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 $|T|=5$?


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).

  • $\begingroup$ 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

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