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Theorem: A partially ordered set is totally ordered if it obeys the law of trichotomy.

Things I know:

  • A relation on some set $A$ is said to be a partially ordered set if the relation is reflexive, anti-symmetric and transitive.
  • The law of trichotomy states: for all $a$ and $b$ in the poset $A$, the law of trichotomy holds if exactly one of the following is true: $a < b$, $a = b$ or $b < a$.
  • A partially ordered set is said to be totally ordered if given any two elements in A, either $a \leq b$ (less than/equal to) or $b < a$.

I restructured the theorem to match the form of $P$ implies $Q$. If a partially ordered set $A$ obeys the law of trichotomy, then it is totally ordered.

Proceed by direct proof. Let $A$ be any arbitrary but fixed partially ordered set that follows the law of trichotomy. Let $a$ and $b$ be ABF elements in $A$, such that exactly one of the following is true: $a < b$, $a = b$ or $b < a$. By the definition of the partially ordered set, the relation on $A$ is reflexive, anti-symmetric, and transitive.

How do I continue?

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1 Answer 1

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You want to show that if $\langle A,\le\rangle$ is a partial order that satisfies the law of trichotomy, then $\le$ is a total order on $A$. You need to focus on the goal of showing that $\le$ totally orders $A$: you need to show that for any $a,b\in A$, either $a\le b$, or $b<a$. You should therefore be starting with arbitrary $a,b\in A$, period: there’s no need to specify that exactly one of $a<b$, $a=b$, or $b<a$ holds, because that’s built into the assumption that $\langle A,\le\rangle$ satisfies the trichotomy law.

Now simply apply trichotomy to $a$ and $b$. If $a<b$ or $a=b$, then certainly $a\le b$. The only remaining possibility is that $b<a$. Thus, in every case either $a\le b$, or $b<a$. Since $a$ and $b$ were arbitrary elements of $A$, this shows that $\langle A,\le\rangle$ is a total order. That’s all there is to it.

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  • $\begingroup$ Thank you very much - I appreciate your help! $\endgroup$
    – Ethan
    Commented Jun 21, 2015 at 4:14
  • $\begingroup$ @Ethan: You're welcome! $\endgroup$ Commented Jun 21, 2015 at 4:21

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