Understanding descending intersections of generic extensions

Question

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.

Answer 1

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. $\square$