Help on a proof about Ramsey ultrafilters I was studying in this proof the implication $1\Rightarrow3$. I don't understand why $\{x_n,x_m\}\in[X]^2$ and $x_m>x_n$ implies $x_m\in X_{x_n}$. I believe it is not true, I think, first of all, we have to take $X^\prime_n=\bigcap_{i\leq n,i\in Y}X_i\cap Y$, but I can't fix the rest. Can anyone help me please?

Begin with a definition: $\mathcal{F}$ is called Ramsey if for every descending sequence $\{X_n:n\in\omega\}\subset\mathcal{F}$ there exists $\{x_n:n\in\omega\}\in\mathcal{F}$ such that $x_n\in X_n$ for every $n\in\omega$.
I want to prove that:
If $\mathcal{F}$ is Ramsey then for every set $A\subset[\omega]^2$ (not-ordered pairs of elements in $\omega$) there exists $X\in\mathcal{F}$ such that $[X]^2\subset A$ or $[X]^2\cap A=\emptyset$.
Here is the proof: for $n\in\omega$ define $X_n:=\{k>n:\{n,k\}\in A\}$, there exists $Y\in\mathcal{F}$ such that $X_n\in\mathcal{F}$ $\forall n\in Y$ or $\omega\backslash X_n\in\mathcal{F}$ $\forall n\in Y$. Assume that the first case holds. For $n\in Y$ define $X^\prime_n:=\bigcap_{i\leq n,i\in Y}X_i$. Apply the hypothesis to $\{X^\prime_n:n\in Y\}$ to get a set $X\in\mathcal{F}$. Then if $n,m\in Y$ and $x_n<x_m$ we have $x_m\in X_{x_n}$ and so $\{x_n,x_m\}\in A$.
I don't understand at the end, why $x_m\in X_{x_n}$.
 A: This isn't an answer, but it's too long for a comment.
Let's say that a non-principal ultrafilter $\mathcal{F}$ on $\omega$ is:  


*

*Ramsey(BJ) iff for every decreasing sequence $X_n$ of sets in $\mathcal{F}$, there exists a set $\{ x_n | n \in \omega \} \in \mathcal{F}$ such that for each $n$, $x_n \in X_n$;

*Ramsey(BJ)${}^+$ iff for such a decreasing sequence $X_n$, the set of $x_n$ that we get satisfy the stronger condition that $x_0 \in X_0$ and $x_{n+1} \in X_{x_n}$; 

*Ramsey(J) iff for every partition $\omega = \bigsqcup _n Z_n$ into "small" sets (i.e. each $Z_n \not\in \mathcal{F}$), there exists $X \in \mathcal{F}$ such that for each $n$, $|X \cap Z_n| \leq 1$;

*Ramsey(R) iff for every colouring $c : [\omega]^2 \to 2$ there exists $X \in \mathcal{F}$ such that $|c''[X]^2| = 1$; and

*Ramsey(R)${}^+$ iff for every colouring $c : [\omega]^n \to k$ (for any finite $n, k$) there exists $X \in \mathcal{F}$ such that $|c''[X]^n| = 1$.


The "BJ" is supposed to make you think of Bartoszynski and Judah, the "J" for Jech, and the "R" for Ramsey, since the Ramsey(R) definitions look most evidently like generalizations of Ramsey's theorem.
Now, Bartoszynski and Judah define a Ramsey ultrafilter to be a Ramsey(BJ) ultrafilter.  And Theorem 4.5.2 in that text is essentially proving that Ramsey(BJ), Ramsey(J), and Ramsey(R) are equivalent (ignoring clause 4 of that theorem).  The proof they give of "Ramsey(BJ) implies Ramsey(R)" seems to use Ramsey(BJ)${}^+$, however.  So at first glance, there appears to be a problem.
On the other hand, Jech, as you might guess, defines a Ramsey ultrafilter to be a Ramsey(J) ultrafilter.  Then in Lemma 9.2 he proves that an ultrafilter is Ramsey(J) iff it's Ramsey(R)${}^+$.  He proves the forward (harder) direction in two steps, first he proves that Ramsey(J) implies Ramsey(BJ)${}^+$, and then that Ramsey(BJ)${}^+$ implies Ramsey(R)${}^+$.
So (assuming Jech's proof is correct, as is the rest of the proof in Bartoszynski and Judah), we know that the last four definitions above are all equivalent, and the first is weaker, but perhaps not strictly weaker.  I doubt the definition given in Bartoszynski and Judah is a mistake (although I suppose it's possible), so how can we see that Ramsey(BJ) implies one of the other four characterizations?
A: This is not an answer, but it would probably be too long for a comment.
First some definitions and results I copied from Bartoszynski-Judah:
Definition 4.4.1: p-point = ultrafilter with pseudointersection property
Lemma 4.4.3: $\mathcal F$ is a p-point $\Leftrightarrow$ for every partition of $\omega$, $\{Y_n; n\in\omega\}$, 
either there exists $n\in\omega$ such that $Y_n\in\mathcal F$ or there exists $X\in\mathcal F$ such that $X\cap Y_n$ is finite for $n\in\omega$.
Definition 4.5.1: A filter $\mathcal F$ is called Ramsey if for every descending sequence $\{X_n; n\in\omega\}\subseteq\mathcal{F}$ of sets
there exists a sequence $\{x_n; n\in\omega\}\in\mathcal F$ such that $x_n\in X_n$ for $n\in\omega$.
$\mathcal F$ is called a q-point if for every partition of $\omega$ into finite pieces $\{I_n: n\in\omega\}$
there exists $X \in \mathcal F$ such that $|X \cap I_n| \le 1$ for $n\in\omega$.
Theorem 4.5.2
Let $\mathcal F$ be an ultrafilter on  $\omega$. The following conditions are equivalent:


*

*$\mathcal F$ is Ramsey,

*for every partition of $\omega$, $\{Y_n : n\in\omega\}$, either $Y_n \in \mathcal F$ for some $n\in\omega$ or
there exists $X\in \mathcal F$ such that $|X_n\cap Y_n| \le 1$ for $n\in\omega$,

*for every set $A\subseteq[\omega]^2$ there exists $X\in\mathcal F$ such that $[X]^2\subseteq A$ or $[X]^2\cap A=\emptyset$,

*$\mathcal F$ is a p-point and a q-point.

When I tried to search in various literature (books, papers); I found quite frequently that Ramsey is equivalent to p-point and q-point. Most authors defined Ramsey ultrafilters using 3 (colorings). But I also found 2, sometimes under name selective ultrafilter.
I did not find the condition 1 in literature, the closest what I found was:
Lemma I.1.4. A nonprincipal ultrafilter $\mathcal U$ on $\omega$ is Ramsey iff for every sequence $\{M_i, i\in\omega\}\subseteq \mathcal U$  there exists $M \in \mathcal U$ such that $j \in M_i$ for all $i < j$ in $M$. (In Spiros A. Argyros, Stevo Todorcevic: Ramsey Methods in Analysis).
Several authors define selective ideals using diagonalization, which is very similar to condition from Lemma I.1.4.
Jech (Set Theory, Millenium Edition) in proof of Lemma 9.2 show as an auxiliary result that a decreasing system $X_n$ of sets from a Ramsey ultrafilter $D$ there exists $\{a_0<a_1<\dots\}\in D$ such that $a_0\in X_0$ and $a_{n+1}\in X_{a_n}$.

I believe that after replacing their definition of Ramsey ultrafilter by some of the above conditions, the proof from Bartoszynski-Judah might work; but I am still not convinced that their definition is incorrect.
