# Are these two definitions of $\mathcal{U}$-Ramsey set equivalent?

Let $$\mathcal{U}$$ be an ultrafilter over $$\omega$$, and let $$\mathcal{X} \subseteq [\omega]^\omega$$. In two separate texts, there are two possible interpretations of a $$\mathcal{U}$$-Ramsey set, as described below (Definition 7.37 and Definition 3). My question is:

Do these two definitions coincide? What if we restrict $$\mathcal{U}$$ to be a Ramsey ultrafilter?

Interpretation 1. The first interpretation is found in Stevo Todorcevic's book Introduction to Ramsey Spaces.

Definition 7.29. For any ultrafilter $$\mathcal{U}$$ over $$\omega$$ (not necessarily Ramsey), we define a $$\mathcal{U}$$-tree to be a subtree $$T$$ of $${}^{<\omega}\omega$$ (finite subsets of $$\omega$$, not sequences) with the property that for every finite subset $$t \subseteq \omega$$ such that $$\operatorname{stem}(T) \subseteq t$$, we have: $$\{n \in \omega : t \cup \{n\} \in T\} \in \mathcal{U}$$ Definition 7.30. For two $$\mathcal{U}$$-trees $$T'$$ and $$T$$, we say that $$T'$$ is a pure refinement of $$T$$ if $$\operatorname{stem}(T') = \operatorname{stem}(T)$$ and $$T' \subseteq T$$.

Definition 7.37. We then say that $$\mathcal{X}$$ is $$\mathcal{U}$$-Ramsey if for every $$\mathcal{U}$$-tree $$T$$, there is a pure refinement $$T'$$ of $$T$$ such that $$[T'] \subseteq \mathcal{X}$$ or $$[T'] \subseteq \mathcal{X}^c$$.

Interpretation 2. I first saw this interpretation in the paper Happy and mad families in $$L(\mathbb{R})$$, but I believe this is quite a standard definition.

Definition 1. If $$y_0 \supseteq y_1 \supseteq y_2 \supseteq \cdots$$ is a decreasing sequence of subsets of $$\omega$$, then we call a set $$y_\infty \in [\omega]^\omega$$ a diagonalisation of the sequence $$\langle{y_n : n < \omega}\rangle$$ iff $$f(n+1) \in y_{f(n)}$$ for every $$n < \omega$$, where $$f : \omega \to \omega$$ is the increasing enumeration of $$y_\infty$$.

Definition 2. A $$\mathcal{H} \subseteq [\omega]^\omega$$ is called a coideal if is satisfies the conditions:

1. (Upward-closure) If $$x \in H$$ and $$y \supseteq x$$, then $$y \in H$$.

2. (Pigeonhole) If $$x_0 \cup \cdots \cup x_n \in H$$, then $$x_k \in H$$ for some $$k$$.

Furthermore, $$\mathcal{H}$$ is said to be selective if it satisfies the condition:

1. (Selectivity) Every decreasing sequence $$y_0 \supseteq y_1 \supseteq \cdots$$ of members of $$\mathcal{H}$$ has a diagonalisation in $$H$$.

A selective coideal is also called a happy family.

Definition 3. If $$H$$ is a coideal (typically a happy family) and $$\mathcal{X} \subseteq [\omega]^\omega$$ is a set of reals, then we say $$\mathcal{X}$$ is $$\mathcal{H}$$-Ramsey if there is $$M \in \mathcal{H}$$ such that $$[M]^\omega \subseteq X$$ or $$[M]^\omega \subseteq X^c$$.

Note that if $$\mathcal{H}$$ is an ultrafilter, then it is a happy family iff it is a Ramsey ultrafilter.

If I understand your question correctly, you ask if Defintion 7.37. and Definition 3. are equivalent for a Ramsey ultrafilter $$\mathcal{U}(=\mathcal{H})$$. The short answer is no, but let me elaborate.
Since for a $$\mathcal{U}$$-tree $$T$$, the set $$[T]$$ is technically a subset of $$\omega^\omega$$, we will only consider $$\mathcal{U}$$-trees $$T$$ such that $$\text{stem}(T)$$ is increasing. For such a tree $$T$$ there exists a pure refinement $$T' \subseteq T$$ such that every $$x \in [T']$$ is increasing, hence $$\text{ran}(x) \in [\omega]^\omega$$ can uniquely be identified with $$x \in T'$$.
Fact: If $$\mathcal{U}$$ is a Ramsey ultrafilter, then $$\mathbb{M}_\mathcal{U}$$ (Mathias forcing with an ultrafilter $$\mathcal{U}$$) is dense in $$\mathbb{L}_\mathcal{U}$$ (Laver forcing with an ultrafilter $$\mathcal{U}$$ such that for every $$T \in \mathbb{L}_\mathcal{U}$$ we have that $$\text{stem}(T)$$ is increasing). In particular, for every $$T \in \mathbb{L}_\mathcal{U}$$ there exists $$A \in \mathcal{U}$$ such that $$\{\text{ran}(\text{stem}(T)) \cup x \,\, \colon \,\ x \in [A]^\omega \} \subseteq [T]$$.
Now Defintion 7.37. obviously implies Definition 3. : Let $$\mathcal{X} \subseteq [\omega]^\omega$$ be arbitrary such that $$\mathcal{X}$$ is $$\mathcal{U}$$-Ramsey. Let $$T$$ be a pure refinement of $$\omega^\omega$$ such that either $$[T] \subseteq \mathcal{X}$$ or $$[T] \cap \mathcal{X} = \emptyset$$. By the above Fact we can find $$A \in \mathcal{U}$$ such that $$\{\emptyset \cup x \,\, \colon \,\ x \in [A]^\omega \} \subseteq [T]$$. Hence $$\mathcal{X}$$ is also $$\mathcal{H}$$-Ramsey.
On the other hand Definition 3. does not imply Defintion 7.37. : By an induction of length $$\mathfrak{c}$$ we can construct a $$\mathcal{X} \subseteq [\omega]^\omega$$ which is not $$\mathcal{H}$$-Ramsey. W.l.o.g. we can assume $$0 \notin x$$ for every $$x \in \mathcal{X}$$. Now define $$\mathcal{X}':=\{\{0\} \cup x \,\, \colon \,\, x \in \mathcal{X}\}$$, which is obviously $$\mathcal{H}$$-Ramsey. Now if $$T$$ is a $$\mathcal{U}$$-tree such that $$\text{stem}(T)=\langle 0 \rangle$$, there cannot exist a pure refinement $$T' \subseteq T$$ such that $$[T'] \subseteq \mathcal{X}'$$ or $$[T'] \cap \mathcal{X}' = \emptyset$$, because there does not exist an $$A \in \mathcal{U}$$ such that $$\{\{0\} \cup x \,\, \colon \,\ x \in [A]^\omega \} \subseteq \mathcal{X}'$$ or $$\{\{0\} \cup x \,\, \colon \,\ x \in [A]^\omega \} \cap \mathcal{X}' = \emptyset$$.
• May I know where can I find a text which discusses the fact that $\mathbf{M}_\mathcal{U}$ is dense in $\mathbf{L}_\mathcal{U}$? Nov 3, 2021 at 14:10