Devlin K.: The Joy of Sets (Springer, Undergraduate Texts in Mathematics)
In naive set theory we assume the existence of some
given domain of 'objects', of which we may build sets. Just what
these objects are is of no interest to us. Our only concern is the
behavior of the 'set' concept. This is, of course, a very common
situation of mathematics. For example, in algebra, when we discuss
a group, we are (usually) not interested in what
the elements of the group are, but rather in the way the group operation
acts upon those elements.
The above quote is mentioned in connection with "definition" of sets, but it shows that this situation is quite common in mathematics.
It is not important how the real numbers are represented, the important thing are their properties.
In the case of Dedekind cuts the starting point is that we suppose we already have defined the rational numbers $\mathbb Q$, and we what somehow get a new set $\mathbb R$, which will have nicer properties. This means that we want somehow define a set $\mathbb R$ together with operations $+$ and $\cdot$ and relation $\le$, such that
- they have "all familiar properties"; i.e. $(\mathbb R,+,\cdot)$ is an ordered field;
- they "contain" rational numbers; which formally means that there is an injective map $e:\mathbb Q\to\mathbb R$, which preserves addition, multiplication and inequality;
- they "improve" the set of rational numbers in the sense that it contains all "missing" numbers; every non-empty subset of $\mathbb R$ which is bounded from above has a supremum, see wikipedia: Least-upper-bound-property.
Note that rational numbers do not have least-upper-bound-property, the set $\{x\in\mathbb Q; x^2<2\}$ does not have a supremum in $\mathbb Q$.
We can give many different definitions which will fulfill the above properties; theoretically they are all equally good; for practical purposes some of them might be easier to work with.
The construction of reals using Cauchy sequences has a similar spirit, in this case the property which we want to add is completeness as a metric space. (Rational numbers do not have this property.)
Let me mention two books, which deal with this topic:
Artmann B: The concept of number (Ellis Horwood, 1988).
This books mentions several constructions of reals (Dedekind cuts, Cauchy sequences, decimal representation, continued fractions). Advantages and disadvantages of various approaches are mentioned in this book. (Although all construction lead to "the same" - isomorphic - set of reals, some properties of $\mathbb R$ are easy to prove and some might be more difficult, depending on the chosen approach.)
Ethan D. Bloch: The Real Numbers and Real Analysis, Springer, 2001. This book is intended as a textbook for a course in real analysis, but it discusses the two most usual definitions of real numbers in detail in the first two chapters.
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:
Lemma: Given a Dedekind cut $\alpha$ and any $r\in\Bbb Q$ with $r>0,$ there exist $a,b\in\Bbb Q$ with $a\in\alpha$ and $b$ a non-least element of $\Bbb Q\setminus\alpha$ such that $0<b-a<r$.
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.
Best Answer
Just to clarify the definition of the Dedekind cut you gave, $\alpha$ is the actual Dedekind cut and $\beta$ is its complement in $\mathbb{Q}$. That wasn't clearly stated in your formulation and it is important to know that to proceed.
Now let $x\in\mathbb{Q}$ and define $\alpha=\{q\in\mathbb{Q}|q<x\}$. Now your goal is to show that $\alpha$ satisfies the three axioms you spelled out in your question. Here are a few hints:
I was intentionally vague to promote thought. Let me know in a comment if you need more of a hint.