Timeline for Does totally proper forcing imply countable distributivity?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 7, 2015 at 12:36 | comment | added | Stefan Mesken | Right, I completely ignored the fact that we were talking about totally proper forcings. Thank you. | |
| Sep 7, 2015 at 11:25 | comment | added | Joel David Hamkins | 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. | |
| Sep 7, 2015 at 11:19 | comment | added | Stefan Mesken | I have OP's definition in mind. A forcing is countable distributive iff the intersection of countably many open dense sets remains to be dense. This follows from not adding countable subsets if the forcing is separative. | |
| Sep 7, 2015 at 11:14 | comment | added | Joel David Hamkins | 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? | |
| Sep 7, 2015 at 11:03 | comment | added | Stefan Mesken | Sure, but from this we can conclude its countable distributivity only if the forcing is separative, right? | |
| Sep 6, 2015 at 23:54 | comment | added | Joel David Hamkins | @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. | |
| Sep 6, 2015 at 22:48 | comment | added | Stefan Mesken | Are you assuming that $Q$ is separative? | |
| Sep 4, 2015 at 12:52 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |