Suppose $\mathbb P$ is a countably closed forcing, and $\mathbb Q$ is a c.c.c. forcing that adds reals. Is $\mathbb P$ still proper in $V^{\mathbb Q}$?



Preservation of properness
Here is the short version of a negative answer. Let $P$ be the collapse of $\omega\_2$ to $\omega\_1$ with countable conditions. Fix a tree $(N\_\eta:\eta\in 2^{<\omega})$ of quite different models, increasing along each branch. Specifically, make sure that along each branch in $2^\omega$ the union of the models will get a different intersection with $\omega\_2$. Now go to any extension $V^Q$ where there is a new real. The new real will give a new branch, yielding a new model; no condition in $P$ can be generic for this model, since a generic condition has to know the model. 


*(I could not find a counterexample for a week. But since I am in Jerusalem at the moment, I eventually gave up and asked Shelah. It took him only a minute or so to come up with this answer. He also said that the reason he is using this specific $P$ is just to understand generic conditions better; many others would work as well.)*


Here is the more explicit version. (Perhaps too explicit? Perhaps too complicated? It is quite possible that I have overlooked a simpler solution.) 


Let $P$ be the set of countable partial 1-1 functions from $\omega\_1$ to $\omega\_2$. Note that if a condition $q$ is $N$-generic it must satisfy the following properties:


* $dom(q) \supseteq \omega\_1\cap N $. (Otherwise $q\cup \{(\alpha,\beta)\}$ for some $\alpha\in \omega\_1\cap N$, $\beta\in \omega\_2\setminus N$ will force that $N[G]$ has the new ordinal $\beta$.)
* $q[\omega\_1\cap N]\subseteq \omega\_2\cap N$. Obviously.
* $q[\omega\_1\cap N] \supseteq \omega\_2 \cap N$. (If $\beta\in \omega\_2\cap N$ is outside the range of $q[\omega\_1\cap N]$, then $q$ would force that $g^{-1}(\beta)\in \omega\_1\cap N$ is an ordinal outside $N$. Here I write $g$ for the generic function, i.e. the union of all conditions in the filter.)
* Summarizing: $q\restriction(\omega\_1\cap N)$ is a bijection between $\omega\_1\cap N$ and $\omega\_2\cap N$. (If I have not overlooked anything, this is also sufficient for being generic.)


So a generic condition must really know a lot about $N$. We will use this to get a counterexample. 


Let $(N\_\eta:\eta\in 2^{<\omega})$ be a family of countable elementary submodels with the following properties:


* For alle $\eta\subseteq \nu$ we have $N\_\eta \subseteq N\_\nu$.
* For all $\eta,\nu\in 2^\omega$ with $\eta\not=\nu$ we have $N\_\eta \cap \omega\_2 \not=N\_\nu\cap \omega\_2$. ($N\_\eta$ is defined naturally as $\bigcup\_n N\_{\eta\restriction n}$.)
* Moreover: For every $\eta\in 2^{<\omega}$ there is an ordinal $\alpha\_\eta\in \omega\_2$ witnessing that the two branches that split at $\eta$ will yield models whose intersections with $\omega\_2$ are different: $\alpha\_\eta\in N\_{\eta^\frown 0}$, but not in $N\_{\eta^\frown1}$, and also $\alpha\_\eta\notin N\_\nu$ for any $\nu$ extending $\eta^\frown1$.
* There is a $\delta$ such that $\omega\_1\cap N\_\eta=\delta$ for all $\eta\in 2^\omega$. (We will in fact have $\omega\_1\cap N\_\eta=\delta$ also for all $\eta\in 2^{<\omega}$.)


Now let $Q$ be a (ccc, or just proper) forcing adding a new real $\nu\in 2^\omega$. In the extension $V^Q$ the family $(N\_{\nu\restriction n}[G]:n\in \omega)$ is an increasing sequence of elementary submodels; its union $N\_\nu[G]$ is again an elementary submodel, and $N\_\nu[G]\cap \omega\_1=\delta$. 


Assume that $q$ is generic for this model. Then $S:=q[\delta]=\omega\_2 \cap N\_\nu[G]$ is a set in the ground model. But now $\nu$ can be computed from $S$: $\eta(0)=1 $ iff $\alpha\_{\langle\rangle}\in S$, $\eta(1)=1 $ iff $\alpha\_{\langle\eta(0)\rangle}\in S$, etc. Contradiction. 


It remains to find a tree as described above. (I took some of the following arguments from chapter XV of "Proper and Improper Forcing". *I suspect there might be a much easier construction.*)


* Start with a Namba tree $(N\_s:s\in \omega\_2^{<\omega})$ of countable models which are increasing along each branch and satisfy $s\in N\_s$ for all $s\in \omega\_2^{<\omega}$.
* Thin out the tree to make all $N\_s\cap\omega\_1$ equal to some $\delta$ for all $s\in \omega\_2^\omega$, still keeping $\aleph\_2$ many successors of each node. (Proper and improper, XV 2.12, based on XI 3.5, going back to a partition theorem in Rudin+Shelah, APAL 1987, RuSh:117. Use determinacy of a low level Borel game.)
* Replace each model $N\_s$ ($s\in \omega\_2^{<\omega}$) by the Skolem closure of $N\_s\cup\delta$.
* Now all $\omega\_1\cap N\_s=\delta$ for all $s\in \omega\_2^{\le \omega}$.


Now find a binary tree $(s\_\eta:\eta\in 2^{<\omega})$ with corresponding models $M\_\eta:=N\_{s\_\eta}$ as follows.


* Start with $s\_{\langle\rangle}:=\langle\rangle$, $M\_{\langle\rangle}=N\_{\langle\rangle}$.
* Given $M\_\eta=N\_{s\_\eta}$, first pick an ordinal $\alpha\_\eta$ which appears in one of the successors of $M\_\eta$ in the $N$-tree (i.e., in some $N\_{s\_\eta^\frown i}$, but not in coboundedly many of them).
* Let $M\_{\eta^\frown 0}$ be a successor of $M\_{\eta}$ containing $\alpha\_\eta$.
* Let $M\_{\eta^\frown 1}$ be a successor of $M\_{\eta}$ not containing $\alpha\_\eta$, but containing some larger ordinal $\beta\_\eta<\omega\_2$.
* Note that no model $M\supseteq M\_{\eta^\frown1}$ with $M\cap \omega\_1=\delta$ can contain $\alpha\_\eta$ as an element, since already $M\_{\eta^\frown1}$ sees a bijection between $\omega\_1$ and $\beta\_\eta$; if a new element appeared in the range of this bijection, a new element would have to appear also in the domain.


This finishes the inductive step, and thus completes the construction of the tree of models. 


