Partial order which forces finite intersections with sets not in an ultrafilter The following is from the Paper ,,Canonical models for fragments of the Axiom of Choice" written by Paul Larson and Jindrich Zapletal, in Claim 3.4:
Let $U$ be a nonprincipal ultrafilter on $\omega$ with empty intersection with $A$, where $A \subseteq \mathcal{P}(\omega)$ is an infinite $MAD$ family. Let $P$ be the usual $c.c.c.$ poset adding a set $\dot x_{gen} \subseteq \omega$ which has finite intersection with every set not in $U$.
We work with ZFC and the hypothesis, that Woodin cardinals exist.
My question now is: What is such an usual partial order $P$, that forces the above?
 A: One possible poset would be as follows: Let $\mathbb{P}=[\omega]^{<\omega}\times U$, and $(s,X)\ge (t,Y)$ if and only if

*

*$s\subseteq t$,

*$X\supseteq Y$, and

*$t\setminus \max s \subseteq X$.

Let $G$ be a $\mathbb{P}$-generic filter, and let $x_{gen}=\bigcup\{s\mid \exists X\in U: (s,X)\in G\}$. For each $Z\subseteq \omega$ that is not in $U$, consider
$$\mathcal{D}_Z = \{(s,X)\in\mathbb{P}\mid X\subseteq \omega\setminus Z\}.$$
Then we can see that $\mathcal{D}_Z$ is dense. Hence there is $s$ such that $(s,\omega\setminus Z)\in G$. From this, we can see that $x_{gen}\setminus \max s\subseteq \omega\setminus Z$. Especially, $x_{gen}\cap Z\subseteq \max s$.

It remains to show that $\mathbb{P}$ has c.c.c. Let $\{(s_\alpha,X_\alpha)\mid \alpha<\omega_1\}\subseteq \mathbb{P}$. Since $[\omega]^{<\omega}$ is countable, we may assume that $s_\alpha$ is constant with value $s$. Hence it suffices to show that $(U,\subseteq)$ satisfies c.c.c., and in fact, every pair of elements of $U$ is compatible.
A: Given a family of co-infinite sets, $X$, we define the forcing that avoids all the sets in $X$, $\Bbb P_X$ as the collection of pairs $(p,S)$ such that:

*

*$p\colon\omega\to 2$ is a partial function with finite domain.

*$S\subseteq X$ is a finite family of sets.

Now define the order as follows: $(q,T)\leq(p,S)$ iff:

*

*$p\subseteq q$,

*$S\subseteq T$, and

*for all $A\in S$, if $n\in (A\cap\operatorname{dom}(q))\setminus\operatorname{dom}(p)$, then $q(n)=0$.

In other words, once $A$ goes into the $S$ component, any further extension must avoid it.
Now it is easy to see that $\Bbb P_X$ is c.c.c. since any two conditions with the same stem (the $p$ coordinate) must be compatible and there are only countably many possible stems.
And if $g$ is the the generic subset given by the generic filter $G$, then for all $A\in X$, there is some condition $(p,S)\in G$ such that $A\in S$, and therefore any further extension beyond $p$ must force that $g\cap A$ is finite.
