Timeline for Dominating reals: another low-level Q
Current License: CC BY-SA 3.0
8 events
when toggle format | what | by | license | comment | |
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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 |
added 2 characters in body
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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 |