# An iteration of proper forcing without proper iterands

Famously, a countable support iteration with proper iterands is again proper. This is mostly stated as follows: Suppose $$(\mathbb{P}_{\alpha},\dot{\mathbb{Q}}_{\alpha})_{\alpha<\lambda}$$ is a countable support iteration such that for each $$\alpha<\lambda$$, $$\mathbb{P}_{\alpha}\Vdash$$ "$$\dot{\mathbb{Q}}_{\alpha}$$ is proper". Then the countable support limit $$\mathbb{P}_{\lambda}$$ is proper. I am interested in what happens if we merely require each $$\mathbb{P}_{\alpha}$$ to be proper (note that $$\mathbb{P}_{\alpha}*\dot{\mathbb{Q}}_{\alpha}$$ being proper does not imply that $$\mathbb{P}_{\alpha}$$ forces $$\dot{\mathbb{Q}}_{\alpha}$$ to be proper by Properness of quotient forcing). The question could probably be resolved by answering the following:

Suppose $$(\mathbb{P}_n,\dot{\mathbb{Q}}_n)_{n\in\omega}$$ is an iteration such that $$\mathbb{P}_n$$ is proper for each $$n\in\omega$$. Is the fully supported limit $$\mathbb{P}_{\omega}$$ proper as well?

Consider the forcing that adds a generic partition of $$\omega_1$$ into sets $$\omega_1=\sqcup_n A_n$$. One generically chooses for each ordinal which piece of the partitiion it is in. Each $$A_n$$ is stationary. This forcing is equivalent to adding a Cohen subset of $$\omega_1$$, which is countably closed and hence proper.
After doing this at the beginning, we perform subsequent forcing to add a club set $$C_n$$ avoiding the stationary set $$A_n$$. This step is not forced to be proper in the quotient, of course, but the combination of adding the generic sets and then clubs avoiding them is proper, for the same reason as in the other question. The combination of adding a Cohen set and then killing it is isomorphic to Cohen forcing, which is proper.
So every $$\mathbb{P}_n$$ will be proper, but the iteration will collapse $$\omega_1$$, since it will kill every stationary set in the countable partition.