I am looking for an example of a specific element of the Cantor set which isn't a rational number, and how it comes about in the set when constructing it, to understand why the Cantor set isn't a subset of $\mathbb{Q}$. I have seen the proof that there is bijection between all ternary strings having only $0$'s and $2$'s and the Cantor set, however intuitively I still feel that all of the elements of the Cantor set are rationals, as these elements in the cantor set are just the endpoints of the intervals left from when we took away the open intervals from the set to make the Cantor set, all of which endpoints seem to be rationals because we keep on removing $1 \over 3$, a rational amount, from the remaining intervals at each stage.
EDIT: I am now trying to understand why the elements of the Cantor set aren't just the endpoints of the open intervals we removed; this is counter-intuitive as the limit of $(\frac 2 3 )^{n}$ as $n$, the number of intervals we remove, gets arbitrarily large, is $0$, so it seems we would have nothing left over if we indefinitely remove intervals. Is it the case that the reals are so numerous that we cannot finely divide them or something like that?
Best Answer
Let's start with what you know!
So, you know that there is a bijection between all ternary strings with only zeroes and twos and the cantor set. In the proof, what is used is the fact that the Cantor set can be characterized as all numbers that have only zeroes and twos in their ternary representation.
So, let's just find a number that has only zeroes and twos in its ternary representation, and is not rational. Well, here is one:
$$0.20020002000020000020000002\cdots$$