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