# Separative Quotients, and also the Induced Order

Let $P$ be some partial order.

We claim that $x$ and also $y$ are suitable if $\exists r\in P (r\le x \wedge r\le y)$, we represent this by $x \perp y$. Or else, we claim that $x$ and also $y$ are inappropriate .
. An order $P$ is called separative if whenever $\neg (x \le y)$ after that there exists some $r\le x$ which inappropriate with $y$.

In Jech is "Set Theory" he confirms a lemma (14.11) which construct a separative ratio, $Q$, and also a mapping $h$ for a (basic) partly gotten set, $P$, that pleases 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 building and construction is by taking a ratio of $P$ over the equivalence relationship $x \sim y \iff \forall z(z \perp x \leftrightarrow z \perp y)$, and also $[x] \preceq [y] \iff \forall z \le x(z \perp y)$. (And clearly sufficient, $h(x) = [x]$).

Today I was attempting to confirm the adhering to lemma (which does not show up on Jech is publication):

Let $P$ be some partial order, and also $Q$ its separative ratio. If $[x] \preceq [y]$ after that there exist $x' \in [x], y' \in [y]$ such that $x' \le y'$.
(That is to claim that the separative mapping is "virtually" order - preserving.)

With ease it appears right, yet it could quite possibly be a not - usual misbelief. Any kind of tips, partial evidence or full evidence (provided that I have not confirmed it myself yet - in which instance I'll hurry to the local computer system and also upgrade) will certainly be most invited.

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2019-05-18 22:19:21
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Then it is very easy to validate that $[x] \prec [y]$ yet $x \not{<} y$.