Let $P$ be some partial order.
We say that $x$ and $y$ are compatible if $\exists r\in P (r\le x \wedge r\le y)$, we denote this by $x \perp y$. Otherwise, we say that $x$ and $y$ are incompatible.
An order $P$ is called separative if whenever $\neg (x \le y)$ then there exists some $r\le x$ which incompatible with $y$.
In Jech's "Set Theory" he proves a lemma (14.11) which construct a separative quotient, $Q$, and a mapping $h$ for a (general) partially ordered set, $P$, that satisfies the following:
- $x\le y \Rightarrow h(x) \preceq h(y)$
- $x \perp y \iff h(x) \perp h(y)$
And $h$ is onto $Q$.
The construction is by taking a quotient of $P$ over the equivalence relation $x \sim y \iff \forall z(z \perp x \leftrightarrow z \perp y)$, and $[x] \preceq [y] \iff \forall z \le x(z \perp y)$. (And obviously enough, $h(x) = [x]$).
Today I was trying to prove the following lemma (which does not appear on Jech's book):
Let $P$ be some partial order, and $Q$ its separative quotient. If $[x] \preceq [y]$ then there exist $x' \in [x], y' \in [y]$ such that $x' \le y'$.
(That is to say that the separative mapping is "almost" order-preserving.)
Intuitively it seems right, but it might very well be a not-common misbelief. Any hints, partial proofs or complete proofs (granted that I haven't proved it myself yet - in which case I'll rush to the nearest computer and update) will be most welcomed.