Dual Ordering is Ordering

Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $\succeq$ denote the dual ordering of $\preceq$.

Then $\succeq$ is an ordering on $S$.

Proof

By definition of ordering, $\preceq$ is reflexive, transitive and antisymmetric.

By definition, $\succeq$ is the inverse relation to $\preceq$.

By Inverse of Reflexive Relation is Reflexive, $\succeq$ is reflexive.

By Inverse of Antisymmetric Relation is Antisymmetric, $\succeq$ is antisymmetric.

By Inverse of Transitive Relation is Transitive, $\succeq$ is transitive.

Thus by definition $\succeq$ is an ordering on $S$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.2$
• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.8$
• 1967: Garrett Birkhoff: Lattice Theory (3rd ed.): $\S \text{I}.2$: Theorem $2$
• 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 $27 \ \text {(a)}$
• 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.14 \ \text{(b)}$