Mapping from Preordering reflects Ordering
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Theorem
Let $\struct {S, \RR}$ be a preordered set.
Then there exists:
- an ordered set $\struct {T, \preccurlyeq}$
- a mapping $f: S \to T$ such that:
- $\RR: = \set {\tuple {x, y}: \map f x \preccurlyeq \map f y}$
Proof
Let $\sim_\RR$ denote the equivalence on $S$ induced by $\RR$:
- $x \sim_\RR y$ if and only if $x \mathrel \RR y$ and $y \mathrel \RR x$
Let $\preccurlyeq_\RR$ be the ordering on the quotient set $S / {\sim_\RR}$ by $\RR$:
- $\eqclass x {\sim_\RR} \preccurlyeq_\RR \eqclass y {\sim_\RR} \iff x \mathrel \RR y$
where $\eqclass x {\sim_\RR}$ denotes the equivalence class of $x$ under $\sim_\RR$.
Let $f: S \to S / {\sim_\RR}$ be the quotient mapping induced by $\RR$:
- $\forall x \in S: \map f x = \eqclass x \RR$
From Preordering induces Ordering we have that $\struct {T, \preccurlyeq}$ is an set with exactly the properties required.
$\blacksquare$
Sources
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $41 \ \text {(e)}$