# Antisymmetric Preordering is Ordering

## Theorem

Let $\mathcal R \subseteq S \times S$ be a preordering on a set $S$.

Let $\mathcal R$ also be antisymmetric.

Then $\mathcal R$ is an ordering on $S$.

## Proof

By definition, a preordering on $S$ is a relation on $S$ which is:

- $(1): \quad$ reflexive

and:

- $(2): \quad$ transitive.

Thus $\mathcal R$ is a relation on $S$ which is reflexive, transitive and antisymmetric.

Thus by definition $\mathcal R$ is an ordering on $S$.

$\blacksquare$

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 7$: Exercise $7$