Ordering on 1-Based Natural Numbers is Total Ordering

Theorem

Let $\N_{> 0}$ be the $1$-based natural numbers.

Let $<$ be the strict ordering on $\N_{>0}$.


Then $<$ is a (strict) total ordering.


Proof

From Ordering on $1$-Based Natural Numbers is Trichotomy we have that $<$ is trichotomy.

From Ordering on $1$-Based Natural Numbers is Transitive we have that $<$ is transitive.

From Trichotomy is Antireflexive it follows that $<$ is antireflexive.

It follows by definition that $<$ is a strict ordering.

By the trichotomy law it follows that $<$ is a strict total ordering.

$\blacksquare$


Sources

  • 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $2.2$: Theorem $2.10$
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