[bjornsing]

I have an issue with this (albeit parenthesised) line: "It turns out that, in some sense, the real numbers would still look like a line under infinite magnification, but the rational numbers would be dots separated by spaces."

In-between any two rational numbers there's an infinite number of other rational numbers. So, in any reasonable sense and at any level of "magnification", if you can "see" two dots representing two rational numbers then they are connected by a line of other little dots (just like the reals). Perhaps you could argue though that at "infinite magnification" there are no rational numbers to be seen, it's just empty space, whereas the reals of course still make a nice line.

[thaumasiotes]

Well, consider the ruler function[1], which is continuous on the irrationals and discontinuous on the rationals. The real numbers really are denser than the rationals; that's why something like the ruler function is possible (notably, a conceptual reverse, continuous on the rationals and discontinuous on the irrationals, cannot exist -- the rationals are too far apart). I'm pretty sure this is precisely the phenomenon the quote you extract is referring to: if you were standing, infinitely magnified, at a point on the ruler function, then the function would be continuous ("look like a line") if your point was irrational, but if your point was rational, there would be a measurable gulf separating you from the rest of the function.

[1] http://en.wikipedia.org/wiki/Thomae%27s_function

[bjornsing]

Infinity is a pretty strange concept. :) I'm not sure arguing over it in this format is meaningful, but for the fun of it:

Consider that the integral of the ruler function from 0 to 1 is 0 (as is stated in your reference 1). In layman's terms you could express this as "there are infinitely more irrational than rational numbers between 0 and 1". At the same time, "for every two rational numbers there are infinitely many rational numbers in-between them". What sort of "picture" is this compatible with?

I still think that the only picture that really makes any sense is a solid line at any finite magnification, yet empty space at infinite magnification.

[thaumasiotes]

I don't understand the point you're trying to make?

The Cantor set shares the property that "for every two [points in the set] there are infinitely many [points in the set] in between", but no one would describe it as looking like a line. It's rather sparse.

[bjornsing]

What I take issue with is an "image" of two rational numbers as two separate dots, with empty space in-between. That's a very deceiving image IMHO, since I cannot think of a sane way to produce it.

The Cantor set is very different. It's even easy to give an example of two points in the set that can (sanely) be depicted with empty space in-between: 1/3 and 2/3. If I'm not mistaken that example also disproves your stated conjecture... ;)

[thaumasiotes]

First of all, let me point out that 1/3 and 2/3 are both rational numbers, so if you can imagine them with empty space between, you've imagined two rational numbers with empty space between.

> It's even easy to give an example of two points in the [Cantor] set that can (sanely) be depicted with empty space in-between: 1/3 and 2/3. If I'm not mistaken that example also disproves your stated conjecture... [that between any two points in the set, there is a third one] ;)

Fair enough. Consider, then, the intersection of the Cantor set with the irrational numbers (you can think of this as the "open Cantor set"). It is, obviously, a subset of the Cantor set, and really does have the property described.

Since I'm feeling embarrassed about that last time, a proof follows:

-----

The Cantor set consists of all real numbers in the interval [0,1] which have a "decimal" expansion in trinary which does not contain the digit 1. That is to say, they can be expressed in terms of powers of (1/3) such that the coefficient of each power of 1/3 is either 0 or 2. (1/3 would usually be represented in trinary as 0.1, but is in the Cantor set because of its representation as 0.02222222...)

Let a,b be two irrational numbers in the Cantor set, a less than b. There is some decimal place at which they diverge, and since a is smaller, it has a 0 at that point, while b has a 2. Since a is irrational, it also has a 0 at some later point in its expansion (if every digit after that were 2, then a's expansion would be repeating and a would be rational). The number constructed by substituting a 2 for a 0 at that index is greater than a, less than b, and in the Cantor set.

Graphical representation of the proof:

    a = 0.......0......
    b = 0.......2......

then

    a = 0.......0....0.....
    c = 0.......0....2.....
    b = 0.......2..........

[gizmo686]

I don't think that works. The rational numbers are a dense subset of the real numbers. Informally this means every real number is either a rational number, or is arbitrarily close to a rational number. This means that at any magnification, if there was a hole that is filled by a real number, then their would also be a rational number that is arbitrarily close to that real number.

[bjornsing]

Well, I'm not a big fan of this "infinite magnification" idea in the first place, but "arbitrarily close" is typically one of those things that infinity can beat.

(Compare e.g. with the fourier transform of a function. It consists of a sum series which comes "arbitrarily close" to the function, but "at the limit" when the number of terms approaches infinity the function and its fourier transform is one and the same.)

[jfarmer]

Well, "in some sense" gives a lot of wiggle room! I agree the magnifying glass analogy is bad, but here's one "sense" in which the rationals "have more holes" than the irrationals.

There exists a function of the reals which is continuous at every irrational point but discontinuous at every rational point. However, there is no function of the reals which is discontinuous on the irrationals but continuous on the rationals. In this sense, the irrationals are "more continuous" than the rationals.

That's about the best I can do, though, which I admit is a stretch.