The simple dichotomy for omega-one-generated ideals

Work in ZFC. Let $S$ be uncountable. An ideal of countable subsets of $S$ is a family $\mathcal{I}\subseteq [S{]}^{\le \omega }$ that contains every finite subset of $S$ and is closed under taking subsets and finite unions. For $a,b\subseteq S$, write $a{\subseteq }^{\ast }b$ when $a\setminus b$ is finite.

The ideal $\mathcal{I}$ is generated modulo finite by ${\omega }_1$ members if there are $A_{\xi }\in \mathcal{I}$ for $\xi <{\omega }_1$ such that

$a\in \mathcal{I}⟺\mathrm{\exists }u\in [{\omega }_1{]}^{<\omega }a{\subseteq }^{\ast }{\bigcup }_{\xi \in u}{A}_{\xi }$

for every countable $a\subseteq S$. This is equivalent to having an explicit family closed under finite unions: replace the displayed family by the sets ${\bigcup }_{\xi \in u}{A}_{\xi }$ for finite $u\subseteq {\omega }_1$. The resulting family has cardinality at most ${\omega }_1$ in ZFC; if an ${\omega }_1$-indexed family is desired, repeat members to pad the enumeration. Membership in $\mathcal{I}$ then means almost containment in one member of that family. No increasing sequence is asserted: an ideal need not contain the union of countably many of its members.

For $X\subseteq S$:

• $X$ is inside $\mathcal{I}$ when $[X{]}^{\le \omega }\subseteq \mathcal{I}$;
• $X$ is outside, or orthogonal to, $\mathcal{I}$ when $\mathrm{\mid }X\cap a\mathrm{\mid }<\omega$ for every $a\in \mathcal{I}$.

Equivalently, $X$ is outside exactly when its intersection with every chosen generator is finite: one direction uses that generators belong to the ideal; the other uses the finite-union and finite-error formula above. Also $\mathcal{I}\upharpoonright X=\{a\cap X:a\in \mathcal{I}\}$ then consists only of finite sets.

The simple dichotomy for ${\omega }_1$-generated ideals is the assertion that for every such $S$ and $\mathcal{I}$, either some uncountable $X\subseteq S$ is inside $\mathcal{I}$, or some uncountable $X\subseteq S$ is outside $\mathcal{I}$. “Either” is inclusive: different witnesses can in principle satisfy the two clauses. One uncountable $X$ cannot satisfy both in ZFC, because AC gives a countably infinite $a\subseteq X$; inside gives $a\in \mathcal{I}$, while outside says $X\cap a=a$ is finite. This extraction of $a$ is the only choice used in the definitional discussion. Empty and finite $X$ are allowed by the two local predicates but are not witnesses to the dichotomy.