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

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.

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Perhaps you mean an "arithmetic progression"? It is a more common term. $\{1,4,7\}$ would be an arithmetic progression of length $3$ because all its terms can be written in the form $1 + 3k$, and an arithmetic progression in general would look like $a+kb$ with $k$ over some range of integers. Containing such a progression would mean that collecting terms in an arithmetic progression and making them as a set, if that set is a subset of A, then A would contain the arithmetic progression. – Patrick Da Silva Nov 3 '11 at 4:19
you are correct! But, I need each problem separately. Please solve any three. So that, I will try rest of them by my own. Because, I want to learn... – mathew Nov 3 '11 at 5:39
This question is broken, and the OP doesn't seem able to fix it. Voting to close. – TonyK Nov 3 '11 at 8:55

## closed as not a real question by TonyK, Ross Millikan, Asaf Karagila, Mike Spivey, t.b.Nov 7 '11 at 5:15

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

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 $\{1,3,6,7\}$
 ! Could you give little more explanation and prove the all 7 questions.Please... – mathew Nov 3 '11 at 5:37 Patrick Da Silva has explained what an arithmetic progression is. There are some of them in $\{1,2,\ldots,7\}$ like 1,2,3; 3,5,7; etc. If A is too small, it doesn't have to meet any of these. Unless we know more about A, there is no question here. – Ross Millikan Nov 3 '11 at 14:58 Hello experts! I am waiting for your valuable response on my post. Please... – mathew Nov 4 '11 at 8:03 @mathew: I think I showed there is not a solvable question here. I can't even guess what it is, as making A have 3 or 4 elements doesn't work either. If you can't find the original, I don't know where to go. – Ross Millikan Nov 4 '11 at 12:59 Thank you for your response sir. – mathew Nov 5 '11 at 8:49