let $\pi:\omega\to\omega$ be permutation and $\mathcal{F}$ is Ramsey selective ultrafilter on $\omega$. There are uncountable many increasing subsequences of $\pi$. Can one proof that one of them has domain in $\mathcal{F}$ ?

12$\begingroup$ Assuming that "growing" means monotonically increasing, the answer is yes. The Ramsey property implies that, on some set in $\mathcal F$, $\pi$ will be either increasing or decreasing. But there is no infinite decreasing sequence in $\omega$, so "increasing" is the only possibility. $\endgroup$ – Andreas Blass Dec 9 '17 at 17:54

$\begingroup$ Sorry for my bad English. "growing" == "increasing". But how to proof the fact ? $\endgroup$ – ar.grig Dec 10 '17 at 20:03

2$\begingroup$ @ar.grig: define a coloring $f: [\omega]^2 \to 2$ such that for $i<j$, f(i,j)=0 if $\pi(i)<\pi(j)$, otherwise it gets $1$. By the Ramsey property, there is a large set such that $f$ is monochromatic restricted on the set. It must then be 0. $\endgroup$ – Jing Zhang Dec 11 '17 at 15:42

2$\begingroup$ I'm not sure what remains to be explained. Just apply the definition of "Ramsey ultrafilter" to the partition of $[\omega]^2$ into the pairs that remain in the same order when you apply $\pi$ and the pairs whose order is reversed by $\pi$. $\endgroup$ – Andreas Blass Dec 11 '17 at 18:58

3$\begingroup$ The best detailed text on ultrafliters is the book "Theory of ultrafilters" by Comfort and Negrepontis. The specific result, relating selective to Ramsey, is due to Kunen and was published in David Booth's thesis; I think the title is "Ultrafilters on countable sets" and it's in one of the early volumes of the Annals of Mathematical Logic, but I don't have time to check right now. $\endgroup$ – Andreas Blass Dec 12 '17 at 14:01
Thanks to @AndreasBlass and @JingZhang for the answer. First of all saying "Ramsey ultrafilter" I meant "Every partition of $\omega$ into sets not in ultrafilter admits a selector in ultrafilter". Sodefined ultrafilter better to be called selective. And Ramsey ultrafilter can be defined as follows "Every partition of $[\omega]^2$ into two pieces has a homogeneous set in ultrafilter". This two definitions are eqivalent, so termins "Ramsey" and "selective" are used as synonyms in literature, which led me to some misunderstanding. If we know that this two definitions are equivalent, the answer is trivial, what was noted by @AndreasBlass and @JingZhang.
But proof of equivalence is not trivial in part "selective" $\implies$ "Ramsey". Proof can be found in the book "Theory of ultrafilters" by Comfort and Negrepontis cited by @AndreasBlass.

1$\begingroup$ While the proof of "selective implies Ramsey" is not trivial, it is quite straightforward to prove (assuming CH) that Ramsey ultrafilters exist, so that (since the implication "Ramsey implies selective" is trivial) at least some selective ultrafilters are Ramsey. $\endgroup$ – bof Dec 20 '17 at 9:56