Let’s take a closer look at the conditions defining a Dedekind cut. A subset $D$ of $\Bbb Q$ is a Dedekind cut if:
- whenever $r\in D$ and $s<r$, then $s\in D$;
- there is a number $q\in\Bbb Q$ such that $r\le q$ for all $r\in D$; and
- for each $r\in D$ there is an $s\in D$ such that $r<s$.
Think of the rationals in the usual pictorial fashion, laid out as a line extending infinitely far in both directions, negative rationals to the left and positive rationals to the right. Condition (1) says that if $D$ contains some rational $r$, it contains every rational to the left of $r$ as well. One set that satisfies this condition is $\Bbb Q$ itself. Another is $(\leftarrow,0)$, the set of negative rationals, containing every rational strictly to the left of $0$. Yet another is $(\leftarrow,1]$, the set of rationals at or to the left of $1$. On the other hand, the set $(\leftarrow,0)\cup(2,4)$ does not satisfy condition (1): it contains $3$, but it doesn’t contain $1$ even though $1<3$. It’s not an initial segment of $\Bbb Q$.
Condition (2) is the simplest: it just says that $D$ cannot extend infinitely far to the right. There must be some rational number $q$ such that every member of $D$ is at or to the left of $q$. Almost all of the sets that satisfy condition (1) satisfy condition (2) as well; the only exception is $\Bbb Q$ itself, and condition (2) is designed specifically to rule out $\Bbb Q$.
Condition (3) is perhaps the hardest to get a grip on, but what it says is actually quite simple: it says that $D$ must not have a largest element. No matter what $r$ you choose in $D$, $D$ contains some bigger number $s$. This rules out sets like $(\leftarrow,1]=\{q\in\Bbb Q:q\le 1\}$, that have a maximum element.
A Dedekind cut, therefore, is a subset of the rational numbers that looks more or less like $(\leftarrow,2)$, say: it’s an initial segment of the rational number line, it’s not the whole line, and it has no largest element. In fact, every $r\in\Bbb Q$ defines a Dedekind cut in just this way, namely, the cut $$(\leftarrow,r)=\{q\in\Bbb Q:q<r\}$$ consisting of every rational to the left of (smaller than) $r$.
However, these aren’t the only Dedekind cuts. For example, let $$D=\{q\in\Bbb Q:q<0\text{ or }q^2<2\}\;.$$ It’s not hard to check that this $D$ satisfies conditions (1)-(3); if you’ve not already seen this, you probably will soon. But $D$ is not $(\leftarrow,r)$ for any rational number $r$, because if it were, we could show that $r^2=2$, when in fact we know that $\sqrt2$ is irrational. It’s these ‘extra’ Dedekind cuts, the ones not of the form $(\leftarrow,r)$ for any rational number $r$, that make Dedekind cuts useful and interesting: they ‘fill in’ the gaps in $\Bbb Q$ corresponding to the irrational numbers and allow us to construct the real numbers rigorously starting with just the rationals. In the end the cuts $(\leftarrow,r)$ for $r\in\Bbb Q$ are going to correspond to the rationals, and the ‘extra’ Dedekind cuts are going to correspond to the irrationals. But in order to make that work, we have to be able to define the various arithmetic operations on these Dedekind cuts in such a way that they behave the way they should. Your exercise here is part of showing how to define multiplication of Dedekind cuts.
I’ll say nothing about the exercise itself, as breeden has already covered that in some detail.
Added: In view of the comments, I think that I should emphasize that in what I’ve written above, $(\leftarrow,q)$ and so forth are to be understood as subsets of $\Bbb Q$. That is, I’m writing $(\leftarrow,q)$ as an abbreviation for $\{r\in\Bbb Q:r<q\}$, not for $\{r\in\Bbb R:r<q\}$.
As another example to show how a set can be bounded above (as is required by (2)) and still have no largest element, let $D$ be the set of negative rationals. Every member of $D$ is less than $0$ (and hence less than every positive rational as well), but $D$ has no largest element. If $m$ and $n$ are positive integers, so that $-\frac{m}n\in D$, then $-\frac{m}n<-\frac{m}{2n}<0$, so $-\frac{m}{2n}$ is a member of $D$ that is bigger than $-\frac{m}n$. This shows that $D$ has no largest element: give me any element of $D$, and I’ve just shown you how to find a bigger one.
