# Ordering is Preordering

## 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.

$\blacksquare$