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Is it consistent (from suitable large cardinals) that there is a ccc poset which forces PFA?

This seems quite implausible to me. If we could force PFA via ccc forcing, the ground model would have to be quite close to a model of PFA (having the correct continuum, no squares, SCH holding etc.). However, the ground model cannot be a model of full PFA (or even BPFA), since it follows from a result of Caicedo and Veličković that any ccc forcing over such models destroys BPFA.

The reference for the Caicedo-Veličković result is

Andrés Eduardo Caicedo and Boban Veličković, The bounded proper forcing axiom and well orderings of the reals, Math. Res. Lett. 13 (2006), no. 3, 393--408. (link)

They show that if $V$ and an inner model $M$ agree on $\omega_2$ and both satisfy BPFA then they have the same subsets of $\omega_1$. The conclusion above follows since any ccc forcing adds a subset of $\omega_1$.

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  • $\begingroup$ Do you have a reference for the Caicedo-Velickovic result? $\endgroup$
    – Paul McKenney
    Commented Jul 11, 2016 at 14:22
  • $\begingroup$ @PaulMcKenney I've added a reference. $\endgroup$
    – Miha Habič
    Commented Jul 11, 2016 at 14:36
  • $\begingroup$ The only "nontrivial" that a ccc forcing can really do and not destroy PFA is to shoot branches through Suslin trees. (Or so I think, anyway.) Which is of course consistent with your observation from the paper of Andres and Boban. But it seems really weird if that might happen. E.g. if you add a Cohen real to a model of PFA and somehow revive PFA by shooting branches through the Suslin trees you've added and restoring all sort of cardinal invariants to their rightful size. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 11, 2016 at 16:43
  • $\begingroup$ (And yes, that does seem a bit implausible!) $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 11, 2016 at 16:43
  • $\begingroup$ @AsafKaragila Forcing with a Suslin tree forces $\mathfrak{t}= \omega_1$. This is Lemma 2 in Farah, 'OCA and towers in P(N)/fin', but there it is mentioned that Booth would have known of this result. $\endgroup$
    – user3462
    Commented Jul 12, 2016 at 17:41

1 Answer 1

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I think a negative answer can be derived from the following observations.

One. A nontrivial c.c.c. forcing adds a subset of $\omega_{1}$ (consider the least cardinal $\kappa$ for which it adds a subset of $\kappa$, and the tree of possible initial segments for this subset; the splitnodes in the tree give rise to a name for a new subset of $\omega_{1}$).

Two. A c.c.c. partial order $P$ has the property that for any ordinal $\delta$, if in a $P$-extension one has a continuous $\subseteq$-increasing chain $\bar{N} = \langle N_{\alpha} : \alpha < \omega_{1} \rangle$ of countable subsets of $\delta$ with union $\delta$, then for some club $C \subseteq \omega_{1}$ in the ground model, $\langle N_{\alpha} : \alpha \in C \rangle$ exists already in the ground model. Now if we look at Justin Moore's MRP coding for subsets of $\omega_{1}$ as in Section 4 of this paper (http://arxiv.org/pdf/math/0501526v1.pdf) we see that every subset of $\omega_{1}$ coded in a $P$-extension by such a sequence $\bar{N}$ would be coded already by restriction of $\bar{N}$ in the ground model.

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  • $\begingroup$ So we can even say more, that MRP cannot be forced with a ccc forcing (unless it held true in the ground model maybe?). $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 14, 2016 at 16:16
  • $\begingroup$ Yes, I think that's right. $\endgroup$
    – Paul Larson
    Commented Jul 14, 2016 at 16:23
  • $\begingroup$ Excellent! Thank you, this was very helpful. $\endgroup$
    – Miha Habič
    Commented Aug 30, 2016 at 6:22

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