# The completion of a Boolean algebra is unique up to isomorphism

Jech defines a completion of a Boolean algebra $B$ to be a complete Boolean algebra $C$ such that $B$ is a dense subalgebra of $C$.

I am trying to prove that given two completions $C$ and $D$ of $B$, then the mapping $\pi: C \rightarrow D$ given by $\pi(c) = \sum^D \{ u \in B \ \vert \ u \le c \}$ is an isomorphism. I understand that $c \not= 0 \implies \pi(c) \not= 0$, and that given any $d \in D$, I can write it as $d = \sum^D \{ u \in B \ \vert \ u \le d \}$, but I am unsure how to proceed. Any help would be appreciated.

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I am afraid that it doesn't make sense that $\pi (c)\leq c$ because both elements live in different lattices $C$ and $D$, so they are not comparable. So it's better to denote differently both orders as $\leq_C$ and $\leq_D$.

So let $\pi (c)= \sum^D${$u\in B:u\leq_C c$}, and let $\rho: D\to C$ be such that $\rho(d)=\sum^C${$u\in B:u\leq_D d$}.

For any $c\in C$ and $u\in B$ we have $u\leq_C c\Rightarrow u\leq_D\pi(c)$, by the property of the supremum. Conversely, let $u\in B$ be such that $u\leq_D\pi(c)$. Then $u=u\wedge\pi(c)=u\wedge\sum^D${$b\in B:b\leq_C c$}$=\sum^D${$u\wedge b:b\leq_C c$}, since $D$ is a complete Boolean algebra, and so it has the general distributivity. Now, if $b\leq_C c$ then $u\wedge b\leq_Cu\wedge c\leq_C c$. Therefore $u\leq_C c$. We have shown that $u\leq_C c\Leftrightarrow u\leq_D\pi(c)$.

Therefore $c=\sum^C${$u\in B:u\leq_C c$}$=\sum^C${$u\in B:u\leq_D \pi(c)$} $=\rho(\pi(c))$.

Similarly $\pi(\rho(d))=d$ for each $d\in D$. So $\rho=\pi^{-1}$, and $\pi$ is an isomorphism.

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Edit: Ignore this answer. See Loronegro's.

Apparently I haven't done enough on Math.SE before, so I have to post this as an answer instead of a comment.

Hopefully this is helpful: first notice that $\pi c \le c$ for every $c\in C$ (the supremum of a bunch of things that are $\le c$ is also $\le c$). From this, you can deduce that, for every $u\in B$, $u\le c$ iff $u\le \pi c$. Therefore the sets $\{u\in B : u\le c\}$ and $\{u\in B : u\le \pi c\}$ are identical, so $\pi^{-1}\pi = 1$. By starting the argument with $\pi^{-1}c \le c$ instead of $\pi c \le c$, you can also conclude that $\pi\pi^{-1} = 1$.

You should feel cheated now, as I have avoided any mention of density. I guess proving that $\sum^C \{u\in B: u\le c\} = c$ requires the density of $B$.

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Also, hi Paul!! – Zach N Feb 5 '12 at 15:38
Thanks Zach. I basically needed $\pi(c) \le c$ which I stupidly did not see! See you on Wednesday, hopefully I can get to chapter 9 haha :p – Paul Slevin Feb 5 '12 at 15:49
You didn't see it because it is false! AHHH!!! See Loronegro's answer. – Zach N Feb 7 '12 at 22:19

Note that $\pi$ has an inverse, defined in the same way, $\pi^{-1}(d)=\sum^C\{u\in B: u\leq d\}$ and the maps are order-preserving.

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But how do I use density to show those 2 facts? – Paul Slevin Feb 5 '12 at 10:56
Well, I think I can see why each is order-preserving, but I am having some difficulty showing that $\pi \pi^{-1} = 1$. – Paul Slevin Feb 5 '12 at 13:04