12
$\begingroup$

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 $\omega_1$.

$\endgroup$
  • $\begingroup$ Do you have a reference for the Caicedo-Velickovic result? $\endgroup$ – Paul McKenney Jul 11 '16 at 14:22
  • $\begingroup$ @PaulMcKenney I've added a reference. $\endgroup$ – Miha Habič Jul 11 '16 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 Jul 11 '16 at 16:43
  • $\begingroup$ (And yes, that does seem a bit implausible!) $\endgroup$ – Asaf Karagila Jul 11 '16 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. $\endgroup$ – tci Jul 12 '16 at 17:41
12
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Jul 14 '16 at 16:16
  • $\begingroup$ Yes, I think that's right. $\endgroup$ – Paul Larson Jul 14 '16 at 16:23
  • $\begingroup$ Excellent! Thank you, this was very helpful. $\endgroup$ – Miha Habič Aug 30 '16 at 6:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.