PFA implies the simple ideal dichotomy

Put $T={\bigcup }_{\xi <{\omega }_1}{A}_{\xi }$ and $R=S\setminus T$. Generation modulo finite makes $R$ outside $\mathcal{I}$. AC chooses an enumeration of each countable $A_{\xi }$, so $T$ injects into ${\omega }_1\times \omega$ and [F6] gives $\mathrm{\mid }T\mathrm{\mid }\le {\mathrm{\aleph }}_1$. If $R$ is uncountable, it already supplies the outside branch of Form 1. Otherwise $\mathrm{\mid }S\mathrm{\mid }\le {\mathrm{\aleph }}_1$, and uncountability gives $\mathrm{\mid }S\mathrm{\mid }={\mathrm{\aleph }}_1$; transport $S,\mathcal{I}$, and the generators along a bijection with ${\omega }_1$. Thus, for the nontrivial Form-1 case, we may work on $S={\omega }_1$.

Assume that $S$ is not a countable union of sets inside $\mathcal{I}$. Define $P_1$ as follows. A condition $p=(x_p,d_p,{\mathcal{N}}_p)$ has finite $x_p,d_p\subseteq {\omega }_1$ and a finite membership chain ${\mathcal{N}}_p$ of countable elementary submodels of a fixed well-ordered expansion of $H({\mathrm{\aleph }}_2)$ containing $\mathcal{I}$ and the generator map. Require that whenever $\alpha <\beta$ are in $x_p$, some $N\in {\mathcal{N}}_p$ satisfies $\alpha \in N$ and $\beta \mathrm{\notin }N$, equivalently $\alpha <N\cap {\omega }_1\le \beta$; this is the meaning of “the models separate distinct points of $x_p$.” If $\eta \in x_p$ lies above $N\cap {\omega }_1$ for $N\in {\mathcal{N}}_p$, then $\eta$ belongs to no $Y\in N$ that is inside $\mathcal{I}$. A stronger $q\le p$ enlarges all three finite coordinates and, for each $\xi \in d_p$, freezes $x_q\cap A_{\xi }=x_p\cap A_{\xi }$.

For every $\gamma <{\omega }_1$, conditions putting a point above $\gamma$ into $x_p$ are dense. Given $p$, append a countable model $N$ containing $p$ and $\gamma$, so $x_p\subseteq N$ and $\gamma <N\cap {\omega }_1$. The union of the countably many inside sets belonging to $N$ cannot cover $S$ by the assumption in step 2.1. A point outside that union is outside $N\cap {\omega }_1$ because every singleton from $N$ is an inside set in $N$; append that point to $x_p$. For every $\xi <{\omega }_1$, the set of conditions with $\xi \in d_p$ is dense by simply enlarging $d_p$.

To prove properness, take a large countable $M\prec H(\kappa )$ containing $P_1$ and a condition $p_0\in M\cap P_1$. Append $N^{\ast }=M\cap H({\mathrm{\aleph }}_2)$ to the side chain. Because every finite coordinate of $p_0$ lies in $M$, this is a condition $p\le p_0$. Fix $r\le p$ and dense $D\in M$, and first strengthen $r$ into $D$. The model $N^{\ast }$ cuts the increasing enumeration $x_r=\{{\alpha }_0<\cdots <{\alpha }_k\}$ after some ${\alpha }_i$; the lower part $r\upharpoonright M$ belongs to $M$.

Let $E\in M$ be the set of $(k+1)$-tuples end-extending the $x$-coordinate of $r\upharpoonright M$ that occur as the $x$-coordinate of some condition in $D$ extending that lower part. It contains $({\alpha }_0,\dots ,{\alpha }_k)$. We use the following fibre observation at each side model: if $N$ is countable, $b\mathrm{\notin }N$ avoids every inside set in $N$, $a\in N$, and $H=\{z<{\omega }_1:\phi (z,a)\}\in N$ contains $b$, then $H$ is not inside $\mathcal{I}$; otherwise the condition's avoidance clause would exclude $b$. Starting at ${\alpha }_k$ and moving down to ${\alpha }_{i+1}$, apply this observation to the definable successive fibres of $E$. The intervening side model contains $E$ and all earlier coordinates but lies below the current coordinate. We obtain in $M$ nested non-inside candidate sets $Y_{i+1},\dots ,Y_k$ such that every successive choice from them completes to a tuple in $E$.

