3
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Are you assuming that $Q$ is separative? $\endgroup$ Sep 6, 2015 at 22:48
  • $\begingroup$ @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. $\endgroup$ Sep 6, 2015 at 23:54
  • $\begingroup$ Sure, but from this we can conclude its countable distributivity only if the forcing is separative, right? $\endgroup$ Sep 7, 2015 at 11:03
  • $\begingroup$ 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? $\endgroup$ Sep 7, 2015 at 11:14
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 7, 2015 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.