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To be precise, let the dominating forcing $D$ consist of all pairs $(m,f)$, where $m<\omega$ and $f\in\omega^\omega$, with the order $(m,f)\le(m',f')$ ($(m',f')$ is stronger) iff $m\le m'$, $f(k)\le f'(k)$ for all $k$, and $f'|m=f|m$.

Assume that $a\in\omega^\omega$ is dominating-generic over a CTM $M$ (that is, ($M\cap D$)-generic) and $b\in\omega^\omega$ is dominating-generic over $M[a]$ (that is, ($M[a]\cap D$)-generic).

Q1. Is $b$ then dominating-generic over $M$, a submodel of $M[a]$?

A1. Yes.

Q2. Is it true that $a\in M[b]$?

A2. No, and in fact $M[b]\cap M[a]=M$.

Q3, motivated by A2. Is $a$ in any way generic over $M[b]$?

A3. IDK, and surprisingly this Q does not seem to be an easy one.

Q4. Clearly $a+b$ (termwise addition) is generic over $M[a]$. Is it true that $M[a+b]\cap M[b]=M$? Or weaker, $M[a+b]\cap M[b]\cap2^\omega\subseteq M$?

A4. IDK, ditto.

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  • $\begingroup$ Your definition of the ordering of conditions makes $(m',f')$ stronger, contrary to your stated intention. $\endgroup$
    – Andreas Blass
    Commented Aug 29, 2016 at 16:01
  • $\begingroup$ 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) $\endgroup$
    – Noah Schweber
    Commented Aug 29, 2016 at 16:56
  • $\begingroup$ . . . 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. $\endgroup$
    – Noah Schweber
    Commented Aug 29, 2016 at 16:56
  • $\begingroup$ 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. $\endgroup$
    – Vladimir Kanovei
    Commented Aug 29, 2016 at 17:03

2 Answers 2

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Regarding question 3, it is a general and nontrivial fact that if $M\subseteq M[G]$ is any forcing extension and $N$ is an intermediate transitive model of ZFC, with $M\subseteq N\subseteq M[G]$, then $N$ is a forcing extension of $M$ and $M[G]$ is a forcing extension of $N$. This is proved, for example, in Corollary 15.43 in Jech's book (see also Fact 11 in my article, Set-theoretic geology, where we give a proof).

In your case, you have the two-step forcing iteration $M\subseteq M[a][b]$, which has $M[b]$ as an intermediate model. So $a$ is definitely generic over $M[b]$, and the forcing is a quotient of the two-step forcing giving rise to $a*b$.

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  • $\begingroup$ 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 $M[b]$? $\endgroup$
    – Vladimir Kanovei
    Commented Aug 29, 2016 at 20:03
  • $\begingroup$ 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? $\endgroup$
    – Vladimir Kanovei
    Commented Aug 29, 2016 at 20:34
  • $\begingroup$ 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 $V$. $\endgroup$
    – Joel David Hamkins
    Commented Aug 29, 2016 at 22:28
  • $\begingroup$ 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. $\endgroup$
    – Joel David Hamkins
    Commented Sep 3, 2016 at 22:13
  • $\begingroup$ 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 $M[b]$. $\endgroup$
    – Vladimir Kanovei
    Commented Sep 4, 2016 at 4:41
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Q4 is solved in the positive: $M[a+b]\cap M[b]\cap 2^\omega\subseteq M$. A point of dissatisfaction is that the most natural way to establish the result, that is, prove that $(a+b,b)$ is $(D\times D)$-generic (product-generic) over $M$, still does not go through. OOps - this fails because if $(a+b,b)$ is $(D\times D)$-generic then $a=(a+b)-b$ is Cohen-generic, contrary to the choice of $a$.

By the way, the result $M[a+b]\cap M[b]\cap 2^\omega\subseteq M$ is a key lemma in my proof that it is true in the dominating-generic extension that every countable OD set of reals consists of OD reals, arxived http://arxiv.org/abs/1609.01032.

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