Trichotomy Law (Ordering)
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Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Then $\preceq$ is a total ordering if and only if:
- $\forall a, b \in S: \paren {a \prec b} \lor \paren {a = b} \lor \paren {a \succ b}$
That is, every element either strictly precedes, is the same as, or strictly succeeds, every other element.
In other words, if and only if $\prec$ is a trichotomy.
Proof
\(\ds \forall a, b \in S: \, \) | \(\ds \) | \(\) | \(\ds a \preceq b \lor b \preceq a\) | Definition of Total Ordering | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall a, b \in S: \, \) | \(\ds \) | \(\) | \(\ds a \preceq b \lor a \succeq b\) | Definition of Dual Ordering | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall a, b \in S: \, \) | \(\ds \) | \(\) | \(\ds \paren {a = b \lor a \prec b} \lor \paren {a = b \lor a \succ b}\) | Strictly Precedes is Strict Ordering | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall a, b \in S: \, \) | \(\ds \) | \(\) | \(\ds a \prec b \lor a = b \lor a \succ b\) | Rules of Commutation, Association and Idempotence |
$\blacksquare$
Also known as
The Trichotomy Law can also be seen referred to as the trichotomy principle.
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order