Rudin gives the definition of a Dedekind Cut to be:
A set of rational numbers is said to be a cut if
(I) $\alpha$ contains at least one rational, but not every rational;
(II) if $p\in\alpha$ and $q<p$ (q rational), then $q\in\alpha$;
(III) $\alpha$ contains no largest rational.
I’m confused as to how a set of rationals can contain no largest rational, yet not contain every rational.
This is very possible. Consider $\{r \in \mathbb Q \mid r < 0\}$. Suppose this set contained some largest rational $p$. But $p$ is negative, so $\frac p2$ is also negative and greater than $p$ (i.e. “less negative”).
This can easily be adapted into an argument that for all $q \in \mathbb Q$ the set $L_q = \{r \in \mathbb Q \mid r < q\}$ is a Dedekind cut. But it’s not all of them as the standard “$\sqrt2$” example shows.
Sorry for the answer, I can’t comment yet, but as a hint: which is the largest rational of the set $\alpha = \{p \in \mathbb{Q}: p^2 < 2\}$? If the set in the reals (I know this is kinda cheating) was something like $[-\infty, \sqrt{2}]$ then in the rationals it would contain no largest element.
$$ \left\{\frac{9}{10}, \frac{99}{100}, \frac{999}{1000}, \frac{9999}{10000},\ldots \right\}$$
Consider all the rationals which are strictly less than $0$. This set has no maximal element, but is very far from being the entire set of rational numbers.
There are enough examples shows that, a set which satisfies all the conditions exists, I’m not gonna repeat them. Just some additions to the properties of that set:
Of course it is infinite. Because as
$ \forall p \in \alpha, \, q < p \,\land q\in Q=>\, q \in \alpha $ and as a property of rational numbers, $ \forall q \in Q, \exists r $ such that $ r<q \land r \in Q$, then the set must contain every rational number $q$ such that $q<r$ if it contains such a number $p$, that is: $$p\in \alpha \land q\in Q<p => q \in Q$$
The meaning of having no largest rational is that, the set have a maximum value, but does not contain it; that is it should be an open set.
That’s why $ \alpha = \{p\,| p<a, p \in Q, a \in R\} $ is a great example for such a set. Also remember that if you choose $a$ as an irrational number, then the set $\{b \,| -b \in Q – \alpha \}$ will also be a Dedekind Cut but if you choose $a$ as a rational number, then such a set would have a largest rational.
If you already have the real numbers $\mathbb{R}$ a Dedekind Cut is a set $(-\infty,a) \cap \mathbb{Q}$,use where $a$ is the real number that is represented by the Dedekind Cut. But of course you go the other way, you use Dedekind Cuts to define the elements of $\mathbb{R}$ without assuming the existence of $\mathbb{R}$.