Timeline for Dominating reals: another low-level Q
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2016 at 7:40 | comment | added | Joel David Hamkins | Ah, I understand now. | |
| Sep 4, 2016 at 4:41 | comment | added | Vladimir Kanovei | I am very sorry about this, I was not able to find a more appropriate way to express my inner expectations to have the forcing notion in $M$, not in $M[b]$. | |
| Sep 3, 2016 at 22:13 | comment | added | Joel David Hamkins | I don't understand your comment about "tiny bit off the point". You asked whether $a$ is "in any way generic over $M[b]$," and my answer is that yes, it is generic over $M[b]$, and I furthermore identified the partial order for which it is generic. | |
| Aug 29, 2016 at 22:28 | comment | added | Joel David Hamkins | The subalgebra $\mathbb{B}_0$ is in $V$ because it is defined in $V$ as the Boolean algebra generated by a certain collection of Boolean values which is in $V$, defined from a certain name that is in $V$. | |
| Aug 29, 2016 at 20:34 | comment | added | Vladimir Kanovei | By the way both 15.43 and your 11 lack clearness in an important detail: does $D$ in Jech, or $B_0$ in Fact 11, belong to V? | |
| Aug 29, 2016 at 20:03 | comment | added | Vladimir Kanovei | It's tiny bit off the point. Yes $M[a][b]$ is a generic extension of $M[b]$ by Solovay's $\Sigma$-method, that is, in this case, via a subforcing $\Sigma\in M[b]$ of the two-step forcing mentioned, and in fact $(a,b)$ is the $\Sigma$-generic object extending $M[b]$ to $M[a][b]$, not $a$ itself. But I wonder, if $\Sigma'$ is the projection of $\Sigma$ on its $a$-domain, will $a$ be $\Sigma'$-generic over $M[b]$? | |
| Aug 29, 2016 at 18:49 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |