Strategy for proving that Dedekind cuts satisfy the property of trichotomy I'm trying to prove that given Dedekind cuts $\alpha, \beta$, exactly one of $\alpha < \beta$, $\beta < \alpha$, and $\alpha = \beta$ holds.
I'm only interested in the proof strategy, because once I have the strategy in mind, I think I can prove it. The idea I have is the following. I need to show:
(i) at least one of $\alpha < \beta$, $\beta < \alpha$, and $\alpha = \beta$ holds.
(ii) only one holds, i.e., two cannot hold simultaneously.
To prove (i), I believe I can proceed as follows. I can prove that if $\alpha \nless \beta$ and $\beta \nless \alpha$ then $\alpha = \beta$. Then the argument has the form:
\begin{align*}
& (\neg (\alpha < \beta) \wedge \neg (\beta < \alpha)) \implies (\alpha = \beta) \\
& \equiv \neg (\neg (\alpha < \beta) \wedge \neg (\beta < \alpha)) \lor (\alpha = \beta) \\
& \equiv (\alpha < \beta) \lor (\beta < \alpha) \lor (\alpha = \beta). 
\end{align*}
To prove (ii), I think I have to prove that if $\alpha < \beta$ holds, then $\beta < \alpha$ and $\alpha = \beta$ doesn't hold; if $\beta < \alpha$, then $\alpha < \beta$ and $\alpha = \beta$ cannot hold; and if $\alpha = \beta$, then neither $\alpha < \beta$ or $\beta < \alpha$ hold.
I'm interested in whether there is some symmetry to the above argument. If I prove the first statement when $\alpha < \beta$, that seems exactly symmetric to proving the second statement where $\beta < \alpha$, interchanging labels of $\alpha$ and $\beta$. So I could do in this two instead of three steps.
I'd appreciate any thoughts on this proof strategy.
 A: A relation $<$ on set $S$ is trichotomous $\iff{}$ for all $\alpha{},\beta{}\in{}S$, one and only one of the following is true:

*

*$\alpha{}=\beta{}$

*$\alpha{}<\beta{}$

*$\beta{}<\alpha{}$
In your case, $<$ is the proper subset relation. It is clear that no more than one of 1, 2, 3 can be true. If you need clarification on this point do not hesitate to ask.
How can we prove that at least one of 1, 2, 3 is true?
Suppose 1 and 2 are false. Then if we can prove that 3 must be true, we are done*.
Note that $\alpha{},\beta{}\neq{}\varnothing{}$ by the definition of cut. Given that 1 and 2 are false, we have $\alpha{}\not\subset{}\beta{}$. Thus there exists $q\in{}\mathbb{Q}$ with $q\in{}\alpha{}$ and $q\notin{}\beta{}$. Now consider $q'\in{}\beta{}$. If $q'>q$ then $q\in{}\beta{}$ by the definition of cut, which is a contradiction. Therefore $q'<q$, which forces $q'\in{}\alpha{}$ by the same part of the definition of cut. As $q'$ was an arbitrary element of $\beta{}$, this shows that $\beta{}\subset{}\alpha{}$. The fact that $q\in{}\alpha{}$ and $q\notin{}\beta{}$ forces $\beta{}<\alpha{}$.
*This is a general proof technique; in two-valued logic, if we have $n$ statements with $n\in{}\mathbb{N}$, $n\geq{}2$, then to show that at least one of the $n$ statements is true it suffices to assume $n-1$ of the statements are false and then to prove that the last statement must be true.
