My comment in response to Andy Steadman's comment summarizes the answer to the challenge in Part 2 of "Cantor set":
Andy, you got it. The Cantor set is the numbers whose ternary representation contains only 0s and 2s, no 1s. Of course, if you then convert all the 2s to 1s and interpret the result as binary, you see at once that the Cantor set is just as numerous as the whole real interval [0,1].
By "ternary representation" I mean what we mean by "decimal representation," only base 3:
1/3=0.1, 2/3=0.2, 1/2=0.111..., 2/13=0.011011011..., etc. using digits 0,1,2.
Binary, of course, is the corresponding representation base 2:
1/2=0.1, 1/3=0.010101..., 2/3=0.101010..., 3/7=0.011011011..., etc. using digits 0,1.
These are members of the Cantor set (base 3):
0.222..., 0.002220220002202, 0.000220002200022...
This is not:
The "compatible representation" for this predicate--"ternary has only 0s and 2s"--is of course ternary representation. In Part 3 of "Cantor set" I asked, given a number in another, noncompatible representation, can it be tested by the given predicate? For example, suppose you define a number by some algebraic equation of high degree or by some transcendental equation like e^x=x; and suppose you calculate the first 10,000 ternary digits (tits?) and find them all to be 0 or 2. How do you know if a 1 appears later? This is hard. A problem of equivalent difficulty is: Do 777 consecutive 7s occur in the decimal expansion of pi? Let's see... 3.1415926535... Well, not in the first 10 places, anyway.