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$\mathbb{R}$ is real numbers set, $\mathbb{Q}$ denotes rational numbers set.

This is quoted from Rudin's mathematical analysis book page 17 about Dedekind' s construction.

Why the members of $\mathbb{R}$ will be certain subsets of $\mathbb{Q}$?

There are two levels' confusing, one is that in Mathematics, another is this English sentence, or the expression of this fact.

Maybe I'd like to comprehend like this: members of $\mathbb{R}$ are some thing decided by some certain subsets of of $\mathbb{Q}$.

At first glance, it seems like members of $\mathbb{R}$ are are equal to some subsets of $\mathbb{Q}$. But which subset is $\sqrt{2}$ correspoinding to ? This maybe not so obvious. IMO

As @Hagen von Eitzen's answer mentioned, it means that $\mathbb{Q}$ is obtained from Integers. And this is just one construction, I agree.

That is obvious.

And the same to $\mathbb{C}$, Complex number is a pair of real numbers, we do accept the fact quikly.

But if you say,

$\mathbb{C}$ is some certain subsets of $\mathbb{R}$

$\mathbb{C}$ is some certain subsets of $\mathbb{Q}$

$\mathbb{C}$ is some certain subsets of $\mathbb{Z}$

There will also be some confusions at first glance in my point of view.

@Robert Israel 's answer is more about the fact what Real Number is.

HyperGroups
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  • There is a difference between an ordered pair of real numbers and a two member subset of the real numbers. – Henry Jun 23 '13 at 13:16
  • @Henry: But the difference is only one level: An ordered pair or real numbers is (modelled by) a set of sets of real numbers. – celtschk Jun 23 '13 at 13:36
  • @celtschk: Indeed: ${{x},{x,y}}$ is not a subset of the set of real numbers, but is a subset of the powerset of real numbers. The latter is a much bigger set. – Henry Jun 23 '13 at 13:39
  • So a complex number "is" a pair of real numbers, that "is" a set of sets or real numbers, that "is" a set of sets of sets of rational numbers, that "is" a set of sets of sets of pairs of integers, that "is" a set of sets of sets of sets of sets of integers, that "is" a set of sets of sets of sets of sets of pairs of natural numbers, that "is" a set of sets of sets of sets of sets of sets of sets of natural numbers. And of course the natural numbers "are" just special sets of sets … well, let's just say, complex numbers are very specific sets of sets. ;-) – celtschk Jun 23 '13 at 13:49
  • @Henry: BTW, we could also model pairs of real numbers by subsets of real numbers: If $a<b$, represent the pair $(a,b)$ by ${a,(a+b)/2,b}$, otherwise represent the pair by ${a,b}$. – celtschk Jun 23 '13 at 14:00
  • @celtschk: Indeed. We could model a complex number as a subset of the integers (as a hint, consider $123.45 - 9.0123i$ and then ${1201, 802, 403, -404, -805, 609, -600, -1001, -1402, -1803}$ where the right hand digits correspond to the digits in the complex number, the next digit is $0$ and the remaining parts identify the power of ten and whether the term is positive or negative and real or imaginary). But I was making a small pedantic point to the OP. – Henry Jun 23 '13 at 16:31

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There are many ways to define the real numbers, which in the end turn out to be equivalent. There is no point in arguing about which are the "real" real numbers: all that matters is what you can do with them.

Robert Israel
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For the same reason that rationals "are" pairs of integers (or are they?). It is just a construction. If you want to communicate the notion of "real number" to someone who only knows rational numbers, you might in simple terms try to say that real numbers are obtained by adding objects "between" rationals. So a real number is characterized by how it is between rationals (and if $xy$ are between the same rationals, then $x=y$). However, you can't say that a real number is specified by being between two specific rationals (after all there are also infinitely many rationals between these two rationals). Instead you have to refer to all rationals; they are divided into those rationals greater than your real number and those smaller (and there is one rational equal to your real number if it is in fact rational). Dedekind cuts is a formalization of this idea.

jkn
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Hagen von Eitzen
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Real numbers correspond to certain subsets of rational numbers. If you take the real numbers as given, and consider for any real number $r$ the set $D_r:=\{x\in\mathbb Q: x < r\}$, you'll find that it uniquely characterizes the real number, that is, $r=s\iff D_r=D_s$. Therefore you can specify a real number $r$ just by giving the set $D_r$. However, since $D_r$ only contains rational numbers, you don't need to take the real numbers as given to speak about those sets, provided that you find a way to define those sets without speaking about real numbers. That's exactly what the Dedekind cuts are about: Defining sets which uniquely specify a real number, without actually using the real numbers. After having done that, you can just declare that every such set defines a real number. Of course you then also need to explain which set correspond e.g. to the real number $a+b$ if you've given the sets for the real numbers $a$ and $b$, and so on.

The advantage to do so is that if you have proven that rational numbers are a sound concept, then you also know that those subsets of rational numbers are a sound concept. Since each such set corresponds to a real, this means that also reals are a sound concept (because those subsets of rational numbers are a model of the real numbers).

Note that Dedekind cuts are not the only way to define the real numbers; there's also the definition using Cauchy sequences, for example. So are now real numbers certain subsets of rational numbers, or equivalence classes of Cauchy sequences? Well, they correspond to both, and both give rise to the same concept because there's, again, a correspondence between the equivalence classes of Cauchy sequences and Dedekind cuts, as for each Dedekind cut there exists an equivalence class of Cauchy sequences, and vice versa.

celtschk
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It says certain subsets of $\mathbb{Q}$. It does not say certain elements of $\mathbb{Q}$. If you have a set A:={1, 3, 7, 14}, then some certain subsets of A are {1, 3}, {1}, and {7, 1, 14}. But, none of those are elements of A. Only 1, 3, 7, and 14 are elements of A.

By "certain subsets" I feel confident in saying that Rudin doesn't mean the entire set $\mathbb{Q}$, but specific subsets of $\mathbb{Q}$.

Doug Spoonwood
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  • ah, thanks for pointing out this, that is one expression error of mine, the same like a+bI is the certain subset {a,b} of R, but still not so obvious in the case posted, maybe the subsets have infinite numbers of elements. – HyperGroups Jun 23 '13 at 13:22
  • @HyperGroups Yes, in this case they do have an infinity of elements. – Doug Spoonwood Jun 23 '13 at 13:30