4
$\begingroup$

This question refers to the construction of $\mathbb{R}$ from $\mathbb{Q}$ using Dedekind cuts, as presented in Rudin's "Principles of Mathematical Analysis" pp. 17-21.

More specifically, in the last paragraph of step 4, Rudin says that for $\alpha$ a fixed cut, and given $v \in 0^*$, setting $w=- v / 2$, there exists an integer $n$ such that $nw \in \alpha$ but $(n+1)w$ is not inside $\alpha$. Rudin says that this depends on the Archimedean property of the rationals, however he has not proved it.

Could somebody prove the existence of the integer $n$?

$\endgroup$

2 Answers 2

Reset to default
8
$\begingroup$

By definition there is some rational $q$ that is not in $\alpha$. Let $p=|q|+1$; then $p>q$, so $p\notin\alpha$, and $p>0$. The rational $w$ is also positive, so by the Archimedean property of the rationals there is a positive integer $m$ such that $mw>p$, and therefore $mw\notin\alpha$.

We also know that there is some rational $r\in\alpha$. Let $s=-|r|-1$; then $s<r$, so $s\in\alpha$, and $-s>0$, so again by the Archimedean property there is a positive integer $k$ such that $kw>-s$. But then $(-k)w<s$, so $(-k)w\in\alpha$. We now have positive integers $k$ and $m$ such that $-kw\in\alpha$ and $mw\notin\alpha$.

Let $S=\{i\in\Bbb N:(i-k)w\notin\alpha\}$; $m+k\in S$, so $S\ne\varnothing$. Since $\Bbb N$ is well-ordered, $S$ has a smallest element, $j$. Note that $0\notin S$, since $-kw\in\alpha$, so $j>0$. Let $n=(j-1)-k$. Then $nw\in\alpha$, and $(n+1)w\notin\alpha$.

$\endgroup$
4
  • $\begingroup$ Concerning the last part... I've checked, and think Rudin does not mention that $\mathbb N$ is well-ordered in chapter 1 or anywhere else in the book. Might this have been his omission, or is the well-ordering of $\mathbb N$ a fact the reader is supposed to know before reading Rudin? $\endgroup$
    – Anthony Pulido
    Commented Dec 1, 2014 at 20:23
  • 2
    $\begingroup$ @Anthony: Probably the latter: Walter was extremely careful about such things. Still, I’d not bet a large amount either way. $\endgroup$
    – Brian M. Scott
    Commented Dec 1, 2014 at 20:25
  • $\begingroup$ @Brian, I have seen a few proofs on Math StackExchange for proving that the rationals have the Archimedean property, the simplest being basically this: Let $m/n \in Q$. Then $n(m/n) + 1 \gt m/n$ (math.stackexchange.com/q/61632). I'm wondering if this is really rigorous, and if there are any underlying assumptions that I am missing. I am working through Rudin's construction of $R$ using cuts right now as well. $\endgroup$
    – user137044
    Commented Oct 2, 2016 at 16:27
  • 1
    $\begingroup$ @Addison: It’s rigorous as the term is normally used. To make it any more rigorous, one would have to go back to the construction of the rationals and the verification that they are an ordered field and recast the argument in that context, verifying that it still works. I’d say that you’re probably not missing anything. $\endgroup$
    – Brian M. Scott
    Commented Oct 2, 2016 at 19:01
0
$\begingroup$

If $v\in 0^*$, then $v<0$, so $w>0$. Let $\gamma=\sup \alpha$. The Archimedean property says that for any $\gamma$ there is some integer $m$ such that $mw\geq\gamma$. Since $\mathbb{N}$ is well-ordered, there is a smallest $m$ such that $mw\geq\gamma$, call this $n+1$. This means that $nw<\gamma$, and hence $nw\in \alpha$. But also $(n+1)w\notin \alpha$ since it is larger than $\gamma$ (OK, so there might be some subtleties with endpoints you should think about, but essentially this is the idea).

$\endgroup$
6
  • $\begingroup$ How do you know that $\gamma=\sup \alpha$ exists? $\endgroup$
    – Manos
    Commented Aug 26, 2012 at 17:10
  • $\begingroup$ Sorry. I don't have the book in front of me, but isn't this step 3? $\endgroup$
    – Matt
    Commented Aug 26, 2012 at 18:26
  • $\begingroup$ As i understand $\alpha$ is a cut, and as such it has no upper bound. $\endgroup$
    – Manos
    Commented Aug 26, 2012 at 20:25
  • 1
    $\begingroup$ @ Matt: I think step 3 proves that $R$, the set of cuts, has the least upper bound property, not that a cut has a least upper bound, which it does not have to have since it is a subset of the rationals. Supposing you were right about that, I'm also not sure of what follows in your argument. Suppose the cut $\alpha$ represents $-20$, so that $\gamma = -20$, and say, $v = -2 \in 0^*$, so that $w = -v/2 = 1$. (continued.) $\endgroup$
    – Anthony Pulido
    Commented Dec 1, 2014 at 23:24
  • 1
    $\begingroup$ The archimedean property says that there is a positive integer $m$ such that $mw > \gamma$, or $m > -20$, as you said. Just to be sure, by the second fact to "be used freely" (written just before step 2): if $r \notin \alpha$ and $r < s$ then $s \notin \alpha$, it follows that $m \notin \alpha$, although we already knew this since $m$ is bigger than the supremum. Since $m \in \mathbb N$ and you are using its well-ordering, the smallest possible value for $m$ is $1$. If $m = n + 1$, then $n = 0$. It follows that $nw = 0 \notin \alpha$ and $(n+1)w \notin \alpha$. $\endgroup$
    – Anthony Pulido
    Commented Dec 1, 2014 at 23:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .