I am trying to prove that given any two Dedekind cuts, exactly one of $\alpha < \beta$, $\beta < \alpha$, or $\alpha = \beta$ hold.
To show that at least one holds, Rudin proceeds by supposing $\alpha < \beta$ and $\alpha = \beta$ fail, then proving that $\beta < \alpha$. From a propositional logic standpoint, this checks out:
\begin{align*}
& \neg (\alpha < \beta \lor \alpha = \beta) \implies \beta < \alpha \\
& \equiv \neg \neg (\alpha < \beta \lor \alpha = \beta) \lor \beta < \alpha \\
& \equiv \alpha < \beta \lor \alpha = \beta \lor \beta < \alpha.
\end{align*}
But intuitively, this proof strategy doesn't make any sense to me. In fact, it seems like I would need to prove that given any to of these statements failing, the third holds. Why I'm allowed to just pick two, doesn't make any sense to me.
Of course, there's a separate proof that only one can hold simultaneously, though I'm not too concerned with that, as it's clear to me by the definition of subsets of $\mathbb{Q}$.
Best Answer
"At least one of the statements holds" is the negation of "None of the statements hold". The idea here is that "None of the statements hold" must be false, since if neither $\alpha<\beta$, nor $\alpha=\beta$, then $\alpha>\beta$, so all three statements can't be false.
And then you also need to show that no two of the statements can hold simultaneously, but as you said, that step is not hard.