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Recall the following equivalent definitions of a Ramsey ultrafilter over $\omega$:



> 
> **Theorem (Ramsey Ultrafilter).** Let $\U$ be a non-principal ultrafilter over $\omega$. TFAE:
> 
> 
> 1. For any partition $F : [\omega]^n \to k$, there is a homogeneous set $H \in \U$.
> 2. For any partition $F : [\omega]^2 \to 2$, there is a homogeneous set $H \in \U$.
> 3. For every partition of $\omega$, $\{A\_n : n \in \omega\}$ such that $A\_n \notin \U$ for all $n$, there is a sequence $\langle{x\_n : n \in \omega}\rangle$ such that $x\_n \in A\_n$ and $\{x\_n : n \in \omega\} \in U$.
> 4. Given a decreasing sequence of sets $A\_0 \supseteq A\_1 \supseteq \cdots$ in $\U$, there exists a set $\{x\_i\}\_{i \in \omega} \in \U$ such that for all $n \in \omega$, $x\_{n+1} \in A\_{x\_n}$.
> 
> 
> 


I wish to generalise this notion to countable posets, and am in particular more interested in criterion (4). More specifically, let $(\P,\leq)$ be a countable poset, and let $\U$ be an ultrafilter over $\P$. Let's suppose we say that $\U$ is a *Ramsey ultrafilter* over $\P$ if for any partition $F : [\P]^n \to k$, there is a homogeneous set $H \in \U$.


I wish to attain an equivalence that is of something like the following:



> 
> **Conjecture.** Let $\U$ be a non-principal ultrafilter over a countable poset $(\P,\leq)$ (with possibly more assumptions required on $(\P,\leq)$). TFAE:
> 
> 
> 1. $\U$ is Ramsey. That is, for any partition $F : [\P]^n \to k$, there is a homogeneous set $H \in \U$.
> 2. (?) Let $\F = \{A\_p : p \in \P\}$ be a family of sets in $\U$ such that $p < q \implies A\_p \supseteq A\_q$. Then there exists a set $\Q \subseteq \P$, $\Q \in \U$, such that for all $p,q \in \Q$, if $p < q$ then $q \in A\_p$.
> 
> 
> 


If this is false in full generality, what additional assumptions should we impose on $(\P,\leq)$? Several assumptions I'm more than willing to impose are:


1. Every chain in $\P$ is well-ordered.
2. For all $p,q \in \P$, there are only finitely many $r$ such that $p < r < q$.


But I wish to stay away from the linear order case.




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**EDIT**: Allow me to explain the main obstacle I face in generalising the equivalence: The most accessible proof of (3) $\implies$ (4) appears to be Jech's book, Lemma 9.2. His proof is as follows:



> 
> [Let] $X\_0 \supseteq X\_1 \supseteq \cdots$ be sets in $D$ [where $D$ is a Ramsey ultrafilter]. Since $D$ is a p-point, there exists $Y \in D$ such that each $Y - X\_n$ is finite. Let us define a sequence $y\_0 < y\_1 < \cdots$ in $Y$ as follows:
> 
> 
> * $y\_0 = $ the least $y\_0 \in Y$ such that $\{y \in Y : y > y\_0\} \subseteq X\_0$.
> * $y\_1 = $ the least $y\_1 \in Y$ such that $\{y \in Y : y > y\_2\} \subseteq X\_{y\_0}$.
> * $\dots$
> * $y\_n = $ the least $y\_n \in Y$ such that $\{y \in Y : y > y\_{n-1}\} \subseteq X\_{y\_{n-1}}$.
> 
> 
> For each $n$, let $A\_n = \{y \in Y : y\_n < y \leq y\_{n+1}\}$. Since $D$ is Ramsey, there exists a set $\{z\_n\}\_{n=0}^\infty$ such that $z\_n \in A\_n$ for all $n$.
> 
> 
> We observe that for each $n$, $z\_{n+2} \in X\_{z\_n}$: Since $z\_{n+2} > y\_{n+2}$, we have $z\_{n+2} \in X\_{y\_{n+2}}$, and since $y\_{n+1} \geq z\_n$, we have $X\_{y\_{n+1}} \subseteq X\_{z\_n}$ and hence $z\_{n+2} \in X\_{z\_n}$.
> 
> 
> Thus if we let $a\_n = z\_{2n}$ and $b\_n = z\_{2n+1}$, for all $n$, then either $\{a\_n\}\_{n=0}^\infty \in D$ or $\{b\_n\}\_{n=0}^\infty \in D$; and in either case we get a sequence that satisfies [the property (4)].
> 
> 
> 


For general posets, we cannot so simply split to two subsets $a\_n = z\_{2n}$ and $b\_n = z\_{2n+1}$. If there are infinitely many branches, then we cannot guarantee one such subset will be in $D$. It also makes no sense to consider $\sigma$-complete Ramsey ultrafilters.



Ramsey ultrafilters on partial order
Presumably in (2) you meant to assume the sets $A\_p$ belong to $\mathcal U$.
The nontrivial thing is to show that (1) implies (2). The main point is that the Ramsey property of $\mathcal U$ implies that $\mathbb P$ has a $\mathcal U$-large suborder isomorphic either to $\omega$, $\omega^\*$, or an infinite discrete order. Since (2) only depends on the $\mathcal U$-almost everywhere structure of $\mathbb P$, one only needs to check that (2) holds for these three orders. The case $\mathbb P = \omega$ is standard and the other two cases are basically trivial. Here are the details.


Enumerate $\mathbb P$ as $(p\_n)\_{n < \omega}$ and for $n < m$, set $f(p\_n,p\_m) = 0$ if $p\_n < p\_m$, $f(p\_n,p\_m) = 1$ if $p\_n > p\_m$, and $f(p\_n,p\_m) = 2$ if $p\_n$ and $p\_m$ are incomparable.


If $f$ has a $\mathcal U$-large 0-homogeneous set $H$, then you are essentially in the standard situation since $H \cong \omega$, so you finish using the standard argument.


If $f$ has a $\mathcal U$-large 1-homogeneous set $H$, then letting $n = \min\{k : p\_k\in H\}$, $p\_n$ is the maximum element of $H$. So if $(A\_p)\_{p\in \mathbb P}$ is a sequence with $A\_p\in \mathcal U$ and $p < q$ implies $A\_p\supseteq A\_q$, then for $p < q$ in $H\cap A\_{p\_n}$, $q\in A\_p$ since $p \leq p\_n$ and hence $A\_{p\_n}\subseteq A\_p$.


If $f$ has a $\mathcal U$-large 2-homogenous set, then there is a $\mathcal U$-large set of incomparables, and so (2) holds vacuously.


