Let us say that a partial order is "countably closed with infima" if every descending $\omega$-sequence has an infimum.


Suppose $P$ and $Q$ are posets that are countably closed with infima, and for some dense $D \subseteq Q$, there is a projection $\pi : D \to P$, i.e. $\pi$ is an order-preserving map such that whenever $p \leq \pi(q)$, then there is $q' \leq q$ in $D$ such that $\pi(q') = p$. Assume also that $D$ is closed under descending $\omega$-sequences, and $\pi$ is continuous on such sequences.


Whenever $G \subseteq P$ is generic, we define the quotient $Q/G$ as $\{ q : (\exists d \in D) d \leq q \wedge \pi(d) \in G \}$.


It is a well-known fact that under these assumptions, if $D = Q$ (i.e. $\pi$ is defined everywhere), then $Q/G$ is forced to be countably closed with infima. Does this necessarily hold when $D$ is a proper subset of $Q$?



Countable closure of quotient forcing
This does (unfortunately?) not hold in general.


Consider the following (trivial) forcings $P$ and $Q$:


The only thing $Q$ does is generically pick out some $N\leq\omega$, it does this in the following way. $Q$ has two types of conditions:


* atoms deciding $N$, one atom $a\_m$ representing "$N=m$" for $m\leq\omega$,
* conditions $q\_m$ representing "$N\geq m$" for integers $m<\omega$.


$Q$ is ordered in the obvious way and is countably closed with infima: Essentially the only nontrivial decreasing chain is $(q\_m)\_{m<\omega}$ and it converges to $a\_\omega$. Let $D$ be the dense set of the first type, i.e. the atoms.


Now $P$ has only two (incompatible) conditions, $p\_{\mathrm{fin}}$ representing "$N$ is finite" and $p\_{\mathrm{inf}}$ for "$N$ is $\omega$" and this suggests how to define the projection $\pi\colon D\rightarrow P$. For $G=\{p\_{\mathrm{fin}}\}$, we get $Q/G=Q-\{a\_\omega\}$ which is not countably closed.


