# Understanding descending intersections of generic extensions

Let $$B_{0}\supseteq B_{1}\supseteq\dots\supseteq B_{\alpha}\supseteq\dots\,\,\left(\alpha<\kappa\right)$$ be a descending sequence of complete Boolean algebras, $$B_{\kappa}:=\bigcap_{\alpha<\kappa}B_{\alpha}$$, $$G_0$$ a $$V$$-generic filter on $$B_0$$ and for every $$\alpha\leq\kappa$$, $$G_{\alpha}:=G\cap B_{\alpha}$$. I want to understand the intersection $$\bigcap_{\alpha<\kappa}V\left[G_{\alpha}\right]$$, which clearly contains $$V[G_{\kappa}]$$.

In Iterating ordinal definability, Zadrożny surveys without proof some results. First are general results:

1. (Grigorief) $$\bigcap_{\alpha<\kappa}V\left[G_{\alpha}\right]\vDash ZF$$
2. (Jech) If $$B_0$$ is $$\kappa$$-distributive, then $$\bigcap_{\alpha<\kappa}V\left[G_{\alpha}\right]=V[G_\kappa]$$ (so in particular satisfies $$ZFC$$)

[The latter appears as lemma 26.6 in the 1978 edition of Jech's Set Theory, but curiously I haven't found it in the 3rd Millennium edition].

Then he gives a more concrete characterization of $$V\left[G_{\kappa}\right]$$, attributed to Sakarovitch:

For $$p,q\in B_0^+$$, let $$p\sim q$$ iff $$\exists \alpha<\kappa$$ such that $$\inf\{d\in B_\alpha \mid d\geq p\}=\inf\{d\in B_\alpha \mid d\geq q\}$$

Then the separative part of $$B_0/{\sim}$$ is isomorphic to $$B_\kappa^+$$. In particular, from $$G_0$$ one can define a $$B_0/{\sim}$$ generic $$G_0/{\sim}$$, and if $$B_0$$ is $$\kappa$$-distributive then $$\bigcap_{\alpha<\kappa}V\left[G_{\alpha}\right]=V[G_\kappa]=V[G_0/{\sim}]$$

I want to understand this result more, however Zadrożny references Sakarovitch's PhD thesis, which is in French, not available online and as far as I can tell, has no adaptation to a paper. So my questions are:

1. Is there some other source where this result is presented?
2. Can someone provide a proof or at least a sketch?
3. What more can be said on this intersection, given the properties of the descending sequence?

A particular case I'm interested in, is when the sequence is given by "tails" of an iteration (or even product): assume $$\langle P_\alpha \mid\alpha\leq\kappa\rangle$$ is an iteration, and set for every $$\alpha$$ $$P_\kappa=P_\alpha * \dot{P}^\alpha$$ (so $$\dot{P}^\alpha$$ is the "tail" of the iteration), and "$$B_\alpha=ro(\dot{P}^\alpha)$$" (I guess that for this to make perfect sense we should say something like $$B_\alpha=ro(P_\kappa)/ro(P_\alpha)$$). Can something more concrete be said in this case?

Edit: note that if $$P_\kappa$$ is a direct limit (and $$\kappa$$-distributive), we'd get that all elements are eventually equivalent so $$B_0/{\sim}$$ is trivial. So on one hand this shows that the intersection is $$V$$. And on the other hand, the question is interesting only when non-direct limit is taken.

• Regarding your edit: You need to additionally assume that $P_0$ is $\kappa$-distrubutive. Otherwise the intersection can differ from $V$. Commented Jun 21, 2022 at 21:47
• @AndreasLietz yes ofcourse, in my head I was still under that assumption which is present in Sakarovitch's characterization. I'll add it. Commented Jun 23, 2022 at 7:43
• When you say "direct limit", in what sense do you mean this? (I've always found this terminology confusing when it comes to iterations; we have a notion of supports, just tell me the ideal/support system you're using.) Commented Jun 23, 2022 at 8:46
• Also, the result of Grigorieff should probably be in his paper about intermediate models of ZF set theory from 1975 (Annals of Mathematics). Commented Jun 23, 2022 at 8:56
• So after digging a bit more on Google scholar I managed to find a note by Sakarovitch summarizing the results, which is even in the public domain (although still in French) - gallica.bnf.fr/ark:/12148/bpt6k5619035q/f13.item Commented Jun 27, 2022 at 9:32

The theorem should state that the separative quotient of $$B_{0}/{\sim}$$ is isomorphic to $$B_{\kappa}^+$$.

