# Is there a forcing closure?

The main theorem of forcing says that for any c.t.m of $ZFC$ like $M$ and for all partial order $\mathbb{P}$ and $\mathbb{P}$-generic $G$ over $M$, there is a c.t.m of $ZFC$, like $N$ such that $N$ is the least (by inclusion order) c.t.m of $ZFC$ which $M\subseteq N$ and $G\in N$. Now the question is:

Question (1): Let $I$ be an index set and $\lbrace M_{i}\rbrace_{i\in I}$ a family of c.t.m s of $ZFC$, $\lbrace \mathbb{P}_{i}\rbrace_{i\in I}$ a family of partial orders and $\lbrace G_{i}\rbrace_{i\in I}$ a family of sets such that for all $i\in I$, $G_{i}$ is a $\mathbb{P}_{i}$ - generic filter over $M_{i}$.

Is there a c.t.m $N$ of $ZFC$ such that $N$ be the least c.t.m of $ZFC$ (by inclusion order) with the property: $\forall i\in I~~~~~M_{i}\subseteq N~~\wedge~~G_{i}\in N$?

We call $N$ the forcing closure of $\lbrace (M_{i},\mathbb{P}_{i},G_{i})~|~i\in I\rbrace$.

Question (2): Is any c.t.m of $ZFC$, a forcing closure of some family $\lbrace (M_{i},\mathbb{P}_{i},G_{i})~|~i\in I\rbrace$?

In general, there may be no such $N$, even when $I$ has only two elements. To see this, suppose that we have a countable transitive model $M$, and suppose there is no countable transitive model with more ordinals. Let $z$ be a real coding the ordinals of $M$ as a relation on $\omega$, so that $z$ is a forbidden real, which cannot be added to $M$ without collapsing all the ordinals. Now, build two $M$-generic Cohen reals $c$ and $d$, in stages. First, enumerate the dense sets in $M$ as $D_0,D_1,\ldots$, and let $c_0$ be any element of $D_0$. Let $d_0$ be all-zeros, of exactly the same length, followed by $1$, followed by the digit $z(0)$, followed by an extension putting it into $D_0$. Now extend $c_0$ by adding $0$s until it has the same length as $d_0$, followed by $1$, and then make an extension putting it into $D_1$. Now extend $d_0$ to $d_1$ by adding $0$s, then a $1$, then $z(1)$, and then extend into $D_1$. Continue in this way. Each of $c=\bigcup c_n$, $d=\bigcup d_n$ is an $M$-generic Cohen real, so we have the forcing extensions $M[c]$ and $M[d]$, but if $c$ and $d$ appear together in a model $N$, then by inspecting the blocks of zeros, we can define the number $z$, which collapses all the ordinals of $M$. In this case, $N$ would have to be taller than $M$, and this would violate our assumption.
Basically, the example shows that without a mutual-genericity hypothesis, the forcing extensions of a given model $M$ are not upward directed. I find it more natural to look for an extension $N$ with the same ordinals as $M$, and furthermore, to look for an $N$ that is itself a forcing extension of $M$. Thus, one is looking at the generic multiverse of $N$ (the collection of all models that are obtainable from $M$ by going to a forcing extension or a ground). And here there are some interesting things to say.
Theorem. Suppose that $W$ is a countable model of ZFC, and $$W[G_0]\of W[G_1]\of\cdots\of W[G_n]\of\cdots$$ is an increasing sequence of forcing extensions of $W$, with $G_n\of\Q_n\in W$ being $W$-generic. If the cardinalities of the $\Q_n$'s in $W$ are bounded in $W$, then there is a set-forcing extension $W[G]$ with $W[G_n]\of W[G]$ for all $n<\omega$.
The bounded cardinality assumption is very natural, since otherwise one could collapse more and more cardinals of $M$, or add more and more Cohen reals over $M$, and these extensions could clearly not be amalgamated.