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