Mapping on Total Ordering reflects Transitivity

Theorem
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.

Let $f: S \to T$ be a mapping to an arbitrary set $T$.

Let $\RR$ be a relation on $T$ defined such that:
 * $\RR: = \set {\tuple {\map f x, \map f y}: x \preccurlyeq y}$

That is, $a$ is related to $b$ in $T$ they have preimages $x$ and $y$ under $f$ such that $x$ precedes $y$.

Then $\RR$ is transitive.

Proof
Let $a, b, c \in T$ such that:
 * $a \mathrel \RR b$
 * $b \mathrel \RR c$

Then there exist $x, y, z \in S$ such that:


 * $a = \map f x$
 * $b = \map f y$
 * $c = \map f z$

such that:
 * $x \preccurlyeq y$
 * $y \preccurlyeq z$

As $\preccurlyeq$ is a total ordering, it follows that:
 * $x \preccurlyeq z$

and so by definition of $\RR$:
 * $\map f x \mathrel \RR \map f z$

That is:
 * $a \mathrel \RR c$

As $a$, $b$ and $c$ were arbitrary, it follows that $\RR$ is transitive.