# Ramsey ultrafilters on partial order

$$\newcommand{\U}{\mathcal{U}}$$ $$\newcommand{\P}{\mathbb{P}}$$ $$\newcommand{\Q}{\mathbb{Q}}$$ $$\newcommand{\F}{\mathcal{F}}$$ 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.

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.

• In the characterization of Ramsey ultrafilters that you quoted, items 3 and 4 are mis-stated. In 3, the sets $A_n$ must not be in $\mathcal U$. In 4, the sets $A_n$ must be in $\mathcal U$. Commented Oct 20, 2021 at 15:54
• @AndreasBlass corrected, thanks for highlighting. Commented Oct 20, 2021 at 15:57
• What is the definition of an ultrafilter over a poset? What is the definition of $[\mathbb P]^n$, is it the set of all $n$-element chains in $\mathbb P$, or is it the set of all $n$-element subsets of $\mathbb P$? In statement 1 of the Conjecture, why do you have $F:[\omega]^n\to k$ instead of $F:[\mathbb P]^n\to k$?
– bof
Commented Oct 21, 2021 at 4:03
• @bof 1) Ultrafilter over a poset is just the usual ultrafilter but treating the poset as just a set. 2) $n$-element subsets of $\mathbb{P}$. 3) Typo, I've corrected it. Thanks for highlighting. Commented Oct 21, 2021 at 5:22

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.