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

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  • $\begingroup$ 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 $\mathcal U$. $\endgroup$ Commented Oct 20, 2021 at 15:54
  • $\begingroup$ @AndreasBlass corrected, thanks for highlighting. $\endgroup$ Commented Oct 20, 2021 at 15:57
  • $\begingroup$ 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 $F:[\mathbb P]^n\to k$? $\endgroup$
    – bof
    Commented Oct 21, 2021 at 4:03
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    $\begingroup$ @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. $\endgroup$ Commented Oct 21, 2021 at 5:22

1 Answer 1

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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.

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