MacNeille completion

In a first course on real analysis, one is generally introduced to the concept of a Dedekind cut. It is a way of constructing the set of real numbers from the rationals. This is a process commonly known as the completion of the rationals. Three key features of this completion are:

• the rationals can be embedded in its completion (the reals)

• every subset with an upper bound has a least upper bound

• every subset with a lower bound has a greatest lower bound

If we extend the reals by adjoining $+\mathrm{\infty }$ and $-\mathrm{\infty }$ and define the appropriate ordering relations on this new extended set (the extended real numbers), then it is a set where every subset has a least upper bound and a greatest lower bound.

When we deal with the rationals and the reals (and extended reals), we are working with linearly ordered sets. So the next question is: can the procedure of a completion be generalized to an arbitrary poset? In other words, if $P$ is a poset ordered by $\le$, does there exist another poset $Q$ ordered by ${\le }_Q$ such that

1. $P$ can be embedded in $Q$ as a poset (so that $\le$ is compatible with ${\le }_Q$), and

2. every subset of $Q$ has both a least upper bound and a greatest lower bound

In 1937, MacNeille answered this question in the affirmative by the following construction:

Given a poset $P$ with order $\le$, define for every subset $A$ of $P$, two subsets of $P$ as follows:

$$A^u=\{p\in P\mid a\le p\text{ for all }a\in A\}\text{ and }{A}^{\mathrm{\ell }}=\{q\in P\mid q\le a\text{ for all }a\in A\}.$$

Then $M(P):=\{A\in 2^P\mid {(A^u)}^{\mathrm{\ell }}=A\}$ ordered by the usual set inclusion is a poset satisfying conditions (1) and (2) above.

This is known as the MacNeille completion $M(P)$ of a poset $P$. In $M(P)$, since lub and glb exist for any subset, $M(P)$ is a complete lattice. So this process can be readily applied to any lattice, if we define a completion of a lattice to follow the two conditions above.