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..........