# Why Cantor set is uncountable despite each element is rational.

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?

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

• Not every element of the Cantor set is rational. – Lord Shark the Unknown Oct 29 '17 at 6:22
• Said differently, your set is not a Cantor set. – quasi Oct 29 '17 at 6:23
• Your construction is definitely not the usual one. – user99914 Oct 29 '17 at 6:23
• Why do you call the set $C$ you have constructed the "Cantor set", and why do you suppose that it is uncountable? – Lord Shark the Unknown Oct 29 '17 at 6:24
• Nothing is wrong with it. It's just not the Cantor set. Your set is definitely a subset of the set of rational numbers. – quasi Oct 29 '17 at 6:25