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:
-
(Upward-closure) If $x \in H$ and $y \supseteq x$, then $y \in H$.
-
(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:
- (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.
Best Answer
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 $\{\{0\} \cup x \,\, \colon \,\ x \in [A]^\omega \} \cap \mathcal{X}' = \emptyset$.