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.

• $\{r\in\mathbb Q: r<1\}$ – Matemáticos Chibchas Oct 8 '13 at 6:34
• Followup: So every such cut must be an infinite set? – Danny Oct 8 '13 at 6:44
• If $x$ is in a cut, then so is $x-1$. – Hurkyl Oct 8 '13 at 7:23

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.

• I think you mean $\alpha = \{p \in \mathbb Q : p^2 < 2 \lor p < 0\}$. If you forget to put the "$\lor p < 0$" there, you won't get a cut because the set won't be downward closed. – kahen Oct 8 '13 at 6:37
• @kahen indeed you are right and I thought about adding it, but I was answering only to the I'm confused as to how a set of rationals can contain no largest rational, yet not contain every rational. part. I thought that adding the "$\vee p < 0$" could have led to more confusion. – bernatguillen Oct 8 '13 at 6:46

$$\left\{\frac{9}{10}, \frac{99}{100}, \frac{999}{1000}, \frac{9999}{10000},\ldots \right\}$$

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 $\mathbb{R}$.