Mapping from Preordering reflects Ordering

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 $\RR$:
 * $x \sim_\RR y$ $x \mathrel \RR y$ 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.