Union of Orderings is not necessarily Ordering
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Theorem
Let $A$ be a set.
Let $\RR$ and $\SS$ be orderings on $A$.
Then $\RR \cup \SS$ is not necessarily an ordering on $A$.
Proof
Let $\RR$ be an ordering as asserted.
Let $\SS$ be the dual ordering of $\RR$.
From Dual Ordering is Ordering, $\SS$ is an ordering.
By definition, the dual of $\RR$ is the inverse of $\RR$:
- $\SS = \RR^{-1}$
From Union of Antisymmetric Relation with Inverse is Antisymmetric iff Diagonal, it is generally not the case that $\RR \cup \RR^{-1}$ is antisymmetric.
Hence, in general, $\RR \cup \RR^{-1}$ is not an ordering.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.27 \ \text {(c)}$