# Can there be a p-point ultrafilter that is 'aggressively non-Ramsey'?

These are fairly standard terms, but for the sake of completeness: An ultrafilter $$\mathcal{U}$$ on $$\omega$$ is a p-point if whenever $$(A_n)_{n<\omega}$$ is a partition of $$\omega$$ such that $$A_n \notin \mathcal{U}$$ for all $$n$$, there is an $$X \in \mathcal{U}$$ such that $$X \cap A_n$$ is finite for all $$n$$. $$\mathcal{U}$$ is Ramsey if the same holds but with $$|X\cap A_n| = 1$$ for all $$n$$. Clearly Ramsey ultrafilters are p-point ultrafilters. The name Ramsey comes from the fact that if $$\mathcal{U}$$ is a Ramsey ultrafilter, then any graph on $$\omega$$ has a homogeneous set in $$\mathcal{U}$$.

Fix an edge relation $$E$$ on $$\omega$$ making it into a copy of the random graph. (For concreteness, we could say that $$nEm$$ if and only if the $$\min(n,m)$$th binary bit of $$\max(n,m)$$ is $$1$$.) A remarkable property of the random graph is that for any finite partition $$X_0,X_1,\dots,X_{n-1}$$ of $$\omega$$, there is an $$i such that $$(X_i,E)$$ is isomorphic to $$(\omega,E)$$. (See Proposition 3 here.) This implies the following.

Proposition. There is an ultrafilter $$\mathcal{U}$$ on $$\omega$$ with the property that for any $$X \in \mathcal{U}$$, there is a $$Y \subseteq X$$ such that $$(Y,E)$$ is isomorphic to $$(\omega,E)$$.

Proof. For each finite partition $$P$$ of $$\omega$$, let $$F_P$$ be the (clopen) set of ultrafilters $$\mathcal{U}$$ in $$\beta \omega \setminus \omega$$ satisfying that for the unique $$X$$ in $$\mathcal{U} \cap P$$, there is a $$Y \subseteq X$$ such that $$(Y,E)$$ is isomorphic to $$(\omega,E)$$. By the above fact, each $$F_P$$ is non-empty. The family $$\\{F_P:P\text{ a finite partition of }\omega\\}$$ has the finite intersection property. To see this, note that if $$P_0,P_1,\dots,P_{n-1}$$ are finite partitions of $$\omega$$, then for any mutual refinement $$Q$$ of the $$P_i$$'s, $$F_Q \subseteq \bigcap_{i. Therefore, by compactness, there is an ultrafilter $$\mathcal{U}$$ such that for every $$P$$, $$\mathcal{U} \in F_P$$. For any set $$X \in \mathcal{U}$$, we have that $$\mathcal{U} \in F_{ \\{X,\omega\setminus X\\} }$$, so $$X$$ has the required property. $$\square$$

Clearly, any such ultrafilter cannot be Ramsey, since an $$E$$-homogeneous set will contain no subset isomorphic to the random graph. Call such an ultrafilter aggressively non-Ramsey.

Question. Is it consistent with $$\mathsf{ZFC}$$ that there is an aggressively non-Ramsey p-point?

## 1 Answer

No, such a filter cannot exist. Suppose $$\mathcal U$$ is a $$p$$-point. For $$s\in 2^{<\omega}$$, let $$A_s$$ consist of all $$m<\omega$$ such that $$\forall n\in\mathrm{dom}(s)\ (nEm\Leftrightarrow s(n)=1).$$ Note that there is a unique $$x\in 2^\omega$$ so that $$A_{x\upharpoonright n}\in \mathcal U$$ for all $$n<\omega$$. As $$\mathcal U$$ is a $$p$$-point, there is $$B\in\mathcal U$$ with $$B\subseteq^\ast A_{x\upharpoonright n}$$ for all $$n<\omega$$. The graph $$(B, E)$$ has the following property: For all $$n\in B$$, all but maybe finitely many $$m\in B$$ satisfy $$n E m\Leftrightarrow x(n)=1.$$ This easily implies that there is no $$C\subseteq B$$ so that $$(C,E)$$ is (isomorphic to) the random graph, so $$\mathcal U$$ is not aggresively non-Ramsey.