I’ll show that the sum of two rational cuts is rational. Suppose that $A$ and $B$ are rational cuts; then there are $a,b\in\Bbb Q$ such that $A=\{q\in\Bbb Q:q<a\}$ and $B=\{q\in\Bbb Q:q<b\}$. By definition $$A+B=\{p+q:p\in A\text{ and }q\in B\}\;;$$ I’ll prove that $A+B=\{q\in\Bbb Q:q<a+b\}$, the Dedekind cut corresponding to the rational number $a+b$.
Suppose that $r\in A+B$. Then there are $p\in A$ and $q\in B$ such that $r=p+q$. Since $p\in A$, we know that $p<a$, and similarly, since $q\in B$, we know that $q<b$, so $r=p+q<a+b$. This shows that $A+B\subseteq\{q\in\Bbb Q:q<a+b\}$.
Now suppose that $r\in\Bbb Q$ and $r<a+b$. Let $d=\frac12(a+b-r)>0$, and note that $d$ is rational. Let $p=a-d$ and $q=b-d$; then $p,q\in\Bbb Q$. Moreover $p<a$ (since $d>0$), so $p\in A$, and $q<b$, so $q\in B$, and $$p+q=(a-d)+(b-d)=a+b-2d=a+b-(a+b-r)=r\;.$$ Thus, $r\in A+B$, and we’ve shown that $\{q\in\Bbb q:q<a+b\}\subseteq A+B$. Putting the pieces together, we conclude that $A+B=\{q\in\Bbb Q:q<a+b\}$. $\dashv$
Here I didn’t have to prove directly that $A+B$ was a Dedekind cut, because I could show that it was equal to something known to be a Dedekind cut. You should try to prove that for all Dedekind cuts $A$ and $B$, $A+B=\{p+q:p\in A\text{ and }q\in B\}$ is a Dedekind cut.
- It’s easy to show that $A+B\ne\varnothing$.
- Suppose that $r\in A+B$, $s\in\Bbb Q$, and $s\le r$; you need to show that $s\in A+B$. You know that $r=p+q$ for some $p\in A$ and $q\in B$. Let $d=\frac12(r-s)$, and consider the rational numbers $p-d$ and $q-d$.
- Suppose that $r\in A+B$; you need to show that there is an $s\in A+B$ such that $r<s$. Start by writing $r=p+q$ for some $p\in A$ and $q\in B$, and use the fact that $A$ and $B$ have no largest elements.
Once you’ve done that, you can show that if $A$ is a rational cut and $B$ is an irrational cut, then $A+B$ is irrational. You already know that $A+B$ is a Dedekind cut, so you just have to show that it’s not a rational Dedekind cut: for each $r\in\Bbb Q$, $A+B\ne\{q\in\Bbb Q:q<r\}$. This can be done by contradiction: show that if $A+B=\{q\in\Bbb Q:q<r\}$ for some $r\in\Bbb Q$, then the cut $B$ is rational.
When both $A$ and $B$ are irrational cuts, $A+B$ can be either rational or irrational, depending on exactly which cuts $A$ and $B$ are; you can’t prove any general conclusion here.
Best Answer
For (1): As Dylan Moreland points out in a comment, the problems lies in (b). It's not true that a rational smaller than an element of your $S$ is also in $S$.
For (2) and (3), I have the same comment in both: A common and understandable confusion when being first exposed to Dedekind cuts is that you have to completely forget that real numbers exist. It doesn't really make sense to say that a cut is on an irrational number, since there are no such thing as rational numbers when you're defining Dedekind cuts. (Of course, most of the things you're saying in this respect are "morally correct" in the sense that they end up being true once you've defined real numbers, but in the long run, this is circular.)
Similarly, for your $2^{1/3}$ Dedekind cut, it may be helpful while you're going through this to never ever use notations like $2^{1/3}$ or $\sqrt[3]{2}$, except possibly as shorthand for the Dedekind cut $S$ you describe (and even then, only once you've defined multiplication of cuts and checked that $S^3=2$), since in the rational world, this number does not exist. So no, you can't just say it's that set, because you don't know that set is a Dedekind cut. You need to prove that it has no maximal element, and if you think about it, checking that there is no rational number whose cube equals 2 is a little shy of that stronger conclusion.