Intersection of Orderings is Ordering

Theorem

Let $A$ be a set.

Let $\RR$ and $\SS$ be orderings on $A$.

Then $\RR \cap \SS$ is an ordering on $A$.

Proof

By definition of ordering:

$\RR$ and $\SS$ are reflexive

$\RR$ and $\SS$ are transitive

$\RR$ and $\SS$ are antisymmetric.

We have:

Intersection of Reflexive Relations is Reflexive

Intersection of Transitive Relations is Transitive

Intersection of Antisymmetric Relations is Antisymmetric

and the result follows.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.27 \ \text {(c)}$