In Chapter 7 of Jech's Set Theory, he proves that every Boolean algebra has a completion (see this post here for the definition of a Boolean algebra and its partial ordering). The proof is as follows:
Theorem 7.13. Every Boolean algebra has a completion.
Proof. We use a construction similar to the method of Dedekind cuts. Let $A$ be a Boolean algebra. Let us call a set $U\subseteq A^+$ a cut if
$$(7.24) \qquad p\leq q \text{ and } q\in U\text{ implies } p\in U.$$
For every $p\in A^+$, let $U_p$ denote the cut $\{x:x\leq p\}$.A cut $U$ is regular if
$$\text{whenever } p\not\in U, \text{ then there exists }q\leq p \text{ such that }U_q\cap U=\varnothing.$$
Note that every $U_p$ is regular, and that every cut includes some $U_p$.We let $B$ be the set of all regular cuts in $A^+$. We claim that $B$, under the partial ordering by inclusion, is a complete Boolean algebra. note that the intersection of any collection of regular cuts is a regular cut, and hence each cut $U$ is included in a least regular cut $\overline{U}$. In fact,
$$\bar U=\{p:(\forall q\leq p) U\cap U_q\neq \varnothing\}.$$
Thus for $u,v\in B$ we have
$$u\cdot v=u\cap v,\qquad u+v=\overline{u+v}.$$
The complement of $u\in B$ is the regular cut
$$-u=\{p:U_p\cap u=\varnothing\}.$$
And, of course, $\varnothing$ and $A^+$ are the zero and the unit of $B$. It is not difficult to verify that $B$ is a complete Boolean algebra, and we leave the verification to the reader. Furthermore, for all $p,q\in A^+$ we have $U_p+U_q=U_{p+q}$, $U_p\cdot U_q=U_{p\cdot q}$, and $-U_p=U_{-p}$. Thus $A$ embeds in $B$ as a dense subalgebra.
I don't think I understand what a regular cut is, or in particular why it is the case that every $U_r$ (where $r\in A^+$; I changed the letter for convenience) is regular.
I'm trying to use Dedekind cuts in the rationals to build intuition, but it seems like no Dedekind cuts are regular, given that any two cuts intersect eventually.
Is there any good intuition behind what it means for a cut to be regular? At this point I'm not really even convinced that regular cuts even exist, much less why the ones highlighted here are regular.
Best Answer
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 couple closing notes