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.
Best Answer
Well, for the first, suppose that $x_0\in\alpha^*.$ Then by definition, there is some rational $r>0$ such that $-x_0-r\notin\alpha.$ But then $\frac r2$ is also a positive rational, and $x_1=x_0+\frac r2$ is rational and greater than $x_0,$ and $-x_1-\frac r2=-x_0-r\notin\alpha.$ Thus, for any $x\in\alpha^*,$ we can find a $y\in\alpha^*$ with $x<y$.
For the second, we will need the following result:
To prove this we let $z_0$ be the least element of $\Bbb Q\setminus\alpha$--if there is such an element. We define a function $m:\Bbb Q\times\Bbb Q\to\Bbb Q$ by $$m(x,y)=\frac{x+y}2,$$ we fix any $a_0\in\alpha,$ and any non-least $b_0\in\Bbb Q\setminus\alpha.$ Then we define sequences $\langle a_n\rangle_{n=0}^\infty$ and $\langle b_n\rangle_{n=0}^\infty$ recursively as follows:
(1) If $m(a_n,b_n)=z_0$ for some $n,$ then for all integers $k\ge n$ we let $a_{k+1}=m(a_k,z_0)$ and $b_{k+1}=m(z_0,b_k).$
(2) If $m(a_n,b_n)\in\alpha,$ let $a_{n+1}=m(a_n,b_n)$ and let $b_{n+1}=b_n.$
(3) If $m(a_n,b_n)$ is a non-least element of $\Bbb Q\setminus\alpha,$ then let $a_{n+1}=a_n$ and $b_{n+1}=m(a_n,b_n).$
It can be shown that $m$ is well-defined, that $\langle a_n\rangle_{n=0}^\infty$ is a well-defined sequence of elements of $\alpha,$ that $\langle b_n\rangle_{n=0}^\infty$ is a well-defined sequence of non-least elements of $\Bbb Q\setminus\alpha,$ and that for all integers $n\ge 0$ we have $$0<b_0-a_0=2^n\cdot(b_n-a_n).$$ By the Archimedean Property of the rationals, there exists some positive integer $n$ such that $n\cdot r>b_0-a_0,$ so since $2^n>n>0,$ we have $$2^n\cdot r>n\cdot r>b_0-a_0=2^n\cdot(b_n-a_n)>0,$$ whence $$0<b_n-a_n<r,$$ as desired.
I leave the details to you (unless, of course, you already have that result). Now, to show that $0^*\subseteq\alpha+\alpha^*,$ take any $q\in0^*,$ and put $r=-q,$ so $r\in\Bbb Q$ and $r>0.$ By the Lemma, there exist some $a\in\alpha$ and some non-least $b\in\Bbb Q\setminus\alpha$ such that $0<b-a<r.$ Then $q=-r<a-b=a+-b.$ It remains only to show that $-b\in\alpha^*,$ which again I leave to you.