2
$\begingroup$

The author uses $\subset$ as $\subseteq$. Also notice the difference by font that $\mathscr{C}$ is a class, while $C$ a set.

Definition. A cut in $\Bbb{Q}$ is a pair of subsets $A,B$ of $\Bbb{Q}$ such that

  1. $A\cup B=\Bbb{Q},\quad A\ne\emptyset,\quad B\ne\emptyset,\quad A\cap B=\emptyset.$
  2. $a\in A \land b \in B \implies a\lt b.$
  3. $A$ contains no largest element.

Definition. The cut $x=A|B$ is less than or equal to the cut $y=C|D$ if $A\subset C$.

Theorem. If $S$ is a non-empty subset of $\Bbb{R}$ and is bounded above then in $\Bbb{R}$ there exists a least upper bound for $S$.

Proof: Let $\mathscr{C}\subset\Bbb{R}$ be any non-empty collection of cuts which is bounded above, say by the cut $X|Y$. Define $$C=\{a\in\Bbb{Q}: \stackrel{?_1}{\text{for some cut }}A|B\in\mathscr{C},a\in A\},D=\text{the rest of }\Bbb{Q}.$$ It is easy to see that $z=C|D$ is a cut. Clearly, it is an upper bound for $\mathscr{C}$ since the "$A$" for every element of $\mathscr{C}$ is contained in $C$. Let $z'=C'|D'$ be any upper bound for $\mathscr{C}$. By the assumption that $A|B\le C'|D'$ for all $A|B\in\mathscr{C}$, we see that the "$A$" for every member of $\mathscr{C}$ is contained in $C'$. Hence $C\subset C'$, so $z\le z'$. That is, among all upper bounds for $\mathscr{C}$, $z$ is $\stackrel{?_2}{\text{least}}$.

Is the cut $X|Y$ not important since it's never used in the proof? Why we need it?

$?_1$. Does that for some cut mean that $\forall A|B$?

Is the result equivalent to the union of all the "A"s in $\mathscr{C}$? It's not clear for me...

I'm reading Pugh's Real Mathematical Analysis.

Is the definition of Dedekind cut of Rudin's more rigor/better?

  • 3
    Anyway, those classes are certainly sets, aren't they? – 2017-02-23
  • 0
    Sorry about that, and thanks... – 2017-02-23
  • 2
    Definition of Rudin has advantage that it deals with the lower half of the cut, so that you have to deal with only one set for one real number, but then everytime you have to show (or convince yourself) that this one set is not the set $\mathbb{Q}$. Dedekind used two sets in his pamphlet "Stetigkeit und irrationale zahlen" (continuity and irrational numbers) and I prefer that. Rigor wise both approaches are same. – 2017-02-23
  • 0
    @ParamanandSingh: I don't understand that why $C \subset C'$? Can you give me a hint? Is $z\in\mathscr{C}$ true? – 2017-10-01
  • 2
    @H.R.: this is not difficult if you clearly note the definition of $C$ and $C'$. If $c\in C$ then there is some cut $A\mid B\in\mathscr{C}$ such that $c\in A$. By definition $A\mid B\leq C'\mid D'$ and hence $A\subset C'$ and thus $c\in C'$. Therefore $C\subset C'$. Also $z$ does not necessarily belong to $\mathscr{C}$. – 2017-10-01
  • 1
    @H.R.: you should read Dedekind's original pamphlet "Continuity and irrational numbers". It is much easier to comprehend than the exposition provided by modern authors who perhaps simply want to downplay this beautiful aspect of construction of real numbers. – 2017-10-01
  • 0
    @ParamanandSingh: I was read this proof of Pugh many times and it really seemed unclear to me! :D There are two cases, my mathematical understanding is really poor or the proof is not well written! :( Maybe I should do what you say. Is the pamphlet available on the net? Can you provide any link. :) – 2017-10-01
  • 1
    @H.R.: Dedekind's pamphlet is available [here](https://archive.org/details/essaysintheoryof00dedeuoft). I have not read Pugh and don't personally know you so can't comment on either Pugh's book or your mathematical understanding, but Rudin's exposition of this topic is very very boring and poor. – 2017-10-01
  • 0
    @ParamanandSingh: Thanks. :) The proof by Pugh is exactly the one mentioned in the question. :) – 2017-10-01
  • 0
    To @ParamanandSingh: Ohh man I finally understand what you said! Especially that *$z$ does not necessarily belong to $\mathscr{C}$*. Thank you! – 2018-03-17

2 Answers 2

3

The cut $X\vert Y$ is used to ensure that $D$ is nonempty - the point being that $D$ contains $Y$, since $X\vert Y$ is an upper bound of $\mathscr C$. This was not written out explicitly, but is implicit in the sentence "It is easy to see that $z=C\vert D$ is a cut."

So the proof is completely rigorous, it just wasn't explicit about one step.

As to $?_1$, I'm not sure what the "$\forall$" is doing there, but your conclusion - that $C=\bigcup_{A\vert B\in \mathscr C} A$ - is correct.

  • 0
    It is not really clear that $C|D$ is a cut! At least to me! What about checking properties 2 and 3? – 2017-10-07
3

Apparently, "for some cut $A|B\in\mathscr C$, $a\in A$" is supposed to mean $\exists A|B\in\mathscr C\colon a\in A$. In other words, $$C=\bigcup_{A|B\in\mathscr C}A $$

The cut $X|Y$ better be used to show that $D\ne \emptyset$.

  • 0
    What else could "for some cut $A\vert B\in\mathcal{C}$, $a\in A$" mean? – 2017-02-23
  • 0
    Is it also true that $D=\bigcap_{A|B\in\mathscr C}B$? – 2017-10-07