Why Cantor set is uncountable despite each element is rational.

Question

Note I am not asking about a proof that Cantor set is uncountable. I want to get some "natural" answer on my question (see below).

A Construction of the Cantor set

Let us change a bit the famous construction of Cantor set. Instead of removing intervals we will add points at each iteration. So we build Cantor set by induction:

1. At the first iteration $n=0$ there is a set $A_0 = \{0,1\}$
2. After $A_n$ has been constructed $A_{n+1}$ is obtained as the folllowing: $$A_{n+1} = A_n\cup \{a_i+\frac{(-1)^{3^{n-1} a_i}}{3^n}\,|\, a_i \in A_n\}$$

By the above costruction Cantor set is the union of all $A_n$'s: $$\mathcal{C} = \bigcup_{n=0}^{\infty}A_n$$

Question

From this construction the following statement arises $$\xi\in\mathcal{C}\Rightarrow \xi \in \mathbb{Q}$$ So if each element in $\mathcal{C}$ is rational how it (set) is uncountable?

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P.S. I am not sure that my construction of Cantor set is absolutely correct. I will appreciate any ideas, advices, corrections etc.

Answer 1

You seem to think the only elements in the Cantor set are the endpoints of the intervals you have deleted. This is not true.