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Is it known whether Todorcevic's Open Coloring Axiom implies $2^{\aleph_0}=\aleph_2$?

The only consistency proofs for OCA that I know are the following:

1) PFA implies OCA (and also $2^{\aleph_0}=\aleph_2$).

2) Constructing a finite support ccc iteration of length $\omega_2$ starting from a model satisfying CH. (CH at intermediate stages is needed to ensure the ccc of the forcing notions, so we end with $2^{\aleph_0}=\aleph_2$).

3) Starting from a model of OCA and forcing with a Suslin tree (which does not add reals).

On the other hand, I know that MA+OCA imply $2^{\aleph_0}=\aleph_2$, as well as OCA plus the Abraham-Rubin-Shelah version of OCA.

The last fact is proved in

Moore, J.T. Open colorings, the continuum, and the second uncountable cardinal, Proceedings of the American Mathematical Society, 130 (2002), n 9, pp. 2753-2759.

where it is also asked whether OCA implies $2^{\aleph_0}=\aleph_2$. So my question is if this question has been answered since 2002.

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    $\begingroup$ It might help you get more answers if you add the reference-request tag @Carlos. $\endgroup$
    – Jose Arnaldo Bebita Dris
    Aug 14, 2014 at 10:07
  • $\begingroup$ I was going to add the tag, but David has already done it. Thanks! $\endgroup$
    – Carlos
    Aug 14, 2014 at 12:12

1 Answer 1

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Whether $OCA$ implies $2^{\aleph_0}=\aleph_2$ appears as problem 7.2 in J.T. Moore´s The Proper Forcing Axiom in the Proceedings of the ICM, 2010. Apparently it was still open by the end of 2012 according to T. Yorioka´s abstract for the Fields Institute Set Theory seminar.

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    $\begingroup$ To add to Ramiro's answer, I spoke with Justin Moore at the end of 2013, and he told me the problem was still open then. $\endgroup$
    – Garrett Ervin
    Aug 14, 2014 at 19:38
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    $\begingroup$ More information: Ilijas Farah has a proof (unpublished) that the restriction of $OCA$ to sets of size $\aleph_1$ is consistent with a large continuum. $\endgroup$
    – Paul McKenney
    Aug 17, 2014 at 3:16

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