Real Analysis/Dedekind's construction

Historically Dedekind gave the first construction of the real numbers. Like Cantor's construction, Dedekind's method constructs the real numbers from the set rational numbers. Dedekind's construction gives a more geometric picture of the real numbers.

The idea of the construction is that every real number $$r$$ should cut the number line into two subsets, the numbers less than $$r$$ and the numbers greater than or equal to $$r$$. These two sets are different for every real number, and given these sets we could determine $$r$$. In fact, because each real number divides the number line, it also divides the rational numbers into two. And if we knew these two sets of rational numbers, then we could also determine $$r$$. Hence when looking for a way to define the real numbers, we could define them as the collection of ways to partition the rational numbers into two sets. Today when discussing Dedekind cuts one usually only keeps track of one of these two sets. Our definition of a cut could informally be thought of as the numbers less than $$r$$.

Definition
We say that $$\alpha\subset\Q$$ is a cut if and only if the following statements are true:
 * 1) The set $$\alpha$$ is nonempty. That is, there exists some a rational number $$p$$ such that $$p\in\alpha$$.
 * 2) If $$p\in\alpha$$ and $$q$$ is a rational number so that $$q<p$$, then $$q\in\alpha$$ . So for any element of the cut, every rational number smaller than it is also an element of the cut.
 * 3) There exists a rational number $$a$$ such that $$p<a$$ for all $$p\in\alpha$$ . So there's a rational number that's larger than any element in a particular cut.
 * 4) For every $$p\in\alpha$$, there exists an $$r\in\alpha$$ such that $$p<r$$ . So every element of the cut has yet another element larger than it. If that seems impossible, imagine a series of rational numbers that approximate the real number we're trying to define. For example, the cut $$x^2<2$$ contains a series of countably infinitely many rational numbers like 1, 1.2, 1.4, 1.41, 1.414, 1.4142, etc. each of which is larger than every previous one and a bit closer to $$\sqrt2$$.

The Construction
We now outline how to make the set of Dedekind cuts form an ordered field that satisfies the least upper bound axiom.

Order
First we discuss how to compare the relative size of two cuts.

Definition Given two cuts $$\alpha,\beta$$, we say $$\alpha=\beta$$ if they are the same subset of the rational numbers. That is if $$\alpha\sube\beta$$ and $$\beta\sube\alpha$$.

Exercise Show equality defined as in the previous definition is an equivalence relation.

Definition Given two cuts $$\alpha,\beta$$, we say $$\alpha<\beta$$ if $$\alpha\subset\beta$$.

Exercise Show the above order satisfies the order axioms and that $$\R$$ is totally ordered.

We now also assume the usual definitions to $$\ge,<,>$$, we leave this to the reader to supply the correct definitions.

Exercise Given two cuts $$\alpha,\beta$$, show that if $$\alpha < \beta$$ then there is a rational number $$p\in\beta$$ so that $$p\notin\alpha$$.

Exercise (Trichotomy) Show that with this order, every cut satisfies exactly one of the following three inequalities: $$\alpha<0,\alpha=0,\alpha>0$$.

Field Axioms
Let $$\alpha,\beta$$ be two cuts

Define addition as $$\alpha+\beta=\{p+q:p\in\alpha,q\in\beta\}$$.

Define negation as $$-\alpha=\{r\in\Q:r<-p\text{ for some }p\not\in\alpha\}$$

Let 0 be the cut defined by $$\mathbf0=\{r:r<0\}$$.


 * Exercise For two cuts α, β verify that α+β, &minus;α, and 0 are cuts. Furthermore show that the set of cuts forms an abelian group with this definition of addition.  What is the identity element?

Defining multiplication requires a bit more care. We define
 * $$\alpha\cdot\beta=\begin{cases}\{p\cdot q\mid p\in \alpha, p>0, q\in\beta\}&\text{if }\alpha\geq\mathbf{0},\beta\geq\mathbf{0};\\

\{p\cdot q\mid p\notin \alpha, q\in\beta, q>0\}&\text{if }\alpha\geq\mathbf{0}, \beta<\mathbf{0};\\ \{p\cdot q\mid p\in \alpha, q\notin\beta, q>0\}&\text{if }\alpha<\mathbf{0}, \beta\geq\mathbf{0};\\ \{r\mid r\mathbf{0};\\

\{r\mid r<\frac{1}{p}, p\not\in \alpha,\text{ and }p<0\}&\text{if }\alpha<\mathbf{0}. \end{cases}$$

Exercise For two cuts α, β verify that α&middot;β, α-1 (assuming α &ne; 0), and 1 are cuts. Show that with these definitions the set of Dedekind cuts forms an ordered field.

Least Upper Bound Property
We will now show the set of Dedekind cuts satisfies the least upper bound axiom.

Let A be a non-empty collection of cuts, and suppose that there is a cut β such that α&lt;β for all α. In other words, A is a non-empty collection of cuts that has an upper bound given by β.

Now we define a subset of the rational numbers, δ, by taking the unions of all of the subsets given by the cuts in A. That is $$\delta=\displaystyle\bigcup_{\alpha\in A}\alpha$$

Now we wish to show that δ is a cut. Suppose p&isin;δ, if follows that p&isin;α0 for some α0 in A. Then if q&lt;p then q&isin;α0 and hence q&isin;δ. This shows the first property. Also, there is some r&isin;α0 with r&gt;p, but then r&isin;δ. Hence the third property holds. To see that the second property holds, notice that δ&sube;β because each α&sube;β. By the cut properties for β there is a rational number b so that b&gt;p for all p&isin;β. Therefore, b&gt;p for all p&isin;δ. Hence δ is a cut.

Now, by definition of the order, δ is an upper bound of A. If η any cut with η&lt;δ, then there is a rational number p&isin;δ and p&notin;η. But p&isin;α0 for some α0 in A. Hence η is not greater then α0, and therefore η is not an upperbound for A. Therefore δ is the least upper bound. This shows that the set of Dedekind cuts satisfies the least upper bound axiom.

Reference

 * W. Rudin, Principles of Mathematical Analysis, McGraw-Hill International