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Given a separative poset $P$, we can form it's Boolean completion $B(P)$. This is a Boolean algebra whose elements are regular cuts on $P$, as defined here. Also, $P$ embeds densely into $B(P)$ by mapping $p$ to the principal ideal $U_p$.

Let's now take a poset $Q$ and a $P\subseteq Q$, both of which are separative. The inclusion $P\to Q$ is monotone. What I want to know is, can we somehow induce a complete embedding $f\colon B(P)\to B(Q)$ from this that makes the obvious diagram commute?

It seems to me there is only one thing that should work. Since $P$ is densely embedded in $B(P)$ and we know $f$ on $P$, the only reasonable choice seems to be $$f(U)=\bigvee_{p\in U}U_p$$

But this doesn't even seem to be a Boolean algebra homomorphism. Let's look at the meets: $$f(U\cap V)=\bigvee_{p\in U\cap V} U_p \stackrel{?}{=} \bigvee_{p\in U} U_p\cap \bigvee_{q\in V} U_q= f(U)\cap f(V)$$

Obviously, the left hand set is contained in the right hand set, but do we have equality? I just don't see it.

Have I missed something obvious? Is this even the right way of defining the map $f$?

Added: Apostolos dealt with the meets in the comments, just using distributivity. But I'm having trouble with joins now. We have $$f(U\vee V)=\bigvee_{p\in U\vee V} U_p^Q \stackrel{?}{=} \bigvee_{p\in U\cup V} U_p^Q= f(U)\vee f(V)$$ (I'll be using $U_p^P$ and $U_p^Q$ to distinguish principal ideals in $P$ and $Q$ respectively)

To focus on the nontrivial inclusion, take $x\in f(U\vee V)$ and $y\leq x$. We wish to find a $p_0\in U\cup V$ such that $U_y^Q\cap U_{p_0}^Q\neq\emptyset$. According to the definition of the join, there exists a $p\in U\vee V$ such that $U_y^Q\cap U_p^Q\neq\emptyset$. Taking an element $q$ of this set, we get $q\leq y$ and $q\leq p$. Now, if we could find an element of $P$ below this $q$, we would be done (since $p\in U\vee V$), but I don't see any reason for such an element to exist. In particular, in the application I have in mind, $P$ is very far from being dense in $Q$. How do I proceed here?

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For every complete Boolean algebra you have $(\bigvee_{i\in I}a_i)\land(\bigvee_{j\in J}b_j)=\bigvee_{(i,j)\in I\times J}(a_i\land b_j)$. Thus your right hand side is $\bigvee_{(p,q)\in U\times V}U_p\cap U_q$. So if $a$ is in your right hand side it is in some $U_p\cap U_q$ for $p\in U$ and $q\in V$. Then $a\in U$ and $a\in V$ because they are downwards closed (and they contain $p$ and $q$ and $a$ is below them). Hence $a\in f(U\cap V)$. – Apostolos Jun 6 '12 at 19:41
@Apostolos You need to be a bit careful, since the join in $B(P)$ isn't just the union but the "regular closure" of the union. But you're basically right. I was missing distributivity. If you see an elegant way to show this is a complete embedding, please write something up. Otherwise, I'll think about this more in the morning. – Miha Habič Jun 6 '12 at 21:03
If you want the diagram induced by the inclusion map to commute there is really only one way to define your map. Whether or not it is complete, I'm not sure. Note that $P\subseteq Q$ means that $B(P)$ is realized inside $B(Q)$ as the regular cuts defined by the elements of $P$. I suppose it might be possible to create a nice counterexample of two separative posets $P\subseteq Q$ such that $B(P)$ is not a complete subalgebra of $B(Q)$ by using an example of $B'\subseteq B$ which are both complete BA, but $B'$ is not a complete sub-alg. of $B$, then taking $B^+$ and $B'^+$ as your posets. – Asaf Karagila Jun 13 '12 at 17:24
@AsafKaragila : what is it that you denote by $B^{+}$ ? – Ewan Delanoy Jun 14 '12 at 4:55
@EwanDelanoy I expect $B^+$ is the poset of positive elements of $B$. – Miha Habič Jun 14 '12 at 5:21

One trivial counterexample comes to mind:

Take $Q$ to be the power set of $\mathbb R$ and $P$ the power set of $\mathbb N$, both without the zero element, i.e. only non-empty sets. The order on both is inclusion.

Trivially $P$ and $Q$ are separative and $P\subseteq Q$. However since both are the positive elements of a complete Boolean algebra, and the completion is unique [up to isomorphism], the completion is both is simply adding the zero element.

However an embedding of a Boolean algebra into another must respect the $0,1$ elements, but $1_{B(P)}=\mathbb N\neq\mathbb R=1_{B(Q)}$. So there is no obvious way of closing the diagram.

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Wow, I didn't expect this to fail this badly. Thanks for the counterexample Asaf! Although this doesn't quite apply in the particular case I'm considering (there, $Q$ is a "lower extension" of $P$, i.e. if an element of $Q$ is above an element of $P$, it was already in $P$), it's enough for me to start to be very careful. – Miha Habič Jul 13 '12 at 6:33

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