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This is consequence of my earlier post. I am not very happy with earlier replies of my post. I hope, I will get good response for this post. With lot of hope, I am sending the following problem and I want to learn the best from the experts. Please discuss/prove the following one. Thank you.

Let $A$ be a subset of $\{1,2, \dots,8\}$ containing no Arithmetic Progression. Then one of the following Two must be hold:

  1. $|A \cap\{3, 5\}|\le 1$ and $|A \cap\{1, 2, 4, 6, 7, 8\}|\le 1$.
  2. $|A \cap \{4, 6\}|\le 1$ and $|A \cap\{1, 2, 3, 5, 7, 8\}|\le 1$.

Let A be a subset of $\{1, 2, \dots,10\}$ containing no Arithmetic Progression. Then one of $|A \cap \{1, 2, \dots,7\}|$ and $|A \cap \{4, 5, \dots, 10\}|$ is at most $3$.

Let $A$ be a subset of $\{1,2,\dots,13\}$ containing no Arithmetic Progression. Then one of the following Two must be hold:

  1. $|A \cap \{1, 2, 3, 4, 6\}|\le 3$ and $|A \cap \{5, 7, 8, 9, 10, 11, 12, 13\}|\le 4$.
  2. $|A \cap\{8, 10, 11, 12, 13\}|\le 3$ and $|A \cap \{1, 2, 3, 4, 5, 6, 7, 9\}|\le 4$.

Thank you all.

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    This is not set theory. I will re-tag.2011-11-05
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    In the last part of your question, don't you mean for $A$ to be a subset of $\lbrace1,2,\dots,13\rbrace$?2011-11-05
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    By the way, you should include a link to your earlier question so we don't reinvent too many wheels here.2011-11-05
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    It is generally a bad idea to use (non standard!) abbreviations in titles: it is impossible to know what the question is about from its title... You saved an infinitessimal amount of time writing «A.P.» and as a consequence lots of people (among who are those that will probably help you!) to lose time: when writing for others, one should be very mindful of *their* convenience...2011-11-05
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    I am giving my previous post link below: http://math.stackexchange.com/questions/78474/prove-that-a-subset-of-1-2-ldots-7-which-contains-no-arithmetic-series-do2011-11-05

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