{2n:n≥0\}\ $,then $B=\{1,3,5,...\},2B = \mathbb Z_{\ge 2}$ for n=2? $\endgroup$ – Mike Jun 23 '14 at 6:08
  • $\begingroup$ Certainly not, as the sum of two odd numbers is even. $\endgroup$ – Christian Remling Jun 23 '14 at 6:12
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    $\begingroup$ 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. $\endgroup$ – S. Carnahan Jun 23 '14 at 7:44
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    $\begingroup$ What is a "consecutive set"? $\endgroup$ – S. Carnahan Jun 23 '14 at 8:17
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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.

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  • $\begingroup$ Thanks,How about $A=\{x\ |\ x\in\mathbb Z_{\ge 0},\ x=\ $some forms$\ \}$? $\endgroup$ – Mike Jun 23 '14 at 8:54
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    $\begingroup$ As I read this, I get that one picks c_j and d_j and then puts into A and B integers to ensure that c_j ends up in 2A and d_j ends up in 2B. However, it is also needed for the conclusion that (something like) c_j does not end up in 2B and d_j does not end up in 2A. How do we get this guarantee? $\endgroup$ – The Masked Avenger Jun 23 '14 at 16:07
  • $\begingroup$ @TheMaskedAvenger: hmm, I think you're exactly right. Thanks for pointing out the error. S. Carnahan's comment on the OP seems more correct. (At least this gave me the opportunity to learn that one can't delete an accepted answer....) $\endgroup$ – Greg Martin Jun 23 '14 at 17:47
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    $\begingroup$ At least you can edit it, give Scott's answer with attribution, and assuage your guilt by making the answer CW. $\endgroup$ – The Masked Avenger Jun 23 '14 at 17:51
  • $\begingroup$ Well, I didn't feel all that guilty :) but I still followed your good suggestion. $\endgroup$ – Greg Martin Jun 24 '14 at 22:14

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