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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.

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Presumably, $\perp$ is supposed to stand for "incompatible"? –  Arturo Magidin Oct 5 '10 at 19:40
    
Oh right. I forgot to mentioned that $x \perp y$ means they are compatible (it was there at first and somehow through revising my question before submitting it got away). –  Asaf Karagila Oct 5 '10 at 19:45

2 Answers 2

up vote 4 down vote accepted

My friend came up with a good counter-example:

Counter example

Then it is easy to verify that $[x] \prec [y]$ but $x \not{<} y$.

However, if anyone has any further insights into the lemma or this whole idea, I'd very like to hear.

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Another example, perhaps more transparent, because you can see how I arrived at it.

Let $P$ be the set of all infinite subsets of the natural numbers $\omega$. Then $A \subseteq^* B$ means that $A\setminus B$ is finite.

Take your favorite subsets $A \subseteq B$ with $A \not=^* B$. Let $P'$ be the set of all infinite subsets $S \subseteq \omega$ which satisfy one of the following properties:

  1. $S \not=^* A$ and $S \not=^* B$.

  2. $S=^* A$, and $22\in S$.

  3. $S=^* B$, but $22\notin S$.

Now $P'$ is essentially the same as $P$ (it inherits the $\le$ and $\le^*$ relations). The sets of type 2 form a $=^*$-class with is $\subseteq^*$-below the class of sets of type 3, but there is no pair of representatives where the true $\subseteq$-relations holds.

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