[proofplan] The proof is a direct verification from the definition of the separative preorder. First we show that $\leq_s$ is reflexive and transitive, and then pass to the quotient by its symmetric part. The main point is that compatibility is preserved exactly by the quotient map: one direction follows from ordinary common refinements, while the other uses the definition of $\leq_s$ to turn quotient common refinements into genuine common refinements in $\mathbb P$. Separativity and the dense-set assertions then follow by translating every statement back through the quotient map. [/proofplan] [step:Verify that the separative relation is a preorder] We first prove that $\leq_s$ is reflexive. Let $p \in \mathbb P$. If $r \in \mathbb P$ satisfies $r \leq p$, then $r \parallel p$, witnessed by $r$ itself. Hence $p \leq_s p$. We next prove transitivity. Let $p,q,t \in \mathbb P$ satisfy $p \leq_s q$ and $q \leq_s t$. To prove $p \leq_s t$, let $r \in \mathbb P$ satisfy $r \leq p$. Since $p \leq_s q$, the condition $r$ is compatible with $q$, so choose $u \in \mathbb P$ such that $u \leq r$ and $u \leq q$. Since $q \leq_s t$ and $u \leq q$, the condition $u$ is compatible with $t$, so choose $v \in \mathbb P$ such that $v \leq u$ and $v \leq t$. Then $v \leq r$ and $v \leq t$, so $r \parallel t$. Since this holds for every $r \leq p$, we have $p \leq_s t$. Thus $\leq_s$ is reflexive and transitive, hence a preorder on $\mathbb P$. [guided] We need to check the two defining properties of a preorder. For reflexivity, fix $p \in \mathbb P$. The statement $p \leq_s p$ means: every strengthening of $p$ is compatible with $p$. So let $r \in \mathbb P$ satisfy $r \leq p$. Compatibility asks for a common strengthening of $r$ and $p$. The condition $r$ itself works, because $r \leq r$ by reflexivity of the original preorder and $r \leq p$ by assumption. Hence $r \parallel p$, and therefore $p \leq_s p$. For transitivity, assume $p \leq_s q$ and $q \leq_s t$. We must prove $p \leq_s t$, so take an arbitrary $r \in \mathbb P$ with $r \leq p$. The first hypothesis, $p \leq_s q$, says that every strengthening of $p$ is compatible with $q$; applying this to $r$, choose $u \in \mathbb P$ with $u \leq r$ and $u \leq q$. Now $u$ is a strengthening of $q$, so the second hypothesis, $q \leq_s t$, says that $u$ is compatible with $t$. Thus there exists $v \in \mathbb P$ such that $v \leq u$ and $v \leq t$. Since $v \leq u \leq r$, the same $v$ is a common strengthening of $r$ and $t$. Hence $r \parallel t$. Because every $r \leq p$ is compatible with $t$, we have $p \leq_s t$. Together with reflexivity, this proves that $\leq_s$ is a preorder. [/guided] [/step] [step:Pass from the preorder to a partial order on equivalence classes] Define $p =_s q$ iff $p \leq_s q$ and $q \leq_s p$. Since $\leq_s$ is reflexive, symmetric in this definition, and transitive, the relation $=_s$ is an [equivalence relation](/page/Equivalence%20Relation). Let $\mathbb Q := \mathbb P/{=_s}$, and let $\pi:\mathbb P \to \mathbb Q$ be the quotient map $p \mapsto [p]_{=_s}$. Define a relation on $\mathbb Q$ by \begin{align*} [p]_{=_s} \leq [q]_{=_s} \quad \Longleftrightarrow \quad p \leq_s q. \end{align*} This relation is well-defined: if $p =_s p'$ and $q =_s q'$, then $p' \leq_s p \leq_s q \leq_s q'$ by transitivity of $\leq_s$, so $p' \leq_s q'$. Reflexivity and transitivity on $\mathbb Q$ follow from reflexivity and transitivity of $\leq_s$. For antisymmetry, suppose $[p]_{=_s} \leq [q]_{=_s}$ and $[q]_{=_s} \leq [p]_{=_s}$. Then $p \leq_s q$ and $q \leq_s p$, so $p =_s q$, and therefore $[p]_{=_s} = [q]_{=_s}$. Hence $\leq$ is a partial order on $\mathbb Q$. [/step] [step:Show that compatibility is exactly preserved by the quotient map] Let $p,q \in \mathbb P$. First suppose $p \parallel q$ in $\mathbb P$. Choose $r \in \mathbb P$ such that $r \leq p$ and $r \leq q$. We claim that $r \leq_s p$ and $r \leq_s q$. Indeed, if $u \leq r$, then $u \leq p$ and $u \leq q$, so $u \parallel p$ and $u \parallel q$, each witnessed by $u$. Thus \begin{align*} \pi(r) \leq \pi(p) \end{align*} and \begin{align*} \pi(r) \leq \pi(q). \end{align*} Therefore $\pi(p) \parallel \pi(q)$ in $\mathbb Q$. Conversely, suppose $\pi(p) \parallel \pi(q)$ in $\mathbb Q$. Then there exists $a \in \mathbb Q$ such that $a \leq \pi(p)$ and $a \leq \pi(q)$. Choose $t \in \mathbb P$ with $a = \pi(t)$. Then $t \leq_s p$ and $t \leq_s q$. Since $t \leq t$, the