Prove that a subset of $\{1,2,\ldots,7\}$ which contains no arithmetic series does not intersect with certain subsets [closed]

Question

I have seen the following problem on the website few days ago. I forgot the site name. Any how, this problem is very interesting and fascinating for me. I have tried and What I felt, I am giving, under the problem. Please could you solve this?

Let $A$ be a subset of $\{1,2,\ldots,7\}$ containing no arithmetic series, then prove that:

1. $A\cap\{1, 4, 7\} = \varnothing$

2. $A \cap \{3, 4, 5\} = \varnothing$

3. $|A \cap T| \le 1$ for $T\in \{\{1, 2, 3\}, \{1, 3, 5\}\}$.

4. $|A \cap T| \le 1$ for $T\in \{\{3, 5, 7\}, \{5, 6, 7\}\}$.

5. $|A \cap T| \le 1$ for $T\in \{\{1, 3, 5\}, \{1, 4, 7\}, \{3, 4, 5\}, \{4, 5, 6\}\}$.

6. $|A \cap T| \le 1$ for $T\in \{\{1, 4, 7\}, \{3, 5, 7\}, \{2, 3, 4\}, \{3, 4, 5\}\}$.

7. $|A \cap T| \le 1$ for $T\in \{\{2, 3, 4\}, \{3, 4, 5\}, \{4, 5, 6\}, \{2, 4, 6\}\}$.

Work done: I'm not sure what 'containing an arithmetic series' means. Surely that doesn't contain an arithmetic series. Or $\{1,2,4\}$. $1,2$ could be part of an AP, as could any other 2 numbers. I notice that all the subsets in the question have 3 elements. Perhaps this is a requirement that I have missed some thing. At the same time, I feel may be I am wrong in understanding the problem. Please explain.

Answer 1

You need something about the size of $A$. By "arithmetic series" do they mean a 3-term arithmetic progression like $2,4,6$ or three elements $a,b,c$ with $a+b=c$? If $A=\{4\}$ it contains no arithmetic series and has non-null intersection with $\{1,4,7\}$ and $\{3,4,5\}$. If $A$ is required to have $4$ elements, item $1$ says the only $A$ is $\{2,3,5,6\}$, which does not include an arithmetic progression, but neither does $\{1,3,6,7\}$