Let ${\cal P}(\omega)/({\rm fin})$ be the quotient of the Boolean algebra ${\cal P}(\omega)$ where two sets are considered to be equivalent if they differ by a finite number of elements.


It turns out that ${\cal P}(\omega)/({\rm fin})$ is an atomless Boolean algebra. Is the Dedekind-MacNeille completion isomorphic to ${\cal P}(\omega)$?



Dedekind-MacNeille completion of ${\cal P}(\omega)/({\rm fin})$
No. $\overline{P(\omega)/\text{fin}}$ is not isomorphic to $P(X)$ for any set $X$ because $P(\omega)/\text{fin}$ is atomless.


We shall write $\overline{B}$ for the Dedekind-MacNeille completion of a Boolean algebra $B$, and we shall write $B^{+}$ for $B\setminus\{0\}$. A minimal element in $B^{+}$ is said to be an atom of the Boolean algebra $B$. A Boolean algebra $B$ is said to be atomic if for each $b\in B^{+}$, there is an $a\in B^{+}$ with $a\leq b$.


Proposition: If $B$ is a complete Boolean algebra, then the following are equivalent.


1. $B$ is atomic.
2. $B\simeq P(X)$ for some set $X$.
3. $B$ satisfies the complete distributivity identity
$$\bigwedge\_{i\in I}\bigvee\_{j\in A\_{i}}a\_{i,j}=\bigvee\_{f}\bigwedge\_{i\in I}a\_{i,f(i)}$$ where $f$ ranges over the collection of all choice functions $f:I\rightarrow\bigcup\_{i\in I}A\_{i}$ with $f(i)\in A\_{i}$ for $i\in I$.


To prove $1\rightarrow 2$, let $p$ be the set of all atoms of $B$. Then $p$ is a partition of the Boolean algebra $B$. The mapping $\phi:P(p)\rightarrow B$ defined by letting $\phi(R)=\bigvee R$ is your required Boolean algebra isomorphism.


To prove $3\rightarrow 1$, we observe that from complete distributivity, we can show that the lattice of all partitions of the Boolean algebra $B$ is complete, and the least element $p$ in the lattice of all partitions of $B$ is actually the set of all atoms in the Boolean algebra $B$. One can show that for each $b\in B^{+}$, there must be an $a\in p$ with $a\leq b$.


Fact: A Boolean algebra $B$ is atomic if and only if the completion $\overline{B}$ is atomic, and if $e:B\rightarrow\overline{B}$ is the canonical embedding, then the mapping $e$ restricts to a bijection between the set of all atoms of $B$ and the set of all atoms of $\overline{B}$. Therefore, if $B$ is a Boolean algebra, then $\overline{B}\simeq P(X)$ if and only if $B$ is atomic, and $B$ has $|X|$ many atoms.


