Ordering is Preordering
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Theorem
Let $S$ be a set.
Let $\RR$ be an ordering on $S$.
Then $\RR$ is also a preordering on $S$.
Proof
By definition of ordering:
$\RR$ is an ordering on $S$ if and only if $\RR$ satisfies the ordering axioms:
\((1)\) | $:$ | $\RR$ is reflexive | \(\ds \forall a \in S:\) | \(\ds a \mathrel \RR a \) | |||||
\((2)\) | $:$ | $\RR$ is transitive | \(\ds \forall a, b, c \in S:\) | \(\ds a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c \) | |||||
\((3)\) | $:$ | $\RR$ is antisymmetric | \(\ds \forall a, b \in S:\) | \(\ds a \mathrel \RR b \land b \mathrel \RR a \implies a = b \) |
By definition of preordering:
$\RR$ is a preordering on $S$ if and only if $\RR$ satifies the preordering axioms:
\((1)\) | $:$ | $\RR$ is reflexive | \(\ds \forall a \in S:\) | \(\ds a \mathrel \RR a \) | |||||
\((2)\) | $:$ | $\RR$ is transitive | \(\ds \forall a, b, c \in S:\) | \(\ds a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c \) |
Thus an ordering is a preordering which is antisymmetric.
$\blacksquare$
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