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I think the following question is due to Prikry:

Question. Is it consistent that any non-trivial c.c.c forcing notion adds a Cohen real or a Random real?

Is the question still open? What partial results are known about this question (with references, please).

Remark. I have the following simple observation which might be well-know.

Theorem. Souslin hypothesis (SH) hols iff any non-trivial c.c.c forcing notion adds a new real.

Proof. One direction is trivial, since a Souslin tree, considered as a forcing notion, is c.c.c and adds no new reals.

For the other direction suppose there is a non-trivial c.c.c forcing notion $P$ which adds no new reals. Let $B=R.O(P)$ be the completion of $P$. B is a c.c.c. complete Boolean algebra which is $(\omega, \omega)$-distributive, hence it is in fact $(\omega, \infty)-$distributive, thus it is a Souslin algebra, which implies the existence of a Souslin tree, and hence SH fails.

Update. Can anyone say the first place where the above question has appeared?

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  • $\begingroup$ What about "Prikry reals"? Fix an ultrafilter U on $\omega$, and let $(s,A) \in \mathbb{P}$ when $s$ is finite, $A \in U$, etc., same as Prikry forcing. This is obviously c.c.c., so do you know if it is consistent that Prikry real forcing adds a Cohen real or Random real? $\endgroup$
    – Monroe Eskew
    Commented Nov 3, 2013 at 7:20
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    $\begingroup$ One can show that Mathias forcing relative to an ultrafilter U adds Cohen reals exactly when U is not selective. $\endgroup$
    – Mohammad Golshani
    Commented Nov 3, 2013 at 7:47
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    $\begingroup$ The question whether every Suslin ccc forcing notion adding a real must add a Cohen real or add a random real is problem 4.7 in Shelah's "On what I do not understand": shelah.logic.at/files/666.pdf $\endgroup$
    – Haim
    Commented Nov 3, 2013 at 11:22
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    $\begingroup$ Check also the paper "Ccc forcing and splitting reals" by Velickovic, according to which every Suslin ccc forcing adds a Cohen real or is a Maharam algebra. It's also worth mentioning that the question whether every Maharam algebra is a measure algebra was answered negatively by Talagrand. $\endgroup$
    – Haim
    Commented Nov 3, 2013 at 11:26
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    $\begingroup$ I believe the theorem in your question was proved by Jim Baumgartner. I know that it was proved by me, quite some time ago, but then I was told that Jim had proved it considerably earlier yet. $\endgroup$
    – Andreas Blass
    Commented Nov 19, 2013 at 18:39

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