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.
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
I assume that $A$ is supposed to be a Dedekind cut. Start by making a sketch:
The line as a whole represents $\Bbb Q$, and everything to the left of the parenthesis is in the cut $A$. Now look at the definition of $-A$:
$$-A=\{-(q+b):q\in\mathbb{Q}^+, b\in\mathbb{Q}\setminus A\}$$
It depends on two subsets of $\Bbb Q$, $\Bbb Q^+$, and $\Bbb Q\setminus A$, so we should figure out where those are in the picture. $\Bbb Q\setminus A$ is easy:
it’s everything to the right of $A$. And we know just what $\Bbb Q^+$ is: it’s the set of positive rationals. What happens when you add a positive rational $q$ to every member of $\Bbb Q\setminus A$? You shift the set $\Bbb Q\setminus A$ to the right by $q$ units:
That last picture shows $\{q+b:b\in\Bbb Q\setminus A\}$ for a particular $q\in\Bbb Q^+$. Now what happens when you look at the negatives of these rational numbers, $\{-(q+b):b\in\Bbb Q\setminus A\}$? You simply flip the line $180$° around $0$ to get a picture more or less like this:
The plus signs mark the set $-(q+\Bbb Q\setminus A)$, and the o’s mark the set $\{-a:a\in A\}$. The gap in the middle has length $q$. $\{-a:a\in A\}$ is always in the same place, but the location of $-(q+\Bbb Q\setminus A)$ depends on $q$: when $q$ is large, it’s far to the left of $\{-a:a\in A\}$, and when $q$ is small, it’s very close to $\{-a:a\in A\}$.
The set $-A$ that you’re to prove is a Dedekind cut is just the union of all these sets $-(q+\Bbb Q\setminus A)$ as $q$ ranges over the positive rationals, so it’s the union of all possible sets like those marked with plus signs in the pictures below:
What do you think that union will look like in a sketch of this kind? Won’t it look something like this?
That looks an awful lot like a Dedekind cut. Now you just have to prove it. For instance, you have to show that there is some rational that is not in the set. From the picture it appears that any $-a$ with $a\in A$ should work, so you should try to prove that this is the case. (Remember, the $q$’s that are being added are all strictly positive.)
You need to show that if $p$ is rational and $p<r\in-A$, then $p\in -A$. The picture certainly makes that look plausible, and it’s not hard to show. If $r\in-A$, then $r=-(q+a)$ for some $q\in\Bbb Q^+$ and $a\in A$. Can you find another positive rational $q'$ such that $p=-(q'+a)$? The number $r-p$ may be useful.