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In Rudin's Principles of Mathematical Analysis, it is said that real numbers will be certain subsets of rational numbers called cut. I am having trouble accepting this fact, because if we consider real numbers as those having members that are subsets of rationals, then how do one account for gaps left by rationals? I.e. irrationals.

Please help me understand the actual idea being conveyed here. Thank you.

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  • $\begingroup$ I explained about Dedekind cuts here. Maybe it helps. $\endgroup$ Feb 10, 2019 at 11:08
  • $\begingroup$ It might help to see the picture located in the following wiki page <en.wikipedia.org/wiki/Dedekind_cut>. A cut at an irrational number is the collection of all rationals strictly below it; however there are rational numbers that can be found arbitrarily close to any irrational. This "fills" in the gaps by thinking of real numbers as collections. $\endgroup$ Feb 11, 2019 at 3:41

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What you're overlooking is that there aren't any "gaps." The rationals form a "dense" subset of the reals, which means that any open set in the reals contains at least one rational number.

The usual way in which Dedekind cuts are used to define the reals is to define a Dedekind cut as $\{R \subset \Bbb Q ~ \vert ~ x \in R \Rightarrow \forall y \lt x ~ (y \in R) \}, \text{ where } R \neq \emptyset, \Bbb Q$. We think of a real number as simply all rationals (strictly) less than that number. So, for example, in this construction we think of $\pi$ as $\{x \in \Bbb Q~ \vert ~ x \lt \pi ~ \}$.

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  • $\begingroup$ Is it done so that, each irrational is uniquely determined? But still it doesn't give a feel for magnitude of irrationals, by defining the cuts in this way. Any intuitive ways to understand cuts , maybe an example...please $\endgroup$
    – sam soft
    Feb 10, 2019 at 7:45
  • $\begingroup$ @fleablood and I each provided examples. And yes, this method uniquely determines irrational numbers as well as rational numbers. The order relationship is simply set containment: $R_1 \lt R_2 \iff R_1 \subset R_2.$ $\endgroup$ Feb 10, 2019 at 15:25
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Sets are not the same things as numbers and the sets actually "fit" in the "gaps".

And as a cut is a set that has no maximum element it need match up with any rational number. (That's actually the entire point.)

Consider this set: $A=\{q\in \mathbb Q|q < \frac 23\} $. This set is a "cut" (it fits all the criteria). And it sits directly "on top of" The rational number $\frac 23$ as clearly $\sup A=\frac 23$. In this way set $A$ is linked to and represents the real (and rational) number $\frac 23$.

Now consider the set $B=\{q\in \mathbb Q|q <0$ or $q^2 <2\} $. This set is also a cut. But it does not "sit on" any rational number as there is no rational number that is $\sup B $. This set falls into a "gap" between rational numbers. Set $B $ is linked to and represents the real (and irrational) number $\sqrt 2$.

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  • $\begingroup$ what exactly do you mean by sit on top of? $\endgroup$
    – sam soft
    Feb 10, 2019 at 9:25
  • $\begingroup$ is this approach similar to that of a one-one function, each cut corresponding to each real number? $\endgroup$
    – sam soft
    Feb 10, 2019 at 9:50
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    $\begingroup$ No cut has a maximum element. But we can order cuts by which cuts are subsets of each other. Thus every cut "fits" somewhere in relation to other cuts. Some of these cuts have least upper rational bounds. These cuts are in one to one correspondence with the rational numbers AND this 1-1 corespondence preserves order. So they "fit" or "sit on top of" the rational numbers. But other cuts do not have leat upper rational bounds. These cuts "fill" the gaps. $\endgroup$
    – fleablood
    Feb 10, 2019 at 16:19
  • $\begingroup$ To Rudin ' s thinking the cuts ARE the real numbers. The cuts with rational least upper bounds are in 1-1 order preserving, operation preserving corespondence with field of the rational. That makes the two fields "equivalent". There is no difference between them.The cuts with rational least upper bounds are proper subfield of the field of all cuts. The rational numbers themselves are a proper subfield of... something. But we can't define them except by fiat. if there is a universe of concepts. They are the preimage of all cuts while the rationals are the preimage of some cuts. $\endgroup$
    – fleablood
    Feb 10, 2019 at 16:40
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Intuitively the idea behind the Dedeking cuts is the following. Assuming that you "know" what the real numbers are, then you can associate to each real number $a$ the set $$( - \infty ,a ) \cap \mathbb Q=: C_a$$

Now, a set $C_a$ satisfies the four axioms of definition of a Dedekind cut, and moreover, each Dedekind cut is exactly one and only one of the sets $C_a$. Indeed, if $C$ is a Dedekind cut, and $a= \sup(C)$ then $C=C_a$.

Because of this, the correspondence $a \leftrightarrow C_a$ is a bijection between the real numbers and the set of Dedekind cuts.

Now, the issue is what to do if you don't know the real numbers, but you want to construct them. Well, you can construct the Dedekind cuts without assuming the existence of a real number. This way we can give an axiomatic construction for the reals.

And always remember, when we think about Dedekind cuts as "real numbers", what we mean is that, once we formally define $\mathbb R$, for each cut $C$ there exists an unique real number $a$ such that the cut $C$ is exactly $$C= (- \infty, a) \cap \mathbb Q$$

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