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 $\mathcal{F}$ ?



Permutation on $\omega$ and Ramsey ultrafilter
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". So-defined 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.


