## Set theory seminar -Forcing axioms and inner models

Today I started a series of talks on “Forcing axioms and inner models” in the Set Theory Seminar. The goal is to discuss a few results about strong forcing axioms and to see how these axioms impose a certain kind of rigidity’ to the universe.

I motivated forcing axioms as trying to capture the intuition that the universe is wide’ or saturated’ in some sense, the next natural step after the formalization via large cardinal axioms of the intuition that the universe is tall.’

The extensions of ${\sf ZFC}$ obtained via large cardinals and those obtained via forcing axioms share a few common features that seem to indicate their adoption is not arbitrary. They provide us with reflection principles (typically, at the level of the large cardinals themselves, or at small cardinals, respectively), with regularity properties (and determinacy) for many pointclasses of reals, and with generic absoluteness principles.

The specific format I’m concentrating on is of axioms of the form ${\sf FA}({\mathcal K})$ for a class ${\mathcal K}$ of posets, stating that any ${\mathbb P}\in{\mathcal K}$ admits filters meeting any given collection of $\omega_1$ many dense subsets of ${\mathbb P}$. The proper forcing axiom ${\sf PFA}$ is of this kind, with ${\mathcal K}$ being the class of proper posets. The strongest axiom to fall under this setting is Martin’s maximum ${\sf MM}$, that has as ${\mathcal K}$ the class of all posets preserving stationary subsets of $\omega_1$.

Of particular interest is the `bounded’ version of these axioms, which, if posets in ${\mathcal K}$ preserve $\omega_1$, was shown by Bagaria to correspond precisely to an absoluteness statement, namely that $H_{\omega_2}\prec_{\Sigma_1}V^{\mathbb P}$ for any ${\mathbb P}\in{\mathcal K}$.

In the next meeting I will review the notion of properness, and discuss some consequences of ${\sf BPFA}$.

I want now to briefly sketch the following result from the Martin’s Maximum paper (Foreman, Magidor, Shelah. Martin’s Maximum, saturated ideals, and non-regular ultrafilters. Part I, Annals of Mathematics, 127 (1988), 1-47):

Say that a forcing ${\mathbb P}$ preserves stationary subsets of $\omega_1$ (we will usually say that ${\mathbb P}$ is stationary set preserving) iff there is some $p\in{\mathbb P}$ such that $p\Vdash \check S\mbox{ is stationary}$ for any stationary set $S\subseteq\omega_1$

Theorem: Suppose that ${\mathbb P}$ does not preserve stationary subsets of $\omega_1$. Then ${\sf FA}(\{{\mathbb P}\})$ fails.

Proof: Assume ${\mathbb P}$ does not preserve subsets of $\omega_1$. Begin by fixing ${\mathbb P}$-names $\dot S$ and $\dot C$ that are respectively interpreted in any ${\mathbb P}$-generic extension as a ground model-stationary subset of $\omega_1$, and as a club subset of $\omega_1$ disjoint from $S$.

Let $D=\{p\in{\mathbb P}: \mbox{there is }S\in V\mbox{ such that }p\Vdash \dot S=S\}$. If $G\subseteq{\mathbb P}$ is any filter and $p\in G\cap D$, then $p\Vdash S\cap\dot C=\emptyset$ for some stationary set $S$ in the ground model. We now argue that if we could find such a $G$ meeting enough dense sets, then we could define (from $G$) a club subset $C$ (in the ground model) such that $C\cap S=\emptyset$ for some such $p$ and $S$, which of course is a contradiction. We will ensure this by ensuring that if $\alpha\in C\cap S$, then there is some $q\in G$ which we may assume is below $p$, and such that $q\Vdash\alpha\in\dot C$.

The set $C=C_G$ is easy to define: Simply take $C=\{\alpha\in\omega_1:\mbox{ there is }q\in G\mbox{ such that }q\Vdash\alpha\in\dot C\}$. We now proceed to exhibit $\omega_1$ many dense subsets of ${\mathbb P}$ such that if $G$ were a filter meeting all of them (in addition to $D$), then $C$ indeed would be a club set. This proves that the forcing axiom for ${\mathbb P}$ fails, as we want.

We are required to ensure that $C$ is unbounded in $\omega_1$, and that it is closed. For $\alpha<\omega_1$ let $D_\alpha=\{q:\mbox{ either }q\Vdash\alpha\in\dot C\mbox{ or }q\Vdash\alpha\notin\dot C,\mbox{ in which case}$ $\mbox{there are }\beta<\alpha<\gamma\mbox{ such that }q\Vdash(\beta,\alpha)\cap\dot C=\emptyset\mbox{ and }$ $q\Vdash\gamma\in\dot C\}$. Any filter $G$ meeting all the sets $D_\alpha$ ensures that $C$ is unbounded.

But it also ensures that it is closed: Supposed otherwise. Then for some $0<\alpha\notin C$ we have that $\sup(C\cap\alpha)=\alpha$. However, there is some $q\in D_\alpha\cap G$ such that $q\Vdash\alpha\notin \dot C$ and, for some $\beta<\alpha$, $q\Vdash(\beta,\alpha)\cap\dot C=\emptyset$. However, there must be unboundedly many $\delta\in(\beta,\alpha)\cap C$ and for any of them there is some $r\in D_\delta\cap G$ such that $r\Vdash \delta\in \dot C$. It follows that $r$ and $q$ are incompatible, which contradicts that $G$ is a filter. This contradiction shows that $C$ is indeed closed, and completes the proof. $\mbox{QED}$