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:

An ordering on $S$ is a relation $\RR$ on $S$ such that:

\((1)\)   $:$   $\RR$ is reflexive      \(\displaystyle \forall a \in S:\) \(\displaystyle a \mathrel \RR a \)             
\((2)\)   $:$   $\RR$ is transitive      \(\displaystyle \forall a, b, c \in S:\) \(\displaystyle a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c \)             
\((3)\)   $:$   $\RR$ is antisymmetric      \(\displaystyle \forall a \in S:\) \(\displaystyle a \mathrel \RR b \land b \mathrel \RR a \implies a = b \)             


By definition of preordering:

$\RR$ is a preordering on $S$ if and only if:

\((1)\)   $:$   $\RR$ is reflexive      \(\displaystyle \forall a \in S:\) \(\displaystyle a \mathrel \RR a \)             
\((2)\)   $:$   $\RR$ is transitive      \(\displaystyle \forall a, b, c \in S:\) \(\displaystyle a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c \)             


Thus an ordering is a preordering which is antisymmetric.

$\blacksquare$


Sources