Reflexive Closure of Strict Ordering is Ordering

Theorem

Let $S$ be a set.

Let $\prec$ be a strict ordering on $S$.

Let $\preceq$ be the reflexive closure of $\prec$.

Then $\preceq$ is an ordering.

Proof

Since $\prec$ is a strict ordering, it is by definition transitive and asymmetric.

By Asymmetric Relation is Antisymmetric, $\prec$ is antisymmetric.

Thus by Reflexive Closure of Transitive Antisymmetric Relation is Ordering, $\preceq$ is an ordering.

$\blacksquare$

Also see

• Reflexive Reduction of Ordering is Strict Ordering

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.7$
• 1967: Garrett Birkhoff: Lattice Theory (3rd ed.) ... (next): $\S \text I.1$: Lemma 1
• 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $1 \ \text {(b)}$