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The intuition for cuts presumably comes from the standard experience of approximation by terminating decimals. For example, we can approximate $\sqrt{2}$ by the sequence $1,1.4,1.41,1.414,1.4142, \dots$. Now, why a cut set for $\sqrt{2}$ requires "all" rationals less than $\sqrt{2}$. Intuitively, only these rationals $1,1.4,1.41,1.414,1.4142, \dots$ uniquely identifies $\sqrt{2}$. Are not they sufficient ?

Related question : Why the addition of two cut set is defined as the set of all possible summation of elements of the two cutsets ?

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You're trying to define $\sqrt 2$. How do you know how to define the sequence $1,1.4,1.41,1.414,1.4142, \dots$ before defining $\sqrt 2$? –  Git Gud Jun 23 '13 at 11:11
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Presumably the same way you would construct its actual cut; by considering $a_n$ to be the largest $n$-decimal-digit number whose square is less than 2? –  Sharkos Jun 23 '13 at 11:19
    
Actually, my question is what was the motivation of Dedekind behind defining cut set with the property that "if a rational number belongs to a cut set, then "all" the rational numbers less than it also belongs to it." I guess the approximation by terminating decimals was the motivation. Then why "all" rational ? Are not the terminating decimals sufficient ? I hope I made my point. –  podu Jun 23 '13 at 11:28
    
I've tried to answer your deposited specific new question in that comment in my answer - I stress that whilst you can use lots and lots of different constructions, Dedekind gives a remarkably simple one. –  Sharkos Jun 23 '13 at 12:13
    
Yes, now I think that my question is a silly one ! –  podu Jun 26 '13 at 5:09

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The answer to your question is somewhat complicated - the successive decimal approximations define a number whose square is $2$ in a context in which every increasing sequence which is bounded above has a limit.

There is a good discussion here, and more related topics here.

But there is a bigger issue. The square root of $2$ can be taken to be the limit of many different sequences of rational numbers (try working in base $9$ or $11$ instead of base $10$ for starters). If we take sections of "all" the rationals, it becomes easier to work with real numbers because we don't have to keep proving that two representations are equivalent (eg for the purposes of multiplying numbers).

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Now, why a cut set for $\sqrt{2}$ requires "all" rationals less than $\sqrt{2}$.

This is Dedekind's idea (these cut sets are often called "Dedekind cuts", after him), and it works. Simple as that. Your idea works too - there's more than one way to skin a cat.

But as soon as you write down the statement "$\mathbb{R} = \{\text{Dedekind cuts of }\mathbb{Q}\}$", a few things are very obvious:

  • you get $\mathbb{Q}$ sitting inside $\mathbb{R}$
  • you can define addition (just add all the elements in two cut sets, and you get another cut set) and multiplication (similarly)
  • you can define <, and $\mathbb{R}$ is ordered
  • $\mathbb{R}$ (by definition) has the least-upper-bound property: any non-empty subset bounded above has a supremum
  • etc.

You can check all of this with your definition too, and it comes down to the same thing. It's just a little messier.

Notice that I haven't asked you to write down a Dedekind cut for, say, $\pi$ explicitly. (How would you do this?) That's because we don't really want to know that $\pi$ exists at this stage - all we want to know right now is that the number system $\mathbb{R}$ exists and has all the nice properties we think it does. And, of course, it does. Once we've done that, we can worry about all the individual numbers that we need.

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For various reasons, mainly of convenience. One good reason is given by considering the embedding of the rationals in the reals; the decimal approximation of $1$ for example looks a bit silly - one could conceivably write $0.9,0.99,0.999,\cdots$ but otherwise one obtains a finite sequence, which looks strange. Another reason is simply to avoid tying the construction to the decimal counting system. The usual construction is completely unambiguous, more consistent, and has nice operations.

Speaking of the operations - what else would you define addition as? Clearly you need to sum infinitely many pairs because the result must be an infinite set, and there is no natural way to choose only certain pairs in the sets $x$ and $y$ when computing $x+y$. Indeed, if the resulting set must contain all rationals below some cut, you need to consider many possible additions or you may miss some.

Think about multiplication for your sequence now - how would you define $\sqrt 2\times \sqrt 2$? You don't get a sequence with one more digit every time just by multiplying together a particular series of elements. Therefore you need more manipulation to fix this to be of the correct form. This is a big pain!


Are not the terminating decimals sufficient ?

Yes, you could also define things only in terms of terminating decimals, but again there is work to do, and since termination depends on which base you work in this isn't very natural. (It's tied to base 10 again and for no good reason; we're already familiar with non terminating expressions like $1/3$.)

The thing this makes annoying is division of reals. Dividing two terminating rationals doesn't always give you another way ($\frac 1 3$ again) so this requires special attention.

Basically, the point is that the construction should encompass the (ordered) field properties of $\mathbb Q$ if it is to be "nice". Terminating decimals are not a field.

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There's more than one way of modelling the real numbers. Using Dedekind cuts is one way. Or we can use Cauchy sequences. A Cauchy sequence is a sequence of rationals whose elements become arbitrarily close to each other as the sequence progresses (and your example of progressive decimal approximations from below for $\sqrt{2}$ is a nice example of a Cauchy sequence, though in general we need not require Cauchy sequences to be monotonic).

Now of course, there can be many Cauchy sequences which, intuitively, converge to the same place, so for most purposes it will be more convenient to model a real number, not with a particular Cauchy sequence, but rather than with an equivalence class of Cauchy sequences. But your intuition was right: Dedekind cuts aren't the only way to go here.

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The idea behind Dedekind cuts also shows up elsewhere. For example, it can be used to prove that every generalized ordered space can be embedded in a linearly ordered topological space. –  dfeuer Jun 24 '13 at 22:20

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