A question about separative orders. Let $\mathcal{P}=\langle P, \leq \rangle$ be a p.o. set. Two elements $p$ and $q$ of it are called
    compatible if there is an $r \in \mathbb{P}$ such that $r\leq p$ and $r \leq q$; otherwise they
    are called incompatible.
A partial ordering $\leq $ is said to be separative if for any two elements $p$ and $q$ of $P$ either $q\leq p$ or there is an $r\leq q$ that is not compatible with $p$.
Now, from a stationary set $S$ of $\omega_{1}$ we construct a p.o. set $\mathcal{P}_{S}$ of conditions:


*

*a condition $p\in \mathcal{P}_{S}$ is a countable subset of $S$ that is closed in the order topology of $\omega_{1}$. In particular each member $p$ of $\mathcal{P}_{S}$ has maximum.


$$\mathcal{P}_{S} =\{p\subseteq S : |p|\leq\aleph_{0} \hspace{0.1cm}\mbox{and}\hspace{0.1cm} p \hspace{0.1cm}\mbox{is closed in }\omega_{1}    \}    $$


*

*If $p, q\in \mathcal{P}_{S}$, then $$p\leq q \hspace{0.3cm}\text{iff} \hspace{0.3cm}q\subseteq p \hspace{0.1cm}\text{and}\hspace{0.1cm} (p\setminus q) \cap \bigcup q=\emptyset,$$ which is equivalent to the fact that $\alpha>\beta$ for all $\alpha\in p\setminus q$ and $\beta\in q$.


My question is if anyone have any idea how to show that $\leq$ is separative. 
I tried it as follows, let $p,q \in\mathcal{P}_{S}$, and suppose that $\bigcup p +1< \bigcup q$, then consider $r=p\cup \{\bigcup p +1 \} \in \mathcal{P}_{S}$, and note that $r\leq p$. But I can't come to a contradiction in the case that $r$ and $q$ are compatible. 
Thanks
 A: As you noted, the only way to extend a condition $p$ is by taking some closed subset of $p\cup (S\setminus \bigcup p)$ and adding it to $p$, in particular, by extending $p$ with ordinals larger than all ordinals in $p$.
Let $q\not\leq_{\mathcal P_S} p$ for some $p,q\in\mathcal P_S$. Then $p\not\subseteq q$ or $(q\setminus p)\cap \bigcup p$ is nonempty.
In the first case, there is some $\alpha\in p\setminus q$. Pick some $\beta\in S\setminus\bigcup (p\cup q)$, then it is easy to see that $q\cup\{\beta\}$ is a condition extending $q$. However, no extension $r\leq_{\mathcal P_S} q\cup\{\beta\}$ can be an extension of $p$, since any extension of $p$ contains $\alpha$ and since $\alpha<\beta\leq\bigcup r$ we have $\alpha\notin r$, thus $\alpha\notin s$ for any $s\leq_{\mathcal P_S} r$. Hence $p$ and $r$ are incompatible.
In the second case, let $\alpha\in (q\setminus p)\cap\bigcup p$, then $\alpha\in r$ for any $r\leq_{\mathcal P_S} q$, and thus $\alpha\in(r\setminus p)\cap\bigcup p$ for any $r\leq_{\mathcal P_S} q$, thus $p$ and $q$ are incompatible.
A: The poset $\mathcal{P}_S$ satisfies the following property: $p$ and $q$ are compatible iff $p\le q$ or $q\le p$:
one direction is easy. For the other direction, assume that $r\le p,q$. Without loss of generality, we may take $\bigcup p\le \bigcup q$. Since $(r\setminus p)\cap\bigcup p=\varnothing$ and $p\subseteq r$, $r\cap \bigcup p = p$. Similarly, we have $r\cap \bigcup q=q$.
Since $\bigcup p\le\bigcup q$, $$q=(r\cap\bigcup p)\cup (r\cap(\bigcup q\setminus\bigcup p))=p\cup (r\cap(\bigcup q\setminus\bigcup p)).$$ Hence $p\subseteq q$. Moreover,
$$(q\setminus p)\cap\bigcup p \subseteq (r\cap(\bigcup q\setminus\bigcup p))\cap\bigcup p = \varnothing.$$
Therefore, $q\le p$. Similiarly, we have $p\le q$ if $\bigcup q\le\bigcup p$.
As a corollary, $q\not\le p$ implies $p\perp q$ or $p\le q$. In the former case, just take $r=q$. In the latter case, take $\alpha\in S$ such that $\alpha>\bigcup p$ and take $r=q\cup\{\alpha\}$.
