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](https://pascal-francis.inist.fr/vibad/index.php?action=getRecordDetail&idt=PASCAL7830080080), 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.



Understanding descending intersections of generic extensions
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 $i<j$ 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. $\square$


