In Chow's "beginner's guide to forcing", why is $\bigcup U$ a function? I'm reading Timothy Chow's A beginner's guide to forcing (in a quest to finally familiarize myself with "boolean-valued model" forcing), and this passage on page 13 threw me for a loop:

... let $P$ be the partially ordered set of all finite partial functions from $\aleph_0^M\times \aleph_2^M$ into $2$, partially ordered by reverse inclusion. (...) There is a standard method, which we shall not go into here, of completing an arbitrary partially ordered set to a complete Boolean algebra; we take the completion of $P$ in $M$ to be our Boolean algebra $\mathbb B$. Now take an $M$-generic ultrafilter $U$, which exists because $M$ is countable. If we blur the distinction between $P$ and its completion $\mathbb B$ for a moment, then we claim that $F:=\bigcup U$ is a partial function from $\aleph_0^M\times \aleph_2^M$ to $2$. To check this, we just need to check that any two elements $x$ and $y$ of $U$ are consistent with each other where they are both defined, but this is easy: Since $U$ is an ultrafilter, $x$ and $y$ have a common lower bound $z$ in $U$, and both $x$ and $y$ are consistent with $z$.

The bolded part (my bolding!) looks quite dodgy to me. I assume we're really supposed to take $\bigcup(U\cap P)$, but even so the subsequent argument doesn't seem to make sense. For all I know, even if $x$ and $y$ are in $U\cap P$, their meet $z$ might lie in $\mathbb B\setminus P$, in which case it is not a partial function (but probably something like an equivalence class of sets of functions under some appropriate relation, given how "completion" usually goes). So why would its existence imply that $x$ and $y$ are consistent?
What am I missing here? Is it somehow an obvious consequence of $U$ being an ultrafilter? But that concept is defined only with reference to $\mathbb B$, and doesn't even know which of the elements of $\mathbb B$ were in $P$ to begin with.
 A: The complete Boolean algebra $\mathbb{B}$ is constructed so that this is not an issue.  Specifically, $\mathbb{B}$ has the property that $P$ is dense in $\mathbb{B}$.  So if $z\in U$, then there is some $w\in U\cap P$ such that $w\leq z$ by genericity, since the set of such $w$ is dense below $z$.
The explicit construction of $\mathbb{B}$ is this: an element of $\mathbb{B}$ is a downward-closed subset $b\subseteq P$ such that if $p\in P$ and $b$ is dense below $p$, then $p\in b$; we embed $P$ into $\mathbb{B}$ by mapping $p$ to $\{q:q\leq p\}$ (or more generally, the least element of $\mathbb{B}$ which contains $\{q:q\leq p\}$, if $\{q:q\leq p\}$ is not an element of $\mathbb{B}$, though that does not happen for the specific poset $P$ that Chow is discussing).  Each $b\in\mathbb{B}$ is the join of its elements (considered as elements of $\mathbb{B}$ via the embedding), so $P$ is dense in $\mathbb{B}$.  (Beware that for an arbitrary poset $P$ this "embedding" may not be injective, or may only weakly preserve the order.  A poset $P$ is called separative if this "embedding" really is an embedding of posets, or equivalently if $\{q:q\leq p\}$ is an element of $\mathbb{B}$ for all $p\in P$.)
