# Does totally proper forcing imply countable distributivity?

For a suitable model $M$ for $Q$ and a condition $q \in Q$ we say that $q$ is $(M,Q)$-generic if whenever $r \leqslant q$, $D \in M$ dense, $D \subset Q$, $r$ is compatible with an element of $D \cap M$.

If $\lbrace p \in Q \cap M \colon q \leqslant p \rbrace$ is an $(M,Q)$-generic filter, then $q$ is called totally $(M,Q)$-generic.

$Q$ is totally proper if whenever $M$ is a suitable model for $Q$ and $q \in Q \cap M$, $q$ has a totally $(M,Q)$-generic extension.

A forcing notation $P$ is $\kappa$-distributive if the intersection of $\kappa$ open dense sets is open dense.

Now let $P$ be a totally proper forcing notation. Does it follow that $P$ is proper and countable distributive?

I know that the way back holds and want to know if it is equivalent.

The answer is yes. Assume $Q$ is totally proper. It follows that $Q$ is proper. To see that it is countably distributive, let $\sigma$ be a name for an $\omega$-sequence of ordinals. Find a suitable $M$ with $\sigma\in M$. By total properness, we get a condition $q$ whose upward cone is an $M$-generic filter for $P$. In particular, $q$ decides every particular value of $\sigma(\check n)$. So $q$ forces that $\sigma$ is in the ground model. So the forcing cannot add any new countable sequences of ordinals, and so it is countably distributive.
• Are you assuming that $Q$ is separative? – Stefan Mesken Sep 6 '15 at 22:48
• @Stefan Does it matter? If the principle filter above $q$ is $M$-generic, it will decide the values of $\sigma(\check n)$ for every $n$, whether the forcing is separative or not. – Joel David Hamkins Sep 6 '15 at 23:54
• My argument shows that no condition can force that $\sigma$ is an $\omega$ sequence of ordinals that is not in the ground model (since we find such a $q$ below any given $p$), and this is one of the usual definitions of what it means to be countably distributive. So I don't think we need separativity for this. What definition of countable separativity are you using? – Joel David Hamkins Sep 7 '15 at 11:14
• Ah, I hadn't actually considered his definition. But it still seems to be no problem, since if you put the open dense sets into $M$, then $q$ will have to be in all of them, since the filter meets them, and so the intersection is not empty. By working below any condition, the intersection is dense. – Joel David Hamkins Sep 7 '15 at 11:25