One of the standard examples of a forcing which is distributive but not proper (equiv. not closed) is a club shooting into a stationary co-stationary set.


But that forcing is $S$-proper for the stationary set into which we shoot the club. So it's at least a little bit proper. I want more. I am looking for an example of a forcing that is not $S$-proper for any stationary set $S$.


Of course, we can be silly and take the lottery sum over all possible club shooting, that is, first we choose a stationary set, then we shoot a club into it. Since no stationary set is guaranteed to survive, this is not $S$-proper for any stationary set. But this is silly, since the forcing is "locally $S$-proper for some $S$". That is, every condition can be extended such that below the extension the forcing is $S$-proper for some $S$.


So what I am really looking for is a forcing that is both distributive and not $S$-proper below any condition for any stationary set $S$.



Example of a distributive forcing which is entirely improper
$\mathbb P$ being $S$-proper means that for all countable $M \prec H\_\theta$ with $M \cap \omega\_1 \in S$, every $p \in \mathbb P \cap M$ can be extended to an $M$-generic condition.


A positive answer to your question is given by [work of Gitik](https://mathscinet.ams.org/mathscinet-getitem?mr=2648158), who showed that relative to a supercompact, there can exist an inaccessible $\kappa$ and a $\kappa^+$-saturated normal ideal concentrating on $\mathrm{cof}(\omega\_1)$. Let $\mathbb P = \mathcal P(\kappa)/I$ for such an ideal $I$.


The reason this $\mathbb P$ doesn't add $\omega$-sequences is that it forces an embedding $j : V \to M \subseteq V[G]$ with $M^\kappa \cap V[G] \subseteq M$. It can't add bounded subsets of $\kappa$, and by the saturation, being $(\omega,\infty)$-distributive is equivalent to being $(\omega,\kappa)$-distributive, which is equivalent to not adding $\omega$-sequences in $\kappa$. Since it forces $\mathrm{cf}(\kappa)=\omega\_1$, adding an $\omega$-sequence in $\kappa$ means adding a bounded one, which it does not.


Now since $\mathbb P$ makes $\kappa$ a singular cardinal, it must kill the stationarity of some $T \subseteq \kappa \cap \mathrm{cof}(\omega)$, since it adds a club of smaller cardinality than some partition into $\kappa$-many stationary sets. Let $\dot C$ be a name for a club in $\kappa$ forced to be disjoint with $T$. Now let $S \subseteq \omega\_1$ be stationary and let $N \prec H\_\theta$ be a countable elementary submodel with $N \cap \omega\_1 \in S$ and $\sup(N\cap\kappa) \in T$, where $\theta \gg \kappa$ and $\dot C \in N$. If $q$ is $N$-generic, then it forces $\sup(N \cap \kappa) \in \dot C$, contradiction. Thus $\mathbb P$ cannot be $S$-proper.


I am curious whether this can be done with less strength or with easier constructions than Gitik's.


**EDIT:** Here’s an example without large cardinals. The argument above shows that if a forcing $\mathbb P$ is $S$-proper for some stationary $S \subseteq \omega\_1$, then it must preserve all stationary $T \subseteq \kappa \cap \mathrm{cof}(\omega)$ for regular $\kappa>\omega\_1$.


There is a two-step iteration $\mathbb P \* \dot{\mathbb Q}$ that is countably closed such that $\mathbb P$ adds a $\square\_{\omega\_1}$-sequence without adding subsets of $\omega\_1$, and then $\dot{\mathbb Q}$ threads it with an ordertype-$\omega\_1$ club in $\omega\_2^V$. So $\dot{\mathbb Q}$ is forced to be countably distributive.


If $\langle C\_\alpha : \alpha < \omega\_2 \rangle$ is the square sequence added by $\mathbb P$, then for unboundedly many $\gamma<\omega\_1$, the set $T\_\gamma = \{ \alpha : \mathrm{ot}(C\_\alpha) = \gamma \}$ is stationary in $\omega\_2$.


Let $D$ be the thread added by $\mathbb P \* \dot{\mathbb Q}$, and let $D’$ be the limit points of $D$. Then $D’$ intersects each $T\_\gamma$ in at most one point. For if $\alpha \in D’ \cap T\_\gamma$, then $D \cap \alpha = C\_\alpha$ and $\mathrm{ot}(D \cap \alpha) = \gamma$, which holds for exactly one point in $D$. Thus the stationarity of some $T\_\gamma$ has been killed by $\mathbb Q$.


