Is there an example of a strongly proper ccc forcing that is not equivalent to Cohen forcing?
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asked Dec 7, 2016 at 13:24
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$\begingroup$ Shooting from the hip, coding a subset of $\omega_1$ into a real? $\endgroup$– Asaf Karagila ♦Commented Dec 7, 2016 at 13:49
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$\begingroup$ I think this one fails. The conditions are $(s,F)$, where $s$ is a finite subset of $\omega$ and $F$ is a finite set of almost-disjoint subsets of $\omega$ from a given collection of $\omega_1$ many. Let $M \prec H_\theta$ be countable, and let $p \in M$ be a condition. For any $q = (s,F) \leq p$ such that there is $A \in F \setminus M$, there cannot be a reduction of $q$ to $M$. Because if $r \in M$, then $r$ there is $(t,G) \leq r$ in $M$ such that $t$ has some elements of $A$ above $\max s$, so is incompatible with $q$. $\endgroup$– Monroe EskewCommented Dec 7, 2016 at 15:43
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1$\begingroup$ I think "strongly proper" might stand for different things across the literature. Monroe means "strongly proper" in Mitchell's sense. I.e. for club many countable $M$ and every $p \in M$ there is a $p' \le p$ which is an $(M,P)$-strong master condition; which in turn means that for every $p'' \le p'$ there is a $p''|M \in M \cap P$ such that all extensions of $p''|M$ in $M \cap P$ are compatible with $p''$. $\endgroup$– Sean CoxCommented Dec 7, 2016 at 18:20
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$\begingroup$ @Sean: I've never heard "strongly proper" before, and the first Google result was your slides from a talk not long ago. And now I also saw that you two posted a paper on arXiv today. in any case, I didn't know that this definition existed or had different variants. But I did say that I shot from the hip with my suggestion (taking your slide about forcing with side conditions as something which can sometimes be strongly proper, and I figured that coding subsets is in some sense a Cohen forcing with side conditions, so it might just work). $\endgroup$– Asaf Karagila ♦Commented Dec 7, 2016 at 18:48
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3$\begingroup$ @MohammadGolshani all (nontrivial) strongly proper forcings add Cohen reals, since there will be countable models $M$ from $V$ such that $G \cap M$ is generic over $V$ for $P \cap M$ (where $P$ is the strongly proper forcing and $G$ is $(V,P)$-generic). $\endgroup$– Sean CoxCommented Dec 8, 2016 at 20:15
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1 Answer
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The answer to your question is yes, and it follows from the following paper of Koppelberg and Shelah Subalgebras of Cohen algebras need not be Cohen.
In this paper, it is shown, for each $\kappa \geq \aleph_2,$ there exists a non-Cohen complete subalgebra of $Add(\omega, \kappa)$.
This subalgebra, is c.c.c and strongly proper, as a projection of Cohen forcing, but is not isomorphic to Cohen forcing.
answered Dec 10, 2016 at 4:35
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1$\begingroup$ Natural follow-up question: Does there exist a strongly proper ccc forcing of uniform density $\aleph_1$ that is not Cohen? $\endgroup$ Commented Dec 10, 2016 at 19:27
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$\begingroup$ @MonroeEskew About the follow-up question: perhaps this paper of Shelah and Zapletal is of some relevance. $\endgroup$– user3462Commented Dec 12, 2016 at 12:00
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$\begingroup$ Also, these examples are all where for club many models $M$, $1_{\mathbb P}$ is a strong master condition. I wonder if this is a necessary feature of strongly proper ccc forcing. $\endgroup$ Commented Dec 12, 2016 at 16:25
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$\begingroup$ The paper in @tci 's comment, Shelah and Zapletal's Embeddings of Cohen algebras, has updated link arxiv.org/abs/math/9502230 $\endgroup$– David Roberts ♦Commented Sep 26, 2021 at 0:56