Antisymmetric Preordering is Ordering
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Theorem
Let $\RR \subseteq S \times S$ be a preordering on a set $S$.
Let $\RR$ also be antisymmetric.
Then $\RR$ 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 $\RR$ is a relation on $S$ which is reflexive, transitive and antisymmetric.
Thus by definition $\RR$ is an ordering on $S$.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 7$: Exercise $7$