# prove (?) the trichotomy property

Let $$(X, \preceq)$$ be a total ordered set, show that

$$x \prec y \leftrightarrow x \preceq y \land x \neq y$$ is a transitive relation, such that for each $$x$$, $$y$$ $$\in X$$ exactly one case holds: 1) $$x \prec y$$ 2) $$x=y$$ 3) $$y \prec x$$

I am really unsure here, doesn't the transitivity and trichotomy follow from X being total ordered? Is this https://proofwiki.org/wiki/Trichotomy_Law_(Ordering) the proof needed here?

They do follow from the fact that $$\preceq$$ is a total ordering, but there’s a bit of work to be done in order to show that they do.

You’re being asked to show two things. The first is that $$\prec$$ is transitive, i.e., that if $$x,y,z\in X$$ and $$x\prec y\prec z$$, then $$x\prec z$$. That $$x\preceq z$$ is immediate from the transitivity of $$\preceq$$, but you still have to show that $$x\ne z$$, which isn’t quite trivial.

The second is that for each $$x,y\in X$$, exactly one of $$x\prec y$$, $$x=y$$, and $$y\prec x$$ holds. You know that either $$x\preceq y$$ or $$y\preceq x$$. If $$x\preceq y$$, then either $$x\ne y$$, in which case $$x\prec y$$ by definition, and it’s easy to verify that $$y\not\prec x$$, or $$x=y$$, in which case by definition $$x\not\prec y$$ and $$y\not\prec x$$. Finishing it off is just more of the same sort of reasoning.

• Hello Brian, thank you so much for your great helping comments each and everytime, the second part is clear but could you tell me for the first part, how I can exactly show $x \neq y$? May 26, 2020 at 8:59
• @Parinn: You mean that $x\ne z$? Suppose that $x=z$. Then $z\prec y$ and $y\prec z$; is that possible? May 26, 2020 at 16:06
• Thank you, so by taking the assumption that $x = y$, and then showing that a contradiction occurs, I would be finished? Or is there something additional to do for proving the first part? May 28, 2020 at 4:15
• @Parinn: If you assume that $x=z$ and arrive at a contradiction, then you know that $x\ne z$. Since at that point you already know that $x\preceq z$, that immediately tells you that $x\prec z$, which is what you had to prove in order to show that $\prec$ is transitive. May 28, 2020 at 4:18
• Thank you for your help May 28, 2020 at 9:33

Show $$\forall x,y$$($$x\preceq y\lor y\prec x)$$ : Suppose not, i.e. $$\exists x,y(x\npreceq y\land y\nprec x)$$. From $$y\nprec x$$, $$y\npreceq x\lor y=x$$ (definition). If $$y=x$$ holds, then $$x\preceq y$$, which contradicts the assumption $$x\npreceq y$$. Therefore $$y\npreceq x$$ holds, however, again this contradicts that $$(X,\preceq)$$ is total ordered set. Therefore $$\forall x,y$$($$x\preceq y\lor y\prec x)$$ holds, which states at least one of 1),2) and 3) holds.

Now show only one of 1),2) and 3) holds for given $$x,y\in X$$ : Suppose not. If 1) and 2) holds in same time, then $$x\preceq y\land x\neq y\land x=y$$, which clearly a contradiction. The case for 2) and 3) can be seen similarly. If 1) and 3) holds in same time, then $$x\preceq y\land y\preceq x\land x\neq y\Rightarrow x=y\land x\neq y$$, a contradiction. Therefore for given $$x,y\in X$$, only one of 1),2), and 3) holds.

• Thank you so much for your answer, the second paragraph is perfectly clear. In the first one I have struggles to understand your argumentation, you first negate the sentence $\neg ( x \preceq y \land x \neq y)$ in the definition ? But then I could not understand your argument chain, if possible could you please further explain? May 26, 2020 at 9:27