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Sep 3, 2016 at 18:16 answer added Vladimir Kanovei timeline score: 1
Aug 29, 2016 at 18:49 answer added Joel David Hamkins timeline score: 2
Aug 29, 2016 at 17:03 comment added Vladimir Kanovei Well, about $D$, I've copied the definition from the Judah-Bartosh book page 104. Your $\mathbb P$ is a set in $M$ and $M[b]$, of course, and your conjecture that $a$ is $\mathbb P$-generic over $M[b]$ would solve Q3.
Aug 29, 2016 at 16:56 comment added Noah Schweber . . . or if not, that $a$ is generic for the version $\mathbb{P}'$ of $\mathbb{P}$ "thinned" by Solovay's $\Sigma$-process to guarantee that $b$ is Hechler generic over $M[c]$ for any $c$ which is $\mathbb{P}'$-generic over $M$. But this is just a wild guess.
Aug 29, 2016 at 16:56 comment added Noah Schweber First, an observation: this notion of forcing is usually called Hechler forcing. Now, a wild guess: Recall that the ground model of any forcing extension is definable in that extension (this is due to Laver and (independently) Woodin, if I recall correctly). That means that in $M[b]$, there is a poset $\mathbb{P}$ which is the ground Hechler forcing: $\mathbb{P}$ consists of all Hechler conditions which are in $M$. I suspect $a$ is $\mathbb{P}$-generic over $M[b]$ (cont'd)
Aug 29, 2016 at 16:25 history edited Vladimir Kanovei CC BY-SA 3.0
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Aug 29, 2016 at 16:01 comment added Andreas Blass Your definition of the ordering of conditions makes $(m',f')$ stronger, contrary to your stated intention.
Aug 29, 2016 at 11:03 history asked Vladimir Kanovei CC BY-SA 3.0