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 pin{mathbb P} such that pVdash check Smbox{ is stationary} for any stationary set Ssubseteqomega_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={pin{mathbb P}: mbox{there is }Sin Vmbox{ such that }pVdash dot S=S}. If Gsubseteq{mathbb P} is any filter and pin Gcap D, then pVdash Scapdot 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 Ccap S=emptyset for some such p and S, which of course is a contradiction. We will ensure this by ensuring that if alphain Ccap S, then there is some qin G which we may assume is below p, and such that qVdashalphaindot C.

The set C=C_G is easy to define: Simply take C={alphainomega_1:mbox{ there is }qin Gmbox{ such that }qVdashalphaindot 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 }qVdashalphaindot Cmbox{ or }qVdashalphanotindot C,mbox{ in which case} mbox{there are }beta<alpha<gammambox{ such that }qVdash(beta,alpha)capdot C=emptysetmbox{ and } qVdashgammaindot 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<alphanotin C we have that sup(Ccapalpha)=alpha. However, there is some qin D_alphacap G such that qVdashalphanotin dot C and, for some beta<alpha, qVdash(beta,alpha)capdot C=emptyset. However, there must be unboundedly many deltain(beta,alpha)cap C and for any of them there is some rin D_deltacap G such that rVdash deltain 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}


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