MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are many consequences of the Proper Forcing Axiom (PFA) which essentially say that $\omega_2$ has a lot of compactness.

For instance, Matteo Viale and Christoph Weiss have a few papers in which the combinatorial properties $\mathrm{SP}(\kappa)$ and $\mathrm{ISP}(\kappa)$ (which can be seen as generalizations of the tree property) are isolated, and the following theorems (which are modification of older results of Jech and Magidor, respectively) are stated:

  • $\kappa$ is strongly compact iff $\kappa$ is inaccessible and $\mathrm{SP}(\kappa)$
  • $\kappa$ is supercompact iff $\kappa$ is inaccessible and $\mathrm{ISP}(\kappa)$

They also show that $\mathrm{PFA} \rightarrow \mathrm{ISP}(\omega_2)$ (and $\mathrm{SP}(\omega_2)$), so one can read this as saying PFA implies $\omega_2$ is as "compact" as a supercompact, minus the inaccessibility.

On the other hand, there are consequences of PFA (including the statement of PFA itself) which seem to say that $\omega_2$ is very much incompact. For instance, consider Rado's Conjecture (RC). RC is the statement: If $T$ is a tree such that every subtree of size $< \omega_2$ can be decomposed into countably many antichains, then so can $T$. It's not hard to see how RC is, in some sense, saying that $\omega_2$ is compact: If we replace $\omega_2$ with a compact cardinal $\kappa$ in the statement of RC, then the statement is true by the "compactness of the language $\mathcal{L}_{\kappa,\kappa}$" characterization of $\kappa$. But PFA contradicts RC. Nonetheless, both PFA and RC are usually obtained by proper forcings which collapse a supercompact to $\omega_2$.

PFA itself seems to say $\omega_2$ is incompact: Let $\mathbb{P}$ be a proper forcing, and consider the language which has a constant symbol for every element of $\mathbb{P}$, a binary relation symbol (for the relation on $\mathbb{P}$), a unary predicate for each dense subset of $\mathbb{P}$, and a unary predicate which will stand for a generic filter. Consider the theory consisting of the positive diagram of $(\mathbb{P},\leq,p\ (p\in \mathbb{P}), D\ (D \subseteq \mathbb{P}\mbox{ dense}))$ together with formulas saying that $G$ is a filter and $G$ meets every $D$ (a new formula for each $D$). PFA says that any subtheory of size $< \omega_2$ has a model, but there's no filter meeting every dense set in the ground model, so loosely speaking, this theory doesn't have a model (although I suppose there could be a model which adds unnamed elements to each dense subset and has a generic meeting each dense subset at an unnamed condition).

My question is:

Is there some way to reconcile the fact that PFA seems to simultaneously say that $\omega_2$ has properties strongly indicative of some sort of compactness, and also has properties strongly indicative of some sort of incompactness?

share|cite|improve this question
$\omega_2$ being a successor, of course it is going to be non-compact. I doubt this has much to do with PFA or anything else, it is just that it is a successor. This is why rather than looking at direct instances of compactness, you search for traces that have survived (and think of them as evidence that $\omega_2$ is genuinely compact in some inner model). – Andrés Caicedo Sep 7 '11 at 1:22
I can't seem to recall the details, but I do recall a definition of being "$\lambda$-thinly ineffable" (or "thinly $\lambda$-ineffable"). I do recall being in a seminar recently about PFA implying $\omega_2$ is $\lambda$-thinly ineffable. This implies some sense of compactness and as Andres says being a successor implies otherwise. – Asaf Karagila Sep 7 '11 at 8:08
@Asaf, yes in the paper of Viale and Weiss, $\mathrm{ITP}(\kappa,\lambda)$ is used to denote that $\kappa$ is $\lambda$-thinly ineffable. $\mathrm{ITP}(\kappa)$ is then used to denote that $\forall\lambda\geq\kappa,\mathrm{ITP}(\kappa,\lambda)$. $\mathrm{ISP}(\kappa,\lambda)$ denotes that $\kappa$ is $\lambda$-"slenderly" ineffable, and $\mathrm{ISP}(\kappa)$ denotes that $\forall\lambda\geq\kappa,\mathrm{ISP}(\kappa,\lambda)$. $\mathrm{ISP}(\kappa,\lambda)\rightarrow\mathrm{ITP}(\kappa,\lambda)$, so the result that $\mathrm{PFA}\rightarrow\mathrm{ISP}(\omega_2)$ subsumes the one you mention. – Amit Kumar Gupta Sep 7 '11 at 9:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.