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by thaumasiotes 4564 days ago
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.
1 comments
bjornsing 4563 days ago
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... ;)

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thaumasiotes 4563 days ago
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:

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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..........
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