# closure of separative quotients

Does there exist a partial order, nontrivial for forcing, that is countably closed, but whose separative quotient is not countably closed? Supposing the answer is yes, then is there a partial order, nontrivial for forcing, that is countably closed, but is not forcing equivalent to any countably closed separative partial order?

For those of you unfamiliar with the separative quotient of a partial order, it is defined as follows. Two elements of a partial order are compatible iff there is some element below both of them. We form the separative quotient of a partial order by taking equivalence classes: x is equivalent to y when x and y are compatible with the exact same things. We then define a new partial order for the separative quotient -- $x \leq y$ iff everything compatible with x is compatible with y.

A partial order is said to be separative if whenever $x \nleq y$, there is $z \leq x$ such that z is incompatible with y. The separative quotient of any partial order is separative.

Some of the ways, order-theoretically speaking, that two partial orders can be forcing equivalent are

(1) They are isomorphic, or more generally, (2) A dense subset of one of them is isomorphic to a dense subset of the other.

• Welcome to MO, Norman! I added the 'forcing' and 'order-theory' tags to your question. Also, since it seems, in principle, that this question might be answerable by someone outside forcing---it is at heart a combinatorial question about partial orders---I would recommend that you might explain a bit more about what you mean by 'separative quotient' and indeed 'separative', in order to assist this possibility. – Joel David Hamkins Jan 12 '10 at 13:40
• Thanks. I thought about adding the forcing tag, but it only had one other occurrence so far, and the site suggests not creating new tags. – Norman Lewis Perlmutter Jan 13 '10 at 5:09
• You need 250 reputation to create tags. But it is encouraged to use all the appropriate tags that do exist, and I had created the forcing tag for precisely this sort of question. I believe that as MO grows enormous, the tags will play an increasingly critical role for people to find the questions that might interest them. – Joel David Hamkins Jan 13 '10 at 13:57

If $P$ is countably closed then the complete Boolean algebra $RO(P)$ is strategically closed. So, by a result of Boban Velickovic (Playful Boolean Algebras, TAMS 296, 1986), if $RO(P)$ has a dense subset of size $2^{\aleph_0}$ (e.g. when $|P| \leq 2^{\aleph_0}$), $RO(P)$ must have a countably closed dense subset. Another result of Matt Foreman (Games Played on Boolean Algebras, JSL 48, 1983) shows that if $RO(P)$ is $(\kappa,\infty)$-distributive and has a dense set of size $\kappa$, then $RO(P)$ has a countably closed dense subset. This puts some constraints on potential counterexamples.