3
$\begingroup$

Assuming that there are uncountably infinitely many irrationals between any two consecutive rationals, how can the Dedekind cuts (defined on the countably infinite rationals) define all the irrationals?

For instance, is it not true that the real interval between 1 and the largest element of the Dedekind cut with all elements < 1 (i.e. largest element of the cut is $\lim_{n\to\infty} {n\over n+1}$ , natural n) contains uncountably infinitely many irrationals?

$\endgroup$
  • 4
    $\begingroup$ Dedekind cuts are sets of rationals. The family of all sets of rationals is uncountable, so there is no a priori reason why the construction should fail for that reason. One can similarly define irrationals as classes of Cauchy sequences of rationals; such a sequence is a function $f:\Bbb N\to \Bbb Q$, and again, the number of such sequences is uncountably infinite, so there is no trouble about whether there are enough. $\endgroup$ – MJD Feb 28 '14 at 16:12
  • 1
    $\begingroup$ $\pi$ is obtained as the limit of the sequence $3,3.1,3.14,3.141,\dots$ This is a sequence of rationals. Similarly, any number can be expressed as a limit of a strictly increasing sequence of rationals. Dedekind cuts can be seen as a variant of this simple idea. $\endgroup$ – Andrés E. Caicedo Feb 28 '14 at 16:14
  • $\begingroup$ But doesn't each subsequent Dedekind set only have a single member more than its previous set? I.e., are the Dedekind cuts the family of all sets of rationals? $\endgroup$ – user132162 Feb 28 '14 at 16:15
  • 1
    $\begingroup$ Between any two Dedekind cuts, there are infinitely many other Dedekind cuts, there are no "subsequent Dedekind cuts". $\endgroup$ – Daniel Fischer Feb 28 '14 at 16:17
  • 4
    $\begingroup$ $\lim_{n\to\infty}\frac{n}{n+1}$ is equal to 1, so it is not a rational preceding 1. There is no such thing as "consecutive rationals", because between any two distinct rationals $r_1$ and $r_2$, there is an infinite family of other rationals, $\frac12(r_1+r_2)$ being a simple example. $\endgroup$ – MJD Feb 28 '14 at 16:24
14
$\begingroup$

You have several misconceptions about the way numbers work; all of these are common.

First you seem to have the idea that the rationals are a discrete sequence, like the integers, so that every rational number has a predecessor rational and a successor rational. This is not the case. For suppose you claim that $a$ and $b$ are consecutive rationals. But $a\lt\frac{a+b}{2}\lt b,$ and $\frac{a+b}{2}$ is rational, so we have just found a rational number between $a$ and $b$, contradicting your claim that $a$ and $b$ were consecutive. But this construction works for any rationals $a$ and $b$. So there is no such thing as consecutive rationals.

Second, you seem to think that $$ \lim_{n\to \infty}\frac{n}{n+1} $$ means something other than what it does. This notation is a mathematical term of art with a particular, fixed meaning. Without going into details, the $\lim$ notation describes a property that a number might or might not have, and in this particular case one can show that the one and only number with that particular property is the number 1. It is not some number that is magically less than 1 by a tiny amount; it is equal to 1. When you write $\lim_{n\to \infty}\frac{n}{n+1}$, you are intending to communicate a certain idea, but what you have actually written is just a complicated notation for the number 1.

What I believe you are trying to do here is to invent a number $y$ so that $y$ is less than 1, but still closer to 1 than any other number less than 1. This is a common desire, but there is no such $y$. There can't be, because $y$ has contradictory properties. Consider $e = 1-y$. This is positive, because $y<1$. Then $0 < \frac e 2 < e$, so $$ y < y + \frac e2 < y+e = 1 $$ which shows that $y+\frac e2$, while still less than 1, is even closer to 1 than $y$ was, contradicting the original definition of $y$. This shows that the idea for $y$ is incoherent; there is no such object with that property.

It is a common error in mathematics (and philosophy!) to assume that just because some set of properties can be stated, that there must be some object with those properties. This is not the case. One can say

“the current Crown Prince of the Ottoman Empire”,

but there is at present no Ottoman Empire and no Crown Prince of the Ottoman Empire, so the phrase, although meaningful, does not refer to any actual thing, although you have to know a bit about the Ottoman Empire to know this. Similarly one can say

“the rational number preceding 1”

or

“the largest real number less than 1”

and it seems at first to make sense. But if you know a bit more about numbers you know that those phrases do not designate any actual object.

I'm not sure this answers your question, but your question is based on several basic misunderstandings of how numbers work, so I hope this helps clear up some of it.

$\endgroup$
  • 1
    $\begingroup$ A really lovely example came up today of a set of properties that sound reasonable, but which are not possessed by any actual object: The querent in Volume of pentahedron having all sides of length 1? appears to be asking about the geometric properties of a polyhedron with five faces, all triangles. But there is no such polyhedron. $\endgroup$ – MJD Jun 18 '14 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.