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Let $\mathbb{P}$ and $\mathbb{Q}$ be two forcing notions. Recall that we say $\mathbb{Q}$ is a subforcing of $\mathbb{P}$ if there exists a regular embedding $\mathbb{Q} \to \text{r.o.}(\mathbb{P}).$

Question. Let $\mathbb{P} \in \{Add(\omega, \kappa), Col(\omega_1, \kappa) \}.$

(1) Is there a subforcing of $\mathbb{P}$ which is not Proper?

(2) Is there a subforcing of $\mathbb{P}$ which is not semi-Proper?

Remark. In Subalgebras of Cohen algebras need not be Cohen, Koppelberg and Shelah show that there is a subalgebra of Cohen forcing which is not Cohen.

François G. Dorais
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Mohammad Golshani
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1 Answers1

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No. A subforcing of a c.c.c. forcing is c.c.c. A subforcing of a countably closed forcing is countably-strategically-closed, which implies proper. (This is easy to see via countable elementary submodels. Use the strategy to construct a generic condition.)

Furthermore, every subforcing of a proper forcing is proper. Properness is equivalent to preserving stationary subsets of $[\kappa]^\omega$ for all $\kappa$. The stationarity of $X \subseteq [ \kappa ]^\omega$ cannot be restored once killed, since killing it is just adding some $f : [\kappa]^{<\omega} \to \kappa$ such that no $x \in X$ is closed under $f$. (See Abraham's chapter of the Handbook.)

Monroe Eskew
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  • Dear Monroe, thanks a lot; I will use this result to give an answer to the question "http://mathoverflow.net/questions/134101/existence-of-a-regular-subposet-which-collapses-everything-except-the-top-cardin" – Mohammad Golshani Dec 14 '13 at 06:11