Suppose that $\cal P$ is a class of forcings, we denote by $\operatorname{FA}_\kappa(\cal P)$ the statement that whenever $\Bbb P\in\cal P$, $\gamma<\kappa$, and $\{D_\alpha\mid\alpha<\gamma\}$ are dense open sets in $\Bbb P$, then there is a filter $G\subseteq\Bbb P$ such that $G\cap D_\alpha\neq\varnothing$ for all $\alpha<\gamma$.

The usual examples are Martin's Axiom where $\kappa=2^{\aleph_0}$, and $\cal P$ is the class of ccc forcings; or the Proper Forcing Axiom where $\cal P$ is the class of proper forcings and $\kappa=2^{\aleph_0}$ (which then implies $\kappa=\aleph_2$).

Let's focus on subclasses of ccc forcings.

We can force Martin's Axiom for only $\sigma$-centered forcings, or other subclasses, and there is a clear hierarchy here. For example, $$\operatorname{FA}_{2^{\aleph_0}}(\textrm{ccc})\implies\operatorname{FA}_{2^{\aleph_0}}(\sigma\textrm{-centered})\implies\operatorname{FA}_{2^{\aleph_0}}(\sigma\textrm{-linked}).$$

Of course, there are many, many other subclasses that one can consider.

Question. Is there any work on separating the various forcing axioms for subclasses of ccc forcings? For example, is it consistent that $\operatorname{FA}_{\aleph_2}(\sigma\textrm{-linked})$ holds, but only $\operatorname{FA}_{\aleph_1}(\sigma\textrm{-centered})$ holds, and $\operatorname{FA}_{\aleph_1}(\textrm{ccc})$ fails altogether?

(And I cannot stress this enough, the choice of subclasses was entirely arbitrary, and any other subclasses of ccc forcings are of interest here.)

  • $\begingroup$ The way you wrote it, PFA is compatible with CH (in particular it doesn't imply $\kappa=\aleph_2$). To make things nontrivial we should just always require that $\kappa$ is at least $\aleph_2$ when talking about $\mathrm{FA}_\kappa$ (this applies to your question too). $\endgroup$ Commented Feb 27, 2017 at 18:37
  • $\begingroup$ You're right. There's also some discrepancy in the example I gave in the question itself. I'll fix this later when I'm at a computer again. Thanks! $\endgroup$
    – Asaf Karagila
    Commented Feb 27, 2017 at 18:56
  • $\begingroup$ Now I'm not sure whether or not the standard notation is that $\operatorname{FA}_\kappa$ means there is a filter meeting every $\kappa$ dense sets, or if I should just add the requirement that $\kappa>\aleph_1$. $\endgroup$
    – Asaf Karagila
    Commented Feb 28, 2017 at 7:37

2 Answers 2


This paper seems to be the sort of thing you're looking for:

J. H. Barnett, "Weak variants of Martin's Axiom," Fundamenta Mathematicae 141 (1992), pp. 61-73.

The abstract is pretty short and to the point: A hierarchy of weak variants of Martin's Axiom is extended and shown to be strict.

There is a nice picture on page 2 showing the hierarchy of axioms considered in the paper. At the top of the author's hierarchy is MA, and at the bottom is MA(sigma-centered).

Along the same lines, there is some related material on page 339 of Kunen's newer Set Theory book (the one from 2011). He defines MAK to mean that MA holds for all posets with Property K, and he defines MAP to mean that MA holds for all posets with $\aleph_1$ as a pre-calibre. He then proves

Theorem: There is a (finite-support iterated) forcing notion that forces MAK and MAP, but does not kill Suslin trees.

In particular, if you iterate over a ground model with Suslin trees then MA must fail in the extension, even though MAK and MAP both hold.

  • $\begingroup$ Thanks! That certainly looks like this is the direction I was thinking about. I will look at it more carefully tomorrow. I was hoping for a richer separation structure, but I am aware that separating more than two levels of cardinal invariants is usually hard. $\endgroup$
    – Asaf Karagila
    Commented Feb 27, 2017 at 16:27

Another reference is the paper "On partial orderings having precalibre-$\aleph_1$ and fragments of Martin's axiom" by Bagaria-Shelah.


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