1
$\begingroup$

This is the excerpt from the book "Ramsey theory on the integers". There are some confusing moments and I would be thankful if anybody helps me to grasp it.enter image description here

The definition of $\nu_k(n)$ I understood. In the second paragraph author proves that $\nu_3(8)\leq 4$.

I have the following questions:

1) When considers the case $|T|\geq 5$ why in that case we are guaranteed to have a monochromatic 3-term AP?

2) Why he also working on the case $|T|\geq 6$? He already had worked on the case $|T|\geq 5$?!

3) Why he separately examining the case $|T|=5$?

$\endgroup$
1
$\begingroup$

First I will admit I do not fully understand this paragraph ... indeed I would have written ... There are $56$ subset of cardinality $5$ in $[1,8]$; I leave as an exercise to the reader to write them down & confirm that each one contains an arithemetic progression of length $3$. Hence blah blah.

This does not answer any of your 3 Questions.

1) This is a statement of what will be established & at this stage it is a claim.

2) He establishes that it suffices only to look at the case $ \mid T \mid =5$ because a larger cardinality would contain this case.

3) I have addressed in my answer to 2).

$\endgroup$
  • $\begingroup$ Dear Donald! I suppose that I understood you right now after reading your post. As I realized 1) it is just claim(statement) which validity he shows in 2) and 3). Firstly, he shows that if $|T|\geq 6$ then it always contains AP. Secondly, he considers the case $|T|=5$. Right? $\endgroup$ – ZFR Mar 9 '17 at 6:44
  • $\begingroup$ @RFZ Yes, You are right, & we are now sure that $\nu_{3}(8)=4$. Good luck mastering the rest of the good Mr Ramsey's Book. $\endgroup$ – Donald Splutterwit Mar 9 '17 at 20:08
  • $\begingroup$ One more question: In the case of $|T|=5$ (since $|R|=3$) why he considers only these $R$ which contain monochromatic AP? What about if $R$ does not contain AP? For example, $R=\{1,2,7\}$?! $\endgroup$ – ZFR Mar 9 '17 at 20:47
  • $\begingroup$ @RFZ This is the killer question : Why when looking for a maximal cardinality set that avoids an arithemetic progression we need only consider those sets whose complement does contain an arithmetic progression? I did consider this before answering your question & believe me if I had cracked that nut I would have given the explanation in my answer. I will continue to think about this & let you know if I ever find the reason. $\endgroup$ – Donald Splutterwit Mar 9 '17 at 22:16
  • $\begingroup$ Thank you! Would be very thankful if you can give answer in the soon future! $\endgroup$ – ZFR Mar 10 '17 at 8:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.