relation $t \leq_s p$ gives $t \parallel p$, so choose $u \in \mathbb P$ with $u \leq t$ and $u \leq p$. Since $u \leq t$ and $t \leq_s q$, we have $u \parallel q$, so choose $v \in \mathbb P$ with $v \leq u$ and $v \leq q$. Then $v \leq p$ and $v \leq q$, so $p \parallel q$. Thus \begin{align*} p \parallel q \quad \Longleftrightarrow \quad \pi(p) \parallel \pi(q). \end{align*} [/step] [step:Prove that the quotient order is separative] We prove separativity of $\mathbb Q := \mathbb P/{=_s}$. Let $a,b \in \mathbb Q$ satisfy $a \nleq b$. Choose $p,q \in \mathbb P$ such that $a = \pi(p)$ and $b = \pi(q)$. Since $a \nleq b$, we have $p \nleq_s q$. By definition of $\leq_s$, there exists $r \in \mathbb P$ such that $r \leq p$ and $r \perp q$. Because $r \leq p$, the same argument as above gives $r \leq_s p$, hence $\pi(r) \leq \pi(p)=a$. If $\pi(r)$ were compatible with $\pi(q)=b$ in $\mathbb Q$, then the compatibility preservation just proved would imply $r \parallel q$ in $\mathbb P$, contradicting $r \perp q$. Therefore $\pi(r) \perp b$. We have found a condition $\pi(r) \leq a$ incompatible with $b$. Hence $\mathbb Q$ is separative. [/step] [step:Push dense subsets of the original poset to dense subsets of the quotient] Let $D \subseteq \mathbb P$ be dense. We prove that $\pi[D]$ is dense in $\mathbb Q$. Let $a \in \mathbb Q$. Choose $p \in \mathbb P$ with $a=\pi(p)$. Since $D$ is dense in $\mathbb P$, there exists $q \in D$ such that $q \leq p$. As before, $q \leq p$ implies $q \leq_s p$, so \begin{align*} \pi(q) \leq \pi(p)=a. \end{align*} Since $\pi(q) \in \pi[D]$, every condition in $\mathbb Q$ has a strengthening in $\pi[D]$. Thus $\pi[D]$ is dense in $\mathbb Q$. [/step] [step:Pull open dense subsets of the quotient back to dense subsets of the original poset] Let $E \subseteq \mathbb Q$ be open dense. We prove that $\pi^{-1}[E]$ is dense in $\mathbb P$. Let $p \in \mathbb P$. Since $E$ is dense in $\mathbb Q$, there exists $a \in E$ such that $a \leq \pi(p)$. Choose $q \in \mathbb P$ with $a=\pi(q)$. Then $\pi(q) \leq \pi(p)$, so $q \leq_s p$. Since $q \leq q$, the definition of $q \leq_s p$ gives $q \parallel p$. Choose $r \in \mathbb P$ such that $r \leq q$ and $r \leq p$. Because $r \leq q$, we have $\pi(r) \leq \pi(q)=a$. Since $E$ is open and $a \in E$, it follows that $\pi(r) \in E$. Therefore $r \in \pi^{-1}[E]$ and $r \leq p$. Thus every $p \in \mathbb P$ has a strengthening in $\pi^{-1}[E]$, so $\pi^{-1}[E]$ is dense in $\mathbb P$. [guided] The point of this step is that density in the quotient gives us a quotient strengthening below $\pi(p)$, but we still need an actual condition below $p$ in the original poset. Fix $p \in \mathbb P$. Since $E$ is dense in $\mathbb Q$, there is some $a \in E$ with $a \leq \pi(p)$. Choose a representative $q \in \mathbb P$ such that $a=\pi(q)$. The inequality $a \leq \pi(p)$ means, by definition of the quotient order, that $q \leq_s p$. Now unpack $q \leq_s p$. This says that every strengthening of $q$ is compatible with $p$. We may apply it to $q$ itself, since $q \leq q$. Hence $q \parallel p$. Choose a common strengthening $r \in \mathbb P$ such that $r \leq q$ and $r \leq p$. We now check that $r$ actually lies over $E$. Since $r \leq q$, every strengthening of $r$ is also a strengthening of $q$, so every strengthening of $r$ is compatible with $q$. Hence $r \leq_s q$, and therefore \begin{align*} \pi(r) \leq \pi(q)=a. \end{align*} The set $E$ is open, meaning downward closed in the forcing order: whenever $b \leq a$ and $a \in E$, then $b \in E$. Applying this with $b=\pi(r)$, we get $\pi(r) \in E$. Therefore $r \in \pi^{-1}[E]$. Since $r \leq p$, we have found below the arbitrary condition $p$ a stronger condition in $\pi^{-1}[E]$. This proves that $\pi^{-1}[E]$ is dense in $\mathbb P$. [/guided] [/step] [step:Pull arbitrary dense subsets back after taking downward closure] Let $E \subseteq \mathbb Q$ be dense, and define \begin{align*} E^\downarrow = \{a \in \mathbb Q : a \leq b \text{ for some } b \in E\}. \end{align*} By definition, $E^\downarrow$ is downward closed. Since $E \subseteq E^\downarrow$ and $E$ is dense, the set $E^\downarrow$ is dense. Hence $E^\downarrow$ is open dense in $\mathbb Q$. Applying the previous step to the open dense set $E^\downarrow$, we conclude that $\pi^{-1}[E^\downarrow]$ is dense in $\mathbb P$. This proves all asserted properties of the separative quotient construction. [/step]