X is totally ordered under ≤ if and only if X follows the law of trichotomy?

Question

I'm having trouble with this proof. I know that this proof requires two parts.

My attempt so far:

Proof: ($\Rightarrow$)

Assume $X$ is a totally ordered set under $\le$.

Let $m,n$ be two arbitrary element of $X$.

I'm not sure where to go from here.

Thanks

Answer 1

HINT: Since $X$ is totally ordered by $\le$, you know that at least one of $m\le n$ and $n\le m$ holds.

• Show that if both hold, then $m=n$.
• Show that if $m\le n$ and $n\not\le m$, then $m<n$.
• Show that if $n\le m$ and $m\not\le n$, then ... ?

Since exactly one of these three cases must occur, this establishes that the relation $m<n$ defined by

$m<n$ if and only if $m\le n$ and $m\ne n$

satisfies the trichotomy condition.

For the other direction, assume that $<$ satisfies the trichotomy condition, and show that $\le$ totally orders $X$; this is just a matter of checking that $\le$ must have the properties required in order for it to be a total order.

Added: Note that I am assuming that $\le$ is known to be a partial order. If that’s not the case, the result is false.