Reflexive Reduction of Ordering is Strict Ordering

Theorem

Let $\RR$ be an ordering on a set $S$.

Let $\RR^\ne$ be the reflexive reduction of $\RR$.

Then $\RR^\ne$ is a strict ordering on $S$.

Proof 1

Antireflexivity

Follows from Reflexive Reduction is Antireflexive.

$\Box$

Transitivity

Suppose $\tuple {x, y}, \tuple {y, z} \in \RR^\ne$.

By antireflexivity $x \ne y$ and $y \ne z$.

We consider the two remaining cases.

Case 1: $x = z$

If $x = z$ then:

$\tuple {x, y}, \tuple {y, x} \in \RR^\ne$

and so:

$\tuple {x, y}, \tuple {y, x} \in \RR$

Then by the antisymmetry of $\RR$:

$x = y$

and:

$\tuple {x, x} \in \RR^\ne$

which contradicts that $\RR^\ne$ is antireflexive.

Case 2: $x \ne z$

By the transitivity of $\RR$:

$\tuple {x, z} \in \RR$

and by $x$ and $z$ being distinct:

$\tuple {x, z} \notin \Delta_S$

It follows by the definition of reflexive reduction:

$\tuple {x, z} \in \RR^\ne$

Hence $\RR^\ne$ is transitive.

$\blacksquare$

Proof 2

By definition, an ordering is both antisymmetric and transitive.

The result then follows from Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering.

$\blacksquare$

Also see

• Reflexive Closure of Strict Ordering is Ordering

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings (passim)
• 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 $1 \ \text {(b)}$