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I have a question regarding the separative quotient featured in this question.

I want to show that for every dense subset D of the separative quotient Q it's preimage under h is also dense.

I have tried to fix an $x \in P$, find a $[y] \leq [x]$ and extract a $z \in [y]$ with $z \leq x$, but as as shown in the referenced topic this not always possible.

On the other hand since x and y are compatible I can find elements beneath x and y, but I don't see any way to make them part of $h^{-1}(D)$.

How do I use the premise more effectively or come up with possible counter-examples?

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up vote 1 down vote accepted

I think I found a counter-example: Start with the counter-example A in the linked topic and inductively identify each $r_i, s_i$ and $t_i$ with the $y_j$ of a new copy of A.

The set X of all $x_i$ is not dense, but it is the preimage of the dense set $h(X)$ in the separative quotient.

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Note that what you write is true if $D$ is dense and also open. Once you have a $y\in D$ which is compatible with $x$, any witness for the compatibility will also be in $D$.

Here is a variant of pseudo-nym's construction: Let $P= 3^{<\omega}$ be the set of all finite ternary sequences. (EDIT: Ordered in the "usual" upside-down way, that is, by extension: $t\le s$ iff $s$ is an initial segment of $t$.)

Let $Q = P \cup \lbrace s': s\in P\rbrace $ (i.e., add a disjoint copy of $P$), and declare $s^\frown 0 \le s'$, $s^\frown 1\le s'$ for all $s\in P$. (Plus transitive closure.) Thus $s'$ is like a formal disjunction of $s^\frown 0$ and $ s^\frown 1$ for each $s\in P$, whereas $s$ is the disjunction of $s^\frown 0$, $s^\frown 1$, $s^\frown 2$.

Hence in the separative quotient, $s'$ is strictly stronger than $s$. The (image of the) set of all $s'$ is dense in the separative quotient, but no $s'$ is stronger than any $t\in P$, so the set of all $s'$ is not dense in $Q$.

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