What you’ve done is correct, but the fact that $h(x_g)\le x'$ for each $x'\in X'$ isn’t enough to show that $h(x_g)=\bigwedge X'$: it shows that $h(x_g)$ is a lower bound for $X'$, but it doesn’t show that $h(x_g)$ is the greatest lower bound for $X'$. For that you need to verify that $b'\le h(x_g)$ whenever $b'$ is a lower bound for $X'$. But that’s straightforward: if $b'\le x'$ for each $x'\in X'$, then $h^{-1}(b')\le h^{-1}(x')$ for each $x'\in X'$, so $h^{-1}(b')\le x_g$, and therefore $b'\le h(x_g)$.
One way to start, is to naively try to make completion by cuts "work" and see how the need for a new concept naturally comes up. So let's consider the set $C$ of all cuts of $A^+$ under the $\subseteq$ order and try to find a Boolean completion in there somewhere.
A couple easy facts make us immediately optimistic. First off, a union of cuts is a cut, as is the intersection of cuts, so we are in good shape for meets and joins. And moreover, we have $U_p\cap U_q=U_{p\cdot q}$ and $U_p\cup U_q = U_{p+q},$ so an embedding of $A$ embedding is shaping up nicely too.
Where things stop working so nicely is when we consider complements: the complement of a cut is not necessarily a cut and it's not generally the case that $A^+-U_p = U_{-p}.$ We have a nice structure (in fact, it is a complete Heyting algebra!), but it falls short of being a Boolean algebra. Nonetheless, let's see what we can do to make it work.
We know that if our embedding is to work, we need $-U_p=U_{-p}$. As we remarked previously, using set-theoretic complement $-U_p \ne A^+-U_p$ is wrong... we don't want $-U_p$ to contain literally everything that is not $\le p,$ rather we want it to consist of things that are incompatible with $p$. This is is made more clear by the identity $U_{-p}=\{q\in A^+: q\cdot p = 0 \}$, which just comes from the fact that $q\le - p$ if and only if $q\cdot p = 0.$ The natural generalization of this operation to arbitrary cuts is to let $-U = \{q\in A^+: \forall p\in U\; q\cdot p = 0\},$ or more succinctly, $-U = \{q\in A^+: U_q \cap U = \emptyset\}.$
We want our join operation to give us $U+ - U = A^+,$ but it not necessarily true that $U\cup -U=A^+,$ so the join operation needs to do more than just take the union. The key observation is that while $U_p\cup U_{-p}$ might not contain all of $A^+,$ there is a weaker sense in which it is big: it is dense in $A^+$. This suggests that the correct definition of addition is that $U+V$ is the set of all $p\in A^+$ such that $U\cup V$ is dense below $p$.
Finally, we need addition to satisfy $U+U= U,$ so since $U+U$ is the set of all $p$ that $U$ is dense below, it better be the case that $U$ contains every point it is dense below. That's not true for arbitrary cuts, so we restrict our set of cuts to just the ones for which it is true. These are the regular cuts! (Check that Jech's definition of regular cut is indeed equivalent to the statement that the cut contains every point it is dense below.)
Now we check that it works. Are the $U_p$ regular? Yes, any $q\nleq p$ has an extension incompatible with $p$ (this is just the statement that $A^+$ is a separative partial order), so $U_p$ is not dense below $q$ for any $q\notin U_p.$ Is $U+V$ necessarily regular? Yes, if $U+V$ is dense below $p,$ so is $U\cup V$ so if $p\notin U+V$ then $U+V$ is not dense below $p$. What about if we just let $U\cdot V=U\cap V$? That works fine since if $U\cap V$ is dense below $p$, so are $U$ and $V.$ We can likewise confirm that negation works like we expect, and that Jech's statement that $\overline U$ (the collection of all points $U$ is dense below) is the least regular cut containing $U$ and use that to show that joins (finite or infinite) exist and match up to our definition.
That was more than I planned to write... in case all that the motivation has the opposite of the intended effect, the headline news is:
A cut is regular if it contains every point it is dense below.
A couple closing notes
- While our motivation made substantial use of the fact that $A$ is a Boolean algebra, this generalizes to show that any separative partial order densely embeds into a complete Boolean algebra. As remarked above, separativity is what guarantees each $U_p$ is regular.
- The construction of the completion is a special case of a more general construction called the regular open algebra. For any topological space $X$, we say that $U\subseteq X$ is regular open if it is equal to the interior of its closure. The collection of all regular open subsets of $X$, ordered by $\subseteq$, is a complete Boolean algebra. Here, the relevant topological space is $P,$ where $U\subseteq P$ is open if and only if $U$ is a cut. The operation $U\mapsto \overline U$ is taking the interior of the closure.
Best Answer
It is closed under finite intersections, but completeness requires closure under arbitrary intersections. Let $X_i = U \setminus \{i\}$, where $i$ ranges over some infinite subset $I$ of $U$ such that $U \setminus I$ is also infinite (e.g., if $U = \Bbb{R}$, you could take $I = \Bbb{Z}$). Then $\bigcap_{i \in I} X_i = U \setminus I$ which is not in $U'$.
More formally, the family $\{X_i \mid i \in I\}$ has no g.l.b. because any finite (resp. cofinite) set $A$ such that $A \subseteq X_i$ for all $i$ must be a proper subset of $\bigcap_{i \in I} X_i = U \setminus I$ and hence can be extended to give a finite (resp. cofinite) set $B \supset A$ such that $B \subseteq X_i$ for all $i$.