Put $Z={\bigcup }_{\xi \in d_r}{A}_{\xi }\in \mathcal{I}$. At a candidate stage, $Y_j$ is not inside, so elementarity gives a countable $C_j\in M$ with $C_j\subseteq Y_j$ and $C_j\mathrm{\notin }\mathcal{I}$. Since $Z\in \mathcal{I}$, choose $a_j\in C_j\setminus Z$; countability of $C_j\in M$ gives $C_j\subseteq M$. Recursing through the nested fibres gives a tuple in $E\cap M$ and hence $q\in D\cap M$ extending $r\upharpoonright M$, with every new $x_q$-point outside $Z$. The union of $q$ and $r$ is a condition: their model chains merge through $N^{\ast }$; upper points of $r$ avoid every inside generator named by $d_q\in M$; and the replacement points of $q$ avoid every generator named by $d_r$. These last two facts verify both directions of the freezing requirement. Thus $q$ is compatible with $r$.

Step 5.1 says that $p$ is an $(M,P_1)$-master, so [F3] makes $P_1$ proper. Apply PFA to the dense sets in step 3.1. For the resulting filter $G$, let $X={\bigcup }_{p\in G}{x}_p$. It is unbounded, hence uncountable. For each generator $A_{\xi }$, a condition in $G$ puts $\xi$ into its finite $d$-coordinate, after which directedness and freezing show that $X\cap A_{\xi }$ is exactly that condition's finite intersection. By [F1], $X$ is outside $\mathcal{I}$. Together with the alternative excluded in step 2.1 and the reduction in step 1.1, this proves Form 1.

Now suppose there is no uncountable set inside $\mathcal{I}$. For every uncountable $Y\subseteq S$, apply Form 1 to $\mathcal{I}\upharpoonright Y$. Its countable-union branch would make some inside piece uncountable by [F6], contrary to the supposition. Hence every uncountable $Y\subseteq S$ contains an uncountable subset outside $\mathcal{I}$.

Retain $T$ and $R$ from step 1.1. The set $R$ is outside. If $T$ is countable, partition it into singletons, which are outside, and add $R$ as one more piece; this proves the countable outside decomposition. Assume henceforth that $T$ is uncountable. Then $\mathrm{\mid }T\mathrm{\mid }={\mathrm{\aleph }}_1$ by [F6].

On $T$ define $P_2$ to consist of pairs $(f_p,d_p)$ with $f_p:T\rightharpoonup \omega$ finite and $d_p\subseteq {\omega }_1$ finite. Put $q\le p$ when $q$ extends both coordinates and, for every $\xi \in d_p$ and every $n\in \mathrm{ran}(f_p)$, $f_q^{-1}\{n\}\cap A_{\xi }=f_p^{-1}\{n\}\cap A_{\xi }\mathrm{.}$ Thus a recorded generator freezes every colour already present, while a new colour may be introduced once.

The forcing $P_2$ is ccc. Given uncountably many conditions, apply [F6] to their function domains and $d$-coordinates, thin to fixed finite sizes and common roots, and make all functions agree on the domain root. Enumerate the disjoint domain petals in a fixed order. Repeatedly use step 7.1 so that, for each petal coordinate, its uncountable set of values is outside $\mathcal{I}$. Their finite union $O$ is outside. For each remaining condition $p_{\eta }$, the set $B_{\eta }=O\cap {\bigcup }_{\xi \in d_{p_{\eta }}}{A}_{\xi }$ is finite. Thin the finite $B_{\eta }$ to a $\mathrm{\Delta }$-system. Its root meets at most one disjoint domain petal, while its disjoint petals and the domain petals each meet only finitely many petals of the other family. Hence choose distinct $\eta ,\zeta$ with each condition's domain petal disjoint from the other's $B$-set. The coordinatewise unions of their functions and side sets then satisfy both freezing clauses and form a common extension.

The following sets are dense in $P_2$: conditions deciding a specified $t\in T$, conditions placing a specified $\xi <{\omega }_1$ into $d_p$, and conditions whose function range contains a specified $n<\omega$. For the first or third demand, if necessary assign a new point a colour not yet in the finite range (the specified $n$ itself when it is absent); this cannot violate a freeze, which only mentions old colours.

By steps 10.1 and [F5], $P_2$ is proper. PFA applied to the at-most-${\omega }_1$ dense sets of step 10.2 gives a filter whose union is a total $f:T\to \omega$. Fix $n$ and $\xi$. Directedness combines a condition recording $\xi$ with one already using colour $n$; below their common extension that intersection is frozen. Consequently $f^{-1}\{n\}\cap A_{\xi }$ is finite. Each colour class is outside $\mathcal{I}$ by [F1], so these classes, together with $R$, form a countable outside decomposition of $S$. This proves Form 2 under the no-inside hypothesis; its other branch is precisely an uncountable inside set.

Finally, Form 1 alone yields the simple dichotomy: its outside branch is already a witness, while in its countable-union branch [F6] makes at least one inside piece uncountable. Form 2 gives the symmetric conclusion as well. Empty finite coordinates, empty roots, a generator-free $T$, and new colours were handled in steps 1.1, 8.1, 9.1, and 10.2. All model, thinning, enumeration, and witness choices are the AC uses recorded by [F7].