Minimal completion of a Boolean algebra

Question

According to Halmos (Lectures on Boolean Algebras, p. 92), the completion of Boolean algebra $A$ is a complete Boolean algebra $B$ together with a monomorphism $h$ from $A$ into $B$ such that

(1) $h$ preserves all suprema.

(2) The complete Boolean algebra generated by $h(A)$ in $B$ is $B$ itself.

Moreover, a completion $(B,h)$ is minimal if, corresponding to every completion $(C,k)$, there exists a complete monomorphism $f$ from $B$ into $C$ such that $f\circ h=k$

Question 1: If $h(A)$ generates B, then, by definition, B is the smallest complete Boolean algebra containing $h(A)$. Why is this fact not a sufficient condition to define minimality?

Question 2: Is there another way to define minimality (than the one provided above)?

Answer 1

You should cite Halmos properly. Halmos does not define the completion of a Boolean algebra", but a completion of a Boolean algebra. Then he defines a minimal completion using a universal property and insists that

It is not obvious that there are any completions at all, let alone minimal ones.

Question 1. The fact that the complete Boolean algebra generated by $h(A)$ in $B$ is $B$ itself does not suffice to insure that $B$ satisfies the universal property.

Question 2. You could use Halmos' Theorem 11 from the same book as a definition, but it is not really simpler. In this case, you would define the completion of a Boolean algebra $A$ as the Boolean algebra of regular open sets of the Stone dual of $A$. But you could also use another result, known as the Glivenko–Stone theorem:

The minimal completion of a Boolean algebra is also its Dedekind-MacNeille completion.

Answer 2

Actually, properties (1) and (2) are sufficient to define a minimal completion (that is, by Halmos's definition, every completion is minimal). Indeed, let $(B,h)$ be a minimal completion of $A$ and let $(C,k)$ be any other completion. By the universal property of $(B,h)$, there exists a complete monomorphism $f:B\to C$ such that $fh=k$. The image of $f$ is then a complete subalgebra of $C$, which contains $f(h(A))=k(A)$. But by property (2) of $(C,k)$, this implies the image of $f$ is all of $C$. Thus $f$ is surjective, and hence an isomorphism. Since $(B,h)$ is minimal, this implies $(C,k)$ is minimal as well.

[Note that this argument depends crucially on the existence of at least one minimal completion. So Halmos's notion of minimality is not useless--it is still useful as a step along the way to prove that a completion (in the sense of (1) and (2)) is unique up to isomorphism. Still, Halmos seems to have been unaware of this redundancy when he wrote the book.]