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 $T$ if and only if 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.

$\blacksquare$

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 $41 \ \text {(a)}$