Proof. Recall that the separative quotient of a poset $$P$$ is the unique (up to isomorphism) separative $$Q$$ such that there is an order preserving $$h:P\to Q$$ such that $$x$$ is compatible with $$y$$ iff $$h(x)$$ is compatible with $$h(y)$$ (see Jech pg. 205). Since $$B_{\kappa}$$ is separative as a Boolean algebra, we want to provide such $$h:B/{\sim}\to B_{\kappa}^{+}$$.

Let $$b\in B_{0}^{+}$$. For every $$\alpha<\kappa$$ let $$b_{\alpha}=\inf \{ d\in B_{\alpha}\mid d\geq b \}$$, and let $$\bar{b}=\sup\{b_{\alpha}\mid\alpha<\kappa\}$$. Note that $$\{b_{\alpha}\mid\alpha<\kappa\}$$ is an ascending sequence, so in fact for every $$\beta$$, $$\bar{b}=\sup\{b_{\alpha}\mid\beta\leq\alpha<\kappa\}$$, and this is an element of $$B_{\beta}$$, so all-in-all $$\bar{b}\in B_{\kappa}$$. Now if $$b'\sim b$$ then for all large enough $$\alpha$$, $$b_{\alpha}=b_{\alpha}'$$ so $$\bar{b}=\bar{b'}$$. So the function $$h([b])=\bar{b}$$ is well defined, and since in particular $$\bar{b}\geq b>0$$, it is into $$B_{\kappa}^{+}$$. It is order preserving since if $$[b]\leq[c]$$ then for all large enough $$\alpha$$, we have $$b_\alpha \leq c_\alpha$$ so also $$\bar{b}\leq\bar{c}$$.

Let $$[b],[c]\in B_{0}/{\sim}$$. We want to show they are compatible iff $$\bar{b}$$ and $$\bar{c}$$ are compatible.

• If $$[b],[c]$$ are compatible, $$[d]\leq[b],[c]$$, then for all large enough $$\alpha$$ $$d_{\alpha}\leq b_{\alpha},c_{\alpha}$$ so $$\bar{d}\leq\bar{b},\bar{c}$$.
• If $$[b],[c]$$ are incompatible, it means that there is an unbounded $$I\subseteq\kappa$$ such that for $$\alpha\in I$$, $$b_{\alpha}$$ and $$c_{\alpha}$$ are incompatible. But this also implies that for every $$\alpha,\beta\in I$$ $$b_{\alpha}$$ and $$c_{\beta}$$ are incompatible (if $$\alpha<\beta$$ and there is e.g. $$d\leq b_{\alpha},c_{\beta}$$ then since $$b_{\alpha}\leq c_{\beta}$$ we get $$d\leq b_{\beta},c_{\beta}$$), so, as we are in a complete Boolean algebra, $$\overline{b}\cdot\bar{c}=\sum_{\alpha\in I}b_{\alpha}\cdot\sum_{\beta\in I}c_{\beta}=\sum_{\alpha,\beta}b_{\alpha}\cdot c_{\beta}=0$$ i.e. $$\bar{b},\overline{c}$$ are incompatible. $$\square$$

Regarding the specific case, my advisor pointed out to me that if we are considering tails of a full support product, say of length $$\kappa$$, then the quotient poset will be $$\kappa$$-closed.

Proof sketch. Let $$P=\prod_{\alpha<\kappa}Q_{\alpha}$$ be a full support product of posets, and for $$\alpha<\kappa$$ let $$P_{\alpha}=\prod_{\alpha\leq\xi<\kappa}Q_{\xi}$$. Let $$\left\langle p_{i}\mid i<\kappa\right\rangle$$ be sequence such that $$i implies $$[p_{i}]>[p_{j}]$$. Then to construct $$q$$ such that $$[q]$$ is a lower bound, we diagonalize - let $$q(\xi)=p_0(\xi)$$ until the coordinate witnessing $$[p_0]>[p_1]$$, then $$q(\xi)=p_1(\xi)$$ until $$p_2$$ is smaller, and so on. I'll spare you the indexing monstrosity. $$\square$$